# Mathematics Curriculum ```New York State Common Core
Mathematics Curriculum
GEOMETRY • MODULE 2
Similarity, Proof, and Trigonometry
Module Overview .................................................................................................................................................. 3
Topic A: Scale Drawings (G-SRT.A.1, G-SRT.B.4, G-MG.A.3) ................................................................................. 9
Lesson 1: Scale Drawings ........................................................................................................................ 11
Lesson 2: Making Scale Drawings Using the Ratio Method.................................................................... 27
Lesson 3: Making Scale Drawings Using the Parallel Method ................................................................ 44
Lesson 4: Comparing the Ratio Method with the Parallel Method ........................................................ 59
Lesson 5: Scale Factors ........................................................................................................................... 72
Topic B: Dilations (G-SRT.A.1, G-SRT.B.4) ........................................................................................................... 88
Lesson 6: Dilations as Transformations of the Plane .............................................................................. 90
Lesson 7: How Do Dilations Map Segments? ....................................................................................... 104
Lesson 8: How Do Dilations Map Lines, Rays, and Circles? .................................................................. 120
Lesson 9: How Do Dilations Map Angles? ............................................................................................ 135
Lesson 10: Dividing the King’s Foot into 12 Equal Pieces ..................................................................... 148
Lesson 11: Dilations from Different Centers ........................................................................................ 162
Topic C: Similarity and Dilations (G-SRT.A.2, G-SRT.A.3, G-SRT.B.5, G-MG.A.1) .............................................. 179
Lesson 12: What Are Similarity Transformations, and Why Do We Need Them? ............................... 181
Lesson 13: Properties of Similarity Transformations............................................................................ 195
Lesson 14: Similarity ............................................................................................................................. 217
Lesson 15: The Angle-Angle (AA) Criterion for Two Triangles to Be Similar ........................................ 229
Lesson 16: Between-Figure and Within-Figure Ratios.......................................................................... 242
Lesson 17: The Side-Angle-Side (SAS) and Side-Side-Side (SSS) Criteria for Two Triangles
to Be Similar........................................................................................................................ 255
Lesson 18: Similarity and the Angle Bisector Theorem ........................................................................ 271
Lesson 19: Families of Parallel Lines and the Circumference of the Earth ........................................... 283
1Each
lesson is ONE day, and ONE day is considered a 45-minute period.
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NYS COMMON CORE MATHEMATICS CURRICULUM
Module Overview
M2
GEOMETRY
Lesson 20: How Far Away Is the Moon? ............................................................................................... 297
Mid-Module Assessment and Rubric ................................................................................................................ 307
Topics A through C (assessment 1 day, return 1 day, remediation or further applications 4 days)
Topic D: Applying Similarity to Right Triangles (G-SRT.B.4) ............................................................................. 334
Lesson 21: Special Relationships Within Right Triangles—Dividing into Two Similar
Sub-Triangles ...................................................................................................................... 335
Lesson 22: Multiplying and Dividing Expressions with Radicals ........................................................... 349
Lesson 24: Prove the Pythagorean Theorem Using Similarity.............................................................. 375
Topic E: Trigonometry (G-SRT.C.6, G-SRT.C.7, G-SRT.C.8) ................................................................................ 387
Lesson 25: Incredibly Useful Ratios ...................................................................................................... 389
Lesson 26: The Definition of Sine, Cosine, and Tangent....................................................................... 404
Lesson 27: Sine and Cosine of Complementary Angles and Special Angles ......................................... 417
Lesson 28: Solving Problems Using Sine and Cosine ............................................................................ 427
Lesson 29: Applying Tangents .............................................................................................................. 440
Lesson 30: Trigonometry and the Pythagorean Theorem .................................................................... 453
Lesson 31: Using Trigonometry to Determine Area ............................................................................. 466
Lesson 32: Using Trigonometry to Find Side Lengths of an Acute Triangle ......................................... 477
Lesson 33: Applying the Laws of Sines and Cosines ............................................................................. 489
Lesson 34: Unknown Angles ................................................................................................................. 502
End-of-Module Assessment and Rubric ............................................................................................................ 515
Topics A through E (assessment 1 day, return 1 day, remediation or further applications 4 days)
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NYS COMMON CORE MATHEMATICS CURRICULUM
Module Overview
M2
GEOMETRY
Geometry • Module 2
Similarity, Proof, and Trigonometry
OVERVIEW
Just as rigid motions are used to define congruence in Module 1, so dilations are added to define similarity in
Module 2.
To be able to define similarity, there must be a definition of similarity transformations and, consequently, a
definition for dilations. Students are introduced to the progression of terms beginning with scale drawings,
which they first study in Grade 7 (Module 1 Topic D), but in a more observational capacity than in high school
Geometry: Students determine the scale factor between a figure and a scale drawing or predict the lengths
of a scale drawing, provided a figure and a scale factor. In Topic A, students begin with a review of scale
drawings in Lesson 1, followed by two lessons on how to systematically create scale drawings. The study of
scale drawings, specifically the way they are constructed under the ratio and parallel methods, gives us the
language to examine dilations. The comparison of why both construction methods (MP.7) result in the same
image leads to two theorems: the triangle side splitter theorem and the dilation theorem. Note that while
dilations are defined in Lesson 2, it is the dilation theorem in Lesson 5 that begins to tell us how dilations
behave (G-SRT.A.1, G-SRT.A.4).
Topic B establishes a firm understanding of how dilations behave. Students prove that a dilation maps a line
to itself or to a parallel line and, furthermore, dilations map segments to segments, lines to lines, rays to rays,
circles to circles, and an angle to an angle of equal measure. The lessons on proving these properties, Lessons
7–9, require students to build arguments based on the structure of the figure in question and a handful of
related facts that can be applied to the situation (e.g., the triangle side splitter theorem is called on frequently
to prove that dilations map segments to segments and lines to lines) (MP.3, MP.7). Students apply their
understanding of dilations to divide a line segment into equal pieces and explore and compare dilations from
different centers.
In Topic C, students learn what a similarity transformation is and why, provided the right circumstances, both
rectilinear and curvilinear figures can be classified as similar (G-SRT.A.2). After discussing similarity in
general, the scope narrows, and students study criteria for determining when two triangles are similar
(G-SRT.A.3). Part of studying triangle similarity criteria (Lessons 15 and 17) includes understanding side
length ratios for similar triangles, which begins to establish the foundation for trigonometry (G-SRT.B.5). The
final two lessons demonstrate the usefulness of similarity by examining how two ancient Greek
mathematicians managed to measure the circumference of the earth and the distance to the moon,
respectively (G-MG.A.1).
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Module Overview
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
In Topic D, students are laying the foundation to studying trigonometry by focusing on similarity between
right triangles in particular (the importance of the values of corresponding length ratios between similar
triangles is particularly apparent in Lessons 16, 21, and 25). Students discover that a right triangle can be
divided into two similar sub-triangles (MP.2) to prove the Pythagorean theorem (G-SRT.B.4). Two lessons are
spent studying the algebra of radicals that is useful for solving for sides of a right triangle and computing
trigonometric ratios.
An introduction to trigonometry, specifically right triangle trigonometry and the values of side length ratios
within right triangles, is provided in Topic E by defining the sine, cosine, and tangent ratios and using them to
find missing side lengths of a right triangle (G-SRT.B.6). This is in contrast to studying trigonometry in the
context of functions, as is done in Algebra II of this curriculum. Students explore the relationships between
sine, cosine, and tangent using complementary angles and the Pythagorean theorem (G-SRT.B.7, G-SRT.B.8).
Students discover the link between how to calculate the area of a non-right triangle through algebra versus
trigonometry. Topic E continues with a study of the laws of sines and cosines to apply them to solve for
missing side lengths of an acute triangle (G-SRT.D.10, G-SRT.D.11). Topic E closes with Lesson 34, which
introduces students to the functions arcsin, arccos, and arctan, which are formally taught as inverse
functions in Algebra II. Students use what they know about the trigonometric functions sine, cosine, and
tangent to make sense of arcsin, arccos, and arctan. Students use these new functions to determine the
unknown measures of angles of a right triangle.
Throughout the module, students are presented with opportunities to apply geometric concepts in modeling
situations. Students use geometric shapes to describe objects (G-MG.A.1) and apply geometric methods to
solve design problems where physical constraints and cost issues arise (G-MG.A.3).
Focus Standards
Understand similarity in terms of similarity transformations.
G-SRT.A.1
Verify experimentally the properties of dilations given by a center and a scale factor:
a.
A dilation takes a line not passing through the center of the dilation to a parallel line,
and leaves a line passing through the center unchanged.
b.
The dilation of a line segment is longer or shorter in the ratio given by the scale factor.
G-SRT.A.2
Given two figures, use the definition of similarity in terms of similarity transformations to
decide if they are similar; explain using similarity transformations the meaning of similarity
for triangles as the equality of all corresponding pairs of angles and the proportionality of all
corresponding pairs of sides.
G-SRT.A.3
Use the properties of similarity transformations to establish the AA criterion for two
triangles to be similar.
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NYS COMMON CORE MATHEMATICS CURRICULUM
Module Overview
M2
GEOMETRY
Prove theorems involving similarity.
G-SRT.B.4
Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle
divides the other two proportionally, and conversely; the Pythagorean Theorem proved using
triangle similarity.
G-SRT.B.5
Use congruence and similarity criteria for triangles to solve problems and to prove
relationships in geometric figures.
Define trigonometric ratios and solve problems involving right triangles.
G-SRT.C.6
Understand that by similarity, side ratios in right triangles are properties of the angles in the
triangle, leading to definitions of trigonometric ratios for acute angles.
G-SRT.C.7
Explain and use the relationship between the sine and cosine of complementary angles.
G-SRT.C.8
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied
problems.★
Apply geometric concepts in modeling situations.
G-MG.A.1
Use geometric shapes, their measures, and their properties to describe objects (e.g.,
modeling a tree trunk or a human torso as a cylinder).★
G-MG.A.3
Apply geometric methods to solve design problems (e.g., designing an object or structure to
satisfy physical constraints or minimize cost; working with typographic grid systems based
on ratios).★
Extension Standards
Apply trigonometry to general triangles.
G-SRT.D.9
(+) Derive the formula 𝐴𝐴 = 1/2 𝑎𝑎𝑎𝑎 sin(𝐶𝐶) for the area of a triangle by drawing an auxiliary
line from a vertex perpendicular to the opposite side.
G-SRT.D.10 (+) Prove the Laws of Sines and Cosines and use them to solve problems.
G-SRT.D.11 (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown
measurements in right and non-right triangles (e.g., surveying problems, resultant forces).
Foundational Standards
Draw, construct, and describe geometrical figures and describe the relationships between
them.
7.G.A.1
Solve problems involving scale drawings of geometric figures, including computing actual
lengths and areas from a scale drawing and reproducing a scale drawing at a different scale.
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Module Overview
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Understand congruence and similarity using physical models, transparencies, or geometry
software.
8.G.A.3
Describe the effect of dilations, translations, rotations, and reflections on two-dimensional
figures using coordinates.
8.G.A.4
Understand that a two-dimensional figure is similar to another if the second can be obtained
from the first by a sequence of rotations, reflections, translations, and dilations; given two
similar two-dimensional figures, describe a sequence that exhibits the similarity between
them.
8.G.A.5
Use informal arguments to establish facts about the angle sum and exterior angle of
triangles, about the angles created when parallel lines are cut by a transversal, and the
angle-angle criterion for similarity of triangles. For example, arrange three copies of the
same triangle so that the sum of the three angles appears to form a line, and give an
argument in terms of transversals why this is so.
Focus Standards for Mathematical Practice
MP.3
Construct viable arguments and critique the reasoning of others. Critical to this module is
the need for dilations in order to define similarity. In order to understand dilations fully, the
proofs in Lessons 4 and 5 to establish the triangle side splitter and the dilation theorems
require students to build arguments based on definitions and previously established results.
This is also apparent in Lessons 7, 8, and 9, when the properties of dilations are being
proven. Though there are only a handful of facts students must point to in order to create
arguments, how students reason with these facts determine if their arguments actually
establish the properties. It is essential to communicate effectively and purposefully.
MP.7
Look for and make use of structure. Much of the reasoning in Module 2 centers around the
interaction between figures and dilations. It is unsurprising, then, that students must pay
careful attention to an existing structure and how it changes under a dilation, for example,
why it is that dilating the key points of a figure by the ratio method results in the dilation of
the segments that join them. The math practice also ties into the underlying idea of
trigonometry: how to relate the values of corresponding ratio lengths between similar right
triangles and how the value of a trigonometric ratio hinges on a given acute angle within a
right triangle.
Terminology
New or Recently Introduced Terms

Cosine (Let 𝜃𝜃 be the angle measure of an acute angle of the right triangle. The cosine of 𝜃𝜃 of a right
triangle is the value of the ratio of the length of the adjacent side (denoted adj) to the length of the
hypotenuse (denoted hyp). As a formula, cos 𝜃𝜃 = adj/hyp.)
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NYS COMMON CORE MATHEMATICS CURRICULUM
Module Overview
M2
GEOMETRY






Dilation (For 𝑟𝑟 > 0, a dilation with center 𝐶𝐶 and scale factor 𝑟𝑟 is a transformation 𝐷𝐷𝐶𝐶,𝑟𝑟 of the plane
defined as follows:
1. For the center 𝐶𝐶, 𝐷𝐷𝐶𝐶,𝑟𝑟 (𝐶𝐶) = 𝐶𝐶, and
2. For any other point 𝑃𝑃, 𝐷𝐷𝐶𝐶,𝑟𝑟 (𝑃𝑃) is the point 𝑄𝑄 on �����⃗
𝐶𝐶𝐶𝐶 so that 𝐶𝐶𝐶𝐶 = 𝑟𝑟 ∙ 𝐶𝐶𝐶𝐶.)
Sides of a Right Triangle (The hypotenuse of a right triangle is the side opposite the right angle; the
other two sides of the right triangle are called the legs. Let 𝜃𝜃 be the angle measure of an acute angle
of the right triangle. The opposite side is the leg opposite that angle. The adjacent side is the leg
that is contained in one of the two rays of that angle (the hypotenuse is contained in the other ray of
the angle).)
Similar (Two figures in a plane are similar if there exists a similarity transformation taking one figure
onto the other figure. A congruence is a similarity with scale factor 1. It can be shown that a
similarity with scale factor 1 is a congruence.)
Similarity Transformation (A similarity transformation (or similarity) is a composition of a finite
number of dilations or basic rigid motions. The scale factor of a similarity transformation is the
product of the scale factors of the dilations in the composition; if there are no dilations in the
composition, the scale factor is defined to be 1. A similarity is an example of a transformation.)
Sine (Let 𝜃𝜃 be the angle measure of an acute angle of the right triangle. The sine of 𝜃𝜃 of a right
triangle is the value of the ratio of the length of the opposite side (denoted opp) to the length of the
hypotenuse (denoted hyp). As a formula, sin 𝜃𝜃 = opp/hyp.)
Tangent (Let 𝜃𝜃 be the angle measure of an acute angle of the right triangle. The tangent of 𝜃𝜃 of a
right triangle is the value of the ratio of the length of the opposite side (denoted opp) to the length
Note that in Algebra II, sine, cosine, and tangent are thought of as functions whose domains are
subsets of the real numbers; they are not considered as values of ratios. Thus, in Algebra II, the
values of these functions for a given 𝜃𝜃 are notated as sin(𝜃𝜃), cos(𝜃𝜃), and tan(𝜃𝜃) using function
notation (i.e., parentheses are included).
Familiar Terms and Symbols 2







2These
Composition
Dilation
Pythagorean theorem
Rigid motions
Scale drawing
Scale factor
Slope
are terms and symbols students have seen previously.
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Module Overview
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Suggested Tools and Representations

Compass and straightedge
Assessment Summary
Assessment Type
Format
Mid-Module
After Topic C
Constructed response with rubric
G-SRT.A.1, G-SRT.A.2,
G-SRT.A.3, G-SRT.B.4,
G-SRT.B.5, G-MG.A.1,
G-MG.A.3
End-of-Module
After Topic E
Constructed response with rubric
G-SRT.B.4, G-SRT.B.5,
G-SRT.C.6, G-SRT.C.7,
G-SRT.C.8
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New York State Common Core
Mathematics Curriculum
GEOMETRY • MODULE 2
Topic A
Scale Drawings
G-SRT.A.1, G-SRT.B.4, G-MG.A.3
Focus Standards:
G-SRT.A.1
Instructional Days:
Verify experimentally the properties of dilations given by a center and a scale factor:
a.
A dilation takes a line not passing through the center of the dilation to a parallel
line, and leaves a line passing through the center unchanged.
b.
The dilation of a line segment is longer or shorter in the ratio given by the scale
factor.
G-SRT.B.4
Prove theorems about triangles. Theorems include: a line parallel to one side of a
triangle divides the other two proportionally, and conversely; the Pythagorean Theorem
proved using triangle similarity.
G-MG.A.3
Apply geometric methods to solve design problems (e.g., designing an object or
structure to satisfy physical constraints or minimize cost; working with typographic grid
systems based on ratios).★
5
Lesson 1:
Scale Drawings (P) 1
Lesson 2:
Making Scale Drawings Using the Ratio Method (P)
Lesson 3:
Making Scale Drawings Using the Parallel Method (P)
Lesson 4:
Comparing the Ratio Method with the Parallel Method (S)
Lesson 5:
Scale Factors (S)
Students embark on Topic A with a brief review of scale drawings and scale factor, which they last studied in
Grades 7 and 8. In Lesson 1, students recall the properties of a well-scaled drawing and practice creating
scale drawings using basic construction techniques. Lessons 2 and 3 explore systematic techniques for
creating scale drawings. With the ratio method, students dilate key points of a figure according to the scale
factor to produce a scale drawing (G-SRT.A.1). Note that exercises within Lesson 2 where students apply the
ratio method to solve design problems relate to the modeling standard G-MG.A.3. With the parallel method,
students construct sides parallel to corresponding sides of the original figure to create a scale drawing.
Lesson 4 is an examination of these two methods, with the goal of understanding why the methods produce
identical drawings. The outcome of this comparison is the triangle side splitter theorem, which states that a
segment splits two sides of a triangle proportionally if and only if it is parallel to the third side (G-SRT.B.4).
1Lesson
Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
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NYS COMMON CORE MATHEMATICS CURRICULUM
Topic A
M2
GEOMETRY
This theorem is then used in Lesson 5 to establish the dilation theorem: A dilation from a center 𝑂𝑂 maps a
𝑃𝑃𝑃𝑃 and 𝑟𝑟 ≠ 1,
segment 𝑃𝑃𝑃𝑃 to a segment 𝑃𝑃′𝑄𝑄′ so that 𝑃𝑃′ 𝑄𝑄 ′ = 𝑟𝑟 ⋅ 𝑃𝑃𝑄𝑄; additionally, if 𝑂𝑂 is not contained in ⃖����⃗
′
′
⃖�������⃗
⃖����⃗
then 𝑃𝑃𝑃𝑃||𝑃𝑃 𝑄𝑄 .
As opposed to work done in Grade 8 on dilations, where students observed how dilations behaved and
experimentally verified properties of dilations by examples, high school Geometry is anchored in explaining
why these properties are true by reasoned argument. Grade 8 content focused on what was going on, while
high school Geometry content focuses on explaining why it occurs. This is particularly true in Lessons 4 and 5,
where students rigorously explain their explorations of dilations using the ratio and parallel methods to build
arguments that establish the triangle side splitter and dilation theorems (MP.3).
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Lesson 1
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Lesson 1: Scale Drawings
Student Outcomes

Students review properties of scale drawings and are able to create them.
Lesson Notes
Lesson 1 reviews the properties of a scale drawing before studying the relationship between dilations and scale drawings
in Lessons 2 and 3. Students focus on scaling triangles using construction tools and skills learned in Module 1. The
lesson begins by exploring how to scale images using common electronics. After students work on scaling triangles given
various pieces of initial information, the lesson is tied together by showing how triangle scaling can be used in
programming a phone to scale a complex image.
Note that students first studied scale drawings in Grade 7 (Module 1 Lessons 16–22). Teachers may need to modify the
exercises to include the use of graph paper, patty paper, and geometry software (e.g., freely available GeoGebra) to
make the ideas accessible.
Classwork
Opening (2 minutes)

A common feature on cell phones and tablets is the ability to scale, that is, to enlarge or reduce an image by
putting a thumb and index finger to the screen and making a pinching (to reduce) or spreading movement (to
enlarge) as shown in the diagram below.

Notice that as the fingers move outward on the screen (shown on the right), the image of the puppy is
enlarged on the screen.

How did the code for this feature get written? What general steps must the code dictate?

Today we review a concept that is key to tackling these questions.
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Lesson 1
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Opening Exercise (2 minutes)
Opening Exercise
Above is a picture of a bicycle. Which of the images below appears to be a well-scaled image of the original? Why?
Only the third image appears to be a well-scaled image since the image is in proportion to the original.
As mentioned in the Lesson Notes, students have seen scale drawings in Grades 4 and 7. The Opening Exercise is kept
brief to reintroduce the idea of what it means to be well scaled without going into great depth yet.
After the Opening Exercise, refer to the Opening, and re-pose the initial intent. Several questions are provided to help
illustrate the pursuit of the lesson. The expectation is not to answer these questions now but to keep them in mind as
the lesson progresses.

How did the code to scale images get written? What kinds of instructions guide the scaling process? What
steps take an original figure to a scale drawing? How is the program written to make sure that images are well
scaled and not distorted?
To help students answer these questions, the problem is simplified by examining the process of scaling a simpler figure:
a triangle. After tackling this simpler problem, it is possible to revisit the more complex images.
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Lesson 1
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Example 1 (8 minutes)
Example 1 provides students with a triangle and a scale factor. Students use a compass and straightedge to create the
scale drawing of the triangle.
Scaffolding:
Example 1
Use construction tools to create a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 with a scale factor of 𝒓𝒓 = 𝟐𝟐.
Before students begin creating the scale drawing, review the following.

Recall that the scale factor 𝑟𝑟 is the ratio of any length in a scale drawing relative
to its corresponding length in the original figure. A scale factor 𝑟𝑟 > 1 results in
an enlargement of the original figure. A scale factor of 0 < 𝑟𝑟 < 1 results in a
reduction of the original figure.
 One way to facilitate
Example 1 is to use graph
paper, with △ 𝐴𝐴𝐴𝐴𝐴𝐴 located
at 𝐴𝐴(−3, −2), 𝐵𝐵(5, −2),
and 𝐶𝐶(−3,6).
 An alternative challenge is
1
2
to use a scale of 𝑟𝑟 = .
For further background information, refer to Grade 7 (Module 1 Lesson 17).

Since we know that a scale drawing can be created without concern for location, a scale drawing can be done
in two ways: (1) by drawing it so that one vertex coincides with its corresponding vertex, leaving some overlap
between the original triangle and the scale drawing and (2) by drawing it completely independent of the
original triangle. Two copies of the original triangle have been provided for you.
Solution 1: Draw ������⃗
𝑨𝑨𝑨𝑨. To determine 𝑩𝑩′, adjust the compass to the length of 𝑨𝑨𝑨𝑨. Then reposition the compass so that the
����; label the intersection with 𝑨𝑨𝑨𝑨
������⃗ as 𝑩𝑩′. 𝑪𝑪′ is determined in a similar manner.
point is at 𝑩𝑩, and mark off the length of 𝑨𝑨𝑨𝑨
Join 𝑩𝑩′ to 𝑪𝑪′.
MP.5
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Lesson 1
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
������⃗. Label one end as 𝑨𝑨′. Adjust the compass to
Solution 2: Draw a segment that will be longer than double the length of 𝑨𝑨𝑨𝑨
����, and mark off two consecutive such lengths along the segment, and label the endpoint as 𝑩𝑩′. Copy ∠𝑨𝑨.
the length of 𝑨𝑨𝑨𝑨
Determine 𝑪𝑪′ along the �����⃗
𝑨𝑨𝑨𝑨 in the same way as 𝑩𝑩′. Join 𝑩𝑩′ to 𝑪𝑪′.
MP.5


Why do both solutions yield congruent triangles? Both triangles begin with the same criteria: Two pairs of
sides that are equal in length and a pair of included angles that are equal in measurement. By SAS, a unique
triangle is determined. Since △ 𝐴𝐴𝐴𝐴𝐴𝐴 was scaled by a factor of 2 in each case, the method in which we scale
does not change the outcome; that is, we have a triangle with the same dimensions whether we position it on
top of or independent of the original triangle.
���� and ������
Regardless of which solution method you used, measure the length of 𝐵𝐵𝐵𝐵
𝐵𝐵′𝐶𝐶′. What do you notice?


������
𝐵𝐵′𝐶𝐶′ is twice the length of ����
𝐵𝐵𝐵𝐵 .
Now measure the angles ∠𝐵𝐵, ∠𝐶𝐶, ∠𝐵𝐵′, and ∠𝐶𝐶′. What do you notice?

The measures of ∠𝐵𝐵 and ∠𝐵𝐵′ are the same, as are ∠𝐶𝐶 and ∠𝐶𝐶′.
Discussion (3 minutes)

What are the properties of a well-scaled drawing of a figure?


What is the term for the constant of proportionality by which all lengths are scaled in a well-scaled drawing?


A well-scaled drawing of a figure is one where corresponding angles are equal in measure, and
corresponding lengths are all in the same proportion.
The scale factor is the constant of proportionality.
If somewhere else on your paper you created the scale drawing in Example 1 but oriented it at a different
angle, would the drawing still be a scale drawing?

Yes. The orientation of the drawing does not change the fact that it is a scale drawing; the properties
of a scale drawing concern only lengths and relative angles.
Reinforce this by considering the steps of Solution 2 in Example 1. The initial segment can be drawn anywhere, and the
steps following can be completed as is. Ensure that students understand and rehearse the term orientation, and record
a student-friendly definition.
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
Return to the three images of the original bicycle. Why are the first two images classified as not well-scaled?

The corresponding angles are not equal in measurement, and the corresponding lengths are not in
constant proportion.
Exercise 1 (4 minutes)
Students scale a triangle by a factor of 𝑟𝑟 = 3 in Exercise 1. Either of the above solution methods is acceptable. As
students work on Exercise 1, take time to circulate and check for understanding. Note that teachers may choose to
provide graph paper and have students create scale drawings on it.
Exercise 1
1.
Use construction tools to create a scale drawing of △ 𝑫𝑫𝑫𝑫𝑫𝑫 with a scale factor of 𝒓𝒓 = 𝟑𝟑. What properties does your
scale drawing share with the original figure? Explain how you know.
By measurement, I can see that each side is three times the length of the corresponding side of the original figure
and that all three angles are equal in measurement to the three corresponding angles in the original figure.
Make sure students understand that any of these diagrams are acceptable solutions.
A solution where �����
𝑬𝑬𝑬𝑬′ and �����
𝑬𝑬𝑬𝑬′ are drawn first.
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A solution where �����
𝑭𝑭𝑭𝑭′ and �����
𝑭𝑭𝑭𝑭′ are drawn first.
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����� and 𝑫𝑫𝑫𝑫′
����� are drawn first.
A solution where 𝑫𝑫𝑫𝑫′
Example 2 (4 minutes)
1
2
Example 2 provides a triangle and a scale factor of 𝑟𝑟 = . Students use a compass to locate the midpoint.
Example 2
Use construction tools to create a scale drawing of △ 𝑿𝑿𝑿𝑿𝑿𝑿 with a scale factor of 𝒓𝒓 =

𝟏𝟏
.
𝟐𝟐
Which construction technique have we learned that can be used in this question that was not used in the
previous two problems?

We can use the construction to determine the perpendicular bisector to locate the midpoint of two
sides of △ 𝑋𝑋𝑋𝑋𝑋𝑋.
Scaffolding:
For students struggling with
constructions, consider having
them measure the lengths of
two sides and then determine
the midpoints.
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GEOMETRY
As the solutions to Exercise 1 showed, the constructions can be done on other sides of the triangle (i.e., the
���� are acceptable places to start.)
perpendicular bisectors of ����
𝑌𝑌𝑌𝑌 and 𝑋𝑋𝑋𝑋
Exercises 2–4 (13 minutes)
Have students complete Exercise 2 and, if time allows, go on to Exercises 3 and 4. In Exercise 2, using a scale factor of
𝑟𝑟 =
1
4
1
1
is a natural progression following the use of a scale factor of 𝑟𝑟 = in Example 1. Prompt students to consider how
4
2
1
relates to . They should recognize that the steps of the construction in Exercise 2 are similar to those in Example 1.
2
Exercises 2–4
2.
𝟏𝟏
𝟒𝟒
Use construction tools to create a scale drawing of △ 𝑷𝑷𝑷𝑷𝑷𝑷 with a scale factor of 𝒓𝒓 = .
What properties do the scale drawing and the original figure share? Explain how you know.
Scaffolding:
For students who are ready for
them to use a scale factor of
3
4
𝑟𝑟 = .
By measurement, I can see that all three sides are each one-quarter the lengths of the corresponding sides of the
original figure, and all three angles are equal in measurement to the three corresponding angles in the original
figure.
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Lesson 1
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GEOMETRY
3.
Triangle 𝑬𝑬𝑬𝑬𝑬𝑬 is provided below, and one angle of scale drawing △ 𝑬𝑬′𝑭𝑭′𝑮𝑮′ is also provided. Use construction tools to
complete the scale drawing so that the scale factor is 𝒓𝒓 = 𝟑𝟑. What properties do the scale drawing and the original
figure share? Explain how you know.
Extend either ray from 𝑮𝑮′. Use the compass to mark off a length equal to 𝟑𝟑𝟑𝟑𝟑𝟑 on one ray and a length equal to 𝟑𝟑𝟑𝟑𝟑𝟑
on the other. Label the ends of the two lengths 𝑬𝑬′ and 𝑭𝑭′, respectively. Join 𝑬𝑬′ to 𝑭𝑭′.
By measurement, I can see that each side is three times the length of the corresponding side of the original figure
and that all three angles are equal in measurement to the three corresponding angles in the original figure.
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4.
Triangle 𝑨𝑨𝑨𝑨𝑨𝑨 is provided below, and one side of scale drawing △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ is also provided. Use construction tools to
complete the scale drawing and determine the scale factor.
Scaffolding:
One possible solution: We can copy ∠𝑨𝑨 and ∠𝑪𝑪 at points 𝑨𝑨′ and 𝑪𝑪′ so that the new rays
intersect as shown and call the intersection point 𝑩𝑩′. By measuring, we can see that
𝑨𝑨′ 𝑪𝑪′ = 𝟐𝟐𝟐𝟐𝟐𝟐, 𝑨𝑨′ 𝑩𝑩′ = 𝟐𝟐𝟐𝟐𝟐𝟐, and 𝑩𝑩′ 𝑪𝑪′ = 𝟐𝟐𝟐𝟐𝟐𝟐. We already know that 𝒎𝒎∠𝑨𝑨′ = 𝒎𝒎∠𝑨𝑨 and
𝒎𝒎∠𝑪𝑪′ = 𝒎𝒎∠𝑪𝑪. By the triangle sum theorem, 𝒎𝒎∠𝑩𝑩′ = 𝒎𝒎∠𝑩𝑩.
 If students struggle with
constructing an angle of
equal measure, consider
allowing them to use a
protractor to measure
angles 𝐴𝐴 and 𝐶𝐶, and draw
angles at 𝐴𝐴′ and 𝐶𝐶′,
respectively, with equal
measures. This alleviates
time constraints; however,
know that constructing an
angle is a necessary skill to
remediation.
 Use patty paper or
geometry software to
allow students to focus on
the concept development.
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Discussion (3 minutes)

In the last several exercises, we constructed or completed scale drawings of
question that began the lesson.

What does the work we did with scaled triangles have to do with
understanding the code that is written to tell a phone or a computer how to
enlarge or reduce an image? Here is one possible way.

Consider the following figure, which represents an image or perhaps a photo.
A single point 𝑃𝑃 is highlighted in the image, which can easily be imagined to
be one of many points of the image (e.g., it could be just a single point of the
bicycle in the Opening Exercise).

If we know by how much we want to enlarge or reduce the image (i.e., the
scale factor), we can use what we know about scaling triangles to locate where
this point ends up in the scale drawing of the image.

For example, if we were to place our thumb and index finger on the points 𝑇𝑇
and 𝐼𝐼, respectively, and make the zoom-in movement so that our thumb
and index finger end up at 𝑇𝑇′ and 𝐼𝐼′, respectively, the scale factor that
corresponds to the zoom-in motion dictates where 𝑃𝑃′ is located.

Therefore, we can generalize how the code to enlarge or reduce an image
on a phone or a tablet is written. For every point 𝑃𝑃 in an image, a triangle
can be formed with vertices 𝑃𝑃, 𝑇𝑇, and 𝐼𝐼. Since we are able to scale triangles,
we are then also able to scale entire images, point by point. In fact, we can
use this process not just for the code to program electronics but also to
scale a complex image by hand if we so wished.
Closing (1 minute)

What are the key properties of a scale drawing relative to its original figure?


If we were to take any of the scale drawings in our examples and place them in a different location or rotate
them on our paper, would it change the fact that the drawing is still a scale drawing?


There are two properties of a scale drawing of a figure: Corresponding angles are equal in
measurement, and corresponding lengths are proportional in measurement.
No, the properties of a scale drawing have to do with lengths and relative angles, not location or
orientation.
Provided a triangle and a scale factor or a triangle and one piece of the scale drawing of the triangle, it is
possible to create a complete scale drawing of the triangle using a compass and straightedge. No matter
which method is used to create the scale drawing, we rely on triangle congruence criteria to ensure that a
unique triangle is determined.
Exit Ticket (5 minutes)
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Name
Date
Lesson 1: Scale Drawings
Exit Ticket
Triangle 𝐴𝐴𝐴𝐴𝐴𝐴 is provided below, and one side of scale drawing △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ is also provided. Use construction tools to
complete the scale drawing and determine the scale factor. What properties do the scale drawing and the original figure
share? Explain how you know.
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Exit Ticket Sample Solutions
Triangle 𝑨𝑨𝑨𝑨𝑨𝑨 is provided below, and one side of scale drawing △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ is also provided. Use construction tools to
complete the scale drawing and determine the scale factor. What properties do the scale drawing and the original figure
share? Explain how you know.
One possible solution: Since the scale drawing will clearly be a reduction, use the compass to mark the number of lengths
������ is determined to be
equal to the length of ������
𝑨𝑨′𝑩𝑩′ along ����
𝑨𝑨𝑨𝑨. Once the length of 𝑨𝑨′𝑪𝑪′
𝟏𝟏
𝟐𝟐
����, use the compass to
the length of 𝑨𝑨𝑨𝑨
����. Construct circles with radii of lengths
���� and half the length of 𝑩𝑩𝑩𝑩
find a length that is half the length of 𝑨𝑨𝑨𝑨
𝟏𝟏
𝟐𝟐
𝑨𝑨𝑨𝑨 and
from the 𝑨𝑨′ and 𝑩𝑩 , respectively, to determine the location of 𝑪𝑪′, which is at the intersection of the two circles.
′
By measurement, I can see that each side is
𝟏𝟏
𝟐𝟐
𝟏𝟏
𝟐𝟐
𝑩𝑩𝑩𝑩
the length of the corresponding side of the original figure and that all three
angles are equal in measurement to the three corresponding angles in the original figure.
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Problem Set Sample Solutions
1.
Use construction tools to create a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 with a scale factor of 𝒓𝒓 = 𝟑𝟑.
2.
Use construction tools to create a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 with a scale factor of 𝒓𝒓 = .
𝟏𝟏
𝟐𝟐
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3.
Triangle 𝑬𝑬𝑬𝑬𝑬𝑬 is provided below, and one angle of scale drawing △ 𝑬𝑬′𝑭𝑭′𝑮𝑮′ is also provided. Use construction tools to
complete a scale drawing so that the scale factor is 𝒓𝒓 = 𝟐𝟐.
4.
Triangle 𝑴𝑴𝑴𝑴𝑴𝑴 is provided below, and one angle of scale drawing △ 𝑴𝑴′𝑻𝑻′𝑪𝑪′ is also provided. Use construction tools
𝟏𝟏
𝟒𝟒
to complete a scale drawing so that the scale factor is 𝒓𝒓 = .
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5.
Triangle 𝑨𝑨𝑨𝑨𝑨𝑨 is provided below, and one side of scale drawing △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ is also provided. Use construction tools to
complete the scale drawing and determine the scale factor.
The ratio of 𝑩𝑩′ 𝑪𝑪′ : 𝑩𝑩𝑩𝑩 is 𝟓𝟓: 𝟏𝟏, so the scale factor is 𝟓𝟓.
6.
Triangle 𝑿𝑿𝑿𝑿𝑿𝑿 is provided below, and one side of scale drawing △ 𝑿𝑿′𝒀𝒀′𝒁𝒁′ is also provided. Use construction tools to
complete the scale drawing and determine the scale factor.
𝟏𝟏
The ratio of 𝑿𝑿′ 𝒁𝒁′ : 𝑿𝑿𝑿𝑿 is 𝟏𝟏: 𝟐𝟐, so the scale factor is .
𝟐𝟐
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7.
Quadrilateral 𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮 is a scale drawing of quadrilateral 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 with scale factor 𝒓𝒓. Describe each of the following
statements as always true, sometimes true, or never true, and justify your answer.
a.
𝑨𝑨𝑨𝑨 = 𝑮𝑮𝑮𝑮
Sometimes true, but only if 𝒓𝒓 = 𝟏𝟏.
b.
𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 = 𝒎𝒎∠𝑮𝑮𝑮𝑮𝑮𝑮
Always true because ∠𝑮𝑮𝑮𝑮𝑮𝑮 corresponds to ∠𝑨𝑨𝑨𝑨𝑨𝑨 in the original drawing, and angle measures are preserved
in scale drawings.
c.
𝑨𝑨𝑨𝑨
𝑮𝑮𝑮𝑮
=
𝑩𝑩𝑩𝑩
𝑯𝑯𝑯𝑯
Always true because distances in a scale drawing are equal to their corresponding distances in the original
drawing times the scale factor 𝒓𝒓, so
d.
𝑨𝑨𝑨𝑨
𝑮𝑮𝑮𝑮
=
𝑨𝑨𝑨𝑨
𝒓𝒓(𝑨𝑨𝑨𝑨)
=
𝟏𝟏
𝒓𝒓
and
𝑩𝑩𝑩𝑩
𝑯𝑯𝑯𝑯
=
𝑩𝑩𝑩𝑩
𝒓𝒓(𝑩𝑩𝑩𝑩)
𝟏𝟏
= .
𝒓𝒓
𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏 (𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮) = 𝒓𝒓 ∙ 𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨)
Always true because the distances in a scale drawing are equal to their corresponding distances in the original
drawing times the scale factor 𝒓𝒓, so
𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏(𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮) = 𝑮𝑮𝑮𝑮 + 𝑯𝑯𝑯𝑯 + 𝑰𝑰𝑰𝑰 + 𝑱𝑱𝑱𝑱
𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏(𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮) = 𝒓𝒓(𝑨𝑨𝑨𝑨) + 𝒓𝒓(𝑩𝑩𝑩𝑩) + 𝒓𝒓(𝑪𝑪𝑪𝑪) + 𝒓𝒓(𝑫𝑫𝑫𝑫)
𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏(𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮) = 𝒓𝒓(𝑨𝑨𝑨𝑨 + 𝑩𝑩𝑩𝑩 + 𝑪𝑪𝑪𝑪 + 𝑫𝑫𝑫𝑫)
𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏(𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮) = 𝒓𝒓 ∙ 𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨).
e.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 (𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮) = 𝒓𝒓 ∙ 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) where 𝒓𝒓 ≠ 𝟏𝟏
Never true because the area of a scale drawing is related to the area of the original drawing by the factor 𝒓𝒓𝟐𝟐 .
The scale factor 𝒓𝒓 > 𝟎𝟎 and 𝒓𝒓 ≠ 𝟏𝟏, so 𝒓𝒓 ≠ 𝒓𝒓𝟐𝟐 .
f.
𝒓𝒓 < 𝟎𝟎
Never true in a scale drawing because any distance in the scale drawing would be negative as a result of the
scale factor and, thus, cannot be drawn since distance must always be positive.
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Lesson 2
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GEOMETRY
Lesson 2: Making Scale Drawings Using the Ratio Method
Student Outcomes

Students create scale drawings of polygonal figures by the ratio method.

Given a figure and a scale drawing from the ratio method, students answer questions about the scale factor
and the center.
Lesson Notes
In Lesson 1, students created scale drawings in any manner they wanted, as long as the scale drawings met the criteria of
well-scaled drawings. Lesson 2 introduces students to a systematic way of creating a scale drawing: the ratio method,
which relies on dilations. Students dilate the vertices of the provided figure and verify that the resulting image is in fact
a scale drawing of the original. It is important to note that we approach the ratio method as a method that strictly
dilates the vertices. After some practice with the ratio method, students dilate a few other points of the polygonal figure
and notice that they lie on the scale drawing. They may speculate that the dilation of the entire figure is the scale
drawing, but this fact is not generalized in Lesson 2.
Note that students need rulers, protractors, and calculators for this lesson.
Scaffolding:
Classwork
Providing an example of a dilation
(such as in the image below) may
dilations.
Opening Exercise (2 minutes)
Opening Exercise
Based on what you recall from Grade 8, describe what a dilation is.
Student responses will vary; students may say that a dilation results in a reduction or an
enlargement of the original figure or that corresponding side lengths are proportional in
length, and corresponding angles are equal in measure. The objective is to prime them for
an in-depth conversation about dilations; take one or two responses and move on.
Discussion (5 minutes)

In Lesson 1, we reviewed the properties of a scale drawing and created scale drawings of triangles using
construction tools. We observed that as long as our scale drawings had angles equal in measure to the
corresponding angles of the original figure and lengths in constant proportion to the corresponding lengths of
the original figure, the location and orientation of our scale drawing did not concern us.

In Lesson 2, we use a systematic process of creating a scale drawing called the ratio method. The ratio method
dilates the vertices of the provided polygonal figure. The details that we recalled in the Opening Exercise are
characteristics that are consistent with scale drawings too. We will verify that the resulting image created by
dilating these key points is in fact a scale drawing.
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
Recall the definition of a dilation:
Definition
For 𝑟𝑟 > 0, a dilation with center 𝑂𝑂 and scale factor 𝑟𝑟 is
a transformation 𝐷𝐷𝑂𝑂,𝑟𝑟 of the plane defined as follows:
Characteristics
- Preserves angles
- Names a center and a scale factor
For the center 𝑂𝑂, 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝑂𝑂) = 𝑂𝑂, and
For any other point 𝑃𝑃, 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝑃𝑃) is the point 𝑃𝑃 on the ray
�����⃗
𝑂𝑂𝑂𝑂 so that |𝑂𝑂𝑂𝑂′| = 𝑟𝑟 ∙ |𝑂𝑂𝑂𝑂|.
Examples
Dilation
Non-Examples
- Rigid motions such as translations, rotations,
reflections
Note that students last studied dilations in Grade 8 Module 3. At that time, the notation
used was not the capital letter 𝐷𝐷, but the full word dilation. Students have since studied
rigid motion notation in Geometry Module 1 and should be familiar with the style of
notation presented here.

A dilation is a rule (a function) that moves points in the plane a specific distance
along the ray that originates from a center 𝑂𝑂. What determines the distance a
given point moves?


What can we tell about the scale factor of a dilation that pulls any point that is
different from the center toward the center 𝑂𝑂?



We know that the scale factor for a dilation where a point is pulled
toward the center must be 0 < 𝑟𝑟 < 1.
 Consider displaying a
poster with the definition
and notation of dilation.
 Ask students to draw some
examples to demonstrate
understanding, such as,
“Draw a segment 𝑂𝑂𝑂𝑂 on
segment using 𝐷𝐷𝑂𝑂,3 .”
What can we tell about the scale factor of a dilation that pushes all points, except the center, away from the
center 𝑂𝑂?


The location of the scaled point is determined by the scale factor and the
distance of the original point from the center.
Scaffolding:
The scale factor for a dilation where a point is pushed away from the center must be 𝑟𝑟 > 1.
A point, different from the center, that is unchanged in its location after a dilation must have a scale factor of
𝑟𝑟 = 1.
Scale factor is always a positive value, as we use it when working with distance. If we were to use negative
values for scale factor, we would be considering distance as a negative value, which does not make sense.
Hence, scale factor is always positive.
Example 1 (8 minutes)
Examples 1–2 demonstrate how to create a scale drawing using the ratio method. In this example, the ratio method is
1
2
used to dilate the vertices of a polygonal figure about center 𝑂𝑂, by a scale factor of 𝑟𝑟 = .
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
To use the ratio method to scale any figure, we must have a scale factor and center in order to dilate the
vertices of a polygonal figure.

In the steps below, we have a figure with center 𝑂𝑂 and a scale factor of 𝑟𝑟 = . What effect should we expect
1
2
this scale factor to have on the image of the figure?

Since the scale factor is a value less than one (but greater than zero), the image should be a reduction
of the original figure. Specifically, each corresponding length should be half of the original length.
Example 1
𝟏𝟏
𝟐𝟐
Create a scale drawing of the figure below using the ratio method about center 𝑶𝑶 and scale factor 𝒓𝒓 = .
Step 1. Draw a ray beginning at 𝑶𝑶 through each vertex of the figure.
𝟏𝟏
𝟐𝟐
Step 2. Dilate each vertex along the appropriate ray by scale factor 𝒓𝒓 = . Use the ruler to find the midpoint between 𝑶𝑶
and 𝑫𝑫 and then each of the other vertices. Label each respective midpoint with prime notation (e.g., 𝑫𝑫′).

Why are we locating the midpoint between 𝑂𝑂 and 𝐷𝐷?

The scale factor tells us that the distance of the scaled point should be half the distance from 𝑂𝑂 to 𝐷𝐷,
which is the midpoint of ����
𝑂𝑂𝑂𝑂 .
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Step 3. Join vertices in the way they are joined in the original figure (e.g., segment 𝑨𝑨′𝑩𝑩′ corresponds to segment 𝑨𝑨𝑨𝑨).

Does 𝐴𝐴′𝐵𝐵′𝐶𝐶′𝐷𝐷′𝐸𝐸′ look like a scale drawing? How can we verify whether 𝐴𝐴′𝐵𝐵′𝐶𝐶′𝐷𝐷′𝐸𝐸′ is really a scale drawing?

MP.5

Yes. We can measure each segment of the original and the scale drawing; the segments of 𝐴𝐴′𝐵𝐵′𝐶𝐶′𝐷𝐷′𝐸𝐸′
appear to be half as long as their corresponding counterparts in 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, and all corresponding angles
appear to be equal in measurement; the image is a reduction of the original figure.
It is important to notice that the scale factor for the scale drawing is the same as the scale factor for the
dilation.
Students may notice that in the triangle formed by the center and the endpoints of any
segment on the original figure, the dilated segment forms the mid-segment of the triangle.
Have students measure and confirm that the length of each segment in the scale drawing
is half the length of each segment in the original drawing and that the measurements of all
corresponding angles are equal. The quadrilateral 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 is a square, and all four angles
are 90° in measurement. The measurement of ∠𝐷𝐷 = 80°, and the measurements of
∠𝐶𝐶 and ∠𝐸𝐸 are both 50°. The measurements of the side lengths are not provided because
they differ from the images that appear in print form.
Scaffolding:
Teachers may want to consider
using patty paper as an
alternate means to measuring
angles with a protractor in the
interest of time.
Exercise 1 (5 minutes)
Scaffolding:
Exercise 1
1.
Create a scale drawing of the figure below using the ratio method about center 𝑶𝑶 and scale
𝟑𝟑
𝟒𝟒
factor 𝒓𝒓 = . Verify that the resulting figure is in fact a scale drawing by showing that
corresponding side lengths are in constant proportion and the corresponding angles are
equal in measurement.
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In preparing for this lesson,
consider whether the class has
time for each example and
exercise. If time is short,
consider moving from Example
1 to Example 2.
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Verification of the reduced figure should show that the length of each segment in the scale drawing is
𝟑𝟑
the length of
𝟒𝟒
𝟑𝟑
each segment in the original figure (e.g., 𝑨𝑨 𝑮𝑮 = (𝑨𝑨𝑨𝑨)). The angle measurements are as follows (when an angle
𝟒𝟒
′
′
could be interpreted as having two possible measurements, the smaller one was selected in all cases): 𝒎𝒎∠𝑨𝑨 = 𝟗𝟗𝟗𝟗°,
𝒎𝒎∠𝑩𝑩 = 𝟒𝟒𝟒𝟒°, 𝒎𝒎∠𝑪𝑪 = 𝟗𝟗𝟗𝟗°, 𝒎𝒎∠𝑫𝑫 = 𝟗𝟗𝟗𝟗°, 𝒎𝒎∠𝑬𝑬 = 𝟗𝟗𝟗𝟗°, 𝒎𝒎∠𝑭𝑭 = 𝟗𝟗𝟗𝟗°, and 𝒎𝒎∠𝑮𝑮 = 𝟒𝟒𝟒𝟒°.
Example 2 (7 minutes)
Example 2
a.
Create a scale drawing of the figure below using the ratio method about center 𝑶𝑶 and scale factor 𝒓𝒓 = 𝟑𝟑.
Step 1. Draw a ray beginning at 𝑶𝑶 through each vertex of the figure.
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Step 2. Use your ruler to determine the location of 𝑨𝑨′ on ������⃗
𝑶𝑶𝑶𝑶; 𝑨𝑨′ should be three times as far from 𝑶𝑶 as 𝑨𝑨.
Determine the locations of 𝑩𝑩′ and 𝑪𝑪′ in the same way along the respective rays.
Step 3. Draw the corresponding line segments (e.g., segment 𝑨𝑨′𝑩𝑩′ corresponds to segment 𝑨𝑨𝑨𝑨).

Does 𝐴𝐴′𝐵𝐵′𝐶𝐶′𝐷𝐷′ look like a scale drawing of 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴?


Yes
How can we verify whether 𝐴𝐴′𝐵𝐵′𝐶𝐶′𝐷𝐷′ is really a scale drawing of 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴?

We can measure each segment of the original and the scale drawing; the segments of 𝐴𝐴′𝐵𝐵′𝐶𝐶′𝐷𝐷′ should
be three times as long as their corresponding counterparts in 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, and all corresponding angles
should be equal in measurement; the image is an enlargement of the original figure.
Have students measure and confirm that the length of each segment in the scale drawing is three times the length of
each segment in the original drawing and that the measurements of all corresponding angles are equal. The
measurements of the angles in the figure are as follows: 𝑚𝑚∠𝐴𝐴 = 17°, 𝑚𝑚∠𝐵𝐵 = 134° (the smaller of the two possible
options of measuring the angle was selected, but either will do), 𝑚𝑚∠𝐶𝐶 = 22°, 𝑚𝑚∠𝐷𝐷 = 23°. Again, the measurements of
the side lengths are not provided because they differ from the images that appear in print form.
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b.
Locate a point 𝑿𝑿 so that it lies between endpoints 𝑨𝑨 and 𝑩𝑩 on segment 𝑨𝑨𝑨𝑨 of the original figure in part (a).
Use the ratio method to locate 𝑿𝑿′ on the scale drawing in part (a).
Sample response:

Consider that everyone in class could have chosen a different location for 𝑋𝑋 between points 𝐴𝐴 and 𝐵𝐵. What
does the result of part (b) imply?
The result of part (b) implies that all the points between 𝐴𝐴𝐴𝐴 are dilated to corresponding points
between points 𝐴𝐴′ and 𝐵𝐵′.


It is tempting to draw the conclusion that the dilation of the vertices is the same as the dilation of each
segment onto corresponding segments in the scale drawing. Even though this appears to be the case here, we
wait until later lessons to definitively show whether this is actually the case.
c.
Imagine a dilation of the same figure as in parts (a) and (b). What if the ray from the center passed through
two distinct points, such as 𝑩𝑩 and 𝑫𝑫? What does that imply about the locations of 𝑩𝑩′ and 𝑫𝑫′?
Both 𝑩𝑩′ and 𝑫𝑫′also lie on the same ray.
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Exercises 2–6 (11 minutes)
Exercises 2–6
2.
△ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ is a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 drawn by using the ratio method. Use your ruler to determine the location of
the center 𝑶𝑶 used for the scale drawing.
3.
Use the figure below with center 𝑶𝑶 and a scale factor of 𝒓𝒓 =
𝟓𝟓
to create a scale drawing. Verify that the resulting
𝟐𝟐
figure is in fact a scale drawing by showing that corresponding side lengths are in constant proportion and that the
corresponding angles are equal in measurement.
Verification of the enlarged figure should show that the length of each segment in the scale drawing is 𝟐𝟐. 𝟓𝟓 times the
length of each segment in the original figure (e.g., 𝑨𝑨′ 𝑩𝑩′ = 𝟐𝟐. 𝟓𝟓(𝑨𝑨𝑨𝑨)). The angle measurements are 𝒎𝒎∠𝑨𝑨 = 𝟗𝟗𝟗𝟗°,
𝒎𝒎∠𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝒎𝒎∠𝑪𝑪 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝒎𝒎∠𝑫𝑫 = 𝟏𝟏𝟏𝟏𝟏𝟏°, and 𝒎𝒎∠𝑬𝑬 = 𝟏𝟏𝟏𝟏𝟏𝟏°.
4.
Summarize the steps to create a scale drawing by the ratio method. Be sure to describe all necessary parameters to
use the ratio method.
To use the ratio method to create a scale drawing, the problem must provide a polygonal figure, a center 𝑶𝑶, and a
scale factor. To begin the ratio method, draw a ray that originates at 𝑶𝑶 and passes through each vertex of the
figure. We are dilating each vertex along its respective ray. Measure the distance between 𝑶𝑶 and a vertex and
multiply it by the scale factor. The resulting value is the distance away from 𝑶𝑶 at which the scaled point will be
located. Once all the vertices are dilated, they should be joined in the same way as they are joined in the original
figure.
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5.
A clothing company wants to print the face of the Statue of Liberty on a T-shirt. The length of the face from the top
of the forehead to the chin is 𝟏𝟏𝟏𝟏 feet, and the width of the face is 𝟏𝟏𝟏𝟏 feet. Given that a medium-sized T-shirt has a
length of 𝟐𝟐𝟐𝟐 inches and a width of 𝟐𝟐𝟐𝟐 inches, what dimensions of the face are needed to produce a scaled version
that will fit on the T-shirt?
a.
What shape would you use to model the face of the statue?
Answers may vary. Students may say triangle, rectangle, or circle.
b.
Knowing that the maximum width of the T-shirt is 𝟐𝟐𝟐𝟐 inches, what scale factor is needed to make the width
of the face fit on the shirt?
Answers may vary. Sample response shown below.
𝟏𝟏
𝟐𝟐𝟐𝟐
=
𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔
The width of the face on the T-shirt will need to be scaled to
c.
𝟏𝟏
𝟔𝟔
the size of the statue’s face.
What scale factor should be used to scale the length of the face? Explain.
Answers may vary. Students should respond that the scale factor identified in part (b) should be used for the
length.
To keep the length of the face proportional to the width, a scale factor of
d.
𝟏𝟏
𝟔𝟔
should be used.
Using the scale factor identified in part (c), what is the scaled length of the face? Will it fit on the shirt?
𝟏𝟏
(𝟐𝟐𝟐𝟐𝟐𝟐) = 𝟑𝟑𝟑𝟑
𝟔𝟔
The scaled length of the face would be 𝟑𝟑𝟑𝟑 inches. The length of the shirt is only 𝟐𝟐𝟐𝟐 inches, so the face will not
fit on the shirt.
e.
Identify the scale factor you would use to ensure that the face of the statue was in proportion and would fit
on the T-shirt. Identify the dimensions of the face that will be printed on the shirt.
𝟏𝟏
Answers may vary. Scaling by a factor of produces dimensions that are still too large to fit on the shirt. The
largest scale factor that could be used is
𝟐𝟐𝟐𝟐. 𝟓𝟓 inches.
f.
, producing a scaled width of 𝟏𝟏𝟏𝟏 inches and a scaled length of
𝟖𝟖
The T-shirt company wants the width of the face to be no smaller than 𝟏𝟏𝟏𝟏 inches. What scale factors could
be used to create a scaled version of the face that meets this requirement?
𝟏𝟏 𝟏𝟏 𝟏𝟏
Scale factors of , ,
,
𝟏𝟏
, or
𝟖𝟖 𝟗𝟗 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
g.
𝟕𝟕
𝟏𝟏
𝟏𝟏
𝟏𝟏𝟏𝟏
could be used to ensure the width of the face is no smaller than 𝟏𝟏𝟏𝟏 inches.
If it costs the company \$𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 for each square inch of print on a shirt, what is the maximum and minimum
costs for printing the face of the Statue of Liberty on one T-shirt?
The largest scaled face would have dimensions 𝟏𝟏𝟏𝟏 × 𝟐𝟐𝟐𝟐. 𝟓𝟓, meaning the print would cost approximately
\$𝟏𝟏. 𝟗𝟗𝟗𝟗 per shirt. The smallest scaled face would have dimensions 𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏, meaning the print would cost
\$𝟎𝟎. 𝟖𝟖𝟖𝟖 per shirt.
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GEOMETRY
6.
Create your own scale drawing using the ratio method. In the space below:
a.
Draw an original figure.
Scaffolding:
b.
Locate and label a center of dilation 𝑶𝑶.
Figures can be made as simple
or as complex as desired—a
triangle involves fewer
segments to keep track of than
a figure such as the arrow in
Exercise 1. Students should
work with a manageable figure
in the allotted time frame.
c.
d.
e.
Choose a scale factor 𝒓𝒓.
Describe your dilation using appropriate notation.
Complete a scale drawing using the ratio method.
Show all measurements and calculations to confirm that the new figure is a scale drawing.
Next, trace your original figure onto a fresh piece of paper. Trade the traced figure with a
partner. Provide your partner with the dilation information. Each partner should complete
the other’s scale drawing. When finished, check all work for accuracy against your answer
key.
MP.3
Answers will vary. Encourage students to check each other’s work and to discover the reason for any discrepancies
Closing (2 minutes)
Ask students to summarize the key points of the lesson. Additionally, consider asking students the following questions
independently in writing, to a partner, or to the whole class.



To create a scale drawing using the ratio method, each vertex of the original figure is dilated about the center
𝑂𝑂 by scale factor 𝑟𝑟. Once all the vertices are dilated, they are joined to each other in the same way as in the
original figure.
The scale factor tells us whether the scale drawing is being enlarged (𝑟𝑟 > 1) or reduced (0 < 𝑟𝑟 < 1).
How can it be confirmed that what is drawn by the ratio method is in fact a scale drawing?


By measuring the side lengths of the original figure and the scale drawing, we can establish whether
the corresponding sides are in constant proportion. We can also measure corresponding angles and
determine whether they are equal in measure. If the side lengths are in constant proportion, and the
corresponding angle measurements are equal, the new figure is in fact a scale drawing of the original.
It is important to note that though we have dilated the vertices of the figures for the ratio method, we do not
definitively know if each segment is dilated to the corresponding segment in the scale drawing. This remains
to be seen. We cannot be sure of this even if the scale drawing is confirmed to be a well-scaled drawing. We
learn how to determine this in the next few lessons.
Exit Ticket (5 minutes)
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Name
Date
Lesson 2: Making Scale Drawings Using the Ratio Method
Exit Ticket
One of the following images shows a well-scaled drawing of △ 𝐴𝐴𝐴𝐴𝐴𝐴 done by the ratio method; the other image is not a
well-scaled drawing. Use your ruler and protractor to make the necessary measurements and show the calculations that
determine which is a scale drawing and which is not.
Figure 1
Figure 2
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Exit Ticket Sample Solutions
One of the following images shows a well-scaled drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 done by the ratio method; the other image is not a
well-scaled drawing. Use your ruler and protractor to make the necessary measurements and show the calculations that
determine which is a scale drawing and which is not.
Figure 1
Figure 2
Figure 1 shows the true scale drawing.
△ 𝑨𝑨𝑨𝑨𝑨𝑨 angle measurements: 𝒎𝒎∠𝑨𝑨 = 𝟐𝟐𝟐𝟐°, 𝒎𝒎∠𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝒎𝒎∠𝑪𝑪 = 𝟓𝟓𝟓𝟓°, which are the same for △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ in Figure 1. The
value of the ratios of 𝑶𝑶𝑨𝑨′ : 𝑶𝑶𝑶𝑶, 𝑶𝑶𝑶𝑶′ : 𝑶𝑶𝑶𝑶, and 𝑶𝑶𝑶𝑶′ : 𝑶𝑶𝑶𝑶 are the same.
△ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ in Figure 2 has angle measurements 𝒎𝒎∠𝑨𝑨′ = 𝟐𝟐𝟐𝟐°, 𝒎𝒎∠𝑩𝑩′ = 𝟗𝟗𝟗𝟗°, 𝒎𝒎∠𝑪𝑪′ = 𝟔𝟔𝟔𝟔°, and the value of the ratios of
𝑶𝑶𝑶𝑶′ : 𝑶𝑶𝑶𝑶, 𝑶𝑶𝑶𝑶′ : 𝑶𝑶𝑶𝑶, and 𝑶𝑶𝑪𝑪′ : 𝑶𝑶𝑶𝑶 are not the same.
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Problem Set Sample Solutions
Considering the significant construction needed for the Problem Set questions, teachers may feel that a maximum of
three questions is sufficient for a homework assignment. It is up to the teacher to assign what is appropriate for the
class.
1.
𝟏𝟏
𝟒𝟒
Use the ratio method to create a scale drawing about center 𝑶𝑶 with a scale factor of 𝒓𝒓 = . Use a ruler and
protractor to verify that the resulting figure is in fact a scale drawing by showing that corresponding side lengths are
in constant proportion and the corresponding angles are equal in measurement.
The measurements in the figure are 𝒎𝒎∠𝑨𝑨 = 𝟖𝟖𝟖𝟖°, 𝒎𝒎∠𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝒎𝒎∠𝑪𝑪 = 𝟗𝟗𝟗𝟗°, and 𝒎𝒎∠𝑫𝑫 = 𝟓𝟓𝟓𝟓°. All side-length
measurements of the scale drawing should be in the constant ratio of 𝟏𝟏: 𝟒𝟒.
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2.
Use the ratio method to create a scale drawing about center 𝑶𝑶 with a scale factor of 𝒓𝒓 = 𝟐𝟐. Verify that the resulting
figure is in fact a scale drawing by showing that corresponding side lengths are in constant proportion and that the
corresponding angles are equal in measurement.
The measurements in the figure are 𝒎𝒎∠𝑩𝑩 = 𝟑𝟑𝟑𝟑° and 𝒎𝒎∠𝑪𝑪 = 𝟑𝟑𝟑𝟑°. All side-length measurements of the scale
drawing should be in the constant ratio of 𝟐𝟐: 𝟏𝟏.
3.
Use the ratio method to create two scale drawings: 𝑫𝑫𝑶𝑶,𝟐𝟐 and 𝑫𝑫𝑷𝑷,𝟐𝟐 . Label the scale drawing with respect to center 𝑶𝑶
as △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ and the scale drawing with respect to center 𝑷𝑷 as △ 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′.
What do you notice about the two scale drawings?
They are both congruent since each was drawn with the same scale factor.
What rigid motion can be used to map △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ onto △ 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′?
Answers may vary. For example, a translation by vector ���������⃗
𝑨𝑨′ 𝑨𝑨′′ is acceptable.
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4.
Sara found a drawing of a triangle that appears to be a scale drawing. Much of the drawing has faded, but she can
see the drawing and construction lines in the diagram below. If we assume the ratio method was used to construct
△ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ as a scale model of △ 𝑨𝑨𝑨𝑨𝑨𝑨, can you find the center 𝑶𝑶, the scale factor 𝒓𝒓, and locate △ 𝑨𝑨𝑨𝑨𝑨𝑨?
Extend ray 𝑨𝑨′𝑨𝑨 and the partial ray drawn from either 𝑩𝑩′ or 𝑪𝑪′. The point where they intersect is center 𝑶𝑶.
𝑶𝑶𝑨𝑨′
𝑶𝑶𝑶𝑶
𝟑𝟑
𝟑𝟑
𝟐𝟐
𝟐𝟐
vertices to show original △ 𝑨𝑨𝑨𝑨𝑨𝑨.
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𝟐𝟐
𝟐𝟐
= ; the scale factor is . Locate 𝑩𝑩 of the distance from 𝑶𝑶 to 𝑩𝑩′ and 𝑪𝑪 of the way from 𝑶𝑶 to 𝑪𝑪′. Connect the
𝟑𝟑
𝟑𝟑
Making Scale Drawings Using the Ratio Method
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5.
Quadrilateral 𝑨𝑨′′′𝑩𝑩′′′𝑪𝑪′′′𝑫𝑫′′′ is one of a sequence of three scale drawings of quadrilateral 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 that were all
constructed using the ratio method from center 𝑶𝑶. Find the center 𝑶𝑶, each scale drawing in the sequence, and the
scale factor for each scale drawing. The other scale drawings are quadrilaterals 𝑨𝑨′𝑩𝑩′𝑪𝑪′𝑫𝑫′ and 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′𝑫𝑫′′.
Note to the teacher: One option is to simplify this diagram by joining vertices 𝑫𝑫′′′ and 𝑩𝑩′′′, forming a triangle.
Each scale drawing is created from the same center point, so the corresponding vertices of the scale drawings should
align with the center 𝑶𝑶. Draw any two of ��������
𝑨𝑨′′′𝑨𝑨′′, ��������
𝑫𝑫′′′𝑫𝑫′, or �������
𝑩𝑩′′′𝑩𝑩 to find center 𝑶𝑶 at their intersection.
The ratio of 𝑶𝑶𝑨𝑨′′ : 𝑶𝑶𝑶𝑶 is 𝟒𝟒: 𝟏𝟏, so the scale factor of figure 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′𝑫𝑫′′ is 𝟒𝟒.
The ratio of 𝑶𝑶𝑶𝑶′: 𝑶𝑶𝑶𝑶 is 𝟐𝟐: 𝟏𝟏, so the scale factor of figure 𝑨𝑨′𝑩𝑩′𝑪𝑪′𝑫𝑫′ is 𝟐𝟐.
The ratio of 𝑶𝑶𝑩𝑩′′′ : 𝑶𝑶𝑶𝑶 is 𝟖𝟖: 𝟏𝟏, so the scale factor of figure 𝑨𝑨′′′𝑩𝑩′′′𝑪𝑪′′′𝑫𝑫′′′ is 𝟖𝟖.
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6.
𝟏𝟏
𝟐𝟐
Maggie has a rectangle drawn in the corner of an 𝟖𝟖 -inch by 𝟏𝟏𝟏𝟏-inch sheet of printer paper as shown in the
diagram. To cut out the rectangle, Maggie must make two cuts. She wants to scale the rectangle so that she can cut
it out using only one cut with a paper cutter.
a.
What are the dimensions of Maggie’s
scaled rectangle, and what is its scale
factor from the original rectangle?
If the rectangle is scaled from the corner of
the paper at which it currently sits, the
maximum height of the rectangle will be
𝟏𝟏
𝟐𝟐
𝟖𝟖 inches.
𝟏𝟏
𝟐𝟐 = 𝟑𝟑𝟑𝟑
𝟏𝟏 𝟐𝟐𝟐𝟐
𝟔𝟔
𝟒𝟒
The scale factor to the enlarged rectangle is
𝒌𝒌 =
𝟖𝟖
𝟑𝟑𝟑𝟑
.
𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑
(𝟒𝟒)
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏
= 𝟓𝟓. 𝟒𝟒𝟒𝟒
𝒚𝒚 =
𝟐𝟐𝟐𝟐
Using the scale factor, the width of the scaled rectangle is 𝟓𝟓. 𝟒𝟒𝟒𝟒 inches.
𝒚𝒚 =
b.
After making the cut for the scaled rectangle, is there enough material left to cut another identical rectangle?
If so, what is the area of scrap per sheet of paper?
The total width of the sheet of paper is 𝟏𝟏𝟏𝟏 inches, which is more than 𝟐𝟐(𝟓𝟓. 𝟒𝟒𝟒𝟒) = 𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖 inches, so yes, there
is enough material to get two identical rectangles from one sheet of paper. The resulting scrap strip
𝟏𝟏
𝟐𝟐
measures 𝟖𝟖 inches by 𝟎𝟎. 𝟏𝟏𝟏𝟏 inches, giving a scrap area of 𝟏𝟏. 𝟎𝟎𝟎𝟎 𝐢𝐢𝐧𝐧𝟐𝟐 per sheet of paper.
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Lesson 3
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GEOMETRY
Lesson 3: Making Scale Drawings Using the Parallel Method
Student Outcomes

Students create scale drawings of polygonal figures by the parallel method.

Students explain why angles are preserved in scale drawings created by the parallel method using the theorem
of parallel lines cut by a transversal.
Lesson Notes
In Lesson 3, students learn the parallel method as yet another way of creating scale drawings. The lesson focuses on
constructing parallel lines with the use of setsquares, although parallel lines can also be constructed using a compass and
straightedge; setsquares reduce the time needed for the construction. Straightedges, compasses, and setsquares are
needed for this lesson (setsquares can be made in class; refer to Grade 7 Module 6 Lesson 7). Rulers are allowed in one
instance in the lesson (Example 1, part (b)), but the question can completed without them as long as compasses (or
other devices for maintaining a fixed distance) are available.
Classwork
Opening Exercise (2 minutes)
The purpose of this Opening Exercise is to get students thinking about how parallel lines can be used to create a dilation.
Accept all ideas, and use responses to segue to Example 1.
Opening Exercise
Dani dilated △ 𝑨𝑨𝑨𝑨𝑨𝑨 from center 𝑶𝑶, resulting in △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′. She says that she completed the drawing using parallel lines.
How could she have done this? Explain.
MP.2
Accept reasonable suggestions from students and proceed to Example 1.
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Example 1 (6 minutes)
Scaffolding:
Example 1 is intended to remind students how to create a parallel line with the use of a
setsquare. Provide students with compasses, or allow the use of measurement for part
(b).
For further practice with
Module 6 Lesson 7. If students
are familiar with using a
Example 2.
Example 1
a.
Use a ruler and setsquare to draw a line through 𝑪𝑪 parallel to ����
𝑨𝑨𝑨𝑨. What ensures that
the line drawn is parallel to ����
𝑨𝑨𝑨𝑨?
Since the setsquare is in the shape of a right triangle, we are certain that the legs of the setsquare are
����, and the other leg is flush against the ruler and can
perpendicular. Then, when one leg is aligned with 𝑨𝑨𝑨𝑨
slide along the ruler, the 𝟗𝟗𝟗𝟗° angle between the horizontal leg and the ruler remains fixed; in effect, there are
corresponding angles that are equal in measure. The two lines must be parallel.
b.
���� and point 𝑪𝑪.
Use a ruler and setsquare to draw a parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 around 𝑨𝑨𝑨𝑨
Step 1
Step 2
Step 3
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Example 2 (10 minutes)
Example 1 demonstrates how to create a scale drawing using the parallel method.

The basic parameters and initial steps to the parallel method are like those of the initial steps to the ratio
method; a ray must be drawn from the center through all vertices, and one corresponding vertex of the scale
drawing must be determined using the scale factor and ruler. However, as suggested by the name of the
method, the following steps require a setsquare to draw a segment parallel to each side of the figure.
Example 2
Use the figure below with center 𝑶𝑶 and a scale factor of 𝒓𝒓 = 𝟐𝟐 and the following steps to create a scale drawing using the
parallel method.
Step 1. Draw a ray beginning at 𝑶𝑶 through each vertex of the figure.
Step 2. Select one vertex of the scale drawing to locate; we have selected 𝑨𝑨′. Locate 𝑨𝑨′ on ������⃗
𝑶𝑶𝑶𝑶 so that 𝑶𝑶𝑨𝑨′ = 𝟐𝟐𝟐𝟐𝟐𝟐.
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Step 3. Align the setsquare and ruler as in the image below; one leg of the setsquare should line up with side ����
𝑨𝑨𝑨𝑨, and the
perpendicular leg should be flush against the ruler.
Scaffolding:
These steps should be modeled
for all students, but depending
on student needs, these
images may need to be made
larger, or teachers may need to
model the steps more
explicitly.
Step 4. Slide the setsquare along the ruler until the edge of the setsquare passes through 𝑨𝑨′. Then, along the
������⃗, and
perpendicular leg of the setsquare, draw the segment through 𝑨𝑨′ that is parallel to ����
𝑨𝑨𝑨𝑨 until it intersects with 𝑶𝑶𝑶𝑶
label this point 𝑩𝑩′.
Ask students to summarize how they created the scale drawing and why they think this method works.
It may happen that it is not possible to draw the entire parallel segment 𝐴𝐴′𝐵𝐵′ due to the position of the setsquare and
the location of 𝐵𝐵′. Alert students that this may happen and that they should simply pick up the setsquare (or ruler) and
complete the segment once it has been started.
It may even happen that the setsquare is not long enough to meet point 𝐴𝐴′. In such a case, a ruler can be placed flush
against the other leg of the setsquare, and then the setsquare can be removed and a line drawn through 𝐴𝐴′ .
In a similar vein, if any of the rays is not long enough, extend it so that the intersection between the parallel segment
and that ray is visible.
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Step 5. Continue to create parallel segments to determine each successive vertex point. In this particular case, the
����. This is done because, in trying to create a parallel segment from ����
𝑩𝑩𝑩𝑩, the parallel
setsquare has been aligned with 𝑨𝑨𝑨𝑨
segment was not reaching 𝑩𝑩′. This could be remedied with a larger setsquare and longer ruler, but it is easily avoided by
working on the segment parallel to ����
MP.1
Step 6. Use your ruler to join the final two unconnected vertices.
Have students show that △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ is a scale drawing; measure and confirm that the length
of each segment in the scale drawing is twice the length of each segment in the original
drawing and that the measurements of all corresponding angles are equal. △ 𝐴𝐴𝐴𝐴𝐴𝐴 angle
measurements are 𝑚𝑚∠𝐴𝐴 = 104°, 𝑚𝑚∠𝐵𝐵 = 44°, and 𝑚𝑚∠𝐶𝐶 = 32°. The measurements of
the side lengths are not provided because they differ from the images that appear in print
form.
Scaffolding:
Patty paper may facilitate
measurements in the Examples
and Exercises, but students
should be prepared to use
measuring tools in an Exit
Ticket or Assessment.

We want to note here that though we began with a scale factor of 𝑟𝑟 = 2 for the
dilation, we consider the resulting scale factor of the scale drawing created by
the parallel method separately. As we can see by trying the dilations out, the scale factor for the dilation and
the scale factor for the scale drawing by the parallel method are in fact one and the same.

There is a concrete reason why this is, but we do not go into the explanation of why the parallel method
actually yields a scale drawing with the same scale factor as the dilation until later lessons.
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GEOMETRY
Exercises 1–2 (10 minutes)
Exercise 1 differs from Example 1 in one way: in Example 1, students had to locate the initial point 𝐴𝐴′, whereas in
Exercise 1, students are provided with the location of the initial point but not told explicitly how far from the center the
point is. Teachers should use their discretion to decide if students are ready for the slightly altered situation or whether
they need to retry a problem as in Example 1.
Exercises
1.
With a ruler and setsquare, use the parallel method to create a scale drawing of 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 by the parallel method. 𝑾𝑾′
has already been located for you. Determine the scale factor of the scale drawing. Verify that the resulting figure is
in fact a scale drawing by showing that corresponding side lengths are in constant proportion and that
corresponding angles are equal in measurement.
The scale factor is 𝟑𝟑. Verification of the enlarged figure should show that the length of each segment in the scale
drawing is three times the length of each segment in the original figure (e.g., 𝑾𝑾′ 𝑿𝑿′ = 𝟑𝟑(𝑾𝑾𝑾𝑾)). The angle
measurements are 𝒎𝒎∠𝑾𝑾 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝒎𝒎∠𝑿𝑿 = 𝟕𝟕𝟕𝟕°, 𝒎𝒎∠𝒀𝒀 = 𝟔𝟔𝟔𝟔°, and 𝒎𝒎∠𝒁𝒁 = 𝟗𝟗𝟗𝟗°.
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2.
With a ruler and setsquare, use the parallel method to create a scale drawing of 𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫 about center 𝑶𝑶 with scale
𝟏𝟏
𝟐𝟐
factor 𝒓𝒓 = . Verify that the resulting figure is in fact a scale drawing by showing that corresponding side lengths
are in constant proportion and that the corresponding angles are equal in measurement.
Verification of the reduced figure should show that the length of each segment in the scale drawing is one half the
𝟏𝟏
𝟐𝟐
length of each segment in the original figure, (e.g., 𝑫𝑫′𝑬𝑬′ = (𝑫𝑫𝑫𝑫)). The angle measurements are
𝒎𝒎∠𝑫𝑫 = 𝟖𝟖𝟖𝟖°, 𝒎𝒎∠𝑬𝑬 = 𝟗𝟗𝟗𝟗°, 𝒎𝒎∠𝑭𝑭 = 𝟗𝟗𝟗𝟗°, and 𝒎𝒎∠𝑮𝑮 = 𝟕𝟕𝟕𝟕°.
Discussion (5 minutes)

So far we have verified that corresponding angles between figures and their scale drawings are equal in
measurement by actually measuring each pair of angles. Instead of measuring each time, we can recall what
we know about parallel lines cut by transversals to verify that corresponding angles are in fact equal in
measurement. How would you explain this? Mark the following figure as needed to help explain.

If a transversal intersects two parallel lines, then corresponding angles are equal in measurement.
���� ∥ �����
𝐴𝐴′𝐶𝐶′ and
Since we have constructed corresponding segments to be parallel, we are certain that 𝐴𝐴𝐴𝐴
������
����
𝐴𝐴𝐴𝐴 ∥ 𝐴𝐴′𝐵𝐵′. We make use of the corresponding angles fact twice to show that corresponding angles ∠𝐴𝐴
and ∠𝐴𝐴′ are equal in measurement. A similar argument shows the other angles are equal in
measurement.
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Exercise 3 (5 minutes)
The center 𝑂𝑂 lies within the figure in Exercise 3. Ask students if they think this affects the resulting scale drawing, and
allow them to confer with a neighbor.
3.
With a ruler and setsquare, use the parallel method to create a scale drawing of pentagon 𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷 about center 𝑶𝑶
𝟓𝟓
𝟐𝟐
with scale factor 𝒓𝒓 = . Verify that the resulting figure is in fact a scale drawing by showing that corresponding side
lengths are in constant proportion and that corresponding angles are equal in measurement.
Verification of the enlarged figure should show that the length of each segment in the scale drawing is two-and-a𝟓𝟓
𝟐𝟐
half times the length of each segment in the original figure (e.g., 𝑷𝑷′ 𝑻𝑻′ = (𝑷𝑷𝑷𝑷)). Each of the angles has a
measurement of 𝟏𝟏𝟏𝟏𝟏𝟏°.
Closing (2 minutes)
Ask students to summarize the key points of the lesson. Additionally, consider having them answer the following
questions independently in writing, to a partner, or to the whole class.

How are dilations and scale drawings related?


Dilations can be used to create scale drawings by the ratio method or the parallel method.
To create a scale drawing using the ratio method, a center, a figure, and a scale factor must be provided. Then
the dilated vertices can either be measured or located using a compass. To use the parallel method, a center,
a figure, and a scale factor or one provided vertex of the dilated figure must be provided. Then use a
setsquare to help construct sides parallel to the sides of the original figure, beginning with the side that passes
through the provided dilated vertex.
Exit Ticket (5 minutes)
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Name
Date
Lesson 3: Making Scale Drawings Using the Parallel Method
Exit Ticket
With a ruler and setsquare, use the parallel method to create a scale drawing of quadrilateral 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 about center 𝑂𝑂 with
3
4
scale factor 𝑟𝑟 = . Verify that the resulting figure is in fact a scale drawing by showing that corresponding side lengths
are in constant proportion and that the corresponding angles are equal in measurement.
What kind of error in the parallel method might prevent us from having parallel, corresponding sides?
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Exit Ticket Sample Solutions
With a ruler and setsquare, use the parallel method to create a scale drawing of quadrilateral 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 about center 𝑶𝑶 with
𝟑𝟑
𝟒𝟒
scale factor 𝒓𝒓 = . Verify that the resulting figure is in fact a scale drawing by showing that corresponding side lengths
are in constant proportion and that the corresponding angles are equal in measurement.
The measurements of the angles in the figure are 𝒎𝒎∠𝑨𝑨 = 𝟖𝟖𝟖𝟖°, 𝒎𝒎∠𝑩𝑩 = 𝟓𝟓𝟓𝟓°, 𝒎𝒎∠𝑪𝑪 = 𝟏𝟏𝟏𝟏𝟏𝟏°, and 𝒎𝒎∠𝑫𝑫 = 𝟏𝟏𝟏𝟏𝟏𝟏°. All sidelength measurements of the scale drawing should be in the constant ratio of 𝟑𝟑: 𝟒𝟒.
What kind of error in the parallel method might prevent us from having parallel, corresponding sides?
If the setsquare is not aligned with the segment of the figure, you will not create a parallel segment. Also, if the setsquare
is not perfectly flush with the ruler, it will not be possible to create a segment parallel to the segment of the figure.
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Problem Set Sample Solutions
1.
With a ruler and setsquare, use the parallel method to create a scale drawing of the figure about center 𝑶𝑶. One
vertex of the scale drawing has been provided for you.
Determine the scale factor. Verify that the resulting figure is in fact a scale drawing by showing that corresponding
side lengths are in constant proportion and that the corresponding angles are equal in measurement.
𝟓𝟓
𝟐𝟐
The scale factor is 𝒓𝒓 = . The measurements of the angles in the figure are 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑩𝑩 = 𝒎𝒎∠𝑫𝑫 = 𝒎𝒎∠𝑬𝑬 = 𝟗𝟗𝟗𝟗°,
𝒎𝒎∠𝑫𝑫𝑫𝑫𝑫𝑫 = 𝟐𝟐𝟐𝟐°, 𝒎𝒎∠𝑪𝑪𝑪𝑪𝑪𝑪 = 𝟗𝟗𝟗𝟗°, and 𝒎𝒎∠𝑪𝑪 = 𝟕𝟕𝟕𝟕°. All side-length measurements of the scale drawing are in the
constant ratio of 𝟓𝟓: 𝟐𝟐.
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2.
With a ruler and setsquare, use the parallel method to create a scale drawing of the figure about center 𝑶𝑶 and scale
𝟏𝟏
𝟑𝟑
factor 𝒓𝒓 = . Verify that the resulting figure is in fact a scale drawing by showing that corresponding side lengths
are in constant proportion and the corresponding angles are equal in measurement.
The measurements of the angles in the figure are 𝒎𝒎∠𝑨𝑨 = 𝟐𝟐𝟐𝟐°, 𝒎𝒎∠𝑩𝑩 = 𝟒𝟒𝟒𝟒°, 𝒎𝒎∠𝑪𝑪 = 𝟑𝟑𝟑𝟑°, 𝒎𝒎∠𝑫𝑫 = 𝟐𝟐𝟐𝟐°, and
𝒎𝒎∠𝑪𝑪𝑪𝑪𝑪𝑪 = 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏°. All side-length measurements of the scale drawing are in the constant ratio of 𝟏𝟏: 𝟑𝟑.
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3.
With a ruler and setsquare, use the parallel method to create the following scale drawings about center 𝑶𝑶: (1) first
𝟐𝟐
use a scale factor of 𝟐𝟐 to create △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′, and (2) then, with respect to △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′, use a scale factor of to create
𝟑𝟑
scale drawing △ 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′. Calculate the scale factor for △ 𝑨𝑨′′ 𝑩𝑩′′𝑪𝑪′′ as a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨. Use angle and side
The scale factor of △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ relative to △ 𝑨𝑨𝑨𝑨𝑨𝑨 is 𝟐𝟐. The measurements of the angles in the figure are 𝒎𝒎∠𝑨𝑨 ≈ 𝟒𝟒𝟒𝟒°,
𝒎𝒎∠𝑩𝑩 ≈ 𝟔𝟔𝟔𝟔°, and 𝒎𝒎∠𝑪𝑪 ≈ 𝟕𝟕𝟕𝟕°. The scale factor from △ 𝑨𝑨𝑨𝑨𝑨𝑨 to △ 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′ is
side-length proportions between △ 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′ and △ 𝑨𝑨𝑨𝑨𝑨𝑨:
4.
𝑨𝑨𝑨𝑨
=
𝑩𝑩′′ 𝑪𝑪′′
𝑩𝑩𝑩𝑩
=
𝟒𝟒
⋅ 𝟐𝟐 = . Solutions should show the
𝟑𝟑
𝟑𝟑
𝑪𝑪′′ 𝑨𝑨′′
𝟒𝟒
𝑪𝑪𝑪𝑪
= .
𝟑𝟑
Follow the direction in each part below to create three scale drawings of △ 𝑨𝑨𝑨𝑨𝑨𝑨 using the parallel method.
a.
b.
c.
𝟑𝟑
With the center at vertex 𝑨𝑨, make a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 with a scale factor of .
𝟐𝟐
𝟑𝟑
With the center at vertex 𝑩𝑩, make a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 with a scale factor of .
𝟐𝟐
𝟑𝟑
With the center at vertex 𝑪𝑪, make a scale drawing of △ 𝑨𝑨𝑨𝑨𝑨𝑨 with a scale factor of .
𝟐𝟐
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𝟐𝟐
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d.
What conclusions can be drawn about all three scale drawings from parts (a)–(c)?
The three scale drawings are congruent.
𝟑𝟑
𝟑𝟑
𝟐𝟐
𝑨𝑨𝑩𝑩′ = 𝑨𝑨′′ 𝑩𝑩 = 𝑨𝑨𝑨𝑨
Dilation using scale factor .
∠𝑩𝑩′ ≅ ∠𝑨𝑨𝑨𝑨𝑨𝑨
Corresponding ∠'s formed by parallel lines are congruent.
∠𝑨𝑨′′ ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩
′ ′
𝟐𝟐
Corresponding ∠'s formed by parallel lines are congruent.
△ 𝑨𝑨𝑩𝑩 𝑪𝑪 ≅ △ 𝑨𝑨′′𝑩𝑩𝑩𝑩′′
ASA
A similar argument can be used to show △ 𝑨𝑨𝑩𝑩′ 𝑪𝑪′ ≅ △ 𝑨𝑨′′′𝑩𝑩′′′𝑪𝑪, and by transitivity of congruence, all three
scale drawings are congruent to each other.
5.
Use the parallel method to make a scale drawing of the line segments in the following figure using the given 𝑾𝑾′, the
image of vertex 𝑾𝑾, from center 𝑶𝑶. Determine the scale factor.
𝟑𝟑
The ratio of 𝑶𝑶𝑶𝑶′: 𝑶𝑶𝑶𝑶 is 𝟑𝟑: 𝟐𝟐, so the scale factor is . All corresponding lengths are in the ratio of 𝟑𝟑: 𝟐𝟐.
𝟐𝟐
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6.
If we switch perspective and consider the original drawing 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 to be a scale drawing of the constructed image
𝑨𝑨′𝑩𝑩′𝑪𝑪′𝑫𝑫′𝑬𝑬′, what would the scale factor be?
If the original figure were the scale drawing, and the scale drawing were the original figure, the scale factor would
𝟐𝟐
be .
𝟓𝟓
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Lesson 4: Comparing the Ratio Method with the Parallel
Method
Student Outcomes

Students understand that the ratio and parallel methods produce the same scale drawing and understand the
proof of this fact.

Students relate the equivalence of the ratio and parallel methods to the triangle side splitter theorem: A line
segment splits two sides of a triangle proportionally if and only if it is parallel to the third side.
Lesson Notes
This lesson challenges students to understand the reasoning that connects the ratio and parallel methods that have been
used in the last two lessons for producing a scale drawing. The Opening Exercises are important to the discussions that
follow and are two important ideas in their own right. The first key idea is that two triangles with the same base that
have vertices on a line parallel to the base are equal in area. The second key idea is that two triangles with different
bases but equal altitudes have a ratio of areas that is equal to the ratio of their bases. Following the Opening Exercises,
students and the teacher show that the ratio method and parallel method are equivalent. The concluding discussion
shows how that work relates to the triangle side splitter theorem.
Classwork
Today, our goal is to show that the parallel method and the ratio method are equivalent; that is, given a figure in the
plane and a scale factor 𝒓𝒓 > 𝟎𝟎, the scale drawing produced by the parallel method is congruent to the scale drawing
produced by the ratio method. We start with two easy exercises about the areas of two triangles whose bases lie on the
same line, which helps show that the two methods are equivalent.
Opening Exercise (10 minutes)
Students need the formula for the area of a triangle. The first exercise is a famous proposition of Euclid’s (Proposition 37
of Book 1). Consider asking students to go online after class to read how Euclid proves the proposition.
Give students two minutes to work on the first exercise in groups, and walk around the room answering questions and
helping students draw pictures. After two minutes, go through the proof on the board, answering questions about the
parallelogram along the way. Repeat the process for Exercise 2.
Do not look for pristine proofs from students on these exercises, merely for confirmation that they understand the
statements. For example, in Exercise 1, they should understand that two triangles between two parallel lines with the
same base must have the same area. These two exercises help avoid the quagmire of drawing altitudes and calculating
areas in the proofs that follow; these exercises help students to simply recognize when two triangles have the same area
or recognize when the ratio of the bases of two triangles is the same as the ratio of their areas.
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It is useful to leave the statements of Opening Exercises 1 and 2 on the board throughout the lesson to refer back to
them.
Opening Exercise
a.
Suppose two triangles, △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑨𝑨𝑨𝑨, share the
same base 𝑨𝑨𝑨𝑨 such that points 𝑪𝑪 and 𝑫𝑫 lie on a line
parallel to ⃖����⃗
𝑨𝑨𝑨𝑨. Show that their areas are equal, that is,
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨) = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨). (Hint: Why are the
altitudes of each triangle equal in length?)
⃖����⃗ through 𝑪𝑪, and label the intersection of both lines 𝑪𝑪′. Then 𝑪𝑪𝑪𝑪′ is an altitude
Draw a perpendicular line to 𝑨𝑨𝑨𝑨
for △ 𝑨𝑨𝑨𝑨𝑨𝑨. Do the same for △ 𝑨𝑨𝑨𝑨𝑨𝑨 to get an altitude 𝑫𝑫𝑫𝑫′.
Quadrilateral 𝑪𝑪𝑪𝑪′𝑫𝑫′𝑫𝑫 is a parallelogram and, therefore, 𝑪𝑪′ = 𝑫𝑫𝑫𝑫′, both of which follow from the properties
����� and �����
𝑫𝑫𝑫𝑫′ are altitudes of the triangles, we get by the area formula for triangles,
of parallelograms. Since 𝑪𝑪𝑪𝑪′
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨) =
𝟏𝟏
𝟏𝟏
𝑨𝑨𝑨𝑨 ⋅ 𝑪𝑪′ = 𝑨𝑨𝑨𝑨 ⋅ 𝑫𝑫′ = 𝐀𝐀𝐀𝐀𝐞𝐞𝐞𝐞(△ 𝑨𝑨𝑨𝑨𝑨𝑨).
𝟐𝟐
𝟐𝟐
Draw the first picture below while reading through Opening Exercise 2 with the class. Ask questions that check for
MP.6 understanding, like, “Are the points 𝐴𝐴, 𝐵𝐵, and 𝐵𝐵′ collinear? Why?” and “Does it matter if 𝐵𝐵 and 𝐵𝐵′ are on the same side
of 𝐴𝐴 on the line?”
b.
Suppose two triangles have different-length bases, 𝑨𝑨𝑨𝑨 and 𝑨𝑨𝑨𝑨′, that lie on the same line. Furthermore,
suppose they both have the same vertex 𝑪𝑪 opposite these bases. Show that the value of the ratio of their
areas is equal to the value of the ratio of the lengths of their bases, that is,
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨)
𝑨𝑨𝑨𝑨
=
.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑩𝑩′ 𝑪𝑪) 𝑨𝑨𝑩𝑩′
⃖����⃗ through 𝑪𝑪, and
Draw a perpendicular line to 𝑨𝑨𝑨𝑨
label the intersection of both lines 𝑪𝑪′ .
Then 𝑪𝑪𝑪𝑪′ is an altitude for both triangles.
By the area formula for triangles,
𝟏𝟏
𝑨𝑨𝑨𝑨 ⋅ 𝑪𝑪′
𝑨𝑨𝑨𝑨
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨)
𝟐𝟐
=
=
.
′
′
𝟏𝟏
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑩𝑩 𝑪𝑪)
𝑨𝑨𝑩𝑩
′
′
𝑨𝑨𝑩𝑩 ⋅ 𝑪𝑪𝑪𝑪
𝟐𝟐
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Ask students to summarize to a neighbor the two results from the Opening Exercises. Use this as an opportunity to
check for understanding.
Discussion (20 minutes)
The next two theorems generate the so-called triangle side splitter theorem, which is the most important result of this
module. (In standard G-SRT.B.4, it is stated as, “A line parallel to one side of a triangle divides the other two
proportionally, and conversely.”) The triangle side splitter theorem is used many times in the next few lessons to
understand dilations and similarity. Note that using the AA similarity criterion to prove the triangle side splitter theorem
is circular: The triangle side splitter theorem is the reason why a dilation takes a line to a parallel line and an angle to
another angle of equal measure, which are both needed to prove the AA similarity criterion. Thus, the triangle side
splitter theorem must be proven in a way that does not invoke these two ideas. (Note that in Grade 8, we assumed the
triangle side splitter theorem and its immediate consequences and, in doing so, glossed over many of the issues we need
to deal with in this course.)
Even though the following two proofs are probably the simplest known proofs of these theorems (and should be easy to
understand), they rely on subtle tricks that students should not be expected to discover on their own. Brilliant
mathematicians constructed these careful arguments over 2,300 years ago. However, that does not mean that this part
of the lesson is a lecture. Go through the proofs with students at a relaxed pace, ask them questions to check for
understanding, and have them articulate the reasons for each step. If done well, the class can take joy in the clever
arguments presented here!
Discussion
To show that the parallel and ratio methods are equivalent, we need only look at one of the simplest versions of a scale
drawing: scaling segments. First, we need to show that the scale drawing of a segment generated by the parallel method
is the same segment that the ratio method would have generated and vice versa. That is,
The parallel method ⟹ The ratio method,
and
The ratio method ⟹ The parallel method.
MP.1 Ask students why scaling a segment is sufficient for showing equivalence of both methods for scaling polygonal figures.
& That is, if we wanted to show that both methods would produce the same polygonal figure, why is it enough to only
MP.3 show that both methods would produce the same segment?
Students should respond that polygonal figures are composed of segments. If we can show that both methods produce
the same segment, then it makes sense that both methods would work for all segments that comprise the polygonal
figure.
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The first implication above can be stated as the following theorem:
⃖����⃗, construct a scale drawing of 𝑨𝑨𝑨𝑨 with scale factor 𝒓𝒓 > 𝟎𝟎
PARALLEL ⟹ RATIO THEOREM: Given 𝑨𝑨𝑨𝑨 and point 𝑶𝑶 not on 𝑨𝑨𝑨𝑨
′
𝑨𝑨𝑨𝑨 that passes through 𝑨𝑨′. Let 𝑩𝑩′ be the point
using the parallel method: Let 𝑨𝑨 = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑨𝑨), and 𝓵𝓵 be the line parallel to ⃖����⃗
′
������⃗
where 𝑶𝑶𝑶𝑶 intersects 𝓵𝓵. Then 𝑩𝑩 is the same point found by the ratio method, that is,
𝑩𝑩′ = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑩𝑩).
ℓ
Discuss the statement of the theorem with students. Ask open-ended questions that lead students through the
following points:


𝐴𝐴′ 𝐵𝐵′ is the scale drawing of 𝐴𝐴𝐴𝐴 using the parallel method. Why?
𝐴𝐴′ = 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐴𝐴) is already the first step in the ratio method. The difference between the parallel method and the
�����⃗, while in
ratio method is that 𝐵𝐵′ is found using the parallel method by intersecting the parallel line ℓ with 𝑂𝑂𝑂𝑂
the ratio method, the point is found by dilating point 𝐵𝐵 at center 𝑂𝑂 by scale factor 𝑟𝑟 to get 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐵𝐵). We need
𝑂𝑂𝑂𝑂, this can be
to show that these are the same point, that is, that 𝐵𝐵′ = 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐵𝐵). Since both points lie on �����⃗
′
done by showing that 𝑂𝑂𝐵𝐵 = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂.
There is one subtlety with the above theorem to consider discussing with students. In it, we assumed—asserted really—
that ℓ and �����⃗
𝑂𝑂𝑂𝑂 intersect. They do, but is it clear that they do? (Pictures can be deceiving!) First, suppose that ℓ did not
⃖����⃗
⃖����⃗ are parallel. Since ℓ is also parallel to ⃖����⃗
⃖����⃗ and ⃖����⃗
𝐴𝐴𝐴𝐴 , then 𝑂𝑂𝑂𝑂
𝐴𝐴𝐴𝐴 are parallel
intersect 𝑂𝑂𝑂𝑂 , and then by definition, ℓ and 𝑂𝑂𝑂𝑂
(by parallel transitivity from Module 1), which is clearly a contradiction since both contain the point 𝐵𝐵. Hence, it must be
𝑂𝑂𝑂𝑂 or the opposite ray from 𝑂𝑂? There are two cases
that ℓ intersects ⃖����⃗
𝑂𝑂𝑂𝑂. But where does it intersect? Does it intersect �����⃗
𝐴𝐴𝐴𝐴 (i.e., as in the picture above when 𝑟𝑟 > 0), then ℓ is
to consider. Case 1: If 𝐴𝐴′ and 𝑂𝑂 are in opposite half-planes of ⃖����⃗
𝐵𝐵𝐵𝐵 ,
contained completely in the half-plane that contains 𝐴𝐴′ by the plane separation axiom. Thus, ℓ cannot intersect �����⃗
𝐴𝐴𝐴𝐴 (when
which means it must intersect �����⃗
𝑂𝑂𝑂𝑂 . Case 2: Now suppose that 𝐴𝐴′ and 𝑂𝑂 are in the same half-plane of ⃖����⃗
0 < 𝑟𝑟 < 1), and consider the two half-planes of ℓ. The points 𝐵𝐵 and 𝑂𝑂 must lie in the opposite half-planes of ℓ. (Why?
Hint: What fact would be contradicted if they were in the same half-plane?) Thus, by the plane separation axiom, the
line intersects 𝑂𝑂𝑂𝑂, and thus ℓ intersects �����⃗
𝑂𝑂𝑂𝑂 .
There is a set of theorems that revolve around when two lines intersect each other as in the paragraph above. That set
falls under the general heading of “crossbar theorems.” Explore these theorems with students by looking the theorems
up on the web. The theorem above is written in a way that asserts that ℓ and �����⃗
𝑂𝑂𝑂𝑂 intersect, and so covers up these
intersection issues in a factually correct way that helps avoid unnecessarily pedantic crossbar discussions in the future.
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PROOF: We prove the case when 𝒓𝒓 > 𝟏𝟏; the case when 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏 is the same but with a different picture. Construct 𝑩𝑩𝑨𝑨′
and 𝑨𝑨𝑩𝑩′ to form two triangles △ 𝑩𝑩𝑩𝑩𝑩𝑩′ and △ 𝑩𝑩𝑩𝑩𝑨𝑨′ , labeled as 𝑻𝑻𝟏𝟏 and 𝑻𝑻𝟐𝟐 , respectively, in the picture below.
The areas of these two triangles are equal,
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚(𝑻𝑻𝟏𝟏 ) = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑻𝑻𝟐𝟐 ),
by Exercise 1. Why? Label △ 𝑶𝑶𝑶𝑶𝑶𝑶 by 𝑻𝑻𝟎𝟎 . Then 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑨𝑨′ 𝑩𝑩) = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑩𝑩′ 𝑨𝑨) because areas add:
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑨𝑨′ 𝑩𝑩) = 𝐀𝐀𝐫𝐫𝐫𝐫𝐫𝐫(𝑻𝑻𝟎𝟎 ) + 𝐀𝐀𝐫𝐫𝐫𝐫𝐫𝐫(𝑻𝑻𝟐𝟐 )
= 𝐀𝐀𝐫𝐫𝐫𝐫𝐫𝐫(𝑻𝑻𝟎𝟎 ) + 𝐀𝐀𝐫𝐫𝐫𝐫𝐫𝐫(𝑻𝑻𝟏𝟏 )
= 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑩𝑩′ 𝑨𝑨).
Next, we apply Exercise 2 to two sets of triangles: (1) 𝑻𝑻𝟎𝟎 and △ 𝑶𝑶𝑨𝑨′ 𝑩𝑩 and (2) 𝑻𝑻𝟎𝟎 and △ 𝑶𝑶𝑶𝑶′𝑨𝑨.
(2) 𝑻𝑻𝟎𝟎 and △ 𝑶𝑶𝑶𝑶′𝑨𝑨 with
bases on ⃖������⃗
𝑶𝑶𝑩𝑩′
(1) 𝑻𝑻𝟎𝟎 and △ 𝑶𝑶𝑨𝑨′ 𝑩𝑩 with
⃖������⃗′
bases on 𝑶𝑶𝑨𝑨
Therefore,
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△𝑶𝑶𝑨𝑨′ 𝑩𝑩)
𝑶𝑶𝑨𝑨′
=
, and
𝑶𝑶𝑶𝑶
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑻𝑻𝟎𝟎 )
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△𝑶𝑶𝑩𝑩′ 𝑨𝑨)
𝑶𝑶𝑩𝑩′
=
.
𝑶𝑶𝑶𝑶
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑻𝑻𝟎𝟎 )
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Since 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑨𝑨′ 𝑩𝑩) = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑩𝑩′ 𝑨𝑨), we can equate the fractions:
dilating 𝑶𝑶𝑶𝑶 to 𝑶𝑶𝑨𝑨′ , we know that
𝑶𝑶𝑨𝑨′
𝑶𝑶𝑶𝑶
= 𝒓𝒓; therefore,
𝑶𝑶𝑩𝑩′
𝑶𝑶𝑶𝑶
𝑶𝑶𝑨𝑨′
𝑶𝑶𝑶𝑶
=
𝑶𝑶𝑩𝑩′
𝑶𝑶𝑶𝑶
. Since 𝒓𝒓 is the scale factor used in
= 𝒓𝒓, or 𝑶𝑶𝑩𝑩′ = 𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶. This last equality implies that 𝑩𝑩′ is the
dilation of 𝑩𝑩 from 𝑶𝑶 by scale factor 𝒓𝒓, which is what we wanted to prove.
Next, we prove the reverse implication to show that both methods are equivalent to each other.
Ask students why showing that “the ratio method implies the parallel method” establishes equivalence. Why isn’t the
MP.3 first implication “good enough”? (Because we do not know yet that a scale drawing produced by the ratio method
would be the same scale drawing produced by the parallel method—the first implication does help us conclude that.)
This theorem is easier to prove than the previous one. In fact, the previous theorem can be used to quickly prove this
one!
⃖����⃗, construct a scale drawing 𝑨𝑨′ 𝑩𝑩′ of 𝑨𝑨𝑨𝑨 with scale factor
RATIO ⟹ PARALLEL THEOREM: Given 𝑨𝑨𝑨𝑨 and point 𝑶𝑶 not on the 𝑨𝑨𝑨𝑨
′
′
𝒓𝒓 > 𝟎𝟎 using the ratio method (i.e., find 𝑨𝑨 = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑨𝑨) and 𝑩𝑩 = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑩𝑩), and draw 𝑨𝑨′ 𝑩𝑩′ ). Then 𝑩𝑩′ is the same as the point
found using the parallel method.
PROOF: Since both the ratio method and the parallel method start with the same first step of setting 𝑨𝑨′ = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑨𝑨), the
only difference between the two methods is in how the second point is found. If we use the parallel method, we
𝑶𝑶𝑶𝑶 by 𝑪𝑪. Then 𝑩𝑩′ is the
construct the line 𝓵𝓵 parallel to ⃖����⃗
𝑨𝑨𝑨𝑨 that passes through 𝑨𝑨′ and label the point where 𝓵𝓵 intersects ������⃗
same as the point found using the parallel method if we can show that 𝑪𝑪 = 𝑩𝑩′ .
C
𝑶𝑶𝑶𝑶
𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶
The Ratio Method
𝑶𝑶𝑶𝑶
𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶
The Parallel Method
By the parallel ⟹ ratio theorem, we know that 𝑪𝑪 = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑩𝑩), that is, that 𝑪𝑪 is the point on ������⃗
𝑶𝑶𝑶𝑶 such that
������⃗ such that 𝑶𝑶𝑩𝑩′ = 𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶. Hence, they must be the same point.
𝑶𝑶𝑶𝑶 = 𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶. But 𝑩𝑩′ is also the point on 𝑶𝑶𝑶𝑶
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Discussion (8 minutes)
The fact that the ratio and parallel methods are equivalent is often
stated as the triangle side splitter theorem. To understand the
triangle side splitter theorem, we need a definition:
SIDE SPLITTER: A line segment 𝑪𝑪𝑪𝑪 is said to split the sides of △ 𝑶𝑶𝑨𝑨𝑨𝑨
����, 𝑫𝑫 is a point on �����
proportionally if 𝑪𝑪 is a point on 𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶, and
(or equivalently,
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
=
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
=
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
). We call line segment 𝑪𝑪𝑪𝑪 a side splitter.
𝑶𝑶𝑶𝑶
TRIANGLE SIDE SPLITTER THEOREM: A line segment splits two sides of a
triangle proportionally if and only if it is parallel to the third side.
Provide students with time to read and make sense of the theorem. Students should be able to state that a line segment
that splits two sides of a triangle is called a side splitter. If the sides of a triangle are split proportionally, then the line
segment that split the sides must be parallel to the third side of the triangle. Conversely, if a segment that intersects
two sides of a triangle is parallel to the third side of a triangle, then that segment is a side splitter.
Ask students to rephrase the statement of the theorem for a triangle 𝑂𝑂𝐴𝐴′ 𝐵𝐵′ and a segment 𝐴𝐴𝐴𝐴 (i.e., the terminology
used in the theorems above). It should look like this:
Restatement of the triangle side splitter theorem:
In △ 𝑶𝑶𝑶𝑶′𝑩𝑩′, ����
𝑨𝑨𝑨𝑨 splits the sides proportionally (i.e.,
����.
if and only if ������
𝑨𝑨′𝑩𝑩′ || 𝑨𝑨𝑨𝑨


What does
𝑶𝑶𝑩𝑩′
𝑶𝑶𝑶𝑶
)
𝑂𝑂𝐴𝐴′
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝐵𝐵′
𝑂𝑂𝑂𝑂
mean in terms of dilations?
This means that there is a dilation with scale factor 𝑟𝑟 =
𝑂𝑂𝐴𝐴′
𝑂𝑂𝑂𝑂
such that 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐴𝐴) = 𝐴𝐴′ and 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐵𝐵) = 𝐵𝐵′ .
���� splits the sides proportionally” correspond to?
Which method (parallel or ratio) does the statement “𝐴𝐴𝐴𝐴


𝑶𝑶𝑶𝑶
=
Ask students to relate the restatement of the triangle side splitter theorem to the two theorems above. In
order for students to do this, they need to translate the statement into one about dilations. Begin with the
implication that 𝐴𝐴𝐴𝐴 splits the sides proportionally.


𝑶𝑶𝑨𝑨′
The ratio method
What does the ratio ⟹ parallel theorem imply about 𝐵𝐵′ ?
This implies that 𝐵𝐵′ can be found by constructing a line ℓ parallel to 𝐴𝐴𝐴𝐴 through 𝐴𝐴′ and intersecting
that line with �����⃗
𝑂𝑂𝑂𝑂.
Since 𝐴𝐴𝐴𝐴 || ℓ, what does that imply about ������
𝐴𝐴′𝐵𝐵′ and ����
𝐴𝐴𝐴𝐴 ?



The two segments are also parallel as in the triangle side splitter theorem.
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
Now, suppose that ������
𝐴𝐴′𝐵𝐵′ || ����
𝐴𝐴𝐴𝐴 as in the picture below. Which method (parallel or ratio) does this statement
correspond to?


What does the parallel ⟹ ratio theorem imply about the point 𝐵𝐵′?


This statement corresponds to the parallel method because in the parallel method, only the endpoint 𝐴𝐴
of line segment 𝐴𝐴𝐴𝐴 is dilated from center 𝑂𝑂 by scale factor 𝑟𝑟 to get point 𝐴𝐴′. To draw ������
𝐴𝐴′𝐵𝐵′, a line is
′
′
�����⃗
drawn through 𝐴𝐴 that is parallel to 𝐴𝐴𝐴𝐴, and 𝐵𝐵 is the intersection of that line and 𝑂𝑂𝑂𝑂 .
This implies that 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐵𝐵) = 𝐵𝐵′ (i.e., 𝑂𝑂𝐵𝐵′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂).
What does 𝑂𝑂𝐵𝐵′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂 and 𝑂𝑂𝐴𝐴′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂 imply about 𝐴𝐴𝐴𝐴?

𝐴𝐴𝐴𝐴 splits the sides of △ 𝑂𝑂𝑂𝑂′𝐵𝐵.
Closing (2 minutes)
Ask students to summarize the main points of the lesson. Students may respond in writing, to a partner, or to the whole
class.

THE TRIANGLE SIDE SPLITTER THEOREM: A line segment splits two sides of a triangle proportionally if and only if it is
parallel to the third side.

Prior to this lesson we have used the ratio method and the parallel method separately to produce a scale
drawing. The triangle side splitter theorem is a way of saying that we can use either method because both
produce the same scale drawing.
Consider asking students to compare and contrast the two methods in their own words as a way of explaining how the
triangle side splitter theorem captures the mathematics of why each method produces the same scale drawing.
Lesson Summary
THE TRIANGLE SIDE SPLITTER THEOREM: A line segment splits two sides of a triangle proportionally if and only if it is
parallel to the third side.
Exit Ticket (5 minutes)
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Name
Date
Lesson 4: Comparing the Ratio Method with the Parallel Method
Exit Ticket
���� ∥ ����
In the diagram, 𝑋𝑋𝑋𝑋
𝐴𝐴𝐴𝐴 . Use the diagram to answer the following:
1.
If 𝐵𝐵𝐵𝐵 = 4, 𝐵𝐵𝐵𝐵 = 5, and 𝐵𝐵𝐵𝐵 = 6, what is 𝐵𝐵𝐵𝐵?
Not drawn to scale
2.
If 𝐵𝐵𝐵𝐵 = 9, 𝐵𝐵𝐵𝐵 = 15, and 𝐵𝐵𝐵𝐵 = 15, what is 𝑌𝑌𝑌𝑌?
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GEOMETRY
Exit Ticket Sample Solutions
��� ∥ ���
In the diagram, 𝑿𝑿𝑿𝑿
𝑨𝑨𝑨𝑨. Use the diagram to answer the following:
1.
If 𝑩𝑩𝑩𝑩 = 𝟒𝟒, 𝑩𝑩𝑩𝑩 = 𝟓𝟓, and 𝑩𝑩𝑩𝑩 = 𝟔𝟔, what is 𝑩𝑩𝑩𝑩?
𝑩𝑩𝑩𝑩 = 𝟕𝟕. 𝟓𝟓
2.
If 𝑩𝑩𝑩𝑩 = 𝟗𝟗, 𝑩𝑩𝑩𝑩 = 𝟏𝟏𝟏𝟏, and 𝑩𝑩𝑩𝑩 = 𝟏𝟏𝟏𝟏, what is 𝒀𝒀𝒀𝒀?
𝒀𝒀𝒀𝒀 = 𝟏𝟏𝟏𝟏
Not drawn to scale
Problem Set Sample Solutions
1.
Use the diagram to answer each part below.
a.
Measure the segments in the figure below to verify that the proportion
is true.
𝑶𝑶𝑨𝑨′ 𝑶𝑶𝑩𝑩′
=
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
Actual measurements may vary due to copying, but
students should state that the proportion is true.
b.
Is the proportion
𝑶𝑶𝑶𝑶
𝑶𝑶𝑨𝑨′
=
𝑶𝑶𝑶𝑶
also true? Explain algebraically.
𝑩𝑩′
also true? Explain algebraically.
𝑶𝑶𝑩𝑩′
The proportion is also true because the reciprocals of equivalent ratios are also equivalent.
c.
Is the proportion
𝑨𝑨𝑨𝑨′
𝑶𝑶𝑨𝑨′
=
𝑶𝑶𝑩𝑩′
True. 𝑶𝑶𝑨𝑨′ = 𝑶𝑶𝑶𝑶 + 𝑨𝑨𝑨𝑨′, and 𝑶𝑶𝑩𝑩′ = 𝑶𝑶𝑶𝑶 + 𝑩𝑩𝑩𝑩′. So, using the equivalent ratios in part (a):
𝑶𝑶𝑨𝑨′ 𝑶𝑶𝑩𝑩′
=
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶 + 𝑨𝑨𝑨𝑨′ 𝑶𝑶𝑶𝑶 + 𝑩𝑩′
=
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
′
𝑶𝑶𝑶𝑶 𝑨𝑨𝑨𝑨
𝑶𝑶𝑶𝑶 𝑩𝑩′
+
=
+
𝑶𝑶𝑶𝑶 𝑶𝑶𝑶𝑶 𝑶𝑶𝑶𝑶 𝑶𝑶𝑶𝑶
𝑩𝑩′
𝑨𝑨𝑨𝑨′
= 𝟏𝟏 +
𝟏𝟏 +
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
𝑨𝑨𝑨𝑨′ 𝑩𝑩′
=
.
𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶
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2.
Given the diagram below, 𝑨𝑨𝑨𝑨 = 𝟑𝟑𝟑𝟑, line 𝓵𝓵 is parallel to ����
𝑨𝑨𝑨𝑨, and the distance from ����
𝑨𝑨𝑨𝑨 to 𝓵𝓵 is 𝟐𝟐𝟐𝟐. Locate point 𝑪𝑪 on
line 𝓵𝓵 such that △ 𝑨𝑨𝑨𝑨𝑨𝑨 has the greatest area. Defend your answer.
l
The distance between two parallel lines is constant and in this case is 𝟐𝟐𝟐𝟐 units. ����
𝑨𝑨𝑨𝑨 serves as the base of all possible
triangles 𝑨𝑨𝑨𝑨𝑨𝑨. The area of a triangle is one-half the product of its base and its height. No matter where point 𝑪𝑪 is
located on line 𝓵𝓵, triangle 𝑨𝑨𝑨𝑨𝑨𝑨 has a base of 𝑨𝑨𝑨𝑨 = 𝟑𝟑𝟑𝟑 and a height (distance between the parallel lines) of 𝟐𝟐𝟐𝟐. All
possible triangles will therefore have an area of 𝟑𝟑𝟑𝟑𝟑𝟑 units2.
3.
Given △ 𝑿𝑿𝑿𝑿𝑿𝑿, ����
𝑿𝑿𝑿𝑿 and ����
𝒀𝒀𝒀𝒀 are partitioned into equal-length segments by the endpoints of the dashed segments as
shown. What can be concluded about the diagram?
The dashed lines joining the endpoints of the equal length segments
are parallel to ����
𝑿𝑿𝑿𝑿 by the triangle side splitter theorem.
4.
Given the diagram, 𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏, 𝑨𝑨𝑨𝑨 = 𝟔𝟔, 𝑩𝑩𝑩𝑩 = 𝟒𝟒, 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 = 𝒙𝒙°, and 𝒎𝒎∠𝑫𝑫 = 𝒙𝒙°, find 𝑪𝑪𝑪𝑪.
Since ∠𝑨𝑨𝑨𝑨𝑨𝑨 and ∠𝑫𝑫 are corresponding angles and are both 𝒙𝒙°, it
���� ∥ 𝑬𝑬𝑬𝑬
����. By the triangle side splitter theorem, ����
follows that 𝑩𝑩𝑩𝑩
𝑩𝑩𝑩𝑩 is a
proportional side splitter so
𝟏𝟏𝟏𝟏 𝟔𝟔
=
𝑪𝑪𝑪𝑪 𝟒𝟒
𝑨𝑨𝑨𝑨
𝑪𝑪𝑪𝑪
=
𝑨𝑨𝑨𝑨
.
𝑩𝑩𝑩𝑩
𝑪𝑪𝑪𝑪 = 𝟖𝟖
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5.
What conclusions can be drawn from the diagram shown to the
right? Explain.
Since
𝟑𝟑.𝟓𝟓
𝟐𝟐
=
𝟕𝟕
𝟒𝟒
and
𝟏𝟏𝟏𝟏.𝟓𝟓
𝟔𝟔
𝟕𝟕 𝑼𝑼𝑼𝑼
= ,
proportional side splitter.
𝟒𝟒 𝑼𝑼𝑼𝑼
=
This provides several conclusions:
1.
𝑼𝑼𝑼𝑼
����� is a
, so the side splitter 𝑽𝑽𝑽𝑽
𝑼𝑼𝑼𝑼
The side splitter �����
𝑽𝑽𝑽𝑽 is parallel to the third side of the triangle
by the triangle side splitter theorem.
2.
∠𝒀𝒀 ≅ ∠𝑼𝑼𝑼𝑼𝑼𝑼 and ∠𝑿𝑿 ≅ ∠𝑼𝑼𝑼𝑼𝑼𝑼 because corresponding angles
formed by parallel lines cut by a transversal are congruent.
3.
△ 𝑼𝑼𝑼𝑼𝑼𝑼 is a scale drawing of △ 𝑼𝑼𝑼𝑼𝑼𝑼 with a scale factor of .
4.
𝟕𝟕
𝟒𝟒
𝟕𝟕
𝑿𝑿𝑿𝑿 = 𝒖𝒖 because corresponding lengths in scale drawings are
𝟒𝟒
proportional.
6.
Parallelogram 𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷 is shown. Two triangles are formed by a diagonal within the parallelogram. Identify those
triangles, and explain why they are guaranteed to have the same areas.
Opposite sides of a parallelogram are parallel and have the same length, so 𝑸𝑸𝑸𝑸 = 𝑷𝑷𝑷𝑷, and the distance between ����
𝑸𝑸𝑸𝑸
𝑷𝑷𝑷𝑷 forms △ 𝑷𝑷𝑷𝑷𝑷𝑷 and △ 𝑷𝑷𝑷𝑷𝑷𝑷 that have the same base length and the same
and ����
𝑷𝑷𝑷𝑷 is a constant, 𝒉𝒉. Diagonal ����
height and, therefore, the same area.
���� forms △ 𝑷𝑷𝑷𝑷𝑷𝑷 and △ 𝑹𝑹𝑹𝑹𝑹𝑹 that have the same base length and the same height and, therefore, the
Diagonal 𝑸𝑸𝑸𝑸
same area.
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7.
In the diagram to the right, 𝑯𝑯𝑯𝑯 = 𝟑𝟑𝟑𝟑 and 𝑮𝑮𝑮𝑮 = 𝟒𝟒𝟒𝟒. If the ratio of the areas of the triangles is
𝑱𝑱𝑱𝑱, 𝑮𝑮𝑮𝑮, 𝑮𝑮𝑮𝑮, and 𝑱𝑱𝑱𝑱.
����
𝑯𝑯𝑯𝑯 is the altitude of both triangles, so the bases
of the triangles will be in the ratio of the areas of
���� and �����
𝑮𝑮𝑮𝑮, so
the triangles. ���
𝑮𝑮𝑮𝑮 is composed of 𝑱𝑱𝑱𝑱
𝑮𝑮𝑮𝑮 = 𝟒𝟒𝟒𝟒 = 𝑱𝑱𝑱𝑱 + 𝑮𝑮𝑮𝑮.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 △𝑮𝑮𝑮𝑮𝑮𝑮
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 △𝑱𝑱𝑱𝑱𝑱𝑱
𝟓𝟓
= , find
𝟗𝟗
𝑮𝑮𝑮𝑮 𝟓𝟓
=
𝑱𝑱𝑱𝑱 𝟗𝟗
𝑮𝑮𝑮𝑮
𝟓𝟓
=
𝟒𝟒𝟒𝟒 − 𝑮𝑮𝑮𝑮 𝟗𝟗
𝟗𝟗(𝑮𝑮𝑮𝑮) = 𝟓𝟓(𝟒𝟒𝟒𝟒 − 𝑮𝑮𝑮𝑮)
𝟗𝟗(𝑮𝑮𝑮𝑮) = 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟓𝟓(𝑮𝑮𝑮𝑮)
𝟏𝟏𝟏𝟏(𝑮𝑮𝑮𝑮) = 𝟐𝟐𝟐𝟐𝟐𝟐
𝑮𝑮𝑮𝑮 = 𝟏𝟏𝟏𝟏, 𝑱𝑱𝑱𝑱 = 𝟐𝟐𝟐𝟐
By the Pythagorean theorem, 𝑱𝑱𝑱𝑱 = 𝟒𝟒𝟒𝟒 and 𝑮𝑮𝑮𝑮 = 𝟑𝟑𝟑𝟑.
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Lesson 5: Scale Factors
Student Outcomes


Students prove the dilation theorem: If a dilation with center 𝑂𝑂 and scale factor 𝑟𝑟 sends point 𝑃𝑃 to 𝑃𝑃′ and 𝑄𝑄 to
𝑄𝑄′, then |𝑃𝑃′ 𝑄𝑄′ | = 𝑟𝑟|𝑃𝑃𝑃𝑃|. Furthermore, if 𝑟𝑟 ≠ 1 and 𝑂𝑂, 𝑃𝑃, and 𝑄𝑄 are the vertices of a triangle, then
⃖����⃗
𝑃𝑃𝑃𝑃 ∥ ⃖�������⃗
𝑃𝑃′ 𝑄𝑄′ .
Students use the dilation theorem to show that the scale drawings constructed using the ratio and parallel
methods have a scale factor that is the same as the scale factor for the dilation.
Lesson Notes
In the previous lesson, students learned about the triangle side splitter theorem, which is now used to prove the dilation
theorem. In Grade 8, students learned about the fundamental theorem of similarity (FTS), which contains the concepts
that are in the dilation theorem presented in this lesson. It is called the dilation theorem at this point in the module
because students have not yet entered into the formal study of similarity. Some students may recall FTS from Grade 8 as
they enter into the discussion following the Opening Exercise. Their prior knowledge of this topic will strengthen as they
prove the dilation theorem.
Classwork
Opening Exercise (5 minutes)
Have students participate in a Quick Write. A Quick Write is an exercise where students
write as much as they know about a particular topic without concern for correct grammar
or spelling. The purpose of a Quick Write is for students to bring to the forefront all of the
information they can recall about a particular topic. Show the prompt below, and then
allow students to write for two minutes. Give students one minute to share Quick Writes
with a partner. Then select students to share their thoughts with the class.
Opening Exercise
Quick Write: Describe how a figure is transformed under a dilation with a scale factor = 𝟏𝟏, 𝒓𝒓 > 𝟏𝟏,
and 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏.
Scaffolding:
An alternate exercise is
described on the following
page. There are visual aids to
help students make sense of
the prompt. Student responses
should be the same for those
who need the visual as for
those who do not.
Sample student responses should include the following points:
A dilation with a scale factor of 𝒓𝒓 = 𝟏𝟏 produces an image that is congruent to the original figure.
A dilation with a scale factor of 𝒓𝒓 > 𝟏𝟏 produces an image that is larger in size than the original, but the angles of the
figure are unchanged, and the lengths of the larger figure are proportional with the original figure.
A dilation with a scale factor of 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏 produces an image that is smaller in size than the original, but the angles of
the figure are unchanged, and the lengths of the smaller figure are proportional with the original figure.
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GEOMETRY
As an alternative exercise, consider showing students the following three diagrams and asking them to describe how
each of the figures has been transformed with respect to the scale factor of dilation. State that in each of the figures,
△ 𝐴𝐴𝐴𝐴𝐴𝐴 has been dilated from center 𝑂𝑂 by some scale factor to produce the image △ 𝐴𝐴′ 𝐵𝐵′ 𝐶𝐶 ′ . Some groups of students
may benefit from seeing only one figure at a time and having a partner discussion prior to sharing their thoughts with
the whole class. The sample responses above apply to this alternative exercise.
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Discussion (20 minutes)
State the dilation theorem for students, and then allow them time to process what the theorem means.
Discussion
DILATION THEOREM: If a dilation with center 𝑶𝑶 and scale factor 𝒓𝒓 sends point 𝑷𝑷 to 𝑷𝑷′ and 𝑸𝑸 to 𝑸𝑸′, then |𝑷𝑷′ 𝑸𝑸′ | = 𝒓𝒓|𝑷𝑷𝑷𝑷|.
′ 𝑸𝑸′ .
⃖�������⃗
Furthermore, if 𝒓𝒓 ≠ 𝟏𝟏 and 𝑶𝑶, 𝑷𝑷, and 𝑸𝑸 are the vertices of a triangle, then ⃖����⃗
𝑷𝑷𝑷𝑷||𝑷𝑷

Describe in your own words what the dilation theorem states.
Provide students time to discuss the meaning of the theorem with a partner, and then select students to share their
thoughts with the class. Shown below is a sample student response. Consider scripting responses that students might
give. Be sure to elicit facts (1) and (2) from students to ensure their understanding of the theorem. Some students may
comment on the lengths of segments 𝑂𝑂𝑂𝑂, 𝑂𝑂𝑃𝑃′ , 𝑂𝑂𝑂𝑂, and 𝑂𝑂𝑄𝑄′ . These comments should be acknowledged, but make it
clear that the dilation theorem focuses on the parts of a diagram that were not dilated; that is, the points 𝑃𝑃 and 𝑄𝑄 were
dilated, so there is an expectation of what that means and where the images of those points will end up. The focus now
with respect to the theorem is on the segments 𝑃𝑃𝑃𝑃 and 𝑃𝑃′𝑄𝑄′ and what happens to them when the points 𝑃𝑃 and 𝑄𝑄 are
dilated.


The dilation theorem states two things: (1) If two points, 𝑃𝑃 and 𝑄𝑄, are dilated from the same center
using the same scale factor, then the segment formed when you connect the dilated points 𝑃𝑃′ and 𝑄𝑄′ is
exactly the length of ����
𝑃𝑃𝑃𝑃 multiplied by the scale factor, and (2) the lines containing the segments 𝑃𝑃′𝑄𝑄′
and 𝑃𝑃𝑃𝑃 are parallel or equal.
3
2
For example, if points 𝑃𝑃 and 𝑄𝑄 are dilated from center 𝑂𝑂 by a scale factor of 𝑟𝑟 = , then the lines
3
2
containing the segments 𝑃𝑃′𝑄𝑄′ and 𝑃𝑃𝑃𝑃 are parallel, and 𝑃𝑃′ 𝑄𝑄′ = ⋅ 𝑃𝑃𝑃𝑃, as shown below.

The dilation theorem is an important theorem. To prove that the theorem is true, we use the triangle side
splitter theorem (from the previous lesson) several times. When we use it, we consider different pairs of
parallel lines.
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GEOMETRY

Consider the dilation theorem for 𝑟𝑟 = 1. What impact does a scale factor of 𝑟𝑟 = 1 have on a figure and its
image? (Encourage students to use what they wrote during the Quick Write activity from the Opening
Exercise.)


Consider the dilation theorem with the scenario of 𝑂𝑂, 𝑃𝑃, and 𝑄𝑄 not being the vertices of a triangle but in fact
collinear points, as shown below. What impact does that have on the figure and its image?


When the scale factor is 𝑟𝑟 = 1, then the figure remains unchanged. The scale factor being equal to 1
means that each point is taken to itself, that is, a congruence.
If the points 𝑂𝑂, 𝑃𝑃, and 𝑄𝑄 are collinear, then the dilated points 𝑃𝑃′ and 𝑄𝑄′ remain on the line, and center
𝑂𝑂 does not move at all.
A dilation of collinear points from a center with scale factor 𝑟𝑟 can be used to construct the number line.
To clarify the statement “can be used to construct the number line,” consider showing the series of diagrams below.
Begin with the center of dilation at zero and the point 𝑥𝑥 being 1 on the number line, and consider dilations of the point 𝑥𝑥
from the center at zero. The successive lines below show how the point 𝑥𝑥 moves when the scale factor of dilation is
increased from 𝑟𝑟 = 2 to 𝑟𝑟 = 4.
MP.8

How is the location of the dilated point 𝑥𝑥 ′ related to 𝑟𝑟 and 𝑥𝑥?

The location of the dilated point was exactly 𝑟𝑟𝑟𝑟.
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
We could continue to dilate the point 𝑥𝑥 by different scale factors to get all of the positive numbers on the
number line. We can construct each fraction on the number line by dilating 𝑥𝑥 by that fraction. A rotation of
180° around the center at zero of the whole numbers and fractions is a way to construct any rational number.
If we considered rational and irrational scale factors, then we get all of the real numbers, that is, the entire real
number line.
Note that we do not state that the rational and real number lines can be achieved using a negative scale factor of a
dilation. Applying a negative scale factor has the same effect as a 180° rotation (or a reflection across zero) composed
with a dilation whose scale factor is the absolute value of the negative scale factor. We do not complicate the issue by
introducing a new formalism “negative dilation” into the lesson and, more broadly, the module. Therefore, all dilations
have a positive scale factor.

MP.3
Assume that points 𝑃𝑃 and 𝑄𝑄 are numbers 𝑥𝑥 and 𝑦𝑦 on a number line, and point 𝑂𝑂, the center of dilation, is the
number zero. The points 𝑃𝑃′ and 𝑄𝑄′ correspond to the numbers 𝑟𝑟𝑟𝑟 and 𝑟𝑟𝑟𝑟. Explain why the distance between
𝑟𝑟𝑟𝑟 and 𝑟𝑟𝑟𝑟 is 𝑟𝑟 times the distance between 𝑥𝑥 and 𝑦𝑦. Use a diagram below if necessary.
Provide students time to discuss with a partner. Then select students to share with the class. A sample student
response is shown below.

The distance between points on the number line is the absolute value of
their difference. For example, the distance between −2 and 6 is
|−2 − 6| = 8 units. Then the distance between 𝑟𝑟𝑟𝑟 and 𝑟𝑟𝑟𝑟 is |𝑟𝑟𝑟𝑟 − 𝑟𝑟𝑟𝑟|.
By the distributive property, |𝑟𝑟𝑟𝑟 − 𝑟𝑟𝑟𝑟| = |𝑟𝑟(𝑥𝑥 − 𝑦𝑦)|, and since 𝑟𝑟 must be
positive, |𝑟𝑟(𝑥𝑥 − 𝑦𝑦)| = 𝑟𝑟|𝑥𝑥 − 𝑦𝑦|. In terms of the numeric example, if
𝑥𝑥 = −2, 𝑦𝑦 = 6, and 𝑟𝑟 is the scale factor of dilation, then
|𝑟𝑟(−2) − 𝑟𝑟(6)| = |𝑟𝑟(−2 − 6)| = 𝑟𝑟|−2 − 6|.
Scaffolding:
Students may need to see one
or more numerical examples
similar to the one in the sample
response to arrive at the
general statement.
Therefore, the distance between 𝑟𝑟𝑟𝑟 and 𝑟𝑟𝑟𝑟 is exactly 𝑟𝑟 times the distance
between 𝑥𝑥 and 𝑦𝑦.
The remaining part of this discussion uses the triangle side splitter theorem to prove the dilation theorem. The goal is to
use knowledge about dilation, the triangle side splitter theorem, and the properties of parallelograms to explain why
′ 𝑄𝑄 ′ . Show each step of the proof, and ask students to provide reasoning for the steps
⃖�������⃗
|𝑃𝑃′ 𝑄𝑄′ | = 𝑟𝑟|𝑃𝑃𝑃𝑃| and ⃖����⃗
𝑃𝑃𝑃𝑃 ||𝑃𝑃
independently, with a partner, or in small groups.
Now consider the dilation theorem when 𝑶𝑶, 𝑷𝑷, and 𝑸𝑸 are the vertices of △ 𝑶𝑶𝑶𝑶𝑶𝑶. Since 𝑷𝑷′ and 𝑸𝑸′ come from a dilation
with scale factor 𝒓𝒓 and center 𝑶𝑶, we have
𝑶𝑶𝑷𝑷′
𝑶𝑶𝑶𝑶
=
𝑶𝑶𝑸𝑸′
𝑶𝑶𝑶𝑶
= 𝒓𝒓.
There are two cases that arise; recall what you wrote in your Quick Write. We must consider the case when 𝒓𝒓 > 𝟏𝟏 and
when 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏. Let’s begin with the latter.
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GEOMETRY
Dilation Theorem Proof, Case 1
Statements
1.
2.
Reasons/Explanations
A dilation with center 𝑶𝑶 and scale factor 𝒓𝒓 sends
point 𝑷𝑷 to 𝑷𝑷′ and 𝑸𝑸 to 𝑸𝑸′ .
𝑶𝑶𝑷𝑷′
𝑶𝑶𝑶𝑶
=
𝑶𝑶𝑸𝑸′
𝑶𝑶𝑶𝑶
= 𝒓𝒓
3.
⃖�������⃗
⃖����⃗
𝑷𝑷𝑷𝑷‖𝑷𝑷′𝑸𝑸′
4.
A dilation with center 𝑷𝑷 and scale factor
𝑷𝑷𝑷𝑷′
𝑷𝑷𝑷𝑷
𝑷𝑷′𝑹𝑹.
point 𝑶𝑶 to 𝑷𝑷 and point 𝑸𝑸 to 𝑹𝑹. Draw �����
′
sends
1.
Given
2.
By definition of dilation: Corresponding lengths are
proportional, and the ratio of the corresponding
lengths are equal to the scale factor 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏.
3.
By the triangle side splitter theorem
4.
By definition of dilation
5.
�����||𝑶𝑶𝑶𝑶′
�����
𝑷𝑷′𝑹𝑹
5.
By the triangle side splitter theorem
6.
𝑹𝑹𝑹𝑹′𝑸𝑸′𝑸𝑸 is a parallelogram.
6.
By definition of parallelogram
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Consider asking students to state what kind of figure is formed by 𝑅𝑅𝑅𝑅′𝑄𝑄′𝑄𝑄 and stating what they know about the
properties of parallelograms before continuing with the next part of the discussion. Some students may need to see a
parallelogram as an isolated figure as opposed to a figure within the triangles.
7.
𝑹𝑹𝑹𝑹 = 𝑷𝑷′𝑸𝑸′
7.
Opposite sides of parallelogram 𝑹𝑹𝑹𝑹′𝑸𝑸′𝑸𝑸 are equal
in length.
8.
𝑹𝑹𝑹𝑹
8.
9.
𝑹𝑹𝑹𝑹
By the triangle side splitter theorem, �����
𝑷𝑷′𝑹𝑹 splits the
sides of △ 𝑷𝑷𝑷𝑷𝑷𝑷 proportionally.
𝑷𝑷𝑷𝑷
𝑷𝑷𝑷𝑷
=
𝑷𝑷′ 𝑶𝑶
𝑷𝑷𝑷𝑷
9.
= 𝒓𝒓
10. 𝑹𝑹𝑹𝑹 = 𝒓𝒓 ⋅ 𝑷𝑷𝑷𝑷
10. Multiplication property of equality
11. 𝑷𝑷′ 𝑸𝑸′ = 𝒓𝒓 ⋅ 𝑷𝑷𝑷𝑷

By substitution (line 2)
11. By substitution (line 7)
This concludes the proof of Case 1 (when the scale factor of dilation is 0 < 𝑟𝑟 < 1) of the dilation theorem
because we have shown that the dilation with center 𝑂𝑂 and scale factor 0 < 𝑟𝑟 < 1 sends point 𝑃𝑃 to 𝑃𝑃′ and 𝑄𝑄
′ 𝑄𝑄 ′ .
⃖�������⃗
𝑃𝑃𝑃𝑃 ||𝑃𝑃
to 𝑄𝑄′, then |𝑃𝑃′ 𝑄𝑄′ | = 𝑟𝑟|𝑃𝑃𝑃𝑃|, and since 𝑂𝑂, 𝑃𝑃, and 𝑄𝑄 are the vertices of a triangle, then ⃖����⃗
Exercises (10 minutes)
Students complete Exercise 1 independently or in pairs. Upon completion of Exercise 1, select students to share their
proofs with the class. Then have students complete Exercises 2–4. Review the work for Exercises 2–4. Note that
Exercise 4 is revisited in the Closing of the lesson as well.
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Exercises
1.
Prove Case 2: If 𝑶𝑶, 𝑷𝑷, and 𝑸𝑸 are the vertices of a triangle and 𝒓𝒓 > 𝟏𝟏, show that
⃖�������⃗ and (b) 𝑷𝑷′ 𝑸𝑸′ = 𝒓𝒓𝒓𝒓𝒓𝒓. Use the diagram below when writing your proof.
(a) ⃖����⃗
𝑷𝑷𝑷𝑷 || 𝑷𝑷′𝑸𝑸′
Scaffolding:
 Consider having students
compare and contrast this
exercise with the proof they
just finished. They should
notice that the only difference
is the scale factor of dilation.
 It may be necessary to ask
students the following
questions to guide their
thinking:
����� and �����
Case 2: If 𝒓𝒓 > 𝟏𝟏, ����
𝑷𝑷𝑷𝑷 proportionally splits the sides 𝑶𝑶𝑶𝑶′
𝑶𝑶𝑶𝑶′ of 𝜟𝜟𝜟𝜟𝜟𝜟′𝑸𝑸′. By the
⃖�������⃗
⃖����⃗
���� that
triangle side splitter theorem, 𝑷𝑷′𝑸𝑸′ || 𝑷𝑷𝑷𝑷. Next, we construct a line segment 𝑷𝑷𝑷𝑷
splits the sides of △ 𝑶𝑶𝑶𝑶′𝑸𝑸′ and is parallel to side �����
𝑶𝑶𝑶𝑶′. By the triangle side splitter
���� splits the sides proportionally and so
theorem, 𝑷𝑷𝑷𝑷
𝑷𝑷′ 𝑸𝑸′
𝑹𝑹𝑸𝑸′
=
𝑶𝑶𝑷𝑷′
𝑶𝑶𝑶𝑶
= 𝒓𝒓. Now
𝑹𝑹𝑸𝑸′ = 𝑷𝑷𝑷𝑷 because the lengths of opposite sides of parallelogram 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹′ are equal.
So
𝑷𝑷′ 𝑸𝑸′
𝑷𝑷𝑷𝑷
= 𝒓𝒓 and |𝑷𝑷′𝑸𝑸′| = 𝒓𝒓|𝑷𝑷𝑷𝑷|.
2.
a.
Produce a scale drawing of △ 𝑳𝑳𝑳𝑳𝑳𝑳 using either the ratio or parallel method
𝟑𝟑
In the last proof, we
����� so that it split
constructed 𝑃𝑃′𝑅𝑅
����
the sides 𝑃𝑃𝑃𝑃 and ����
𝑃𝑃𝑃𝑃 of △ 𝑂𝑂𝑂𝑂𝑂𝑂
����. How
and is parallel to side 𝑂𝑂𝑂𝑂
does that differ for this figure?
If we apply the triangle side
splitter theorem, what do we
What properties of a
parallelogram will be useful in
proving the dilation theorem?
with point 𝑴𝑴 as the center and a scale factor of .
𝟐𝟐
b.
������, and then measure its length directly using a ruler.
Use the dilation theorem to predict the length of 𝑳𝑳′𝑵𝑵′
Lengths of the line segments can vary due to copy production. By the dilation theorem, it should be predicted
𝟑𝟑
𝟐𝟐
that 𝑳𝑳′ 𝑵𝑵′ = 𝑳𝑳𝑳𝑳. This is confirmed using direct measurement.
c.
Does the dilation theorem appear to hold true?
The dilation theorem does hold true.
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3.
𝟏𝟏
Produce a scale drawing of △ 𝑿𝑿𝑿𝑿𝑿𝑿 with point 𝑿𝑿 as the center and a scale factor of . Use the dilation theorem to
𝟒𝟒
predict 𝒀𝒀′𝒁𝒁′, and then measure its length directly using a ruler. Does the dilation theorem appear to hold true?
Lengths of the line segments can vary due to copy production. By the dilation theorem, it should be predicted that
𝟏𝟏
𝟒𝟒
𝒀𝒀′ 𝒁𝒁′ = 𝒀𝒀𝒀𝒀. This is confirmed using direct measurement.
4.
Given the diagram below, determine if △ 𝑫𝑫𝑫𝑫𝑫𝑫 is a scale drawing of △ 𝑫𝑫𝑫𝑫𝑫𝑫. Explain why or why not.
No. If △ 𝑫𝑫𝑫𝑫𝑫𝑫 was a scale drawing of △ 𝑫𝑫𝑫𝑫𝑫𝑫, then
𝟑𝟑. 𝟐𝟐
𝑫𝑫𝑫𝑫
=
≈ 𝟎𝟎. 𝟒𝟒𝟒𝟒
𝑫𝑫𝑫𝑫 𝟔𝟔. 𝟗𝟗𝟗𝟗
𝑫𝑫𝑫𝑫
𝑫𝑫𝑫𝑫
=
𝑬𝑬𝑬𝑬
𝑮𝑮𝑮𝑮
by the dilation theorem.
𝟓𝟓. 𝟗𝟗
𝑬𝑬𝑬𝑬
=
≈ 𝟎𝟎. 𝟓𝟓𝟓𝟓
𝑮𝑮𝑮𝑮 𝟏𝟏𝟏𝟏. 𝟗𝟗
The ratios of the corresponding sides are not equivalent, so the drawing is not a scale drawing.
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Closing (5 minutes)
Revisit Exercise 4 above. Ask students whether or not ����
𝐸𝐸𝐸𝐸 is parallel to ����
𝐺𝐺𝐺𝐺 . Students should respond that because the
side lengths were not in proportion that △ 𝐷𝐷𝐷𝐷𝐷𝐷 is not a scale drawing of △ 𝐷𝐷𝐷𝐷𝐷𝐷, and we would not expect the lines
containing ����
𝐸𝐸𝐸𝐸 and ����
𝐺𝐺𝐺𝐺 to be parallel. Next, ask students to verbally complete the following in order to informally assess
their understanding of the dilation theorem and its proof.

Restate the dilation theorem in your own words.

Explain how the triangle side splitter theorem was used to prove the dilation theorem.
If time permits, ask students the following question.

We discussed how dilation was used to produce a number line. What other everyday objects may have been
created in this manner?

Any everyday object that is divided into equal parts can be produced using dilation, for example, a ruler
or thermometer.
Exit Ticket (5 minutes)
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Name
Date
Lesson 5: Scale Factors
Exit Ticket
1.
describe two facts that are known about �����
𝑅𝑅′𝑌𝑌′.
2.
3
Two different points 𝑅𝑅 and 𝑌𝑌 are dilated from 𝑆𝑆 with a scale factor of , and 𝑅𝑅𝑅𝑅 = 15. Use the dilation theorem to
4
Which diagram(s) below represents the information given in Problem 1? Explain your answer(s).
a.
b.
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Exit Ticket Sample Solutions
1.
𝟑𝟑
Two different points 𝑹𝑹 and 𝒀𝒀 are dilated from 𝑺𝑺 with a scale factor of , and 𝑹𝑹𝑹𝑹 = 𝟏𝟏𝟏𝟏. Use the dilation theorem to
������.
describe two facts that are known about 𝑹𝑹′𝒀𝒀′
𝟑𝟑
𝟒𝟒
𝟒𝟒
𝟑𝟑
𝟒𝟒
������ ∥ 𝑹𝑹𝒀𝒀
���� or ⃖�������⃗
⃖����⃗.
By the dilation theorem, 𝑹𝑹′ 𝒀𝒀′ = 𝑹𝑹𝑹𝑹, so 𝑹𝑹′ 𝒀𝒀′ = (𝟏𝟏𝟏𝟏) = 𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐, and 𝑹𝑹′𝒀𝒀′
𝑹𝑹′𝒀𝒀′ = 𝑹𝑹𝑹𝑹
2.
a.
Diagram (a) can be a dilation with scale factor
𝟑𝟑
𝟒𝟒
since 𝑹𝑹′𝑺𝑺 and 𝒀𝒀′𝑺𝑺 appear to be
𝟑𝟑
𝟒𝟒
of the distances 𝑹𝑹𝑹𝑹 and 𝒀𝒀𝒀𝒀,
������ = 𝑹𝑹𝑹𝑹
����.
respectively. However, because both line segments lie on the same line, 𝑹𝑹′𝒀𝒀′
b.
Diagram (b) could be a dilation with scale factor
𝟑𝟑
𝟒𝟒
since 𝑹𝑹′𝑺𝑺 and 𝒀𝒀′𝑺𝑺 appear to be
������ ∥ 𝑹𝑹𝑹𝑹
����.
𝒀𝒀𝒀𝒀, respectively. Because 𝑺𝑺, 𝒀𝒀, and 𝑹𝑹 are vertices of a triangle, 𝑹𝑹′𝒀𝒀′
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𝟑𝟑
𝟒𝟒
of the distances 𝑹𝑹𝑹𝑹 and
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Problem Set Sample Solutions
1.
△ 𝑨𝑨𝑨𝑨′𝑪𝑪′ is a dilation of △ 𝑨𝑨𝑨𝑨𝑨𝑨 from vertex 𝑨𝑨, and 𝑪𝑪𝑪𝑪′ = 𝟐𝟐. Use the given information in each part and the diagram
to find 𝑩𝑩′𝑪𝑪′.
a.
𝑨𝑨𝑨𝑨 = 𝟗𝟗, 𝑨𝑨𝑨𝑨 = 𝟒𝟒, and 𝑩𝑩𝑩𝑩 = 𝟕𝟕
𝑩𝑩′ 𝑪𝑪′ = 𝟏𝟏𝟏𝟏
b.
𝑨𝑨𝑨𝑨 = 𝟒𝟒, 𝑨𝑨𝑨𝑨 = 𝟗𝟗, and 𝑩𝑩𝑩𝑩 = 𝟕𝟕
𝑩𝑩′ 𝑪𝑪′ = 𝟖𝟖
c.
𝟓𝟓
𝟗𝟗
𝑨𝑨𝑨𝑨 = 𝟕𝟕, 𝑨𝑨𝑨𝑨 = 𝟗𝟗, and 𝑩𝑩𝑩𝑩 = 𝟒𝟒
𝑩𝑩′ 𝑪𝑪′ = 𝟒𝟒
d.
𝟏𝟏
𝟐𝟐
𝟖𝟖
𝟗𝟗
Drawing Not to Scale
𝑨𝑨𝑨𝑨 = 𝟕𝟕, 𝑨𝑨𝑨𝑨 = 𝟒𝟒, and 𝑩𝑩𝑩𝑩 = 𝟗𝟗
𝑩𝑩′ 𝑪𝑪′ = 𝟏𝟏𝟏𝟏
e.
𝑨𝑨𝑨𝑨 = 𝟒𝟒, 𝑨𝑨𝑨𝑨 = 𝟕𝟕, and 𝑩𝑩𝑩𝑩 = 𝟗𝟗
𝑩𝑩′ 𝑪𝑪′ = 𝟏𝟏𝟏𝟏
f.
𝑨𝑨𝑨𝑨 = 𝟗𝟗, 𝑨𝑨𝑨𝑨 = 𝟕𝟕, and 𝑩𝑩𝑩𝑩 = 𝟒𝟒
𝑩𝑩′ 𝑪𝑪′ = 𝟓𝟓
2.
𝟒𝟒
𝟕𝟕
𝟏𝟏
𝟕𝟕
Given the diagram, ∠𝑪𝑪𝑪𝑪𝑪𝑪 ≅ ∠𝑪𝑪𝑪𝑪𝑪𝑪. Find 𝑨𝑨𝑨𝑨.
����
���� because when two lines are cut
𝑭𝑭𝑭𝑭 ∥ 𝑨𝑨𝑨𝑨
by a transversal, such that the
corresponding angles are congruent, then
the lines are parallel.
△ 𝑪𝑪𝑪𝑪𝑪𝑪 is a scale drawing of △ 𝑪𝑪𝑪𝑪𝑪𝑪 by
the parallel method.
𝑭𝑭 is the image of 𝑨𝑨 after a dilation from
𝟖𝟖
𝑪𝑪 with a scale factor of .
𝟓𝟓
𝟖𝟖
𝑭𝑭𝑭𝑭 = (𝑨𝑨𝑨𝑨) by the dilation theorem.
𝟓𝟓
𝟖𝟖
(𝑨𝑨𝑨𝑨)
𝟓𝟓
𝟓𝟓
𝟓𝟓 𝟖𝟖
(𝟏𝟏𝟏𝟏) = � � (𝑨𝑨𝑨𝑨)
𝟖𝟖
𝟖𝟖 𝟓𝟓
𝟏𝟏𝟏𝟏
= 𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐
𝟏𝟏𝟏𝟏 =
𝟕𝟕. 𝟓𝟓 = 𝑨𝑨𝑨𝑨
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3.
Use the diagram to answer each part below.
a.
����.
△ 𝑶𝑶𝑶𝑶′𝑸𝑸′ is the dilated image of △ 𝑶𝑶𝑶𝑶𝑶𝑶 from point 𝑶𝑶 with a scale factor of 𝒓𝒓 > 𝟏𝟏. Draw a possible 𝑷𝑷𝑷𝑷
Placement of the segment will vary; however, by the dilation theorem, ����
𝑷𝑷𝑷𝑷 must be drawn parallel to ������
𝑷𝑷′𝑸𝑸′,
and because scale factor 𝒓𝒓 > 𝟏𝟏, point 𝑷𝑷 must be between 𝑶𝑶 and 𝑷𝑷′, and point 𝑸𝑸 must be between 𝑶𝑶 and 𝑸𝑸′.
b.
△ 𝑶𝑶𝑶𝑶′′𝑸𝑸′′ is the dilated image of △ 𝑶𝑶𝑶𝑶𝑶𝑶 from point 𝑶𝑶 with a scale factor 𝒌𝒌 > 𝒓𝒓. Draw a possible ��������
𝑷𝑷′′𝑸𝑸′′.
Placement of the segment will vary; however, by the dilation theorem, ��������
𝑷𝑷′′𝑸𝑸′′ must be drawn parallel to ����
𝑷𝑷𝑷𝑷,
and because scale factor 𝒌𝒌 > 𝒓𝒓, point 𝑷𝑷′′ must be placed such that 𝑷𝑷′ is between 𝑷𝑷 and 𝑷𝑷′′ and 𝑸𝑸′′ placed
such that 𝑸𝑸′ is between 𝑸𝑸 and 𝑸𝑸′′.
Possible solutions:
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4.
����, 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 (△ 𝑹𝑹𝑹𝑹𝑹𝑹) = 𝟏𝟏𝟏𝟏 units2, and 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑶𝑶𝑶𝑶) = 𝟐𝟐𝟐𝟐 units2, find 𝑹𝑹𝑹𝑹.
Given the diagram to the right, ����
𝑹𝑹𝑹𝑹 ∥ 𝑷𝑷𝑷𝑷
∆𝑹𝑹𝑹𝑹𝑹𝑹 and ∆𝑶𝑶𝑶𝑶𝑶𝑶 have the same altitude, so the lengths
of their bases are proportional to their areas.
𝟏𝟏
(𝑹𝑹𝑹𝑹)𝟔𝟔 = 𝟏𝟏𝟏𝟏
𝟐𝟐
𝑹𝑹𝑹𝑹 = 𝟓𝟓
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑹𝑹𝑹𝑹𝑹𝑹) =
Since the triangles have the same height 𝟔𝟔, the ratio of
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑹𝑹𝑹𝑹𝑹𝑹): 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑶𝑶𝑶𝑶𝑶𝑶) is equal to the ratio
𝑹𝑹𝑹𝑹: 𝑹𝑹𝑹𝑹.
𝟓𝟓
𝟏𝟏𝟏𝟏
=
𝟐𝟐𝟐𝟐 𝑶𝑶𝑶𝑶
𝑶𝑶𝑶𝑶 = 𝟕𝟕
𝑶𝑶𝑶𝑶 = 𝑶𝑶𝑶𝑶 + 𝑷𝑷𝑷𝑷
𝟕𝟕 = 𝑶𝑶𝑶𝑶 + 𝟑𝟑
𝑶𝑶𝑶𝑶 = 𝟒𝟒
����, then △ 𝑶𝑶𝑶𝑶𝑶𝑶 is a scale drawing of △ 𝑶𝑶𝑶𝑶𝑶𝑶
If ����
𝑹𝑹𝑹𝑹 ∥ 𝑷𝑷𝑷𝑷
𝟕𝟕
from 𝑶𝑶 with a scale factor of . Therefore, 𝑹𝑹 is the
𝟒𝟒
image of 𝑷𝑷, and 𝑺𝑺 is the image of 𝑸𝑸 under a dilation
from point 𝑶𝑶. By the dilation theorem:
𝟕𝟕
(𝟓𝟓)
𝟒𝟒
𝟑𝟑𝟑𝟑
𝑹𝑹𝑹𝑹 =
𝟒𝟒
𝟑𝟑
𝑹𝑹𝑹𝑹 = 𝟖𝟖
𝟒𝟒
𝑹𝑹𝑹𝑹 =
𝟑𝟑
𝟒𝟒
The length of segment 𝑹𝑹𝑹𝑹 is 𝟖𝟖 units.
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5.
���� such that ����
Using the information given in the diagram and 𝑼𝑼𝑼𝑼 = 𝟗𝟗, find 𝒁𝒁 on 𝑿𝑿𝑿𝑿
𝒀𝒀𝒀𝒀 is an altitude. Then, find 𝒀𝒀𝒀𝒀
and 𝑿𝑿𝑿𝑿.
���� is perpendicular to 𝑼𝑼𝑼𝑼
����; thus, 𝒎𝒎∠𝒀𝒀𝒀𝒀𝒀𝒀 =
Altitude 𝒀𝒀𝒀𝒀
𝟗𝟗𝟗𝟗°. Since 𝒎𝒎∠𝑾𝑾𝑾𝑾𝑾𝑾 = 𝟗𝟗𝟗𝟗°, then it follows that
���� ∥ 𝑾𝑾𝑾𝑾
����� by corresponding ∠' s converse. So, △ 𝑼𝑼𝑼𝑼𝑼𝑼 is
𝒀𝒀𝒀𝒀
a scale drawing of △ 𝑼𝑼𝑼𝑼𝑼𝑼 by the parallel method.
Therefore, 𝒁𝒁 is the image of 𝑽𝑽, and 𝒀𝒀 is the image of 𝑾𝑾
under a dilation from 𝑼𝑼 with a scale factor of
By the dilation theorem:
𝟏𝟏𝟏𝟏
𝟓𝟓
.
𝟏𝟏𝟏𝟏
𝑾𝑾𝑾𝑾
𝟓𝟓
𝟏𝟏𝟏𝟏
(𝟒𝟒)
𝒀𝒀𝒀𝒀 =
𝟓𝟓
𝟓𝟓𝟓𝟓
𝒀𝒀𝒀𝒀 =
𝟓𝟓
𝟐𝟐
𝒀𝒀𝒀𝒀 = 𝟏𝟏𝟏𝟏
𝟓𝟓
𝒀𝒀𝒀𝒀 =
By the Pythagorean theorem:
𝑼𝑼𝑽𝑽𝟐𝟐 + 𝑽𝑽𝑾𝑾𝟐𝟐 = 𝑼𝑼𝑾𝑾𝟐𝟐
𝑼𝑼𝑽𝑽𝟐𝟐 + (𝟒𝟒𝟐𝟐 ) = (𝟓𝟓𝟐𝟐 )
𝑼𝑼𝑽𝑽𝟐𝟐 + 𝟏𝟏𝟏𝟏 = 𝟐𝟐𝟐𝟐
𝑼𝑼𝑽𝑽𝟐𝟐 = 𝟗𝟗
𝑼𝑼𝑼𝑼 = 𝟑𝟑
Since 𝒁𝒁 is the image of 𝑽𝑽 under the same dilation:
𝟏𝟏𝟏𝟏
𝑼𝑼𝑼𝑼
𝟓𝟓
𝟏𝟏𝟏𝟏
(𝟑𝟑)
𝑼𝑼𝑼𝑼 =
𝟓𝟓
𝟑𝟑𝟑𝟑
𝑼𝑼𝑼𝑼 =
𝟓𝟓
𝟒𝟒
𝑼𝑼𝑼𝑼 = 𝟕𝟕
𝟓𝟓
𝑼𝑼𝑼𝑼 =
𝑿𝑿𝑿𝑿 + 𝑼𝑼𝑼𝑼 = 𝑼𝑼𝑼𝑼
𝑿𝑿𝑿𝑿 + 𝟕𝟕
𝟒𝟒
= 𝟗𝟗
𝟓𝟓
𝑿𝑿𝑿𝑿 = 𝟏𝟏
𝟏𝟏
𝟓𝟓
𝟐𝟐
𝟏𝟏
The length of ����
𝒀𝒀𝒀𝒀 is 𝟏𝟏𝟏𝟏 units, and the length of ����
𝑿𝑿𝑿𝑿 is 𝟏𝟏 units.
𝟓𝟓
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GEOMETRY • MODULE 2
Topic B
Dilations
G-SRT.A.1, G-SRT.B.4
Focus Standards:
G-SRT.A.1
G-SRT.B.4
Instructional Days:
Lesson 6:
Verify experimentally the properties of dilations given by a center and a scale factor:
a.
A dilation takes a line not passing through the center of the dilation to a parallel
line, and leaves a line passing through the center unchanged.
b.
The dilation of a line segment is longer or shorter in the ratio given by the scale
factor.
Prove theorems about triangles. Theorems include: a line parallel to one side of a
triangle divides the other two proportionally, and conversely; the Pythagorean Theorem
proved using triangle similarity.
6
Dilations as Transformations of the Plane (S) 1
Lesson 7:
How Do Dilations Map Segments? (P)
Lesson 8:
How Do Dilations Map Lines, Rays, and Circles? (S)
Lesson 9:
How Do Dilations Map Angles? (E)
Lesson 10:
Dividing the King’s Foot into 12 Equal Pieces (E)
Lesson 11:
Dilations from Different Centers (E)
Topic B is an in-depth study of the properties of dilations. Though students applied dilations in Topic A, their
use in the ratio and parallel methods was to establish relationships that were consequences of applying a
dilation, not directly about the dilation itself. In Topic B, students explore observed properties of dilations
(Grade 8 Module 3) and reason why these properties are true. This reasoning is possible because of what
students have studied regarding scale drawings and the triangle side splitter and dilation theorems. With
these theorems, it is possible to establish why dilations map segments to segments, lines to lines, etc. Some
of the arguments involve an examination of several sub-cases; it is in these instances of thorough
examination that students must truly make sense of problems and persevere in solving them (MP.1).
In Lesson 6, students revisit the study of rigid motions and contrast the behavior of the rigid motions to that
of a dilation. Students confirm why the properties of dilations are true in Lessons 7–9. Students repeatedly
encounter G.SRT.A.1a and b in these lessons and build arguments with the help of the ratio and parallel
1Lesson
Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
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methods (G.SRT.B.4). In Lesson 10, students study how dilations can be used to divide a segment into equal
divisions. Finally, in Lesson 11, students observe how the images of dilations of a given figure by the same
scale factor are related, as well as the effect of a composition of dilations on the scale factor of the
composition.
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Lesson 6: Dilations as Transformations of the Plane
Student Outcomes

Students review the properties of basic rigid motions.

Students understand the properties of dilations and that a dilation is also an example of a transformation of
the plane.
Lesson Notes
Topic A plunged right into the use of dilations to create scale drawings and arguments to prove the triangle side splitter
theorem and dilation theorem. Topic B examines dilations in detail. In Grade 8 (Module 3), students discovered
properties of dilations, such as that the dilation of a line maps onto another line or that the dilation of an angle maps
onto another angle. We now examine how dilations differ from the other transformations and use reasoned arguments
to confirm the properties of dilations observed in Grade 8.
Topic B begins with a review of the rigid motions studied in Module 1 (Lessons 12–16).
Classwork
Discussion (7 minutes)
Scaffolding:
The goal of Lesson 6 is to study dilations as transformations of the plane. Begin with a
review of what a transformation is and the category of transformations studied in
Module 1. The following questions can be asked as part of a whole-group discussion, or
consider asking for them to be written to let students express their thoughts on paper
before discussing them aloud.
Use Module 1 Lesson 17
Exercises 1–5 to provide
students with good visuals and
review the learned
transformations and the
conclusion that they are
distance-preserving.

Recall that our recent study of translations, reflections, and rotations was a
study of transformations. With a partner, discuss what a transformation of the
plane means.
Allow students time to discuss before sharing responses.


When we refer to the image of 𝑃𝑃 by 𝐹𝐹, what does this refer to?



A transformation of the plane is a rule that assigns each point in the plane to a unique point. We use
function notation to denote transformations, that is, 𝐹𝐹 denotes the transformation of a point, 𝑃𝑃, and is
written as 𝐹𝐹(𝑃𝑃). Thus, a transformation moves point 𝑃𝑃 to a unique point 𝐹𝐹(𝑃𝑃).
The point 𝐹𝐹(𝑃𝑃) is called the image of 𝑃𝑃, or 𝑃𝑃′.
Recall that every point 𝑃𝑃′ in the plane is the image of some point 𝑃𝑃, that is, 𝐹𝐹(𝑃𝑃) = 𝑃𝑃′.
In Module 1, we studied a set of transformations that we described as being “rigid”. What does the term rigid
refer to?

The transformations in Module 1—translations, reflections, and rotations—are all transformations of
the plane that are rigid motions, or they are distance-preserving. A transformation is distancepreserving if, given two points 𝑃𝑃 and 𝑄𝑄, the distance between these points is the same as the distance
between the images of these points, that is, the distance between 𝐹𝐹(𝑃𝑃) and 𝐹𝐹(𝑄𝑄).
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As we know, there are a few implied properties of any rigid transformation:
a.
Rigid motions map a line to a line, a ray to a ray, a segment to a segment, and an angle to an angle.
b.
Rigid motions preserve lengths of segments.
c.
Rigid motions preserve the measures of angles.
Exercises 1–6 (12 minutes)
It is at the teacher’s discretion to assign only some or all of Exercises 1–6. Completion of all six exercises likely requires
more than the allotted time.
Exercises 1–6
������. Find the image 𝑷𝑷′ of point 𝑷𝑷 under this
Find the center and the angle of the rotation that takes ����
𝑨𝑨𝑨𝑨 to 𝑨𝑨′𝑩𝑩′
rotation.
1.
𝑩𝑩𝑩𝑩
Step 1. Determine the location of center 𝑶𝑶 as the
intersection of the perpendicular bisectors of �����
𝑨𝑨𝑨𝑨′ and
�����′.
Step 2. Determine the angle of rotation by joining 𝑨𝑨 to 𝑶𝑶
and 𝑶𝑶 to 𝑨𝑨′ ; the angle of rotation is 𝟓𝟓𝟓𝟓°
counterclockwise.
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Step 3. Rotate 𝑷𝑷 by 𝟓𝟓𝟓𝟓°. 𝑷𝑷 should remain a fixed
distance away from 𝑶𝑶.
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2.
In the diagram below, △ 𝑩𝑩′𝑪𝑪′𝑫𝑫′ is the image of △ 𝑩𝑩𝑩𝑩𝑩𝑩 after a rotation about a point 𝑨𝑨. What are the coordinates
of 𝑨𝑨, and what is the degree measure of the rotation?
By constructing the perpendicular bisector of each segment joining a point and its image, I found the center of
dilation 𝑨𝑨 to be 𝑨𝑨(𝟒𝟒, 𝟏𝟏). Using a protractor, the angle of rotation from △ 𝑩𝑩𝑩𝑩𝑩𝑩 to △ 𝑩𝑩′𝑪𝑪′𝑫𝑫′ about point 𝑨𝑨(𝟒𝟒, 𝟏𝟏)
is 𝟔𝟔𝟔𝟔°.
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3.
Find the line of reflection for the reflection that takes point 𝑨𝑨 to point 𝑨𝑨′. Find the image 𝑷𝑷′ under this reflection.
4.
Quinn tells you that the vertices of the image of quadrilateral 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 reflected over the line representing the
𝟑𝟑
𝟐𝟐
equation 𝒚𝒚 = − 𝒙𝒙 + 𝟐𝟐 are the following: 𝑪𝑪′(−𝟐𝟐, 𝟑𝟑), 𝑫𝑫′(𝟎𝟎, 𝟎𝟎), 𝑬𝑬′(−𝟑𝟑, −𝟑𝟑), and 𝑭𝑭′(−𝟑𝟑, 𝟑𝟑). Do you agree or
disagree with Quinn? Explain.
I disagree because under a reflection, an
image point lies along a line through the
pre-image point that is perpendicular to
the line of reflection. The line representing
𝟐𝟐
𝟑𝟑
the equation 𝒚𝒚 = 𝒙𝒙 + 𝟒𝟒 includes 𝑪𝑪 and is
perpendicular to the line of reflection;
however, it does not include the point
(−𝟐𝟐, 𝟑𝟑). Similar reasoning can be used to
show that Quinn’s coordinates for 𝑫𝑫′, 𝑬𝑬′,
and 𝑭𝑭′ are not the images of 𝑫𝑫, 𝑬𝑬, and 𝑭𝑭,
respectively, under a reflection over
𝟑𝟑
𝟐𝟐
𝒚𝒚 = − 𝒙𝒙 + 𝟐𝟐.
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5.
A translation takes 𝑨𝑨 to 𝑨𝑨′. Find the image 𝑷𝑷′ and pre-image 𝑷𝑷′′ of point 𝑷𝑷 under this translation. Find a vector that
describes the translation.
6.
The point 𝑪𝑪′ is the image of point 𝑪𝑪 under a translation of the plane along a vector.
a.
Find the coordinates of 𝑪𝑪 if the vector used for the translation is ������⃗
𝑩𝑩𝑩𝑩.
𝑪𝑪(𝟏𝟏, 𝟔𝟔)
b.
������⃗.
Find the coordinates of 𝑪𝑪 if the vector used for the translation is 𝑨𝑨𝑨𝑨
𝑪𝑪(−𝟓𝟓, 𝟒𝟒)
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Discussion (7 minutes)
Lead a discussion that highlights how dilations are like rigid motions and how they are different from them.

In this module, we have used dilations to create scale drawings and to establish the triangle side splitter
theorem and the dilation theorem. We pause here to inspect how dilations as a class of transformations are
like rigid transformations and how they are different.

What do dilations have in common with translations, reflections, and rotations?


All of these transformations meet the basic criteria of a transformation in the sense that each follows a
rule assignment where any point 𝑃𝑃 is assigned to a unique point 𝐹𝐹(𝑃𝑃) in the plane. Dilations and
rotations both require a center in order to define the function rule.
What distinguishes dilations from translations, reflections, and rotations?

Dilations are not a distance-preserving transformation like the rigid transformations. For every point 𝑃𝑃,
𝑂𝑂𝑂𝑂 so that the distance
other than the center, the point is assigned to 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝑃𝑃), which is the point 𝑃𝑃 on �����⃗
from 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝑃𝑃) to 𝑂𝑂 is 𝑟𝑟 ∙ 𝑂𝑂𝑂𝑂. The fact that distances are scaled means the transformation is not
distance preserving.

From our work in Grade 8, we have seen that dilations, just like the rigid motions, do in fact map segments to
segments, lines to lines, and rays to rays, but we only confirmed this experimentally, and in the next several
lessons, we create formal arguments to prove why these properties hold true for dilations.

One last feature that dilations share with the rigid motions is the existence of an inverse dilation, just as
inverses exist for the rigid transformations. What this means is that composition of the dilation and its inverse
takes every point in the plane to itself.

Consider a 90° clockwise rotation about a center 𝑂𝑂: 𝑅𝑅𝑂𝑂,90 (𝑃𝑃). The inverse rotation would be a 90° counter-

clockwise rotation to bring the image of point 𝑃𝑃 back to itself: 𝑅𝑅𝑂𝑂,−90 �𝑅𝑅𝑂𝑂,90 (𝑃𝑃)� = 𝑅𝑅𝑂𝑂,0 (𝑃𝑃) = 𝑃𝑃.
What would an inverse dilation rely on to bring the image of a dilated point 𝑃𝑃′ back to 𝑃𝑃?

If we were dilating a point 𝑃𝑃 by a factor of 2, the image would be written as 𝑃𝑃′ = 𝐷𝐷𝑂𝑂,2 (𝑃𝑃). In this case,
𝑃𝑃′ is pushed away from the center 𝑂𝑂 by a factor of two so that it is two times as far away from 𝑂𝑂.
1
To bring it back to itself, we need to halve the distance or, in other words, scale by a factor of , which
MP.7
2
is the reciprocal of the original scale factor: 𝐷𝐷𝑂𝑂,1 �𝐷𝐷𝑂𝑂,2 (𝑃𝑃)� = 𝐷𝐷𝑂𝑂,1 (𝑃𝑃) = 𝑃𝑃. Therefore, an inverse
2
dilation relies on the original center 𝑂𝑂 but requires a scale factor that is the reciprocal (or multiplicative
inverse) of the original scale factor.
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Exercises 7–9 (12 minutes)
Exercises 7–9
7.
A dilation with center 𝑶𝑶 and scale factor 𝒓𝒓 takes 𝑨𝑨 to 𝑨𝑨′ and 𝑩𝑩 to 𝑩𝑩′. Find the center 𝑶𝑶, and estimate the scale
factor 𝒓𝒓.
The estimated scale factor is 𝒓𝒓 ≈ 𝟐𝟐.
8.
Given a center 𝑶𝑶, scale factor 𝒓𝒓, and points 𝑨𝑨 and 𝑩𝑩, find the points 𝑨𝑨′ = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑨𝑨) and 𝑩𝑩′ = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑩𝑩). Compare
length 𝑨𝑨𝑨𝑨 with length 𝑨𝑨′ 𝑩𝑩′ by division; in other words, find
𝑨𝑨′ 𝑩𝑩′
𝑨𝑨𝑨𝑨
. How does this number compare to 𝒓𝒓?
𝐀𝐀′ 𝐁𝐁 ′
𝐀𝐀𝐀𝐀
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=
𝟏𝟏𝟏𝟏.𝟔𝟔
𝟒𝟒.𝟐𝟐
= 𝟑𝟑 = 𝐫𝐫
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9.
Given a center 𝑶𝑶, scale factor 𝒓𝒓, and points 𝑨𝑨, 𝑩𝑩, and 𝑪𝑪, find the points 𝑨𝑨′ = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑨𝑨), 𝑩𝑩′ = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑩𝑩), and
𝑪𝑪′ = 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑪𝑪). Compare 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 with 𝒎𝒎∠𝑨𝑨′ 𝑩𝑩′ 𝑪𝑪′ . What do you find?
The angle measurements are equal.
Closing (2 minutes)

We have studied two major classes of transformations: those that are distance-preserving (translations,
reflections, rotations) and those that are not (dilations).

Like rigid motions, dilations involve a rule assignment for each point in the plane and also have inverse
functions that return each dilated point back to itself.

Though we have experimentally verified that dilations share properties similar to those of rigid motions, for
example, the property that lines map to lines, we have yet to establish these properties formally.
Lesson Summary

There are two major classes of transformations: those that are distance preserving (translations,
reflections, rotations) and those that are not (dilations).

Like rigid motions, dilations involve a rule assignment for each point in the plane and also have inverse
functions that return each dilated point back to itself.
Exit Ticket (5 minutes)
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Name
Date
Lesson 6: Dilations as Transformations of the Plane
Exit Ticket
1.
Which transformations of the plane are distance-preserving transformations? Provide an example of what this
property means.
2.
Which transformations of the plane preserve angle measure? Provide one example of what this property means.
3.
Which transformation is not considered a rigid motion and why?
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Exit Ticket Sample Solutions
1.
2.
Which transformations of the plane are distance-preserving transformations? Provide an example of what this
property means.
Rotations, translations, and reflections are distance-preserving transformations of the plane because for any two
different points 𝑨𝑨 and 𝑩𝑩 in the plane, if 𝑭𝑭 is a rotation, translation, or reflection that maps 𝑨𝑨 → 𝑭𝑭(𝑨𝑨) and 𝑩𝑩 → 𝑭𝑭(𝑩𝑩),
𝑨𝑨𝑨𝑨 = 𝑭𝑭(𝑨𝑨)𝑭𝑭(𝑩𝑩).
Which transformations of the plane preserve angle measure? Provide one example of what this property means.
⃖����⃗ and ⃖����⃗
Rotations, translations, reflections, and dilations all preserve angle measure. If lines 𝑨𝑨𝑨𝑨
𝑩𝑩𝑩𝑩 are coplanar and
⃖�������������������⃗ (or ⃖�������⃗
intersect at 𝑩𝑩 to form ∠𝑨𝑨𝑨𝑨𝑨𝑨 with measure 𝒏𝒏°, ⃖�������������������⃗
𝑭𝑭(𝑨𝑨)𝑭𝑭(𝑩𝑩) (or ⃖�������⃗
𝑨𝑨′𝑩𝑩′) and 𝑭𝑭(𝑩𝑩)𝑭𝑭(𝑪𝑪)
𝑩𝑩′𝑪𝑪′) intersect at 𝑭𝑭(𝑩𝑩) to
3.
form ∠𝑭𝑭(𝑨𝑨)𝑭𝑭(𝑩𝑩)𝑭𝑭(𝑪𝑪) (or ∠𝑨𝑨′𝑩𝑩′𝑪𝑪′) that also has measure 𝒏𝒏°.
Which transformation is not considered a rigid motion and why?
A dilation is not considered a rigid motion because it does not preserve the distance between points. Under a
𝑨𝑨′𝑩𝑩′ must have a
dilation 𝑫𝑫𝑶𝑶,𝒓𝒓 where 𝒓𝒓 ≠ 𝟏𝟏, 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑨𝑨) = 𝑨𝑨′ and 𝑫𝑫𝑶𝑶,𝒓𝒓 (𝑩𝑩) = 𝑩𝑩′, 𝑨𝑨′ 𝑩𝑩′ = 𝒓𝒓(𝑨𝑨𝑨𝑨), which means that ������
length greater or less than ����
𝑨𝑨𝑩𝑩.
Problem Set Sample Solutions
1.
In the diagram below, 𝑨𝑨′ is the image of 𝑨𝑨 under a single transformation of the plane. Use the given diagrams to
show your solutions to parts (a)–(d).
a.
Describe the translation that maps 𝑨𝑨 → 𝑨𝑨′, and then use the translation to locate 𝑷𝑷′, the image of 𝑷𝑷.
The translation that maps 𝑨𝑨 → 𝑨𝑨′ is along horizontal vector �������⃗
𝑨𝑨𝑨𝑨′
b.
Describe the reflection that maps 𝑨𝑨 → 𝑨𝑨′, and then use the reflection to locate 𝑷𝑷′, the image of 𝑷𝑷.
The reflection that maps 𝑨𝑨 → 𝑨𝑨′ is across the
perpendicular bisector of �����
𝑨𝑨𝑨𝑨′.
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c.
Describe a rotation that maps 𝑨𝑨 → 𝑨𝑨′, and then use your rotation to locate 𝑷𝑷′, the image of 𝑷𝑷.
There are many possible correct answers to this part. The center of rotation 𝑪𝑪 must be on the perpendicular
𝟏𝟏
bisector of �����
𝑨𝑨𝑨𝑨′ and the radius 𝑪𝑪𝑪𝑪 ≥ 𝑨𝑨𝑨𝑨′.
𝟐𝟐
d.
Describe a dilation that maps 𝑨𝑨 → 𝑨𝑨′, and then use your dilation to locate 𝑷𝑷′, the image of 𝑷𝑷.
There are many possible correct answers to this part. The center of dilation must be on ⃖�����⃗
𝑨𝑨𝑨𝑨′. If the scale
factor chosen is 𝒓𝒓 > 𝟏𝟏, then 𝑨𝑨 must be between 𝑶𝑶 and 𝑨𝑨′. If the scale factor chosen is 𝒓𝒓 < 𝟏𝟏, then 𝑨𝑨′ must be
between 𝑨𝑨 and 𝑶𝑶, and 𝑶𝑶𝑷𝑷′ = 𝒓𝒓(𝑶𝑶𝑶𝑶). The sample shown below uses a scale factor 𝒓𝒓 = 𝟐𝟐.
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2.
On the diagram below, 𝑶𝑶 is a center of dilation, and ⃖����⃗
𝑨𝑨𝑨𝑨 is a line not through 𝑶𝑶. Choose two points 𝑩𝑩 and 𝑪𝑪 on ⃖����⃗
𝑨𝑨𝑨𝑨
between 𝑨𝑨 and 𝑫𝑫.
a.
b.
c.
d.
𝟏𝟏
𝟐𝟐
Dilate 𝑨𝑨, 𝑩𝑩, 𝑪𝑪, and 𝑫𝑫 from 𝑶𝑶 using scale factor 𝒓𝒓 = . Label the images 𝑨𝑨′ , 𝑩𝑩′ , 𝑪𝑪′ , and 𝑫𝑫′, respectively.
Dilate 𝑨𝑨, 𝑩𝑩, 𝑪𝑪, and 𝑫𝑫 from 𝑶𝑶 using scale factor 𝒓𝒓 = 𝟐𝟐. Label the images 𝑨𝑨′′, 𝑩𝑩′′, 𝑪𝑪′′ , and 𝑫𝑫′′, respectively.
Dilate 𝑨𝑨, 𝑩𝑩, 𝑪𝑪, and 𝑫𝑫 from 𝑶𝑶 using scale factor 𝒓𝒓 = 𝟑𝟑. Label the images 𝑨𝑨′′′ , 𝑩𝑩′′′ , 𝑪𝑪′′′, and 𝑫𝑫′′′, respectively.
Draw a conclusion about the effect of a dilation on a line segment based on the diagram that you drew.
Explain.
Conclusion: Dilations map line segments to line segments.
3.
Write the inverse transformation for each of the following so that the composition of the transformation with its
inverse maps a point to itself on the plane.
a.
𝑻𝑻������⃗
𝑨𝑨𝑨𝑨
������⃗ since this vector
The inverse of a translation along the vector ������⃗
𝑨𝑨𝑨𝑨 would be a translation along the vector 𝑩𝑩𝑩𝑩
has the same magnitude but opposite direction. This translation maps any image point to its pre-image.
b.
c.
𝒓𝒓⃖����⃗
𝑨𝑨𝑨𝑨
The inverse of a reflection over line 𝑨𝑨𝑨𝑨 is the same reflection. The points 𝑷𝑷 and 𝒓𝒓⃖����⃗
⃖����⃗
𝑨𝑨𝑨𝑨, so repeating the reflection takes a point back to itself.
𝑹𝑹𝑪𝑪,𝟒𝟒𝟒𝟒
The inverse of a 𝟒𝟒𝟒𝟒° rotation about a point 𝑪𝑪 would be a rotation about the same point 𝑪𝑪 of −𝟒𝟒𝟒𝟒° , the
opposite rotational direction.
d.
𝑫𝑫𝑶𝑶,𝒓𝒓
The inverse of a dilation with center 𝑶𝑶 and scale factor 𝒓𝒓 would be a dilation from center 𝑶𝑶 with a scale factor
𝟏𝟏
of . Point 𝑨𝑨 in the plane is distance 𝑶𝑶𝑶𝑶 from the center of dilation 𝑶𝑶, and its image 𝑨𝑨′ would, therefore, be
𝒓𝒓
at a distance 𝒓𝒓(𝑶𝑶𝑶𝑶) from 𝑶𝑶. A dilation of 𝑨𝑨′ with scale factor
𝟏𝟏
𝒓𝒓
𝟏𝟏
(𝑶𝑶𝑨𝑨′ ) = �𝒓𝒓(𝑶𝑶𝑶𝑶)� = 𝟏𝟏(𝑶𝑶𝑶𝑶) = 𝑶𝑶𝑶𝑶.
𝒓𝒓
𝟏𝟏
𝒓𝒓
would map the 𝑨𝑨′ to a point that is a distance
By the definition of a dilation, points and their images lie on the
same ray that originates at the center of the dilation. There is only one point on that ray at a distance 𝑶𝑶𝑶𝑶
from 𝑶𝑶, which is 𝑨𝑨.
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To the teacher: Problem 4 reviews the application of dilation on the coordinate plane that was studied in depth in
4.
Given 𝑼𝑼(𝟏𝟏, 𝟑𝟑), 𝑽𝑽(−𝟒𝟒, −𝟒𝟒), and 𝑾𝑾(−𝟑𝟑, 𝟔𝟔) on the coordinate plane, perform a dilation of △ 𝑼𝑼𝑼𝑼𝑼𝑼 from center
𝟑𝟑
𝑶𝑶(𝟎𝟎, 𝟎𝟎) with a scale factor of . Determine the coordinates of images of points 𝑼𝑼, 𝑽𝑽, and 𝑾𝑾, and describe how the
𝟐𝟐
coordinates of the image points are related to the coordinates of the pre-image points.
Under the given dilation, 𝑼𝑼, 𝑽𝑽, and 𝑾𝑾 map to 𝑼𝑼′ , 𝑽𝑽′ , and 𝑾𝑾′,
𝟑𝟑 𝟗𝟗
𝟐𝟐 𝟐𝟐
𝟗𝟗
𝟐𝟐
respectively. 𝑼𝑼′ � , �, 𝑽𝑽′ (−𝟔𝟔, −𝟔𝟔), and 𝑾𝑾′ �− , 𝟗𝟗�. For
each point (𝑿𝑿, 𝒀𝒀) on the coordinate plane, its image point is
𝟑𝟑
𝟐𝟐
𝟑𝟑
𝟐𝟐
𝟑𝟑
� 𝑿𝑿, 𝒀𝒀� under the dilation from the origin with scale
factor .
5.
𝟐𝟐
Points 𝑩𝑩, 𝑪𝑪, 𝑫𝑫, 𝑬𝑬, 𝑭𝑭, and 𝑮𝑮 are dilated images of 𝑨𝑨 from center 𝑶𝑶 with scale factors 𝟐𝟐, 𝟑𝟑, 𝟒𝟒, 𝟓𝟓, 𝟔𝟔, and 𝟕𝟕, respectively.
Are points 𝒀𝒀, 𝑿𝑿, 𝑾𝑾, 𝑽𝑽, 𝑼𝑼, 𝑻𝑻, and 𝑺𝑺 all dilated images of 𝒁𝒁 under the same respective scale factors? Explain why or
why not.
If points 𝒀𝒀, 𝑿𝑿, 𝑾𝑾, 𝑽𝑽, 𝑼𝑼, 𝑻𝑻, and 𝑺𝑺 were dilated images of 𝒁𝒁, the images would all be collinear with 𝑶𝑶 and 𝒁𝒁; however,
the points are not all on a line, so they cannot all be images of point 𝒁𝒁 from center 𝑶𝑶. We also know that dilations
preserve angle measures, and it is clear that each segment meets ������⃗
𝑶𝑶𝑶𝑶 at a different angle.
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6.
Find the center and scale factor that takes 𝑨𝑨 to 𝑨𝑨′ and 𝑩𝑩 to 𝑩𝑩′, if a dilation exists.
𝟑𝟑
The center of the dilation is 𝑶𝑶, and the scale factor is .
𝟐𝟐
7.
Find the center and scale factor that takes 𝑨𝑨 to 𝑨𝑨′ and 𝑩𝑩 to 𝑩𝑩′, if a dilation exists.
After drawing �������⃗
𝑩𝑩′𝑩𝑩 and �������⃗
𝑨𝑨′𝑨𝑨, the rays converge at a supposed center; however, the corresponding distances are not
proportional since
𝑶𝑶𝑨𝑨′
𝑶𝑶𝑶𝑶
= 𝟐𝟐 and
𝑶𝑶𝑩𝑩′
𝑶𝑶𝑶𝑶
≠ 𝟐𝟐. Therefore, a dilation does not exist that maps 𝑨𝑨 → 𝑨𝑨′ and 𝑩𝑩 → 𝑩𝑩′.
���� and ������
It also could be shown that 𝑨𝑨𝑨𝑨
𝑨𝑨′𝑩𝑩′ are not parallel; therefore, the lengths are not proportional by the triangle
side splitter theorem, and there is no dilation.
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Lesson 7: How Do Dilations Map Segments?
Student Outcomes


Students prove that dilation 𝐷𝐷𝑂𝑂,𝑟𝑟 maps a line segment 𝑃𝑃𝑃𝑃 to a line segment 𝑃𝑃′𝑄𝑄′, sending the endpoints to the
⃖������⃗. If the center 𝑂𝑂 does
endpoints so that 𝑃𝑃′ 𝑄𝑄′ = 𝑟𝑟𝑟𝑟𝑟𝑟. If the center 𝑂𝑂 lies in line 𝑃𝑃𝑃𝑃 or 𝑟𝑟 = 1, then ⃖����⃗
𝑃𝑃𝑃𝑃 = 𝑃𝑃′𝑄𝑄′
⃖������⃗.
not lie in line 𝑃𝑃𝑃𝑃 and 𝑟𝑟 ≠ 1, then ⃖����⃗
𝑃𝑃𝑃𝑃 || 𝑃𝑃′𝑄𝑄′
Students prove that if ����
𝑃𝑃𝑃𝑃 and ����
𝑅𝑅𝑅𝑅 are line segments in the plane of different lengths, then there is a dilation
⃖���⃗ or ⃖����⃗
⃖���⃗.
𝑃𝑃𝑃𝑃 || 𝑅𝑅𝑅𝑅
that maps one to the other if and only if ⃖����⃗
𝑃𝑃𝑃𝑃 = 𝑅𝑅𝑅𝑅
Lesson Notes
In Grade 8, students informally showed that a dilation maps a segment to a segment on the coordinate plane. The
lesson includes an opening discussion that reminds students of this fact. Next, students must consider how to prove that
dilations map segments to segments when the segment is not tied to the coordinate plane. We again call upon our
knowledge of the triangle side splitter theorem to show that a dilation maps a segment to a segment. The goal of the
lesson is for students to understand the effect that dilation has on segments, specifically that a dilation maps a segment
to a segment so that its length is 𝑟𝑟 times the original.
To complete the lesson in one class period, it may be necessary to skip the opening discussion and Example 4 and focus
primarily on the work in Examples 1–3. Another option is to extend the lesson to two days so that all examples and
exercises can be given the time required to develop student understanding of the dilation of segments.
Classwork
Opening Exercise (2 minutes)
Scaffolding:
Opening Exercise
a.
MP.3
Is a dilated segment still a segment? If the segment is transformed under a dilation,
explain how.
Accept any reasonable answer. The goal is for students to recognize that a segment
dilates to a segment that is 𝒓𝒓 times the length of the original.
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Use a segment in the
coordinate plane with endpoint
𝑃𝑃(−4,1) and endpoint 𝑄𝑄(3,2).
Show that a dilation from a
center at the origin maps ����
𝑃𝑃𝑃𝑃 to
′
′
������
𝑃𝑃 𝑄𝑄 .
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b.
Dilate the segment 𝑷𝑷𝑷𝑷 by a scale factor of 𝟐𝟐 from center 𝑶𝑶.
MP.3
i.
Is the dilated segment 𝑷𝑷′𝑸𝑸′ a segment?
Yes, the dilation of segment 𝑷𝑷𝑷𝑷 produces a segment 𝑷𝑷′ 𝑸𝑸′ .
ii.
How, if at all, has the segment 𝑷𝑷𝑷𝑷 been transformed?
Segment 𝑷𝑷′𝑸𝑸′ is twice the length of segment 𝑷𝑷𝑷𝑷. The segment has increased in length according to the
scale factor of dilation.
Opening (5 minutes)
In Grade 8, students learned the multiplicative effect that dilation has on points in the coordinate plane when the center
of dilation is the origin. Specifically, students learned that when the point located at (𝑥𝑥, 𝑦𝑦) was dilated from the origin
by scale factor 𝑟𝑟, the dilated point was located at (𝑟𝑟𝑟𝑟, 𝑟𝑟𝑟𝑟). Review this fact with students, and then remind them how to
informally verify that a dilation maps a segment to a segment using the diagram below. As time permits, review what
students learned in Grade 8, and then spend the remaining time on the question in the second bullet point.


Let ����
𝐴𝐴𝐴𝐴 be a segment on the coordinate plane with endpoints at (−2, 1) and (1, −2). If we dilate the segment
from the origin by a scale factor 𝑟𝑟 = 4, another segment is produced.
What do we expect the coordinates of the endpoints to be?

Based on what we know about the multiplicative effect dilation has on coordinates, we expect the
coordinates of the endpoints of ������
𝐴𝐴′𝐵𝐵′ to be (−8, 4) and (4, −8).
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
The question becomes, how can we be sure that the dilation maps the points between 𝐴𝐴 and 𝐵𝐵 to the points
between 𝐴𝐴′ and 𝐵𝐵′? We have already shown that the endpoints move to where we expect them to, but what
about the points in between? Perhaps the dilation maps the endpoints the way we expect, but all other points
form an arc of a circle or some other curve that connects the endpoints. Can you think of a way we can verify
that all of the points of segment 𝐴𝐴𝐴𝐴 map to images that lie on segment 𝐴𝐴′ 𝐵𝐵′ ?


We can verify other points that belong to segment 𝐴𝐴𝐴𝐴 using the same method as the endpoints. For
example, points (−1,0) and (0, −1) of segment 𝐴𝐴𝐴𝐴 should map to points (−4,0) and (0, −4). Using
the 𝑥𝑥- and 𝑦𝑦-axes as our rays from the origin of dilation, we can clearly see that the points and their
dilated images lie on the correct ray and more importantly lie on segment 𝐴𝐴′ 𝐵𝐵′ .
Our next challenge is to show that dilations map segments to segments when they are not on the coordinate
plane. We will prove the preliminary dilation theorem for segments: A dilation maps a line segment to a line
segment sending the endpoints to the endpoints.
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Example 1 (2 minutes)
Example 1
Case 1. Consider the case where the scale factor of dilation is 𝒓𝒓 = 𝟏𝟏. Does a dilation from center 𝑶𝑶 map segment 𝑷𝑷𝑷𝑷 to a
segment 𝑷𝑷′𝑸𝑸′? Explain.
A scale factor of 𝒓𝒓 = 𝟏𝟏 means that the segment and its image are equal. The dilation does not enlarge or shrink the
image of the figure; it remains unchanged. Therefore, when the scale factor of dilation is 𝒓𝒓 = 𝟏𝟏, then the dilation maps
the segment to itself.
Example 2 (3 minutes)
Example 2
Case 2. Consider the case where a line 𝑷𝑷𝑷𝑷 contains the center of the dilation. Does a dilation from the center with scale
factor 𝒓𝒓 ≠ 𝟏𝟏 map the segment 𝑷𝑷𝑷𝑷 to a segment 𝑷𝑷′𝑸𝑸′? Explain.
At this point, students should be clear that a dilation of scale factor 𝑟𝑟 ≠ 1 changes the length of the segment. If
necessary, explain to students that a scale factor of 𝑟𝑟 ≠ 1 simply means that the figure changes in length. The goal in
this example and, more broadly, this lesson is to show that a dilated segment is still a segment not some other figure.
The focus of the discussion should be on showing that the dilated figure is in fact a segment, not necessarily the length
of the segment.
Yes. The dilation sends points 𝑷𝑷 and 𝑸𝑸 to points 𝑷𝑷′ and 𝑸𝑸′. Since the points 𝑷𝑷 and 𝑸𝑸 are collinear with the center 𝑶𝑶, then
both 𝑷𝑷′ and 𝑸𝑸′ are also collinear with the center 𝑶𝑶. The dilation also takes all of the points between 𝑷𝑷 and 𝑸𝑸 to all of the
points between 𝑷𝑷′ and 𝑸𝑸′ (again because their images must fall on the rays 𝑶𝑶𝑶𝑶 and 𝑶𝑶𝑶𝑶). Therefore, the dilation maps
���� to ������
𝑷𝑷′𝑸𝑸′.
𝑷𝑷𝑷𝑷
Example 3 (12 minutes)
Scaffolding:
Example 3
Case 3. Consider the case where ⃖����⃗
𝑷𝑷𝑷𝑷 does not contain the center 𝑶𝑶 of the dilation, and the scale
factor 𝒓𝒓 of the dilation is not equal to 1; then, we have the situation where the key points 𝑶𝑶, 𝑷𝑷,
and 𝑸𝑸 form △ 𝑶𝑶𝑶𝑶𝑶𝑶. The scale factor not being equal to 1 means that we must consider scale
factors such that 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏 and 𝒓𝒓 > 𝟏𝟏. However, the proofs for each are similar, so we focus on
the case when 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏.
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For some groups of students, it
may be necessary for them to
perform a dilation where the
scale factor of dilation is
0 < 𝑟𝑟 < 1 so they have recent
experience allowing them to
question of Example 3.
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When we dilate points 𝑷𝑷 and 𝑸𝑸 from center 𝑶𝑶 by scale factor 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏, as shown,
what do we know about points 𝑷𝑷′ and 𝑸𝑸′?
We know 𝑷𝑷′ lies on ray 𝑶𝑶𝑶𝑶 with 𝑶𝑶𝑷𝑷′ = 𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶, and 𝑸𝑸′ lies on ray 𝑶𝑶𝑶𝑶 with
𝑶𝑶𝑶𝑶′ = 𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶. So,
𝑶𝑶𝑷𝑷′
𝑶𝑶𝑶𝑶
=
of △ 𝑶𝑶𝑶𝑶𝑶𝑶 proportionally.
𝑶𝑶𝑸𝑸′
𝑶𝑶𝑶𝑶
= 𝒓𝒓. The line segment 𝑷𝑷′𝑸𝑸′ splits the sides
���� and
By the triangle side splitter theorem, we know that the lines containing 𝑷𝑷𝑷𝑷
������
𝑷𝑷′𝑸𝑸′ are parallel.
���� and
We use the fact that the line segment 𝑷𝑷′𝑸𝑸′ splits the sides of △ 𝑶𝑶𝑶𝑶𝑶𝑶 proportionally and that the lines containing 𝑷𝑷𝑷𝑷
������ are parallel to prove that a dilation maps segments to segments. Because we already know what happens when
𝑷𝑷′𝑸𝑸′
points 𝑷𝑷 and 𝑸𝑸 are dilated, consider another point 𝑹𝑹 that is on the segment 𝑷𝑷𝑷𝑷. After dilating 𝑹𝑹 from center 𝑶𝑶 by scale
factor 𝒓𝒓 to get the point 𝑹𝑹′, does 𝑹𝑹′ lie on the segment 𝑷𝑷′𝑸𝑸′?
Consider giving students time to discuss in small groups Marwa’s proof shown on the next page on how to prove that a
dilation maps collinear points to collinear points. Also consider providing students time to make sense of it and
paraphrase a presentation of the proof to a partner or the class. Consider also providing the statements for the proof
and asking students to provide the reasoning for each step independently, with a partner, or in small groups.
The proof below relies heavily upon the parallel postulate: Two lines are constructed, ⃖������⃗
𝑃𝑃′𝑄𝑄′ and ⃖������⃗
𝑃𝑃′𝑅𝑅′, both of which are
𝑃𝑃𝑃𝑃 and contain the point 𝑃𝑃′, they must be the same line by the parallel
parallel to ⃖����⃗
𝑃𝑃𝑃𝑃 . Since both lines are parallel to ⃖����⃗
postulate. Thus, the point 𝑅𝑅′ lies on the line 𝑃𝑃′𝑄𝑄′.
Note: This proof on the next page is only part of the reasoning needed to show that dilations map segments to
segments. The full proof also requires showing that points between 𝑃𝑃 and 𝑄𝑄 are mapped to points between 𝑃𝑃′ and 𝑄𝑄′,
and that this mapping is onto (that for every point on the line segment 𝑃𝑃′𝑄𝑄′, there exists a point on segment 𝑃𝑃𝑃𝑃 that
gets sent to it). See the discussion following Marwa’s proof for more details on these steps.
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Marwa’s proof of the statement: Let 𝑂𝑂 be a point not on ⃖����⃗
𝑃𝑃𝑃𝑃 and 𝐷𝐷𝑂𝑂,𝑟𝑟 be a dilation with center 𝑂𝑂 and scale factor
′
⃖����⃗, then 𝐷𝐷𝑂𝑂,𝑟𝑟 (𝑅𝑅) is a point that
0 < 𝑟𝑟 < 1 that sends point 𝑃𝑃 to 𝑃𝑃 and 𝑄𝑄 to 𝑄𝑄′. If 𝑅𝑅 is another point that lies on 𝑃𝑃𝑃𝑃
⃖������⃗
lies on 𝑃𝑃′𝑄𝑄′.
Statement
1.
2.
A dilation 𝐷𝐷𝑂𝑂,𝑟𝑟 with center 𝑂𝑂 and scale factor
𝑟𝑟 sends point 𝑃𝑃 to 𝑃𝑃′ and 𝑄𝑄 to 𝑄𝑄′ .
𝑂𝑂𝑃𝑃′
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑄𝑄′
𝑂𝑂𝑂𝑂
= 𝑟𝑟
3.
⃖������⃗
⃖����⃗
𝑃𝑃𝑃𝑃 ||𝑃𝑃′𝑄𝑄′
4.
Let 𝑅𝑅 be a point on segment 𝑃𝑃𝑃𝑃 between 𝑃𝑃 and 𝑄𝑄.
Let 𝑅𝑅′ be the dilation of 𝑅𝑅 by 𝐷𝐷𝑂𝑂,𝑟𝑟 .
5.
𝑂𝑂𝑃𝑃′
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑅𝑅′
𝑂𝑂𝑂𝑂
= 𝑟𝑟
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Reason
1.
Given
2.
By definition of dilation
3.
By the triangle side splitter theorem
4.
Given
5.
By definition of dilation
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6.
⃖������⃗
⃖����⃗
𝑃𝑃′𝑅𝑅′||𝑃𝑃𝑃𝑃
6.
By the triangle side splitter theorem as applied to
△ 𝑂𝑂𝑂𝑂𝑂𝑂 and the fact that ⃖����⃗
𝑃𝑃𝑃𝑃 and ⃖����⃗
𝑃𝑃𝑃𝑃 are the same
line
7.
⃖������⃗
⃖������⃗
𝑃𝑃′𝑅𝑅′ = 𝑃𝑃′𝑄𝑄′
7.
8.
The point 𝑅𝑅′ lies on ⃖������⃗
𝑃𝑃′𝑄𝑄′.
8.
By the parallel postulate: Through a given
external point 𝑃𝑃′ , there is at most one line parallel
to a given line 𝑃𝑃𝑃𝑃.
By Step 7
There are still two subtle steps that need to be proved to show that dilations map segments to segments when the
center does not lie on the line through the segment. The teacher may decide whether to show how to prove these two
steps or just claim that the steps can be shown to be true. Regardless, the steps should be briefly discussed as part of
what is needed to complete the proof of the full statement that dilations map segments to segments.
The first additional step that needs to be shown is that points on 𝑃𝑃𝑃𝑃 are sent to points on 𝑃𝑃′ 𝑄𝑄′ , that is, if 𝑅𝑅 is between 𝑃𝑃
and 𝑄𝑄, then 𝑅𝑅′ is between 𝑃𝑃′ and 𝑄𝑄′ . To prove this, we first write out what it means for 𝑅𝑅 to be between 𝑃𝑃 and 𝑄𝑄: 𝑃𝑃, 𝑅𝑅,
and 𝑄𝑄 are different points on the same line such that 𝑃𝑃𝑃𝑃 + 𝑅𝑅𝑅𝑅 = 𝑃𝑃𝑃𝑃. By the dilation theorem (Lesson 6), 𝑃𝑃′ 𝑅𝑅′ = 𝑟𝑟 ⋅
𝑃𝑃𝑃𝑃, 𝑅𝑅′ 𝑄𝑄′ = 𝑟𝑟 ⋅ 𝑅𝑅𝑅𝑅, and 𝑃𝑃′ 𝑄𝑄′ = 𝑟𝑟 ⋅ 𝑃𝑃𝑃𝑃. Therefore,
𝑃𝑃′ 𝑅𝑅′ + 𝑅𝑅′ 𝑄𝑄′ = 𝑟𝑟 ⋅ 𝑃𝑃𝑃𝑃 + 𝑟𝑟 ⋅ 𝑅𝑅𝑅𝑅 = 𝑟𝑟(𝑃𝑃𝑃𝑃 + 𝑅𝑅𝑅𝑅) = 𝑟𝑟 ⋅ 𝑃𝑃𝑃𝑃 = 𝑃𝑃′ 𝑄𝑄′
Hence, 𝑃𝑃′ 𝑅𝑅′ + 𝑅𝑅′ 𝑄𝑄′ = 𝑅𝑅′𝑄𝑄′, and therefore 𝑅𝑅′ is between 𝑃𝑃′ and 𝑄𝑄′.
The second additional step is to show that the dilation is an onto mapping, that is, for every point 𝑅𝑅′ that lies on 𝑃𝑃′ 𝑄𝑄′ ,
there is a point 𝑅𝑅 that lies on 𝑃𝑃𝑃𝑃 that is mapped to 𝑅𝑅′ under the dilation. To prove, we use the inverse dilation at
center 𝑂𝑂 with scale factor
1
𝑟𝑟
to get the point 𝑅𝑅 and then follow the proof above to show that 𝑅𝑅 lies on 𝑃𝑃𝑃𝑃.
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Putting together the preliminary dilation theorem for segments with the dilation theorem, we get
DILATION THEOREM FOR SEGMENTS: A dilation 𝑫𝑫𝑶𝑶,𝒓𝒓 maps a line segment 𝑷𝑷𝑷𝑷 to a line segment 𝑷𝑷′𝑸𝑸′ sending the endpoints to
⃖�������⃗.
⃖����⃗ || 𝑷𝑷′𝑸𝑸′
the endpoints so that 𝑷𝑷′ 𝑸𝑸′ = 𝒓𝒓𝒓𝒓𝒓𝒓. Whenever the center 𝑶𝑶 does not lie in line 𝑷𝑷𝑷𝑷 and 𝒓𝒓 ≠ 𝟏𝟏, we conclude 𝑷𝑷𝑸𝑸
Whenever the center 𝑶𝑶 lies in ⃖����⃗
𝑷𝑷𝑷𝑷 or if 𝒓𝒓 = 𝟏𝟏, we conclude ⃖����⃗
𝑷𝑷𝑷𝑷 = ⃖�������⃗
𝑷𝑷′𝑸𝑸′.
As an aside, observe that a dilation maps parallel line segments to parallel line segments. Further, a dilation maps a
directed line segment to a directed line segment that points in the same direction.
If time permits, have students verify these observations with their own scale drawings. It is not imperative that students
do this activity, but this idea is used in the next lesson. At this point, students just need to observe this fact or at least be
made aware of it, as it is discussed in Lesson 8.
Example 4 (7 minutes)
Example 4
Now look at the converse of the dilation theorem for segments: If ����
𝑷𝑷𝑷𝑷 and ����
𝑹𝑹𝑹𝑹 are line segments of different lengths in
⃖����⃗ = ⃖����⃗
⃖����⃗ || 𝑹𝑹𝑹𝑹
⃖����⃗.
𝑹𝑹𝑹𝑹 or 𝑷𝑷𝑷𝑷
the plane, then there is a dilation that maps one to the other if and only if 𝑷𝑷𝑷𝑷
����, so that
Based on Examples 2 and 3, we already know that a dilation maps a segment 𝑷𝑷𝑷𝑷 to another line segment, say 𝑹𝑹𝑹𝑹
⃖����⃗
⃖����⃗ (Example 3). If ⃖����⃗
⃖����⃗, then, because ����
𝑹𝑹𝑹𝑹 (Example 2) or ⃖����⃗
𝑷𝑷𝑷𝑷 || 𝑹𝑹𝑹𝑹
𝑷𝑷𝑷𝑷 || 𝑹𝑹𝑹𝑹
𝑷𝑷𝑷𝑷 and ����
𝑹𝑹𝑹𝑹 are different lengths in the plane,
𝑷𝑷𝑷𝑷 = ⃖����⃗
they are the bases of a trapezoid, as shown.
���� and ����
Since 𝑷𝑷𝑷𝑷
𝑹𝑹𝑹𝑹 are segments of different lengths, then the non-base sides of the trapezoid are not parallel, and the
lines containing them meet at a point 𝑶𝑶 as shown.
����. Explain how to show it.
Recall that we want to show that a dilation maps ����
𝑷𝑷𝑷𝑷 to 𝑹𝑹𝑹𝑹
Provide students time to discuss this in partners or small groups.
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⃖����⃗ || 𝑹𝑹𝑹𝑹
⃖����⃗, then the segment 𝑹𝑹𝑹𝑹 splits the sides of
The triangle formed with vertex 𝑶𝑶, △ 𝑶𝑶𝑶𝑶𝑶𝑶, has 𝑷𝑷𝑷𝑷 as its base. Since 𝑷𝑷𝑷𝑷
the triangle proportionally by the triangle side splitter theorem. Since we know the proportional side splitters of a
triangle are the result of a dilation, then we know there is a dilation from center 𝑶𝑶 by scale factor 𝒓𝒓 that maps points 𝑷𝑷
���� to 𝑹𝑹𝑹𝑹
����.
and 𝑸𝑸 to points 𝑹𝑹 and 𝑺𝑺, respectively. Thus, a dilation maps 𝑷𝑷𝑷𝑷
���� and ����
The case when the segments 𝑷𝑷𝑷𝑷
𝑹𝑹𝑹𝑹 are such that ⃖����⃗
𝑷𝑷𝑷𝑷 = ⃖����⃗
𝑹𝑹𝑹𝑹 is left as an exercise.
Exercises 1–2 (8 minutes)
Students complete Exercises 1–2 in pairs. Students may need support to complete Exercise 2. A hint is shown below
that can be shared with students if necessary.
Exercises 1–2
In the following exercises, you will consider the case where the segment and its dilated image belong to the same line,
���� and ����
𝑹𝑹𝑹𝑹 are such that ⃖����⃗
𝑷𝑷𝑷𝑷 = ⃖����⃗
𝑹𝑹𝑹𝑹.
that is, when 𝑷𝑷𝑷𝑷
1.
Consider points 𝑷𝑷, 𝑸𝑸, 𝑹𝑹, and 𝑺𝑺 on a line, where 𝑷𝑷 = 𝑹𝑹, as shown below. Show there is a dilation that maps ����
𝑷𝑷𝑷𝑷
����. Where is the center of the dilation?
to 𝑹𝑹𝑹𝑹
���� to 𝑹𝑹𝑹𝑹
����, with a scale factor so that 𝒓𝒓 =
If we assume there is a dilation that maps 𝑷𝑷𝑷𝑷
𝑹𝑹𝑹𝑹
, then the center of the
𝑷𝑷𝑷𝑷
dilation must coincide with endpoints 𝑷𝑷 and 𝑹𝑹 because, by definition of dilation, the center maps to itself. Since
points 𝑷𝑷 and 𝑹𝑹 coincide, it must mean that the center 𝑶𝑶 is such that 𝑶𝑶 = 𝑷𝑷 = 𝑹𝑹. Since the other endpoint 𝑸𝑸 of 𝑷𝑷𝑷𝑷
lies on ⃖����⃗
𝑷𝑷𝑷𝑷, the dilated image of 𝑸𝑸 must also lie on the line (draw the ray from the center through point 𝑸𝑸, and the
⃖����⃗). Since the dilated image of 𝑸𝑸 must lie on the line, and point 𝑺𝑺 is to the right of 𝑸𝑸, then a
ray coincides with 𝑷𝑷𝑷𝑷
𝑹𝑹𝑹𝑹.
dilation from center 𝑶𝑶 with scale factor 𝒓𝒓 > 𝟏𝟏 maps ����
𝑷𝑷𝑷𝑷 to �����
2.
Consider points 𝑷𝑷, 𝑸𝑸, 𝑹𝑹, and 𝑺𝑺 on a line as shown below where ����
𝑷𝑷𝑷𝑷 ≠ ����
𝑹𝑹𝑹𝑹. Show there is a dilation that maps ����
𝑷𝑷𝑷𝑷
to ����
𝑹𝑹𝑹𝑹. Where is the center of the dilation?
Students may need some support to complete Exercise 2. Give them enough time to struggle with how to complete the
𝑅𝑅𝑅𝑅′ as
exercise and, if necessary, provide them with the following hint: Construct perpendicular line segments �����
𝑃𝑃𝑃𝑃′ and ����
shown so that 𝑃𝑃𝑄𝑄′ = 𝑃𝑃𝑃𝑃 and 𝑅𝑅𝑆𝑆 ′ = 𝑅𝑅𝑅𝑅.
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Construct perpendicular line segments of lengths 𝑷𝑷𝑷𝑷 and 𝑹𝑹𝑹𝑹 through points 𝑷𝑷 and 𝑹𝑹 respectively. Note the
endpoints of the perpendicular segments as 𝑸𝑸′ and 𝑺𝑺′ . Draw an auxiliary line through points 𝑸𝑸′ and 𝑺𝑺′ that
⃖������⃗||𝑹𝑹𝑹𝑹′
⃖�����⃗ (perpendicular lines),
intersects with ⃖����⃗
𝑷𝑷𝑷𝑷. The intersection of the two lines is the center of dilation 𝑶𝑶. Since 𝑷𝑷𝑷𝑷′
by the triangle side splitter theorem, the segment 𝑷𝑷𝑷𝑷′ splits the triangle △ 𝑶𝑶𝑶𝑶𝑶𝑶′ proportionally, so by the dilation
theorem,
𝑹𝑹𝑺𝑺′
𝑷𝑷𝑸𝑸′
= 𝒓𝒓. This ratio implies that 𝑹𝑹𝑹𝑹′ = 𝒓𝒓 ⋅ 𝑷𝑷𝑷𝑷′. By construction, 𝑷𝑷𝑸𝑸′ = 𝑷𝑷𝑷𝑷 and 𝑹𝑹𝑺𝑺′ = 𝑹𝑹𝑹𝑹. Therefore,
𝑹𝑹𝑺𝑺 = 𝒓𝒓 ⋅ 𝑷𝑷𝑷𝑷. By definition of dilation, since 𝑹𝑹𝑹𝑹 = 𝒓𝒓 ⋅ 𝑷𝑷𝑷𝑷, there is a dilation from center 𝑶𝑶 with scale factor 𝒓𝒓 that
�����
maps ����
𝑷𝑷𝑷𝑷 to 𝑹𝑹𝑹𝑹.
Closing (3 minutes)
Revisit the Opening Exercises. Have students explain their thinking about segments and dilated segments. Were their
initial thoughts correct? Did they change their thinking after going through the examples? What made them change
their minds? Then ask students to paraphrase the proofs as stated in the bullet point below.

Paraphrase the proofs that show how dilations map segments to segments when 𝑟𝑟 = 1 and 𝑟𝑟 ≠ 1 and when
the points are collinear compared to the vertices of a triangle.
Either accept or correct students’ responses accordingly. Recall that the goal of the lesson is to make sense of what
happens when segments are dilated. When the segment is dilated by a scale factor of 𝑟𝑟 = 1, then the segment and its
image would be the same length. When the points 𝑃𝑃 and 𝑄𝑄 are on a line containing the center, then the dilated points
𝑃𝑃′ and 𝑄𝑄′ are also collinear with the center producing an image of the segment that is a segment. When the points 𝑃𝑃
and 𝑄𝑄 are not collinear with the center, and the segment is dilated by a scale factor of 𝑟𝑟 ≠ 1, then the point 𝑃𝑃′ lies on
the ray 𝑂𝑂𝑂𝑂 with 𝑂𝑂𝑃𝑃′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂, and 𝑄𝑄′lies on ������⃗
𝑂𝑂𝑂𝑂 with 𝑄𝑄′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂. Then
𝑂𝑂𝑃𝑃′
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑄𝑄′
𝑂𝑂𝑂𝑂
������ splits the sides of △ 𝑂𝑂𝑂𝑂𝑂𝑂
= 𝑟𝑟. 𝑃𝑃′𝑄𝑄′
𝑃𝑃′𝑄𝑄′ are parallel.
proportionally, and by the triangle side splitter theorem, the lines containing ����
𝑃𝑃𝑃𝑃 and ������
Lesson Summary



When a segment is dilated by a scale factor of 𝒓𝒓 = 𝟏𝟏, then the segment and its image would be the
same length.
When the points 𝑷𝑷 and 𝑸𝑸 are on a line containing the center, then the dilated points 𝑷𝑷′ and 𝑸𝑸′ are also
collinear with the center producing an image of the segment that is a segment.
When the points 𝑷𝑷 and 𝑸𝑸 are not collinear with the center and the segment is dilated by a scale factor
of 𝒓𝒓 ≠ 𝟏𝟏, then the point 𝑷𝑷′ lies on the ray 𝑶𝑶𝑶𝑶′ with 𝑶𝑶𝑷𝑷′ = 𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶, and 𝑸𝑸′ lies on ray 𝑶𝑶𝑶𝑶 with
𝑶𝑶𝑸𝑸′ = 𝒓𝒓 ⋅ 𝑶𝑶𝑶𝑶.
Exit Ticket (5 minutes)
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Name
Date
Lesson 7: How Do Dilations Map Segments?
Exit Ticket
1.
Given the dilation 𝐷𝐷𝑂𝑂,3 , a line segment 𝑃𝑃𝑃𝑃, and that 𝑂𝑂 is not on ⃖����⃗
𝑃𝑃𝑃𝑃 , what can we conclude about the image of ����
𝑃𝑃𝑃𝑃 ?
2.
����, determine which figure has a dilation mapping the
Given figures A and B below, ����
𝐵𝐵𝐵𝐵 ∥ ����
𝐷𝐷𝐷𝐷 , ����
𝑈𝑈𝑈𝑈 ∥ ����
𝑋𝑋𝑋𝑋, and ����
𝑈𝑈𝑈𝑈 ≅ 𝑋𝑋𝑋𝑋
parallel line segments, and locate the center of dilation 𝑂𝑂. For one of the figures, a dilation does not exist. Explain
why.
2
Figure A
Figure B
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Exit Ticket Sample Solutions
1.
����?
Given the dilation 𝑫𝑫𝑶𝑶,𝟑𝟑 , a line segment 𝑷𝑷𝑷𝑷, and that 𝑶𝑶 is not on ⃖����⃗
𝑷𝑷𝑷𝑷, what can we conclude about the image of 𝑷𝑷𝑷𝑷
𝟐𝟐
������ is parallel to 𝑷𝑷𝑷𝑷
����, and 𝑷𝑷′ 𝑸𝑸′ = 𝟑𝟑 (𝑷𝑷𝑷𝑷).
Since 𝑷𝑷 and 𝑸𝑸 are not in line with 𝑶𝑶, 𝑷𝑷′𝑸𝑸′
𝟐𝟐
2.
����, ����
����, and ����
Given figures A and B below, ����
𝑩𝑩𝑩𝑩 ∥ 𝑫𝑫𝑫𝑫
𝑼𝑼𝑼𝑼 ∥ 𝑿𝑿𝑿𝑿
𝑼𝑼𝑼𝑼 ≅ ����
𝑿𝑿𝑿𝑿, determine which figure has a dilation mapping the
parallel line segments, and locate the center of dilation 𝑶𝑶. For one of the figures, a dilation does not exist. Explain
why.
𝑂𝑂
Figure A
Figure B
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���� to 𝑿𝑿𝑿𝑿
����. If the segments are both
There is no dilation that maps 𝑼𝑼𝑼𝑼
parallel and congruent, then they form two sides of a parallelogram,
⃖����⃗. If there was a dilation mapping ����
𝑼𝑼𝑼𝑼 to
which means that ⃖����⃗
𝑼𝑼𝑼𝑼 ∥ 𝑽𝑽𝑽𝑽
����
⃖����⃗
⃖����⃗
𝑿𝑿𝑿𝑿, then 𝑼𝑼𝑼𝑼 and 𝑽𝑽𝑽𝑽 would have to intersect at the center of
dilation, but they cannot intersect because they are parallel. We
also showed that a directed line segment maps to a directed line
segment pointing in the same direction, so it is not possible for 𝑼𝑼 to
map to 𝒀𝒀 and also 𝑽𝑽 map to 𝑿𝑿 under a dilation.
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Problem Set Sample Solutions
1.
Draw the dilation of parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 from center 𝑶𝑶 using the scale factor 𝒓𝒓 = 𝟐𝟐, and then answer the
questions that follow.
a.
Is the image 𝑨𝑨′𝑩𝑩′𝑪𝑪′𝑫𝑫′ also a parallelogram? Explain.
������ and �����
������, and because 𝑨𝑨𝑨𝑨
������. A
���� ∥ 𝑩𝑩𝑩𝑩
�����, it follows that ������
Yes By the dilation theorem, ����
𝑨𝑨𝑨𝑨 ∥ 𝑨𝑨′𝑪𝑪′
𝑩𝑩𝑩𝑩 ∥ 𝑩𝑩′𝑫𝑫′
𝑨𝑨′𝑪𝑪′ ∥ 𝑩𝑩′𝑫𝑫′
������ and ������
������, 𝑨𝑨′𝑩𝑩′𝑪𝑪′𝑫𝑫′ is
𝑨𝑨′𝑪𝑪′ ∥ 𝑩𝑩′𝑫𝑫′
𝑨𝑨′𝑩𝑩′ ∥ 𝑪𝑪′𝑫𝑫′
similar argument follows for the other pair of opposite sides, so with ������
a parallelogram.
b.
What do parallel lines seem to map to under a dilation?
Parallel lines map to parallel lines under dilations.
2.
Given parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 with 𝑨𝑨(−𝟖𝟖, 𝟏𝟏), 𝑩𝑩(𝟐𝟐, −𝟒𝟒), 𝑪𝑪(−𝟑𝟑, −𝟔𝟔), and 𝑫𝑫(−𝟏𝟏𝟏𝟏, −𝟏𝟏), perform a dilation of the plane
centered at the origin using the following scale factors.
a.
Scale factor
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𝟏𝟏
𝟐𝟐
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b.
Scale factor 𝟐𝟐
c.
Are the images of parallel line segments under a dilation also parallel? Use your graphs to support your
𝟏𝟏
The slopes of sides ����
𝑨𝑨𝑨𝑨 and ����
𝑪𝑪𝑪𝑪 in the original graph are − , and the slopes of each of the images of those
𝟏𝟏
𝟐𝟐
𝟐𝟐
𝟐𝟐
���� and ����
sides under the dilations are also − . The slopes of sides 𝑨𝑨𝑨𝑨
𝑩𝑩𝑩𝑩 in the original graph are , and the
𝟐𝟐
𝟓𝟓
slopes of each of the images of those sides under the dilations are also . This informally verifies that the
images of parallel line segments are also parallel line segments.
3.
𝟓𝟓
In Lesson 7, Example 3, we proved that a line segment 𝑷𝑷𝑷𝑷, where 𝑶𝑶, 𝑷𝑷, and 𝑸𝑸 are the vertices of a triangle, maps to
a line segment 𝑷𝑷′𝑸𝑸′ under a dilation with a scale factor 𝒓𝒓 < 𝟏𝟏. Using a similar proof, prove that for 𝑶𝑶 not on ⃖����⃗
𝑷𝑷𝑷𝑷, a
dilation with center 𝑶𝑶 and scale factor 𝒓𝒓 > 𝟏𝟏 maps a point 𝑹𝑹 on 𝑷𝑷𝑷𝑷 to a point 𝑹𝑹′ on line 𝑷𝑷𝑷𝑷.
The proof follows almost exactly the same as the proof of Example 3, but using the following diagram instead:
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4.
������ and ⃖����⃗
������. (Hint: There are 2 cases to
���� to 𝑨𝑨′𝑩𝑩′
On the plane, ����
𝑨𝑨𝑨𝑨 ∥ 𝑨𝑨′𝑩𝑩′
𝑨𝑨𝑨𝑨 ≠ ⃖�������⃗
𝑨𝑨′𝑩𝑩′. Describe a dilation mapping 𝑨𝑨𝑨𝑨
consider.)
���� and ������
Case 1: 𝑨𝑨𝑨𝑨
𝑨𝑨′𝑩𝑩′ are parallel directed line segments oriented in the same direction.
By the dilation theorem for segments, there is a dilation mapping 𝑨𝑨 and 𝑩𝑩 to 𝑨𝑨′ and 𝑩𝑩′, respectively.
���� and ������
Case 2: 𝑨𝑨𝑨𝑨
𝑨𝑨′𝑩𝑩′ are parallel directed line segments oriented in the opposite directions.
We showed that directed line segments map to directed line segments that are oriented in the same direction, so
there is a dilation mapping the parallel segments but only where the dilation maps 𝑨𝑨 and 𝑩𝑩 to 𝑩𝑩′ and 𝑨𝑨′,
respectively.
Note to the teacher: Students may state that a scale factor 𝑟𝑟 < 0 would produce the figure below where the center of a
dilation is between the segments; however, this violates the definition of dilation. In such a case, discuss the fact that a
scale factor must be greater than 0; otherwise, it would create negative distance, which of course does not make
mathematical sense.
5.
Only one of Figures 𝑨𝑨, 𝑩𝑩, or 𝑪𝑪 below contains a dilation that maps 𝑨𝑨 to 𝑨𝑨′ and 𝑩𝑩 to 𝑩𝑩′. Explain for each figure why
the dilation does or does not exist. For each figure, assume that ⃖����⃗
𝑨𝑨𝑨𝑨 ≠ ⃖�������⃗
𝑨𝑨′𝑩𝑩′.
a.
By the dilation theorem, if ����
𝑨𝑨𝑨𝑨 and ������
𝑨𝑨′𝑩𝑩′ are line segments in the plane of different lengths, then there is a
⃖�������⃗. The segments do not lie in the
⃖����⃗ || 𝑨𝑨′𝑩𝑩′
dilation that maps one to the other if and only if ⃖����⃗
𝑨𝑨𝑨𝑨 = ⃖�������⃗
𝑨𝑨′𝑩𝑩′ or 𝑨𝑨𝑨𝑨
same line and are also not parallel, so there is no dilation mapping 𝑨𝑨 to 𝑨𝑨′ and 𝑩𝑩 to 𝑩𝑩′.
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b.
By the dilation theorem, if ����
𝑨𝑨𝑨𝑨 and ������
𝑨𝑨′𝑩𝑩′ are line segments in the plane of different lengths, then there is a
⃖�������⃗. The diagram shows that 𝑨𝑨𝑨𝑨
���� and
dilation that maps one to the other if and only if ⃖����⃗
𝑨𝑨𝑨𝑨 = ⃖�������⃗
𝑨𝑨′𝑩𝑩′ or ⃖����⃗
𝑨𝑨𝑨𝑨 || 𝑨𝑨′𝑩𝑩′
������ do not lie in the same line, and it can also be seen that the line segments are not parallel. Furthermore,
𝑨𝑨′𝑩𝑩′
for the dilation to exist that maps 𝑨𝑨 to 𝑨𝑨′ and 𝑩𝑩 to 𝑩𝑩′, the center of dilation 𝑷𝑷 would need to be between the
segments, which violates the definition of dilation. Therefore, there is no dilation in Figure B mapping 𝑨𝑨 and
𝑩𝑩 to 𝑨𝑨′ and 𝑩𝑩′, respectively.
c.
By the dilation theorem, if ����
𝑨𝑨𝑨𝑨 and ������
𝑨𝑨′𝑩𝑩′ are line segments in the plane of different lengths, then there is a
⃖�������⃗. Assuming that ⃖����⃗
dilation that maps one to the other if and only if ⃖����⃗
𝑨𝑨𝑨𝑨 = ⃖�������⃗
𝑨𝑨′𝑩𝑩′ or ⃖����⃗
𝑨𝑨𝑨𝑨 || 𝑨𝑨′𝑩𝑩′
𝑨𝑨𝑨𝑨 ≠ ⃖�������⃗
𝑨𝑨′𝑩𝑩′, the
segments are shown to lie in the same line; therefore, the dilation exists.
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Lesson 8: How Do Dilations Map Lines, Rays, and Circles?
Student Outcomes

Students prove that a dilation maps a ray to a ray, a line to a line, and a circle to a circle.
Lesson Notes
The objective in Lesson 7 was to prove that a dilation maps a segment to a segment; in Lesson 8, students prove that a
dilation maps a ray to a ray, a line to a line, and a circle to a circle. An argument similar to that in Lesson 7 can be made
to prove that a ray maps to a ray; allow students the opportunity to establish this argument as independently as
possible.
Classwork
Opening (2 minutes)
As in Lesson 7, remind students of their work in Grade 8, when they studied the multiplicative effect that dilation has on
points in the coordinate plane when the center is at the origin. Direct students to consider what happens to the dilation
of a ray that is not on the coordinate plane.

Today we are going to show how to prove that dilations map rays to rays, lines to lines, and circles to circles.

Just as we revisited what dilating a segment on the coordinate plane is like, so we could repeat the exercise
here. How would you describe the effect of a dilation on a point (𝑥𝑥, 𝑦𝑦) on a ray about the origin by scale
factor 𝑟𝑟 in the coordinate plane?
A point (𝑥𝑥, 𝑦𝑦) on the coordinate plane dilated about the origin by scale factor 𝑟𝑟 would then be located
at (𝑟𝑟𝑟𝑟, 𝑟𝑟𝑟𝑟).


Of course, we must now consider what happens in the plane versus the coordinate plane.
Opening Exercise (3 minutes)
Opening Exercise
a.
MP.3
Is a dilated ray still a ray? If the ray is transformed under a dilation, explain how.
Accept any reasonable answer. The goal of this line of questioning is for students to recognize that a segment
dilates to a segment that is 𝒓𝒓 times the length of the original.
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b.
������⃗ by a scale factor of 𝟐𝟐 from center 𝐎𝐎.
Dilate the ray 𝑷𝑷𝑷𝑷
MP.3
i.
���������⃗ a ray?
Is the figure 𝑷𝑷′𝑸𝑸′
Yes. The dilation of ray 𝑷𝑷𝑷𝑷 produces a ray 𝑷𝑷′𝑸𝑸′.
ii.
How, if at all, has the segment 𝑷𝑷𝑷𝑷 been transformed?
The segment 𝑷𝑷′𝑸𝑸′ in ray 𝑷𝑷′𝑸𝑸′ is twice the length of the segment 𝑷𝑷𝑷𝑷 in ray 𝑷𝑷𝑷𝑷. The segment has
increased in length according to the scale factor of dilation.
iii.
Will a ray always be mapped to a ray? Explain how you know.
Students will most likely say that a ray always maps to a ray; they may defend their answers by citing
the definition of a dilation and its effect on the center and a given point.

We are going to use our work in Lesson 7 to help guide our reasoning today.

In proving why dilations map segments to segments, we considered three variations of the position of the
center relative to the segment and value of the scale factor of dilation:
(1) A center 𝑂𝑂 and scale factor 𝑟𝑟 = 1 and segment 𝑃𝑃𝑃𝑃,
(2) A line 𝑃𝑃𝑃𝑃 that does not contain the center 𝑂𝑂 and scale factor 𝑟𝑟 ≠ 1, and
(3) A line 𝑃𝑃𝑃𝑃 that does contain the center 𝑂𝑂 and scale factor 𝑟𝑟 ≠ 1.
Point out that the condition in case (1) does not specify the location of center 𝑂𝑂 relative to the line that contains
segment 𝑃𝑃𝑃𝑃; they can tell by this description that the condition does not impact the outcome and, therefore, is not
more particularly specified.

We use the setup of these cases to build an argument that shows dilations map rays to rays.
Examples 1–3 focus on establishing the dilation theorem of rays: A dilation maps a ray to a ray sending the endpoint to
the endpoint. The arguments for Examples 1 and 2 are very similar to those of Lesson 7, Examples 1 and 3, respectively.
Use discretion and student success with the Lesson 7 examples to consider allowing small-group work on the following
Examples 1–2, possibly providing a handout of the solution once groups arrive at solutions of their own. Again, since the
arguments are quite similar to those of Lesson 7, Examples 1 and 3, it is important to stay within the time allotments of
each example. Otherwise, Examples 1–2 should be teacher led.
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Example 1 (2 minutes)
Encourage students to draw upon the argument from Lesson 1, Example 1. This is intended to be a quick exercise;
consider giving 30-second time warnings or using a visible timer.
Example 1
Does a dilation about center 𝑶𝑶 and scale factor 𝒓𝒓 = 𝟏𝟏 map ������⃗
𝑷𝑷𝑷𝑷 to ���������⃗
𝑷𝑷′𝑸𝑸′? Explain.
A scale factor of 𝒓𝒓 = 𝟏𝟏 means that the ray and its image are equal. That is, the dilation does not enlarge or shrink the
image of the figure but remains unchanged. Therefore, when the scale factor of a dilation is 𝒓𝒓 = 𝟏𝟏, then the dilation
maps the ray to itself.
Example 2 (6 minutes)
Encourage students to draw upon the argument for Lesson 1, Example 3.
Scaffolding:
Example 2
������⃗ does not contain point 𝑶𝑶. Does a dilation 𝑫𝑫 about center 𝑶𝑶 and scale
The line that contains 𝑷𝑷𝑷𝑷
factor 𝒓𝒓 ≠ 𝟏𝟏 map every point of �������⃗
𝑷𝑷𝑷𝑷 onto a point of����������⃗
𝑷𝑷′𝑸𝑸′?

MP.1

A restatement of this problem is as follows: If 𝑅𝑅 is a point on �����⃗
𝑃𝑃𝑃𝑃 , then is 𝐷𝐷(𝑅𝑅) a
′
′
′
′
��������⃗
��������⃗
point on 𝑃𝑃 𝑄𝑄 ? Also, if a point 𝑆𝑆′ lies on the ray 𝑃𝑃 𝑄𝑄 , then is there a point 𝑆𝑆 on
the ray �����⃗
𝑃𝑃𝑃𝑃 such that 𝐷𝐷(𝑆𝑆) = 𝑆𝑆 ′?
Consider the case where the center 𝑂𝑂 is not in the line that contains �����⃗
𝑃𝑃𝑃𝑃 , and the
scale factor is 𝑟𝑟 ≠ 1. Then, points 𝑂𝑂, 𝑃𝑃, and 𝑄𝑄 form △ 𝑂𝑂𝑂𝑂𝑂𝑂.
Draw the following figure on the board.



 If students have difficulty
following the logical
progression of the proof,
rays and dilate them by a
series of different scale
factors and then make
results.
 For students who are
them to attempt to prove
the theorem
independently.
We examine the case with a scale factor 𝑟𝑟 > 1; the proof for 0 < 𝑟𝑟 < 1 is similar.
Under a dilation about center 𝑂𝑂 and 𝑟𝑟 > 1, 𝑃𝑃 goes to 𝑃𝑃′ , and 𝑄𝑄 goes to 𝑄𝑄’; 𝑂𝑂𝑃𝑃′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂 and 𝑂𝑂𝑄𝑄′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂.
What conclusion can we draw from these lengths?
Summarize the proof and result to a partner.
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Draw the following figure on the board.
We can rewrite each length relationship as


𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑄𝑄′
𝑂𝑂𝑂𝑂
= 𝑟𝑟.
By the triangle side splitter theorem, what else can we now conclude?
The segment 𝑃𝑃𝑃𝑃 splits △ 𝑂𝑂𝑂𝑂′𝑄𝑄′ proportionally.

��������⃗ are parallel; ⃖����⃗
The lines that contain �����⃗
𝑃𝑃𝑃𝑃 and 𝑃𝑃′𝑄𝑄′
𝑃𝑃𝑃𝑃 ∥ ⃖������⃗
𝑃𝑃′𝑄𝑄′; therefore, �����⃗
𝑃𝑃𝑃𝑃 ∥ ����������⃗
𝑃𝑃′𝑄𝑄′ .


𝑂𝑂𝑃𝑃′
By the dilation theorem for segments (Lesson 7), the dilation from 𝑂𝑂 of the segment 𝑃𝑃𝑃𝑃 is the segment 𝑃𝑃′𝑄𝑄′,
𝑃𝑃𝑃𝑃 that lies
that is, 𝐷𝐷�𝑃𝑃𝑃𝑃� = 𝑃𝑃′ 𝑄𝑄′ as sets of points. Therefore, we need only consider an arbitrary point 𝑅𝑅 on �����⃗
outside of 𝑃𝑃𝑃𝑃. For any such point 𝑅𝑅, what point is contained in the segment 𝑃𝑃𝑃𝑃?

′
The point 𝑄𝑄

Let 𝑅𝑅 = 𝐷𝐷(𝑅𝑅).

By the dilation theorem for segments, the dilation from 𝑂𝑂 of the segment 𝑃𝑃𝑃𝑃 is the segment 𝑃𝑃′𝑅𝑅′. Also by the
𝑃𝑃′ 𝑄𝑄′ and ��������⃗
𝑃𝑃′ 𝑅𝑅′
dilation theorem for segments, the point 𝐷𝐷(𝑄𝑄) = 𝑄𝑄′ is a point on segment 𝑃𝑃′𝑅𝑅′. Therefore, ��������⃗
′ 𝑄𝑄 ′ , which was what we needed to show.
��������⃗
must be the same ray. In particular, 𝑅𝑅′ is a point on 𝑃𝑃
MP.7
′ 𝑄𝑄 ′ there is a point 𝑆𝑆 on �����⃗
��������⃗
To show that for every point 𝑆𝑆′ on 𝑃𝑃
𝑃𝑃𝑃𝑃 such that 𝐷𝐷(𝑆𝑆) = 𝑆𝑆 ′, consider the dilation from center
1
�����⃗ by the same reasoning as
𝑂𝑂 with scale factor (the inverse of the dilation 𝐷𝐷). This dilation maps 𝑆𝑆′ to a point 𝑆𝑆 on 𝑃𝑃𝑃𝑃
𝑟𝑟
above. Then, 𝐷𝐷(𝑆𝑆) = 𝑆𝑆 ′ .

��������⃗ and, more generally, that dilations map
We conclude that the points of �����⃗
𝑃𝑃𝑃𝑃 are mapped onto the points of 𝑃𝑃′𝑄𝑄′
rays to rays.
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Example 3 (12 minutes)
Encourage students to draw upon the argument for Lesson 1, Example 3.
Example 3
���������⃗?
�������⃗contains point 𝑶𝑶. Does a dilation about center 𝑶𝑶 and scale factor 𝒓𝒓 map �������⃗
The line that contains 𝑷𝑷𝑷𝑷
𝑷𝑷𝑷𝑷 to 𝑷𝑷′𝑸𝑸′

Consider the case where the center 𝑂𝑂 belongs to the line that contains �����⃗
𝑃𝑃𝑃𝑃 .
a.

������⃗ coincides with the center 𝑶𝑶 of the dilation.
Examine the case where the endpoint 𝑷𝑷 of 𝑷𝑷𝑷𝑷
If the endpoint 𝑃𝑃 of �����⃗
𝑃𝑃𝑃𝑃 coincides with the center 𝑂𝑂, what can we say about �����⃗
𝑃𝑃𝑃𝑃 and ������⃗
𝑂𝑂𝑂𝑂 ?
Ask students to draw what this looks like, and draw the following on the board after giving them a head start.
������⃗.
All the points on �����⃗
𝑃𝑃𝑃𝑃 also belong to ������⃗
𝑂𝑂𝑂𝑂 ; �����⃗
𝑃𝑃𝑃𝑃 = 𝑂𝑂𝑂𝑂


By definition, a dilation always sends its center to itself. What are the implications for the dilation of 𝑂𝑂 and 𝑃𝑃?
Since a dilation always sends its center to itself, then 𝑂𝑂 = 𝑃𝑃 = 𝑂𝑂′ = 𝑃𝑃′.

MP.1

Let 𝑋𝑋 be a point on ������⃗
𝑂𝑂𝑂𝑂 so that 𝑋𝑋 ≠ 𝑂𝑂. What happens to 𝑋𝑋 under a dilation about 𝑂𝑂 with scale factor 𝑟𝑟?
The dilation sends 𝑋𝑋 to 𝑋𝑋′ on �����⃗
𝑂𝑂𝑂𝑂, and 𝑂𝑂𝑋𝑋 ′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂.

Ask students to draw what the position of 𝑋𝑋 and 𝑋𝑋′ might look like if 𝑟𝑟 > 1 or 𝑟𝑟 < 1. Points may move farther
away (𝑟𝑟 > 1) or move closer to the center (𝑟𝑟 < 1). They do not draw this for every case; rather, it is a reminder up
front.


𝑟𝑟 > 1
𝑟𝑟 < 1
Since 𝑂𝑂, 𝑋𝑋, and 𝑄𝑄 are collinear, then 𝑂𝑂, 𝑋𝑋′, and 𝑄𝑄 are also collinear; 𝑋𝑋′ is on the ray 𝑂𝑂𝑂𝑂, which coincides with
������⃗
𝑂𝑂𝑂𝑂 by definition of dilation.
������⃗ is �����⃗
Therefore, a dilation of 𝑂𝑂𝑂𝑂
𝑃𝑃𝑃𝑃 (or when the endpoint of 𝑃𝑃 of �����⃗
𝑃𝑃𝑃𝑃 coincides with the center 𝑂𝑂) about center
��������⃗
��������⃗
𝑂𝑂 and scale factor 𝑟𝑟 maps 𝑂𝑂′𝑄𝑄′ to 𝑃𝑃′𝑄𝑄′. We have answered the bigger question that a dilation maps a ray to a
ray.
b.
Examine the case where the endpoint 𝑷𝑷 of ����⃗
𝑷𝑷𝑷𝑷 is between 𝑶𝑶 and 𝑸𝑸 on the line containing 𝑶𝑶, 𝑷𝑷, and 𝑸𝑸.
Ask students to draw what this looks like, and draw the following on the board after giving them a head start.
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

We already know from the previous case that the dilation of the ray ������⃗
𝑂𝑂𝑂𝑂 maps onto itself. All we need to show
is that any point on the ray that is farther away from the center than 𝑃𝑃 maps to a point that is farther away
from the center than 𝑃𝑃′.
����?
Let 𝑋𝑋 be a point on �����⃗
𝑃𝑃𝑃𝑃 so that 𝑋𝑋 ≠ 𝑃𝑃. What can be concluded about the relative lengths of ����
𝑂𝑂𝑂𝑂 and 𝑂𝑂𝑂𝑂
Ask students to draw what this looks like, and draw the following on the board after giving them a head start. The
following is one possibility.
𝑂𝑂𝑂𝑂 < 𝑂𝑂𝑂𝑂


Describe how the lengths 𝑂𝑂𝑂𝑂′ and 𝑂𝑂𝑂𝑂′ compare once a dilation about center 𝑂𝑂 and scale factor 𝑟𝑟 sends 𝑃𝑃 to 𝑃𝑃′
and 𝑋𝑋 to 𝑋𝑋′.
𝑂𝑂𝑃𝑃′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂 and 𝑂𝑂𝑋𝑋 ′ = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂

Multiplying both sides of 𝑂𝑂𝑂𝑂 < 𝑂𝑂𝑂𝑂 by 𝑟𝑟 > 0 gives 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂 < 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂, so 𝑂𝑂𝑃𝑃′ < 𝑂𝑂𝑂𝑂′.

MP.1
Ask students to draw what this might look like if 𝑟𝑟 > 1, and draw the following on the board after giving them a head
start.

Therefore, we have shown that any point on the ray �����⃗
𝑃𝑃𝑃𝑃 that is farther away from the center than 𝑃𝑃 maps to a
point that is farther away from the center than 𝑃𝑃′. In this case, we still see that a dilation maps a ray to a ray.
c.

Examine the remaining case where the center 𝑶𝑶 of the dilation and point 𝑸𝑸 are on the same side of 𝑷𝑷 on the
line containing 𝑶𝑶, 𝑷𝑷, and 𝑸𝑸.
Now consider the relative position of 𝑂𝑂 and 𝑄𝑄 on �����⃗
𝑃𝑃𝑃𝑃 . We use an additional point 𝑅𝑅 as a reference point so the
𝑂𝑂 is between 𝑃𝑃 and 𝑅𝑅.
Draw the following on the board; these are all the ways that 𝑂𝑂 and 𝑄𝑄 are on the same side of 𝑃𝑃.
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

MP.1


By case (a), we know that a dilation with center 𝑂𝑂 maps �����⃗
𝑂𝑂𝑂𝑂 to itself.
Also, by our work in Lesson 7 on how dilations map segments to segments, we know that ����
𝑃𝑃𝑃𝑃 is taken to �����
𝑃𝑃′𝑂𝑂 ,
where 𝑃𝑃′ lies on �����⃗
𝑂𝑂𝑂𝑂 .
The union of the 𝑃𝑃′𝑂𝑂 and �����⃗
𝑂𝑂𝑂𝑂 is ������⃗
𝑃𝑃′𝑅𝑅 . So, the dilation maps �����⃗
𝑃𝑃𝑃𝑃 to ������⃗
𝑃𝑃′𝑅𝑅 .
′ 𝑅𝑅 = ��������⃗
��������⃗
Since 𝑄𝑄′ is a point on ������⃗
𝑃𝑃′𝑅𝑅 and 𝑄𝑄′ ≠ 𝑃𝑃′, we see 𝑃𝑃
𝑃𝑃′𝑄𝑄′. Therefore, in this case, we still see that �����⃗
𝑃𝑃𝑃𝑃 maps to
��������⃗
𝑃𝑃′𝑄𝑄′ under a dilation.
Example 4 (6 minutes)
In Example 4, students prove the dilation theorem for lines: A dilation maps a line to a line. If the center 𝑂𝑂 of the
dilation lies on the line or if the scale factor 𝑟𝑟 of the dilation is equal to 1, then the dilation maps the line to the same
line. Otherwise, the dilation maps the line to a parallel line.
The dilation theorem for lines can be proved using arguments similar to those used in Examples 1–3 for rays and to
Examples 1–3 in Lesson 7 for segments. Consider asking students to prove the theorem on their own as an exercise,
especially if time is an issue, and then provide the following proof:

We have just seen that dilations map rays to rays. How could we use this to reason that dilations map lines to
lines?
MP.3
��������⃗ and
⃖����⃗
𝑃𝑃𝑃𝑃 is the union of the two rays �����⃗
𝑃𝑃𝑃𝑃 and �����⃗
𝑄𝑄𝑄𝑄 . Dilate rays �����⃗
𝑃𝑃𝑃𝑃 and �����⃗
𝑄𝑄𝑄𝑄 ; the dilation yields rays 𝑃𝑃′𝑄𝑄′
�������⃗
𝑄𝑄′𝑃𝑃 ′. ⃖������⃗
𝑃𝑃′𝑄𝑄′ is the union of the two rays ��������⃗
𝑃𝑃′𝑄𝑄′ and �������⃗
𝑄𝑄′𝑃𝑃 ′. Since the dilation maps the rays �����⃗
𝑃𝑃𝑃𝑃 and �����⃗
𝑄𝑄𝑄𝑄 to

′ 𝑃𝑃′ , respectively, then the dilation maps ⃖����⃗
�������⃗
the rays ��������⃗
𝑃𝑃′𝑄𝑄′ and 𝑄𝑄
𝑃𝑃𝑃𝑃 to the ⃖������⃗
𝑃𝑃′𝑄𝑄′.
Example 5 (8 minutes)
In Example 5, students prove the dilation theorem for circles: A dilation maps a circle to a circle and maps the center to
the center. Students need the dilation theorem for circles for proving that all circles are similar in Module 5 (G-C.A.1).
Example 5
Does a dilation about a center 𝑶𝑶 and scale factor 𝒓𝒓 map a circle of radius 𝑹𝑹 onto another circle?
a.

Examine the case where the center of the dilation coincides with the center of the circle.
We first do the case where the center of the dilation is also the center of the circle. Let 𝐶𝐶 be a circle with
Draw the following figure on the board.

If the center of the dilation is also 𝑂𝑂, then every point 𝑃𝑃 on the circle is sent to a point 𝑃𝑃′ on �����⃗
𝑂𝑂𝑂𝑂 so
′
that 𝑂𝑂𝑃𝑃 = 𝑟𝑟 ⋅ 𝑂𝑂𝑂𝑂 = 𝑟𝑟𝑟𝑟; that is, the point goes to the point 𝑃𝑃′ on the circle 𝐶𝐶′ with center 𝑂𝑂 and radius 𝑟𝑟𝑟𝑟.
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
We also need to show that every point on 𝐶𝐶′ is the image of a point from 𝐶𝐶: For every point 𝑃𝑃′ on circle 𝐶𝐶′,
put a coordinate system on ⃖�����⃗
𝑂𝑂𝑂𝑂′ such that �������⃗
𝑂𝑂𝑃𝑃′ corresponds to the nonnegative real numbers with zero
corresponding to point 𝑂𝑂 (by the ruler axiom). Then, there exists a point 𝑃𝑃 on �������⃗
𝑂𝑂𝑂𝑂′ such that
𝑂𝑂𝑂𝑂 = 𝑅𝑅; that is, 𝑃𝑃 is a point on the circle 𝐶𝐶 that is mapped to 𝑃𝑃′ by the dilation.
Draw the following figure on the board.


Effectively, a dilation moves every point on a circle toward or away from the center the same amount, so the
dilated image is still a circle. Thus, the dilation maps the circle 𝐶𝐶 to the circle 𝐶𝐶′.
Circles that share the same center are called concentric circles.
b.
Examine the case where the center of the dilation is not the center of the circle; we call this the general case.

The proof of the general case works no matter where the center of the dilation is. We can actually use this
proof for case (a), when the center of the circle coincides with the center of dilation.

Let 𝐶𝐶 be a circle with center 𝑂𝑂 and radius 𝑅𝑅. Consider a dilation with center 𝐷𝐷 and scale factor 𝑟𝑟 that maps 𝑂𝑂
to 𝑂𝑂′. We will show that the dilation maps the circle 𝐶𝐶 to the circle 𝐶𝐶′ with center 𝑂𝑂′ and radius 𝑟𝑟𝑟𝑟.
Draw the following figure on the board.
𝑅𝑅


If 𝑃𝑃 is a point on circle 𝐶𝐶 and the dilation maps 𝑃𝑃 to 𝑃𝑃′ , the dilation theorem implies that 𝑂𝑂′ 𝑃𝑃′ = 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑟𝑟𝑟𝑟.
So, 𝑃𝑃′ is on circle 𝐶𝐶′.
We also need to show that every point of 𝐶𝐶′ is the image of a point from 𝐶𝐶. There are a number of ways to
prove this, but we will follow the same idea that we used in part (a). For a point 𝑃𝑃′ on circle 𝐶𝐶′ that is not on
⃖������⃗
𝑂𝑂′ 𝑃𝑃′ (the case when 𝑃𝑃′ is on line 𝐷𝐷𝐷𝐷′ is straightforward). Construct line ℓ through 𝑂𝑂 such that
𝐷𝐷𝑂𝑂′ , consider ��������⃗
′ 𝑃𝑃′ , and let 𝐴𝐴 be a point on ℓ that is in the same half-plane of ⃖������⃗
��������⃗
𝐷𝐷𝑂𝑂′ as 𝑃𝑃′ . Put a coordinate system on ℓ
ℓ ‖ 𝑂𝑂
�����⃗ corresponds to the nonnegative numbers with zero corresponding to point 𝑂𝑂 (by the
such that the ray 𝑂𝑂𝑂𝑂
ruler axiom). Then, there exists a point 𝑃𝑃 on �����⃗
𝑂𝑂𝑂𝑂 such that 𝑂𝑂𝑂𝑂 = 𝑅𝑅, which implies that 𝑃𝑃 is on the
circle 𝐶𝐶. By the dilation theorem, 𝑃𝑃 is mapped to the point 𝑃𝑃′ on the circle 𝐶𝐶′.
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
The diagram below shows how the dilation maps points 𝑃𝑃, 𝑄𝑄, 𝑅𝑅, 𝑆𝑆, and 𝑇𝑇 of circle 𝐶𝐶. Ask students to find
point 𝑊𝑊 on circle 𝐶𝐶 that is mapped to point 𝑊𝑊′ on circle 𝐶𝐶′.
Closing (1 minute)
Ask students to respond to the following question and summarize the key points of the lesson.

How are the proofs for the dilation theorems on segments, rays, and lines similar to each other?
•
DILATION THEOREM FOR RAYS: A dilation maps a ray to a ray sending the endpoint to the endpoint.
•
DILATION THEOREM FOR LINES: A dilation maps a line to a line. If the center 𝑂𝑂 of the dilation lies on the line or if the
scale factor 𝑟𝑟 of the dilation is equal to 1, then the dilation maps the line to the same line. Otherwise, the
dilation maps the line to a parallel line.
•
DILATION THEOREM FOR CIRCLES: A dilation maps a circle to a circle and maps the center to the center.
Lesson Summary

DILATION THEOREM FOR RAYS: A dilation maps a ray to a ray sending the endpoint to the endpoint.

DILATION THEOREM FOR LINES: A dilation maps a line to a line. If the center 𝑶𝑶 of the dilation lies on the line
or if the scale factor 𝒓𝒓 of the dilation is equal to 𝟏𝟏, then the dilation maps the line to the same line.
Otherwise, the dilation maps the line to a parallel line.

DILATION THEOREM FOR CIRCLES: A dilation maps a circle to a circle and maps the center to the center.
Exit Ticket (5 minutes)
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Name
Date
Lesson 8: How Do Dilations Map Lines, Rays, and Circles?
Exit Ticket
Given points 𝑂𝑂, 𝑆𝑆, and 𝑇𝑇 below, complete parts (a)–(e):
a.
Draw rays ����⃗
𝑆𝑆𝑆𝑆 and�����⃗
𝑇𝑇𝑇𝑇. What is the union of these rays?
b.
Dilate ����⃗
𝑆𝑆𝑆𝑆 from 𝑂𝑂 using scale factor 𝑟𝑟 = 2. Describe the image of ������⃗
𝑆𝑆𝑆𝑆.
c.
Dilate ����⃗
𝑇𝑇𝑇𝑇 from 𝑂𝑂 using scale factor 𝑟𝑟 = 2. Describe the image of ����⃗
𝑇𝑇𝑇𝑇.
d.
What does the dilation of the rays in parts (b) and (c) yield?
e.
Dilate circle 𝐶𝐶 with radius 𝑇𝑇𝑇𝑇 from 𝑂𝑂 using scale factor 𝑟𝑟 = 2.
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Exit Ticket Sample Solutions
Given points 𝑶𝑶, 𝑺𝑺, and 𝑻𝑻 below, complete parts (a)–(e):
a.
Draw rays �����⃗
𝑺𝑺𝑺𝑺 and ���⃗
𝑻𝑻𝑻𝑻. What is the union of these rays?
�����⃗ and �����⃗
The union of 𝑺𝑺𝑺𝑺
𝑻𝑻𝑻𝑻 is line 𝑺𝑺𝑺𝑺.
b.
�����⃗ from 𝑶𝑶 using scale factor 𝒓𝒓 = 𝟐𝟐. Describe the image of 𝑺𝑺𝑺𝑺
�����⃗.
Dilate 𝑺𝑺𝑺𝑺
�����⃗ is �������⃗
The image of 𝑺𝑺𝑺𝑺
𝑺𝑺′𝑻𝑻′.
c.
���⃗.
Dilate ���⃗
𝑻𝑻𝑻𝑻 from 𝑶𝑶 using scale factor 𝒓𝒓 = 𝟐𝟐. Describe the image of 𝑻𝑻𝑻𝑻
�����⃗ is �������⃗
The image of 𝑻𝑻𝑻𝑻
𝑻𝑻′𝑺𝑺′.
d.
What does the dilation of the rays in parts (b) and (c) yield?
�����⃗ and �����⃗
The dilation of rays 𝑺𝑺𝑺𝑺
𝑻𝑻𝑻𝑻 yields ⃖������⃗
𝑺𝑺′𝑻𝑻′.
e.
Dilate circle 𝑻𝑻 with radius 𝑻𝑻𝑻𝑻 from 𝑶𝑶 using scale factor 𝒓𝒓 = 𝟐𝟐.
See diagram above.
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Problem Set Sample Solutions
1.
In Lesson 8, Example 2, you proved that a dilation with a scale factor 𝒓𝒓 > 𝟏𝟏 maps a ray 𝑷𝑷𝑷𝑷 to a ray 𝑷𝑷′𝑸𝑸′. Prove the
remaining case that a dilation with scale factor 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏 maps a ray 𝑷𝑷𝑷𝑷 to a ray 𝑷𝑷′𝑸𝑸′.
Given the dilation 𝑫𝑫𝑶𝑶,𝒓𝒓 , with 𝟎𝟎 < 𝒓𝒓 < 𝟏𝟏 maps 𝑷𝑷 to 𝑷𝑷′ and 𝑸𝑸 to 𝑸𝑸′, prove that 𝑫𝑫𝑶𝑶,𝒓𝒓 maps ������⃗
𝑷𝑷𝑷𝑷 to ���������⃗
𝑷𝑷′𝑸𝑸′.
By the definition of dilation, 𝑶𝑶𝑷𝑷′ = 𝒓𝒓(𝑶𝑶𝑶𝑶), and likewise,
����.
By the dilation theorem, ������
𝑷𝑷′𝑸𝑸′ ∥ 𝑷𝑷𝑷𝑷
𝑶𝑶𝑷𝑷′
𝑶𝑶𝑶𝑶
= 𝒓𝒓.
⃖�������⃗.
⃖����⃗ ∥ 𝑷𝑷′𝑸𝑸′
Through two different points lies only one line, so 𝑷𝑷𝑷𝑷
������⃗ and ���������⃗
Draw point 𝑹𝑹 on ������⃗
𝑷𝑷𝑷𝑷, and then draw ������⃗
𝑶𝑶𝑶𝑶. Mark point 𝑹𝑹′ at the intersection of 𝑶𝑶𝑶𝑶
𝑷𝑷′𝑸𝑸′.
������ splits 𝑶𝑶𝑶𝑶
���� and ����
By the triangle side splitter theorem, 𝑷𝑷′𝑹𝑹′
𝑶𝑶𝑶𝑶 proportionally, so
𝑶𝑶𝑹𝑹′
𝑶𝑶𝑶𝑶
=
𝑶𝑶𝑷𝑷′
𝑶𝑶𝑶𝑶
= 𝒓𝒓.
𝑷𝑷𝑷𝑷.
Therefore, because 𝑹𝑹 was chosen as an arbitrary point, 𝑶𝑶𝑹𝑹′ = 𝒓𝒓(𝑶𝑶𝑶𝑶) for any point 𝑹𝑹 on ������⃗
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2.
In the diagram below, ��������⃗
𝑨𝑨′𝑩𝑩′ is the image of ������⃗
𝑨𝑨𝑨𝑨 under a dilation from point 𝑶𝑶 with an unknown scale factor; 𝑨𝑨 maps
to 𝑨𝑨′ , and 𝑩𝑩 maps to 𝑩𝑩′. Use direct measurement to determine the scale factor 𝒓𝒓, and then find the center of
dilation 𝑶𝑶.
By the definition of dilation, 𝑨𝑨′ 𝑩𝑩′ = 𝒓𝒓 (𝑨𝑨𝑨𝑨), 𝑶𝑶𝑨𝑨′ = 𝒓𝒓(𝑶𝑶𝑶𝑶), and 𝑶𝑶𝑩𝑩′ = 𝒓𝒓(𝑶𝑶𝑶𝑶).
By direct measurement,
𝑨𝑨′ 𝑩𝑩′
𝑨𝑨𝑨𝑨
=
𝟕𝟕
𝟒𝟒
= 𝒓𝒓.
⃖����⃗ under the dilation, and 𝑨𝑨′ 𝑩𝑩′ > 𝑨𝑨𝑨𝑨, so the center of dilation
The images of 𝑨𝑨 and 𝑩𝑩 are pushed to the right on 𝑨𝑨𝑨𝑨
⃖����⃗
must lie on 𝑨𝑨𝑨𝑨 to the left of points 𝑨𝑨 and 𝑩𝑩.
By the definition of dilation:
𝟕𝟕
(𝑶𝑶𝑶𝑶)
𝟒𝟒
𝟕𝟕
(𝑶𝑶𝑶𝑶 + 𝑨𝑨𝑨𝑨′ ) = (𝑶𝑶𝑶𝑶)
𝟒𝟒
𝑶𝑶𝑶𝑶 + 𝑨𝑨𝑨𝑨′ 𝟕𝟕
=
𝑶𝑶𝑶𝑶
𝟒𝟒
′
𝑶𝑶𝑶𝑶 𝑨𝑨𝑨𝑨
𝟕𝟕
+
=
𝑶𝑶𝑶𝑶 𝑶𝑶𝑶𝑶 𝟒𝟒
𝑨𝑨𝑨𝑨′ 𝟕𝟕
𝟏𝟏 +
=
𝑶𝑶𝑶𝑶 𝟒𝟒
𝑨𝑨𝑨𝑨′ 𝟑𝟑
=
𝑶𝑶𝑶𝑶 𝟒𝟒
𝟑𝟑
𝑨𝑨𝑨𝑨′ = (𝑶𝑶𝑶𝑶)
𝟒𝟒
𝟒𝟒
′
(𝑨𝑨𝑨𝑨 ) = 𝑶𝑶𝑶𝑶
𝟑𝟑
𝑶𝑶𝑨𝑨′ =
3.
Draw a line ⃖����⃗
𝑨𝑨𝑨𝑨, and dilate points 𝑨𝑨 and 𝑩𝑩 from center 𝑶𝑶 where 𝑶𝑶 is not on ⃖����⃗
𝑨𝑨𝑨𝑨. Use your diagram to explain why a
line maps to a line under a dilation with scale factor 𝒓𝒓.
������⃗ and ������⃗
Two rays 𝑨𝑨𝑨𝑨
𝑩𝑩𝑩𝑩 on the points 𝑨𝑨 and 𝑩𝑩 point in opposite directions as shown in the diagram below. The union
��������⃗; likewise, ������⃗
⃖����⃗. We showed that a ray maps to a ray under a dilation, so 𝑨𝑨𝑨𝑨
������⃗ maps to 𝑨𝑨′𝑩𝑩′
of the two rays is 𝑨𝑨𝑨𝑨
𝑩𝑩𝑩𝑩 maps
to ��������⃗
𝑩𝑩′𝑨𝑨′. The dilation yields two rays ��������⃗
𝑨𝑨′𝑩𝑩′ and ��������⃗
𝑩𝑩′𝑨𝑨′ on the points 𝑨𝑨′ and 𝑩𝑩′ pointing in opposite directions. The
union of the two rays is ⃖�������⃗
𝑨𝑨′𝑩𝑩′; therefore, it is true that a dilation maps a line to a line.
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4.
Let ����
𝑨𝑨𝑨𝑨 be a line segment, and let 𝒎𝒎 be a line that is the perpendicular bisector of ����
𝑨𝑨𝑨𝑨. If a dilation with scale factor
������ (sending 𝑨𝑨 to 𝑨𝑨′ and 𝑩𝑩 to 𝑩𝑩′) and also maps line 𝒎𝒎 to line 𝒎𝒎′, show that line 𝒎𝒎′ is the
���� to 𝑨𝑨′𝑩𝑩′
𝒓𝒓 maps 𝑨𝑨𝑨𝑨
perpendicular bisector of ������
𝑨𝑨′𝑩𝑩′.
Let 𝑷𝑷 be a point on line 𝒎𝒎, and let the dilation send 𝑷𝑷 to the point 𝑷𝑷′ on line 𝒎𝒎′. Since 𝑷𝑷 is on the perpendicular
����, 𝑷𝑷𝑷𝑷 = 𝑷𝑷𝑷𝑷. By the dilation theorem, 𝑷𝑷′ 𝑨𝑨′ = 𝒓𝒓𝒓𝒓𝒓𝒓 and 𝑷𝑷′ 𝑩𝑩′ = 𝒓𝒓𝒓𝒓𝒓𝒓. So, 𝑷𝑷′ 𝑨𝑨′ = 𝑷𝑷′ 𝑩𝑩′ , and 𝑷𝑷′ is on the
bisector of 𝑨𝑨𝑨𝑨
������.
perpendicular bisector of 𝑨𝑨′𝑩𝑩′
5.
𝟏𝟏
𝟐𝟐
Dilate circle 𝑪𝑪 with radius 𝑪𝑪𝑪𝑪 from center 𝑶𝑶 with a scale factor 𝒓𝒓 = .
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6.
In the picture below, the larger circle is a dilation of the smaller circle. Find the center of dilation 𝑶𝑶.
Draw ������⃗
𝑪𝑪′𝑪𝑪. Center 𝑶𝑶 lies on ������⃗
𝑪𝑪′𝑪𝑪. Since 𝑨𝑨′ is not on a line with both 𝑪𝑪 and 𝑪𝑪′, I can use the parallel method to find
⃖������⃗.
point 𝑨𝑨 on circle 𝑪𝑪 such that ⃖����⃗
𝑪𝑪𝑪𝑪 ∥ 𝑪𝑪′𝑨𝑨′
Under a dilation, a point and its image(s) lie on a ray with endpoint 𝑶𝑶, the center of dilation. Draw �������⃗
𝑨𝑨′𝑨𝑨, and label
������⃗ ∩ �������⃗
the center of dilation 𝑶𝑶 = 𝑪𝑪′𝑪𝑪
𝑨𝑨′𝑨𝑨.
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Lesson 9: How Do Dilations Map Angles?
Student Outcomes

Students prove that dilations map an angle to an angle with equal measure.

Students understand how dilations map triangles, squares, rectangles, trapezoids, and regular polygons.
Lesson Notes
In this lesson, students show that dilations map angles to angles of equal measure. The Exploratory Challenge requires
students to make conjectures about how dilations map polygonal figures, specifically, the effect on side lengths and
angle measures. The goal is for students to informally verify that dilations map angles to angles of equal measure.
Further, students describe the effect dilations have on side lengths; for example, if side length 𝐴𝐴𝐴𝐴 = 3.6 and is dilated
from a center with scale factor 𝑟𝑟 = 2, then the dilated side 𝐴𝐴′ 𝐶𝐶 ′ = 7.2. The Discussion that follows the Exploratory
Challenge focuses on the effect dilations have on angles. Students should already be familiar with the effect of dilation
on lengths of segments, so the work with polygonal figures extends students’ understanding of the effect of dilations on
figures other than triangles. Consider extending the lesson over two days where on the first day students complete all
parts of the Exploratory Challenge, and on the second day students work to prove their conjectures about how dilations
map angles. The last Discussion of the lesson (dilating a square inscribed in a triangle) is optional and can be completed
if class time permits.
This lesson highlights Mathematical Practice 3: Construct viable arguments and critique the reasoning of others.
Throughout the lesson, students are asked to make a series of conjectures and justify them by experimenting with
diagrams and direct measurements.
Classwork
Exploratory Challenge/Exercises 1–4 (13 minutes)
The Exploratory Challenge allows students to informally verify how dilations map
polygonal figures in preparation for the Discussion that follows. Make clear to students
that they must first make a conjecture about how the dilation affects the figure and then
verify their conjectures by actually performing a dilation and directly measuring angles and
side lengths. Consider having students share their conjectures and drawings with the
class. It may be necessary to divide the class into four groups and assign each group one
exercise to complete. When all groups are finished, they can share their results with the
whole class.
Exploratory Challenge/Exercises 1–4
1.
MP.3
How do dilations map triangles?
a.
Scaffolding:
Some groups of students may
benefit from a teacher-led
model of the first exercise.
explicit examples, such as
dilating triangle 𝐴𝐴𝐴𝐴𝐴𝐴,
𝐴𝐴(3,1), 𝐵𝐵(7,1), 𝐶𝐶(3,5), by a
variety of scale factors to help
students make a conjecture.
Make a conjecture.
A dilation maps a triangle to a triangle with the same angles, and all of the sides of the image triangle are
proportional to the sides of the original triangle.
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b.
Verify your conjecture by experimenting with diagrams and directly measuring angles and lengths of
segments.
𝟕𝟕. 𝟐𝟐 𝟖𝟖. 𝟐𝟐 𝟔𝟔. 𝟐𝟐
=
=
= 𝟐𝟐
𝟑𝟑. 𝟔𝟔 𝟒𝟒. 𝟏𝟏 𝟑𝟑. 𝟏𝟏
The value of the ratios of the lengths of the dilated triangle to the original triangle is equal to the scale factor.
The angles map to angles of equal measure; all of the angles in the original triangle are dilated to angles
equal in measure to the corresponding angles in the dilated triangle.
2.
How do dilations map rectangles?
a.
Make a conjecture.
A dilation maps a rectangle to a rectangle so that the ratio of length to width is the same.
MP.3
b.
Verify your conjecture by experimenting with diagrams and directly measuring angles and lengths of
segments.
The original rectangle has a length to width ratio of 𝟑𝟑: 𝟐𝟐. The dilated rectangle has a length to width ratio
of 𝟏𝟏. 𝟓𝟓: 𝟏𝟏. The values of the ratios are equal. Since angles map to angles of equal measure, all of the right
angles in the original rectangle are dilated to right angles.
3.
How do dilations map squares?
a.
Make a conjecture.
A dilation maps a square with side length 𝑳𝑳 to a square with side length 𝒓𝒓𝒓𝒓 where 𝒓𝒓 is the scale factor of
dilation. Since each segment of length 𝑳𝑳 is mapped to a segment of length 𝒓𝒓𝒓𝒓 and each right angle is mapped
to a right angle, then the dilation maps a square with side length 𝑳𝑳 to a square with side length 𝒓𝒓𝒓𝒓.
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b.
Verify your conjecture by experimenting with diagrams and directly measuring angles and lengths of
segments.
Sample student drawing:
The side length of the original square is 𝟐𝟐 units. The side length of the dilated square is 𝟔𝟔 units. The side
length of the dilated square is equal to the length of the original square multiplied by the scale factor.
Since angles map to angles of equal measure, all of the right angles in the original square are dilated to right
angles.
4.
How do dilations map regular polygons?
a.
Make a conjecture.
A dilation maps a regular polygon with side length 𝑳𝑳 to a regular polygon with side length 𝒓𝒓𝒓𝒓 where 𝒓𝒓 is the
scale factor of dilation. Since each segment of length 𝑳𝑳 is mapped to a segment of length 𝒓𝒓𝒓𝒓 and each angle
is mapped to an angle that is equal in measure, then the dilation maps a regular polygon with side length 𝑳𝑳 to
a regular polygon with side length 𝒓𝒓𝒓𝒓.
MP.3
b.
Verify your conjecture by experimenting with diagrams and directly measuring angles and lengths of
segments.
The side length of the original regular polygon is 𝟑𝟑 units.
The side length of the dilated regular polygon is 𝟏𝟏𝟏𝟏 units.
The side length of the dilated regular polygon is equal to
the length of the original regular polygon multiplied by the
scale factor. Since angles map to angles of equal measure,
all of the angles in the original regular polygon are dilated
to angles of each measure.
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Discussion (7 minutes)
Begin the Discussion by debriefing the Exploratory Challenge. Elicit the information about the effect of dilations on angle
measures and side lengths as described in the sample student responses above. Then, continue with the Discussion
below.


In Grade 8, we showed that under a dilation with a center at the origin and scale factor 𝑟𝑟, an angle formed by
the union of two rays and the image of the angle would be equal in measure.
The multiplicative effect that dilation has on points (when the origin is the center of dilation) was used to show
′ 𝑅𝑅 ′ , respectively. Then, facts about parallel lines cut by a
��������⃗ and 𝑄𝑄
����������⃗
that rays �����⃗
𝑄𝑄𝑄𝑄 and �����⃗
𝑄𝑄𝑄𝑄 map to rays 𝑄𝑄′𝑃𝑃′
transversal were used to prove that 𝑚𝑚∠𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑚𝑚∠𝑃𝑃′ 𝑄𝑄′ 𝑅𝑅′ .
Scaffolding:
For some groups of students, a
simpler example where the
vertex of the angle is on the
𝑥𝑥-axis may aid understanding.

Now that we know from the last two lessons that a dilation maps a segment to a segment, a ray to a ray, and a
line to a line, we can prove that dilations map angles to angles of equal measure without the need for a
coordinate system.
Exercises 5–6 (9 minutes)
Provide students time to develop the proof that under a dilation, the measure of an angle
and its dilated image are equal. Consider having students share their proofs with the
MP.3
class. If necessary, share the proof shown in Exercise 6 with the class as described in the
Scaffolding box to the right.
Exercises 5–6
5.
Recall what you learned about parallel lines cut by a transversal, specifically about the
angles that are formed.
When parallel lines are cut by a transversal, then corresponding angles are equal in
measure, alternate interior angles are equal in measure, and alternate exterior angles are
equal in measure.
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Scaffolding:
 If students struggle,
reference Exercise 5,
particularly the
corresponding angles, as a
hint that guides students’
thinking.
 Consider offering the
proof to students as the
work of a classmate and
having students
paraphrase the statements
in the proof.
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6.
A dilation from center 𝑶𝑶 by scale factor 𝒓𝒓 maps ∠𝑩𝑩𝑩𝑩𝑩𝑩 to ∠𝑩𝑩′ 𝑨𝑨′𝑪𝑪′ . Show that 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝒎𝒎∠𝑩𝑩′ 𝑨𝑨′𝑪𝑪′ .
By properties of dilations, we know that a dilation maps a line to itself or a parallel line. We consider a case where
the dilated rays of the angle are mapped to parallel rays as shown below, and two rays meet in a single point. Then,
𝑨𝑨′ 𝑩𝑩′ . �����⃗
𝑨𝑨𝑨𝑨 is a transversal that cuts the parallel lines. Let 𝑫𝑫
the line containing ������⃗
𝑨𝑨𝑨𝑨 is parallel to the line containing ���������⃗
′
′
������⃗
���������⃗
𝑨𝑨𝑨𝑨. Since corresponding angles
be the point of intersection of 𝑨𝑨𝑨𝑨 and 𝑨𝑨 𝑩𝑩 , and let 𝑬𝑬 be a point to the right of 𝑫𝑫 on �����⃗
𝑨𝑨𝑨𝑨
of parallel lines are congruent and, therefore, equal in measure, then 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝒎𝒎∠𝑩𝑩′ 𝑫𝑫𝑫𝑫. The dilation maps �����⃗
to ��������⃗
𝑨𝑨′ 𝑪𝑪′, and the lines containing those rays are parallel. Then, ��������⃗
𝑨𝑨′𝑩𝑩′ is a transversal that cuts the parallel lines.
Therefore, 𝒎𝒎∠𝑩𝑩′ 𝑫𝑫𝑫𝑫 = 𝒎𝒎∠𝑩𝑩′ 𝑨𝑨′ 𝑪𝑪′ . Since 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝒎𝒎∠𝑩𝑩′ 𝑫𝑫𝑫𝑫 and 𝒎𝒎∠𝑩𝑩′ 𝑫𝑫𝑫𝑫 = 𝒎𝒎∠𝑩𝑩′ 𝑨𝑨′ 𝑪𝑪′ , by the transitive
property 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝒎𝒎∠𝑩𝑩′ 𝑨𝑨′ 𝑪𝑪′ . Therefore, the dilation maps an angle to an angle of equal measure.
MP.7
Discussion (4 minutes)
While the teacher leads the discussion, students should record the information about the dilation theorem and its proof
in their student pages.
Discussion
The dilation theorem for angles is as follows:
DILATION THEOREM: A dilation from center 𝑶𝑶 and scale factor 𝒓𝒓 maps an angle to an angle of equal measure.
We have shown this when the angle and its image intersect at a single point, and that point of intersection is not the
vertex of the angle.

So far, we have seen the case when the angles intersect at a point. What are other possible cases that we
need to consider?
Provide time for students to identify the two other cases: (1) The angles do not intersect at a point, and (2) the angles
have vertices on the same ray. The teacher may need to give students this information if students cannot develop the
cases on their own.

We will cover another case where a dilation maps ∠𝐵𝐵𝐵𝐵𝐵𝐵 to ∠𝐵𝐵′𝐴𝐴′ 𝐶𝐶 ′ . When the angle and its dilated image do
not have intersecting rays (as shown below), how can we show that the angle and its dilated image are equal in
measure?
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If students do not respond, the teacher may need to give the information in the bullet point below.




You can draw an auxiliary line and use the same reasoning to show that the angle and its image are
equal in measure.
Notice now that by drawing the auxiliary line, we have two angles that intersect at a point, much like we had in
the Opening Exercise. Therefore, the same reasoning shows that 𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵 = 𝑚𝑚∠𝐵𝐵′𝐴𝐴′𝐶𝐶′.
The only case left to consider is when the line containing a ray also contains the image of the ray. In this case,
ray 𝐵𝐵𝐵𝐵 and ray 𝐵𝐵′𝐶𝐶′ lie in the same line.
Why does 𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑚𝑚∠𝐴𝐴′𝐵𝐵′𝐶𝐶′?

They are corresponding angles from parallel lines cut by a transversal.
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Discussion (5 minutes)
This Discussion is optional and can be used if class time permits.

Given △ 𝐴𝐴𝐴𝐴𝐴𝐴 with ∠𝐴𝐴 and ∠𝐵𝐵 acute, inscribe a square inside the triangle so that two vertices of the square lie
on side ����
𝐴𝐴𝐴𝐴, and the other two vertices lie on the other two sides.

We begin by drawing a small square near vertex 𝐴𝐴 so that one side is on ����
𝐴𝐴𝐴𝐴 and one vertex is on ����
𝐴𝐴𝐴𝐴 .
Ask students how they can dilate the square so that the other two vertices lie on the other two sides. Provide time for
students to discuss this challenging question in small groups and to share their thoughts with the class. Validate any
appropriate strategies, or suggest the strategy described below.

Next, draw a ray from 𝐴𝐴 through the vertex of the square that does not touch any side of the triangle. Name
the point where the ray intersects the triangle 𝑇𝑇.

The point 𝑇𝑇 can be used as one of the vertices of the desired square. The parallel method can then be used to
construct the desired square.
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Consider trying this on another triangle before proceeding with the question below.

Why does this work?

Three of the four vertices of the square are on two sides of the triangle. The location of the fourth
���� . Since the ray was drawn through the vertex of the small square, a scale
vertex must be on side 𝐵𝐵𝐵𝐵
factor 𝑟𝑟 maps the dilated vertex on the ray. To inscribe it in the desired location, we note the location
where the ray intersects the side of the triangle, giving us the vertex of the desired square. Since
dilations map squares to squares, it is just a matter of locating the point of the vertex along the
opposite side of ∠𝐴𝐴 and then constructing the square.
Closing (2 minutes)

How do dilations map angles?


What foundational knowledge did we need to prove that dilations map angles to angles of equal measure?


Dilations map angles to angles of equal measure.
We needed to know that rays are dilated to rays that are parallel or rays that coincide with the original
ray. Then, using what we know about parallel lines cut by a transversal, we could use what we knew
about corresponding angles of parallel lines being equal to show that dilations map angles to angles of
equal measure. We also needed to know about auxiliary lines and how to use them in order to produce
a diagram where parallel lines are cut by a transversal.
How do dilations map polygonal figures?

Dilations map polygonal figures to polygonal figures whose angles are equal in measure to the
corresponding angles of the original figure and whose side lengths are equal to the corresponding side
lengths multiplied by the scale factor.
Lesson Summary

Dilations map angles to angles of equal measure.

Dilations map polygonal figures to polygonal figures whose angles are equal in measure to the
corresponding angles of the original figure and whose side lengths are equal to the corresponding side
lengths multiplied by the scale factor.
Exit Ticket (5 minutes)
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Name
Date
Lesson 9: How Do Dilations Map Angles?
Exit Ticket
3
4
1.
Dilate parallelogram 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 from center 𝑂𝑂 using a scale factor of 𝑟𝑟 = .
2.
How does 𝑚𝑚∠𝑇𝑇′ compare to 𝑚𝑚∠𝑇𝑇?
3.
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Exit Ticket Sample Solutions
𝟑𝟑
𝟒𝟒
1.
Dilate parallelogram 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 from center 𝑶𝑶 using a scale factor of 𝒓𝒓 = .
2.
How does 𝒎𝒎∠𝑻𝑻′ compare to 𝒎𝒎∠𝑻𝑻?
3.
𝒎𝒎∠𝑻𝑻′ = 𝒎𝒎∠𝑻𝑻 because dilations preserve angle measure.
������; therefore, by
Extend ������⃗
𝑻𝑻𝑻𝑻 such that it intersects �����
𝑺𝑺′𝑻𝑻′ at a point 𝑷𝑷. Dilations map lines to parallel lines, so ����
𝑻𝑻𝑻𝑻 ∥ 𝑻𝑻′𝑼𝑼′
������, ∠𝑺𝑺′𝑷𝑷𝑷𝑷 ≅ ∠𝑻𝑻′. Under the same dilation, �����
����, again, by corresponding
𝑻𝑻𝑻𝑻 ∥ 𝑻𝑻′𝑼𝑼′
corresponding ∠′s, ����
𝑺𝑺′𝑻𝑻′ ∥ 𝑺𝑺𝑺𝑺
����, ∠𝑺𝑺′ 𝑷𝑷𝑷𝑷 ≅ ∠𝑻𝑻. By transitivity, ∠𝑻𝑻′ ≅ ∠𝑻𝑻.
∠'s, �����
𝑺𝑺′𝑻𝑻′ ∥ 𝑺𝑺𝑺𝑺
Problem Set Sample Solutions
1.
𝟓𝟓
𝟐𝟐
Shown below is △ 𝑨𝑨𝑨𝑨𝑨𝑨 and its image △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ after it has been dilated from center 𝑶𝑶 by scale factor 𝒓𝒓 = .
Prove that the dilation maps △ 𝑨𝑨𝑨𝑨𝑨𝑨 to △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ so that 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑨𝑨 , 𝒎𝒎∠𝑩𝑩 = 𝒎𝒎∠𝑩𝑩′, and 𝒎𝒎∠𝑪𝑪 = 𝒎𝒎∠𝑪𝑪′.
′
Locate the center of dilation 𝑶𝑶 by drawing rays through each of the pairs of corresponding points. The intersection
of the rays is the center of dilation 𝑶𝑶. Since dilations map segments to segments, and the dilated segments must
′ 𝑩𝑩′ , ⃖����⃗
′ 𝑪𝑪′ .
⃖������⃗, and ⃖����⃗
⃖�������⃗
⃖����⃗ ∥ 𝑨𝑨
⃖�������⃗
𝑨𝑨𝑨𝑨 ∥ 𝑨𝑨′𝑪𝑪′
𝑩𝑩𝑩𝑩 ∥ 𝑩𝑩
either coincide with their pre-image or be parallel, then we know that 𝑨𝑨𝑨𝑨
′ 𝑩𝑩′ . Then, ∠𝑩𝑩′𝑨𝑨′𝑪𝑪′ is congruent to ∠𝑩𝑩′𝑫𝑫𝑫𝑫 because
���� intersects with side 𝑨𝑨
������
Let 𝑫𝑫 be the point where side 𝑨𝑨𝑨𝑨
⃖������⃗, cut by a transversal, ⃖�������⃗
corresponding ∠'s, ⃖����⃗
𝑨𝑨𝑨𝑨 ∥ 𝑨𝑨′𝑪𝑪′
𝑨𝑨′𝑩𝑩′, are congruent. Then, ∠𝑩𝑩′ 𝑫𝑫𝑫𝑫 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 because
′ 𝑩𝑩′ , cut by a transversal 𝑨𝑨𝑨𝑨
⃖����⃗, are congruent.
⃖�������⃗
𝑨𝑨𝑨𝑨|| 𝑨𝑨
corresponding ∠'s, ⃖����⃗
By the transitive property, ∠𝑩𝑩′ 𝑨𝑨′ 𝑪𝑪′ ≅ ∠𝑩𝑩′ 𝑫𝑫𝑫𝑫 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑩𝑩′ 𝑨𝑨′ 𝑪𝑪′ ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩.
Since congruent angles are equal in measure, 𝒎𝒎∠𝑩𝑩′ 𝑨𝑨′ 𝑪𝑪′ = 𝒎𝒎∠𝑨𝑨′ and 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝒎𝒎∠𝑨𝑨; then, 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑨𝑨′ .
Similar reasoning shows that 𝒎𝒎∠𝑩𝑩 = 𝒎𝒎∠𝑩𝑩′ and 𝒎𝒎∠𝑪𝑪 = 𝒎𝒎∠𝑪𝑪′ .
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2.
Explain the effect of a dilation with scale factor 𝒓𝒓 on the length of the base and height of a triangle. How is the area
of the dilated image related to the area of the pre-image?
⃖����⃗. Thus, the base length of the triangle
���� of △ 𝑨𝑨𝑨𝑨𝑨𝑨, such that 𝑷𝑷 lies on 𝑩𝑩𝑩𝑩
Let 𝑷𝑷 represent the endpoint of altitude 𝑨𝑨𝑨𝑨
𝟏𝟏
is 𝑩𝑩𝑩𝑩, and the height of the triangle is 𝑨𝑨𝑨𝑨. The area of the given triangle then is (𝑩𝑩𝑩𝑩)(𝑨𝑨𝑨𝑨). By the definition of
𝟐𝟐
dilation, 𝑩𝑩′ 𝑪𝑪′ = 𝒓𝒓(𝑩𝑩𝑩𝑩) and 𝑨𝑨′ 𝑷𝑷′ = 𝒓𝒓(𝑨𝑨𝑨𝑨), so the base and height of the dilated image are proportional to the base
and height of the original just as the lengths of the sides of the triangle. The area of the dilated triangle then would
be
𝟏𝟏
𝟐𝟐
�𝒓𝒓(𝑩𝑩𝑩𝑩) ∙ 𝒓𝒓(𝑨𝑨𝑨𝑨)� = 𝒓𝒓𝟐𝟐 ∙
𝟏𝟏
𝟐𝟐
(𝑩𝑩𝑩𝑩)(𝑨𝑨𝑨𝑨).
The ratio of the area of the dilated image to the area of the pre-image is 𝒓𝒓𝟐𝟐 .
3.
𝟏𝟏
𝟐𝟐
Dilate trapezoid 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 from center 𝑶𝑶 using a scale factor of 𝒓𝒓 = .
A dilation maps a trapezoid to a trapezoid so that the ratio of corresponding sides is the same and corresponding
interior angles are the same measure.
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4.
𝟏𝟏
𝟐𝟐
Dilate kite 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 from center 𝑶𝑶 using a scale factor 𝒓𝒓 = 𝟏𝟏 .
A dilation maps a kite to a kite so that the ratio of corresponding sides is the same and corresponding interior angles
are the same measure.
5.
𝟏𝟏
𝟒𝟒
Dilate hexagon 𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫 from center 𝑶𝑶 using a scale factor of 𝒓𝒓 = .
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6.
Examine the dilations that you constructed in Problems 2–5, and describe how each image compares to its preimage under the given dilation. Pay particular attention to the sizes of corresponding angles and the lengths of
corresponding sides.
In each dilation, the angles in the image are the same size as the corresponding angles in the pre-image, as we have
shown that dilation preserves angle measure. We also know that the lengths of corresponding sides are in the same
ratio as the scale factor.
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Lesson 10
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GEOMETRY
Lesson 10: Dividing the King’s Foot into 12 Equal Pieces
Student Outcomes


Students divide a line segment into 𝑛𝑛 equal pieces by the side splitter and dilation methods.
Students know how to locate fractions on the number line.
Materials

Poster paper or chart paper

Yardstick

Compass

Straightedge

Set square
Lesson Notes
Students explore how their study of dilations relates to the constructions that divide a segment into a whole number of
equal-length segments.
Classwork
Opening (2 minutes)
In an age when there was no universal consensus on measurement, the human body was often used to create units of
measurement. One can imagine how a king might declare the length of his foot to be what we know as the unit of a
MP.3
foot. How would we go about figuring out how to divide one foot into twelve equal portions, as the 12 inches that
comprise a foot? Have students write or discuss their thoughts.
Opening Exercise (3 minutes)
Opening Exercise
Use a compass to mark off equally spaced points 𝑪𝑪, 𝑫𝑫, 𝑬𝑬, and 𝑭𝑭 so that 𝑨𝑨𝑨𝑨, 𝑩𝑩𝑩𝑩, 𝑪𝑪𝑪𝑪, 𝑫𝑫𝑫𝑫, and 𝑬𝑬𝑬𝑬 are equal in length.
Describe the steps you took.
���� and then place the point of the compass on 𝑩𝑩 and use the adjustment to make a
I adjust the compass to the length of 𝑨𝑨𝑨𝑨
mark so that it intersects with the ray. This is the location of 𝑪𝑪, and I repeat these steps until I locate point 𝑭𝑭.
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
Marking off equal segments is entirely a matter of knowing how to use the compass.

What if you knew the length of a segment but needed to divide it into equal-length intervals? For example,
suppose you had a segment 𝐴𝐴𝐴𝐴 that was 10 cm in length. How could you divide it into ten 1 cm parts?


We can try to create a perpendicular bisector to locate the midpoint of the segment.

Allow students time to discuss. They may try to use what they know about creating a perpendicular
bisector to locate the midpoint of the segment; however, they quickly see that it does not easily lead to
determining a 1 cm unit.
We can tackle this problem with a construction that relates to our work on dilations.
Exploratory Challenge 1 (12 minutes)
In the Exploratory Challenge, students learn a construction that divides a segment into 𝑛𝑛 equal parts. They understand
that the constructed parallel segments are evenly spaced proportional side splitters of △ 𝐴𝐴𝐴𝐴𝐴𝐴3 .


We are going to use a compass and straightedge to divide a segment of known length by a whole number 𝑛𝑛.
We are going to divide the following segment 𝐴𝐴𝐴𝐴 into three segments of equal length.
Exploratory Challenge 1
Divide segment 𝑨𝑨𝑨𝑨 into three segments of equal lengths.
Draw each step of the Exploratory Challenge so that students can refer to the correct steps whether they are ahead or
working alongside the teacher.

�������⃗1 .
Pick a point 𝐴𝐴1 not on ����
𝐴𝐴𝐴𝐴 . Draw 𝐴𝐴𝐴𝐴

Mark points 𝐴𝐴2 and 𝐴𝐴3 on the ray so that 𝐴𝐴𝐴𝐴1 = 𝐴𝐴1 𝐴𝐴2 = 𝐴𝐴2 𝐴𝐴3 .
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
�����
�����
Use a straightedge to draw 𝐴𝐴
3 𝐵𝐵 . Use a set square to draw segments parallel to 𝐴𝐴3 𝐵𝐵 through 𝐴𝐴1 and 𝐴𝐴2 .

Label the points where the constructed segments intersect ����
𝐴𝐴𝐴𝐴 as 𝐵𝐵1 and 𝐵𝐵2 .
Allow students time to discuss with each other why the construction divides 𝐴𝐴𝐴𝐴 into equal parts.


Did we succeed? Measure segments 𝐴𝐴𝐵𝐵1 , 𝐵𝐵1 𝐵𝐵2 , and 𝐵𝐵2 𝐵𝐵. Are they all equal in measurement?
After the last step of the construction, answer the following: Why does this construction divide the segment
𝐴𝐴𝐴𝐴 into equal parts?
Allow students a moment to jot down their thoughts, and then take responses.

By construction, segments 𝐴𝐴1 𝐵𝐵1 and 𝐴𝐴2 𝐵𝐵2 are parallel to segment 𝐴𝐴3 𝐵𝐵. By the triangle side splitter
������
�������
theorem, 𝐴𝐴
1 𝐵𝐵1 and 𝐴𝐴2 𝐵𝐵2 are proportional side splitters of triangle 𝐴𝐴𝐵𝐵𝐴𝐴3 . So,
MP.2
𝐴𝐴𝐵𝐵2

𝐴𝐴𝐴𝐴
=
𝐴𝐴𝐴𝐴2
𝐴𝐴𝐴𝐴3
2
1
3
2
3
𝐴𝐴𝐵𝐵1
𝐴𝐴𝐴𝐴
=
𝐴𝐴𝐴𝐴1
𝐴𝐴𝐴𝐴3
=
1
3
and
= . Thus, 𝐴𝐴𝐵𝐵1 = 𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐵𝐵2 = 𝐴𝐴𝐴𝐴, and we can conclude that 𝐵𝐵1 and 𝐵𝐵2 divide line
3
segment 𝐴𝐴𝐴𝐴 into three equal pieces.
Would the construction work if you had chosen a different location for 𝐴𝐴1 ?
Try the construction again, and choose a location different from the location in
the first construction.
Scaffolding:
 Consider having students
use a different color for
this construction.
splitting the class so that
one half begins the
construction from 𝐴𝐴, while
the other half begins at 𝐵𝐵.
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Students should retry the construction and discover that the location of 𝐴𝐴1 is irrelevant to dividing ����
𝐴𝐴𝐴𝐴 into three equallength segments. The triangle drawn in this second attempt is different from the one initially created, so though 𝐴𝐴1 is in
a different location, the triangle drawn is also different. Therefore, the proportional side splitters are also different but
achieve the same result, dividing 𝐴𝐴𝐴𝐴 into three equal-length segments. Furthermore, the location of 𝐴𝐴1 was never
specified to begin with, nor was the relative angle of ∠𝐵𝐵𝐵𝐵𝐴𝐴1. So, technically, this question has already been answered.


We call this method of dividing the segment into 𝑛𝑛 equal-length segments the side splitter method in reference
to the triangle side splitter theorem.
SIDE SPLITTER METHOD: If ����
𝐴𝐴𝐴𝐴 is a line segment, construct a ray 𝐴𝐴𝐴𝐴1 , and mark off 𝑛𝑛 equally spaced points using a
compass of fixed radius to get points 𝐴𝐴 = 𝐴𝐴0 , 𝐴𝐴1 , 𝐴𝐴2 , ⋯ , 𝐴𝐴𝑛𝑛 . Construct ������
𝐴𝐴𝑛𝑛 𝐵𝐵 that is a side of △ 𝐴𝐴𝐴𝐴𝐴𝐴𝑛𝑛 . Through
each point 𝐴𝐴1 , 𝐴𝐴2, ⋯, 𝐴𝐴𝑛𝑛−1 , construct line segments ������
𝐴𝐴1 𝐵𝐵1 , �������
𝐴𝐴2 𝐵𝐵2 , …, ������������
𝐴𝐴𝑛𝑛−1 𝐵𝐵𝑛𝑛−1 parallel to ������
𝐴𝐴𝑛𝑛 𝐵𝐵 that connect
two sides of △ 𝐴𝐴𝐴𝐴𝑛𝑛 𝐵𝐵.
Exercise 1 (3 minutes)
Teachers may elect to move directly to the next Exploratory Challenge depending on time. Alternatively, if there is some
time, the teacher may elect to reduce the number of divisions to 3.
Exercise 1
Divide segment 𝑨𝑨𝑨𝑨 into five segments of equal lengths.
Exploratory Challenge 2 (10 minutes)
Now students try an alternate method of dividing a segment into 𝑛𝑛 equal-length
segments.

Let’s continue with our exploration and try a different method to divide a
segment of known length by a whole number 𝑛𝑛. Again, we rely on the use of a
compass, straightedge, and this time, a set square.
Exploratory Challenge 2
Divide segment 𝑨𝑨𝑨𝑨 into four segments of equal length.
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Scaffolding:
 Consider focusing on the
side splitter method alone
for struggling students.
 Using this strategy, select
an early question from the
Problem Set to work on in
class.
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
Use the set square to create a ray 𝑋𝑋𝑋𝑋 parallel to ����
𝐴𝐴𝐴𝐴 . Select the location of the endpoint 𝑋𝑋 so that it falls to the
left of 𝐴𝐴; the location of 𝑌𝑌 should be oriented in relation to 𝑋𝑋 in the same manner as 𝐵𝐵 is in relation to 𝐴𝐴. We
construct the parallel ray below ����
𝐴𝐴𝐴𝐴, but it can be constructed above the segment as well.

From 𝑋𝑋, use the compass to mark off four equal segments along �����⃗
𝑋𝑋𝑋𝑋. Label each intersection as 𝑋𝑋1 , 𝑋𝑋2 , 𝑋𝑋3 ,
and 𝑋𝑋4 . It is important that 𝑋𝑋𝑋𝑋4 ≠ 𝐴𝐴𝐴𝐴. In practice, 𝑋𝑋𝑋𝑋4 should be clearly more or clearly less than 𝐴𝐴𝐴𝐴.

Draw line 𝑋𝑋𝑋𝑋 and line 𝑋𝑋4 𝐵𝐵. Mark the intersection of the two lines as 𝑂𝑂.
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
Construct rays from 𝑂𝑂 through 𝑋𝑋1 , 𝑋𝑋2 , and 𝑋𝑋3 . Label each intersection with ����
𝐴𝐴𝐴𝐴 as 𝐴𝐴1 , 𝐴𝐴2 , and 𝐴𝐴3 .

����
𝐴𝐴𝐴𝐴 should now be divided into four segments of equal length. Measure segments 𝐴𝐴𝐴𝐴1 , 𝐴𝐴1 𝐴𝐴2, 𝐴𝐴2 𝐴𝐴3 , and 𝐴𝐴3 𝐵𝐵.
Are they all equal in measurement?

Why does this construction divide the segment 𝐴𝐴𝐴𝐴 into equal parts?
Allow students time to discuss with each other why the construction divides ����
𝐴𝐴𝐴𝐴 into equal parts.

We constructed ����
𝐴𝐴𝐴𝐴 to be parallel to �����⃗
𝑋𝑋𝑋𝑋. In △ 𝑋𝑋𝑋𝑋𝑋𝑋4 , since the side splitter ����
𝐴𝐴𝐴𝐴 is parallel to �����
𝑋𝑋𝑋𝑋4 , by the
triangle side splitter theorem, it must also be a proportional side splitter. By the dilation theorem, this
means that



𝑋𝑋𝑋𝑋1
=
𝐴𝐴1 𝐴𝐴2
𝑋𝑋1 𝑋𝑋2
=
𝐴𝐴2 𝐴𝐴3
𝑋𝑋2 𝑋𝑋3
=
𝐴𝐴3 𝐵𝐵
𝑋𝑋3 𝑋𝑋4
. Since we know that the values of all four denominators are
the same, the value of all four numerators must also be the same to make the equation true.
Therefore, ����
𝐴𝐴𝐴𝐴 has been divided into four segments of equal length.
We call this method of dividing the segment into equal lengths the dilation method, as the points that divide
𝑂𝑂𝑂𝑂
����
�����⃗.
of the evenly spaced points on 𝑋𝑋𝑋𝑋
𝐴𝐴𝐴𝐴 are by definition dilated points from center 𝑂𝑂 with scale factor 𝑟𝑟 =
𝑂𝑂𝑂𝑂
DILATION METHOD: Construct a ray 𝑋𝑋𝑋𝑋 parallel to ����
𝐴𝐴𝐴𝐴 . On the parallel ray, use a compass to mark off 𝑛𝑛 equally
⃖����⃗ and ⃖������⃗
𝐵𝐵𝑋𝑋𝑛𝑛 intersect at a point 𝑂𝑂. Construct the rays
spaced points 𝑋𝑋1 , 𝑋𝑋2 ,⋯, 𝑋𝑋𝑛𝑛 so that 𝑋𝑋𝑋𝑋𝑛𝑛 ≠ 𝐴𝐴𝐴𝐴. Lines 𝐴𝐴𝐴𝐴
�������⃗
��������⃗
��������⃗
����
𝑂𝑂𝑋𝑋1 , 𝑂𝑂𝑋𝑋2 , …, 𝑂𝑂𝑋𝑋𝑛𝑛 that meet 𝐴𝐴𝐴𝐴 in points 𝐴𝐴1 , 𝐴𝐴2 , …, 𝐴𝐴𝑛𝑛 , respectively.
What happens if line segments 𝑋𝑋𝑋𝑋𝑛𝑛 and 𝐴𝐴𝐴𝐴 are close to the same length?


𝐴𝐴𝐴𝐴1
The point 𝑂𝑂 is very far away.
That is why it is best to make 𝑋𝑋𝑋𝑋𝑛𝑛 clearly more or less than 𝐴𝐴𝐴𝐴. It is also best to keep line segments 𝑋𝑋𝑋𝑋𝑛𝑛 and
𝐴𝐴𝐴𝐴 centered, or the point 𝑂𝑂 will also be far away.
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Exercise 2 (8 minutes)
Students should complete Exercise 2 in pairs. If possible, the teacher should pre-mark each piece of poster paper with a
1-foot mark to allow students to get right to the activity.

Exercise 2
���� into
On a piece of poster paper, draw a segment 𝑨𝑨𝑨𝑨 with a measurement of 𝟏𝟏 foot. Use the dilation method to divide 𝑨𝑨𝑨𝑨
twelve equal-length segments, or into 𝟏𝟏𝟏𝟏 inches.
Closing (2 minutes)

Compare the side splitter method to the dilation method: What advantage does the first method have?


With the side splitter method, we do not have to worry about checking lengths (i.e., the initial point can
be chosen completely arbitrarily), whereas with the dilation method, we have to choose our unit so that
𝑛𝑛 units does not match the original length (or come close to matching).
How does either of the methods help identify fractions on the number line?

We can break up whole units on a number line into any division we choose; for example, we can mimic
a ruler in inches by dividing up a number line into eighths.
Lesson Summary
���� is a line segment, construct a ray 𝑨𝑨𝑨𝑨𝟏𝟏 , and mark off 𝒏𝒏 equally spaced points using a
SIDE SPLITTER METHOD: If 𝑨𝑨𝑨𝑨
������
compass of fixed radius to get points 𝑨𝑨 = 𝑨𝑨𝟎𝟎 , 𝑨𝑨𝟏𝟏 , 𝑨𝑨𝟐𝟐 ,⋯, 𝑨𝑨𝒏𝒏 . Construct 𝑨𝑨
𝒏𝒏 𝑩𝑩 that is a side of △ 𝑨𝑨𝑨𝑨𝑨𝑨𝒏𝒏 . Through each
������
������
point 𝑨𝑨𝟏𝟏 , 𝑨𝑨𝟐𝟐 , ⋯, 𝑨𝑨𝒏𝒏−𝟏𝟏 , construct line segments 𝑨𝑨
𝒊𝒊 𝑩𝑩𝒊𝒊 parallel to 𝑨𝑨𝒏𝒏 𝑩𝑩 that connect two sides of △ 𝑨𝑨𝑨𝑨𝒏𝒏 𝑩𝑩.
DILATION METHOD: Construct a ray 𝑿𝑿𝑿𝑿 parallel to ����
𝑨𝑨𝑨𝑨. On the parallel ray, use a compass to mark off 𝒏𝒏 equally
⃖����⃗ and ⃖�������⃗
𝑩𝑩𝑿𝑿𝒏𝒏 intersect at a point 𝑶𝑶. Construct the rays �������⃗
𝑶𝑶𝑿𝑿𝒊𝒊
spaced points 𝑿𝑿𝟏𝟏 , 𝑿𝑿𝟐𝟐 ,⋯, 𝑿𝑿𝒏𝒏 so that 𝑿𝑿𝑿𝑿𝒏𝒏 ≠ 𝑨𝑨𝑨𝑨. Lines 𝑨𝑨𝑨𝑨
that meet ����
𝑨𝑨𝑨𝑨 in points 𝑨𝑨𝒊𝒊 .
Exit Ticket (5 minutes)
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Name
Date
Lesson 10: Dividing the King’s Foot into 12 Equal Pieces
Exit Ticket
1.
����� into 7 equal-sized pieces.
Use the side splitter method to divide 𝑀𝑀𝑀𝑀
2.
Use the dilation method to divide ����
𝑃𝑃𝑃𝑃 into 11 equal-sized pieces.
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3.
4
If the segment below represents the interval from zero to one on the number line, locate and label .
7
0
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Exit Ticket Sample Solutions
1.
Use the side splitter method to divide ����
𝑴𝑴𝑴𝑴 into 7 equal-sized pieces.
2.
�� into 𝟏𝟏𝟏𝟏 equal-sized pieces.
Use the dilation method to divide ��
𝑷𝑷𝑷𝑷
3.
If the segment below represents the interval from zero to one on the number line, locate and label .
𝟒𝟒
𝟕𝟕
Students may use either the side splitter method or the dilation method and need only find the location of the fourth
equal-sized piece of the segment, as shown in the diagram below.
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Problem Set Sample Solutions
1.
Pretend you are the king or queen and that the length of your foot is the official measurement for one foot. Draw a
line segment on a piece of paper that is the length of your foot. (You may have to remove your shoe.) Use the
method above to find the length of 𝟏𝟏 inch in your kingdom.
���� into twelve equal pieces using the dilation method
I drew ����
𝑨𝑨𝑨𝑨 representing the length of my foot. I then divided 𝑨𝑨𝑨𝑨
as follows:
⃖����⃗ parallel to ����
⃖����⃗, each having
I constructed 𝑫𝑫𝑫𝑫
𝑨𝑨𝑨𝑨 and, using a compass, marked off twelve consecutive segments on 𝑫𝑫𝑫𝑫
���� is a large segment, students will likely choose a length 𝑫𝑫𝑫𝑫 so that the total
length 𝑫𝑫𝑫𝑫. (Note that because 𝑨𝑨𝑨𝑨
⃖����⃗ is noticeably shorter than 𝑨𝑨𝑨𝑨.)
length of all of the segments constructed on 𝑫𝑫𝑫𝑫
I then constructed a ray from 𝑨𝑨 and 𝑩𝑩 through the endpoints of the composed segment, as shown in the diagram, to
���� point 𝑰𝑰. This distance 𝑨𝑨𝑨𝑨 is
form a triangle 𝑨𝑨𝑨𝑨𝑨𝑨. Next, I constructed ������⃗
𝑶𝑶𝑶𝑶, and then I marked its intersection with 𝑨𝑨𝑨𝑨
the length of 𝟏𝟏 inch in my kingdom.
2.
Using a ruler, draw a segment that is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜. This length is referred to as a decimeter. Use the side splitter method
to divide your segment into ten equal-sized pieces. What should be the length of each of those pieces based on
your construction? Check the length of the pieces using a ruler. Are the lengths of the pieces accurate?
Verify that students’ diagrams show the use of the side splitter method. The length of each piece should be 𝟏𝟏 𝐜𝐜𝐜𝐜.
3.
Repeat Problem 2 using the dilation method. What should be the length of each of those pieces based on your
construction? Check the lengths of the pieces using a ruler. Are the lengths of the pieces accurate?
Verify that students’ diagrams show the use of the dilation method. The length of each piece should be 𝟏𝟏 𝐜𝐜𝐜𝐜.
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4.
A portion of a ruler that measured whole centimeters is shown below. Determine the location of 𝟓𝟓
portion of the ruler shown.
𝟐𝟐
𝐜𝐜𝐜𝐜 on the
𝟑𝟑
Responses should show the segment between 𝟓𝟓 and 𝟔𝟔 divided into 𝟑𝟑 equal pieces with the division point closest
𝟐𝟐
𝟑𝟑
to 𝟔𝟔 chosen as the location of 𝟓𝟓 .
5.
Merrick has a ruler that measures in inches only. He is measuring the length of a line segment that is between 𝟖𝟖 𝐢𝐢𝐢𝐢.
and 𝟗𝟗 𝐢𝐢𝐢𝐢. Divide the one-inch section of Merrick’s ruler below into eighths to help him measure the length of the
segment.
Using the side splitter method, I divided the one-inch interval into eighths, labeled as 𝑩𝑩𝟏𝟏 , 𝑩𝑩𝟐𝟐 , etc., on the diagram.
𝟓𝟓
The line segment that Merrick is measuring closely corresponds with 𝑩𝑩𝟑𝟑 , which represents of the distance from
𝟖𝟖
𝟖𝟖 𝐢𝐢𝐢𝐢. to 𝟗𝟗 𝐢𝐢𝐢𝐢. Therefore, the length of the line segment that Merrick is measuring is approximately 𝟖𝟖
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𝟓𝟓
𝐢𝐢𝐢𝐢.
𝟖𝟖
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6.
Use the dilation method to create an equally spaced 𝟑𝟑 × 𝟑𝟑 grid in the following square.
There are several ways to complete this construction. The sample below used the dilation method along two sides of
the square with centers 𝑶𝑶 and 𝑶𝑶𝟐𝟐 to divide the side into three equal-size segments. The sides of the square were
extended such that 𝑫𝑫𝑫𝑫′ = 𝑪𝑪𝑪𝑪, 𝑪𝑪𝑪𝑪′ = 𝑪𝑪𝑪𝑪, 𝑩𝑩′ = 𝑪𝑪𝑪𝑪, and 𝑪𝑪𝑪𝑪′𝟏𝟏 = 𝑪𝑪𝑪𝑪.
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7.
Use the side splitter method to create an equally spaced 𝟑𝟑 × 𝟑𝟑 grid in the following square.
There are several ways to complete this construction. The sample below used the side splitter method along two
sides of the square to divide the side into three equal-size segments.
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Lesson 11: Dilations from Different Centers
Student Outcomes

Students verify experimentally that the dilation of a figure from two different centers by the same scale factor
gives congruent figures. These figures are congruent by a translation along a vector that is parallel to the line
through the centers.

Students verify experimentally that the composition of dilations 𝐷𝐷𝑂𝑂1 ,𝑟𝑟1 and 𝐷𝐷𝑂𝑂2 ,𝑟𝑟2 is a dilation with scale factor
⃖��������⃗
𝑟𝑟1 𝑟𝑟2 and center on 𝑂𝑂
1 𝑂𝑂2 unless 𝑟𝑟1 𝑟𝑟2 = 1.
Lesson Notes
In Lesson 11, students examine the effects of dilating figures from two different centers. By experimental verification,
they examine the impact on the two dilations of having two different scale factors, the same two scale factors, and scale
factors whose product equals 1. Each of the parameters of these cases provides information on the centers of the
dilations, their scale factors, and the relationship between individual dilations versus the relationship between an initial
figure and a composition of dilations.
Classwork
Exploratory Challenge 1 (15 minutes)
In Exploratory Challenge 1, students verify experimentally that the dilation of a figure from two different centers by the
same scale factor gives congruent figures that are congruent by a translation along a vector that is parallel to the line
through the centers.

In this example, we examine scale drawings of an image from different center points.
Exploratory Challenge 1
Drawing 2 and Drawing 3 are both scale drawings of Drawing 1.
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a.
Determine the scale factor and center for each scale drawing. Take measurements as needed.
𝟏𝟏
𝟐𝟐
The scale factor for each drawing is the same; the scale factor for both is 𝒓𝒓 = . Each scale drawing has a
different center.
b.
Is there a way to map Drawing 2 onto Drawing 3 or map Drawing 3 onto Drawing 2?
Since the two drawings are identical, a translation maps either Drawing 2 onto Drawing 3 or Drawing 3 onto
Drawing 2.

What do you notice about a translation vector that maps either scale drawing onto the other and the line that
passes through the centers of the dilations?

A translation vector that maps either scale drawing onto the other is parallel to the line that passes
through the centers of the dilations.
Scaffolding:
 Students can take
responsibility for their own
learning by hand-drawing
the houses; however, if
time is an issue, teachers
can provide the drawings
of the houses located at
the beginning of
Exploratory Challenge 1.



We are not going to generally prove this, but let’s experimentally verify this by
dilating a simple figure (e.g., a segment) by the same scale factor from two
different centers, 𝑂𝑂1 and 𝑂𝑂2 .
Do this twice, in two separate cases, to observe what happens.
In the first case, dilate ����
𝐴𝐴𝐴𝐴 by a factor of 2, and be sure to give each dilation a
������
different center. Label the dilation about 𝑂𝑂1 as 𝐴𝐴
1 𝐵𝐵1 and the dilation about 𝑂𝑂2
�������
as 𝐴𝐴2 𝐵𝐵2 .
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 Use patty paper or
geometry software to help
students focus on the
concepts.
 Consider performing the
dilation in the coordinate
plane with center at the
origin, for example, ����
𝐴𝐴𝐴𝐴
with coordinates 𝐴𝐴(3,1)
and 𝐵𝐵(4, −3).
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
Repeat the experiment, and create a segment, ����
𝐶𝐶𝐶𝐶 , different from ����
𝐴𝐴𝐴𝐴 . Dilate ����
𝐶𝐶𝐶𝐶 by a factor of 2, and be sure
������
������
to give each dilation a different center. Label the dilation about 𝑂𝑂1 as 𝐶𝐶
𝐷𝐷
and
the dilation about 𝑂𝑂2 as 𝐶𝐶
1 1
2 𝐷𝐷2 .
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
What do you notice about the translation vector that maps the scale drawings to each other relative to the line
������
�������
that passes through the centers of the dilations, for example, the vector that maps 𝐴𝐴
1 𝐵𝐵1 to 𝐴𝐴2 𝐵𝐵2 ?
Allow students time to complete this mini-experiment and verify that the translation vector is parallel to the line that
passes through the centers of the dilations.
The translation vector is always parallel to the line that passes through the centers of the dilations.

c.
MP 8
Generalize the parameters of this example and its results.
The dilation of a figure from two different centers by the same scale factor yields congruent figures that are
congruent by a translation along a vector that is parallel to the line through the centers.
Exercise 1 (4 minutes)
Exercise 1
Triangle 𝑨𝑨𝑨𝑨𝑨𝑨 has been dilated with scale factor
𝑩𝑩𝟏𝟏 𝑩𝑩𝟐𝟐 , and 𝑪𝑪𝟏𝟏 𝑪𝑪𝟐𝟐 ?
𝟏𝟏
𝟐𝟐
from centers 𝑶𝑶𝟏𝟏 and 𝑶𝑶𝟐𝟐 . What can you say about line segments 𝑨𝑨𝟏𝟏 𝑨𝑨𝟐𝟐 ,
They are all parallel to the line that passes through 𝑶𝑶𝟏𝟏 𝑶𝑶𝟐𝟐.
Exploratory Challenge 2 (15 minutes)
In Exploratory Challenge 2, students verify experimentally (1) that the composition of dilations is a dilation with scale
⃖��������⃗
factor 𝑟𝑟1 𝑟𝑟2 and (2) that the center of the composition lies on the line 𝑂𝑂
1 𝑂𝑂2 unless 𝑟𝑟1 𝑟𝑟2 = 1. Students may need poster
paper or legal-sized paper to complete part (c).
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Exploratory Challenge 2
If Drawing 2 is a scale drawing of Drawing 1 with scale factor 𝒓𝒓𝟏𝟏 and Drawing 3 is a scale drawing of Drawing 2 with scale
factor 𝒓𝒓𝟐𝟐 , what is the relationship between Drawing 3 and Drawing 1?
a.
Determine the scale factor and center for each scale drawing. Take measurements as needed.
𝟏𝟏
𝟐𝟐
The scale factor for Drawing 2 relative to Drawing 1 is 𝒓𝒓𝟏𝟏 = , and the scale factor for Drawing 3 relative to
𝟑𝟑
𝟐𝟐
Drawing 2 is 𝒓𝒓𝟐𝟐 = .
b.
What is the scale factor going from Drawing 1 to Drawing 3? Take measurements as needed.
𝟑𝟑
𝟒𝟒
The scale factor to go from Drawing 1 to Drawing 3 is 𝒓𝒓𝟑𝟑 = .

Do you see a relationship between the value of the scale factor going from Drawing 1 to Drawing 3 and the
scale factors determined going from Drawing 1 to Drawing 2 and Drawing 2 to Drawing 3?
Allow students a moment to discuss before taking responses.



The scale factor to go from Drawing 1 to Drawing 3 is the same as the product of the scale factors to go
from Drawing 1 to Drawing 2 and then Drawing 2 to Drawing 3. So the scale factor to go from Drawing
1
2
3
2
3
4
1 to Drawing 3 is 𝑟𝑟1 𝑟𝑟2 = � � � � = .
To go from Drawing 1 to Drawing 3 is the same as taking a composition of the two dilations: 𝐷𝐷𝑂𝑂
3
2 ,2
�𝐷𝐷𝑂𝑂 ,1 �.
12
So, with respect to scale factor, a composition of dilations 𝐷𝐷𝑂𝑂2 ,𝑟𝑟2 �𝐷𝐷𝑂𝑂1 ,𝑟𝑟1 � results in a dilation whose scale factor
is 𝑟𝑟1 𝑟𝑟2 .
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c.
MP.1
&
MP.7
Compare the centers of dilations of Drawing 1 (to Drawing 2) and of Drawing 2 (to Drawing 3). What do you
notice about these centers relative to the center of the composition of dilations 𝑶𝑶𝟑𝟑 ?
The centers of each for Drawing 1 and Drawing 2 are collinear with the center of dilation of the composition
of dilations.
Scaffolding:
Students with difficulty in
spatial reasoning can be
to observe what is remarkable
dilations.




From this example, it is tempting to generalize and say that with respect to the centers of the dilations, the
center of the composition of dilations 𝐷𝐷𝑂𝑂2 ,𝑟𝑟2 (𝐷𝐷𝑂𝑂1 ,𝑟𝑟1 ) is collinear with the centers 𝑂𝑂1 and 𝑂𝑂2 , but there is one
situation where this is not the case.
To observe this one case, draw a segment 𝐴𝐴𝐴𝐴 that serves as the figure of a series of dilations.
1
2
For the first dilation 𝐷𝐷1 , select a center of dilation 𝑂𝑂1 and scale factor 𝑟𝑟1 = . Dilate 𝐴𝐴𝐴𝐴 and label the result as
𝐴𝐴′𝐵𝐵′.
For the second dilation 𝐷𝐷2 , select a new center 𝑂𝑂2 and scale factor 𝑟𝑟2 = 2. Determine why the centers of each
of these dilations cannot be collinear with the center of dilation of the composition of dilations 𝐷𝐷2 (𝐷𝐷1 ).
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GEOMETRY
Since Drawing 1 and Drawing 3 are identical figures, the lines that pass through the corresponding
endpoints of the segments are parallel; a translation will map Drawing 1 to Drawing 3.



Notice that this occurs only when 𝑟𝑟1 𝑟𝑟2 = 1.
Also, notice that the translation that maps ����
𝐴𝐴𝐴𝐴 to �������
𝐴𝐴′′𝐵𝐵′′ must be parallel to the line that passes through the
centers of the two given dilations.
d.
Generalize the parameters of this example and its results.
⃖���������⃗
A composition of dilations, 𝑫𝑫𝑶𝑶𝟏𝟏 ,𝒓𝒓𝟏𝟏 and 𝑫𝑫𝑶𝑶𝟐𝟐 ,𝒓𝒓𝟐𝟐 , is a dilation with scale factor 𝒓𝒓𝟏𝟏 𝒓𝒓𝟐𝟐 and center on 𝑶𝑶
𝟏𝟏 𝑶𝑶𝟐𝟐 unless
𝒓𝒓𝟏𝟏 𝒓𝒓𝟐𝟐 = 𝟏𝟏. If 𝒓𝒓𝟏𝟏 𝒓𝒓𝟐𝟐 = 𝟏𝟏, then there is no dilation that maps a figure onto the image of the composition of
dilations; there is a translation parallel to the line passing through the centers of the individual dilations that
maps the figure onto its image.
Exercise 2 (4 minutes)
Exercise 2
Triangle 𝑨𝑨𝑨𝑨𝑨𝑨 has been dilated with scale factor
dilated from center 𝑶𝑶𝟐𝟐 with scale factor
𝟏𝟏
𝟐𝟐
𝟐𝟐
𝟑𝟑
from center 𝑶𝑶𝟏𝟏 to get triangle 𝑨𝑨′ 𝑩𝑩′ 𝑪𝑪′ , and then triangle 𝑨𝑨′ 𝑩𝑩′ 𝑪𝑪′ is
to get triangle 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′. Describe the dilation that maps triangle 𝑨𝑨𝑨𝑨𝑨𝑨 to
triangle 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′. Find the center and scale factor for that dilation.
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The dilation center is a point on the line segment 𝑶𝑶𝟏𝟏 𝑶𝑶𝟐𝟐, and the scale factor is
𝟐𝟐 𝟏𝟏
⋅
𝟑𝟑 𝟐𝟐
=
𝟏𝟏
𝟑𝟑
.
Closing (2 minutes)

In a series of dilations, how does the scale factor that maps the original figure to the final image compare to
the scale factor of each successive dilation?


In a series of dilations, the scale factor that maps the original figure onto the final image is the product
of all the scale factors in the series of dilations.
We remember here that, unlike the previous several lessons, we did not prove facts in general; we made
observations through measurements.
Lesson Summary
In a series of dilations, the scale factor that maps the original figure onto the final image is the product of all the
scale factors in the series of dilations.
Exit Ticket (5 minutes)
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Name
Date
Lesson 11: Dilations from Different Centers
Exit Ticket
3
Marcos constructed the composition of dilations shown below. Drawing 2 is the size of Drawing 1, and Drawing 3 is
8
twice the size of Drawing 2.
1.
Determine the scale factor from Drawing 1 to Drawing 3.
2.
Find the center of dilation mapping Drawing 1 to Drawing 3.
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Exit Ticket Sample Solutions
Marcos constructed the composition of dilations shown below. Drawing 2 is
twice the size of Drawing 2.
1.
𝟑𝟑
𝟖𝟖
the size of Drawing 1, and Drawing 3 is
Determine the scale factor from Drawing 1 to Drawing 3.
Drawing 2 is a 𝟑𝟑: 𝟖𝟖 scale drawing of Drawing 1, and Drawing 3 is a 𝟐𝟐: 𝟏𝟏 scale drawing of Drawing 2, so Drawing 3 is
a 𝟐𝟐: 𝟏𝟏 scale drawing of a 𝟑𝟑: 𝟖𝟖 scale drawing:
𝟑𝟑
𝟖𝟖
Drawing 3 = 𝟐𝟐 � Drawing 1�
Drawing 3 =
𝟑𝟑
(Drawing 1)
𝟒𝟒
𝟑𝟑
The scale factor from Drawing 1 to Drawing 3 is .
2.
𝟒𝟒
Find the center of dilation mapping Drawing 1 to Drawing 3.
See diagram: The center of dilation is 𝑶𝑶𝟑𝟑 .
Problem Set Sample Solutions
1.
In Lesson 7, the dilation theorem for line segments said that if two different-length line segments in the plane were
parallel to each other, then a dilation exists mapping one segment onto the other. Explain why the line segments
must be different lengths for a dilation to exist.
If the line segments were of equal length, then it would have to be true that the scale factor of the supposed dilation
would be 𝒓𝒓 = 𝟏𝟏; however, we found that any dilation with a scale factor of 𝒓𝒓 = 𝟏𝟏 maps any figure to itself, which
implies that the line segment would have to be mapped to itself. Two different line segments that are parallel to
one another implies that the line segments are not one and the same, which means that the supposed dilation does
not exist.
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2.
Regular hexagon 𝑨𝑨′𝑩𝑩′𝑪𝑪′𝑫𝑫′𝑬𝑬′𝑭𝑭′ is the image of regular hexagon 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 under a dilation from center 𝑶𝑶𝟏𝟏 , and
regular hexagon 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′𝑫𝑫′′𝑬𝑬′′𝑭𝑭′′ is the image of regular hexagon 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 under a dilation from center 𝑶𝑶𝟐𝟐 .
Points 𝑨𝑨′ , 𝑩𝑩′ , 𝑪𝑪′ , 𝑫𝑫′ , 𝑬𝑬′ , and 𝑭𝑭′ are also the images of points 𝑨𝑨′′ , 𝑩𝑩′′ , 𝑪𝑪′′ , 𝑫𝑫′′ , 𝑬𝑬′′ , and 𝑭𝑭′′, respectively, under a
translation along vector ����������⃗
𝑫𝑫′′𝑫𝑫′. Find a possible regular hexagon 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨.
Student diagrams will vary; however, the centers of dilation 𝑶𝑶𝟏𝟏 and 𝑶𝑶𝟐𝟐 must lie on a line parallel to vector ����������⃗
𝑫𝑫′′𝑫𝑫′.
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3.
A dilation with center 𝑶𝑶𝟏𝟏 and scale factor
factor
𝟑𝟑
𝟐𝟐
maps figure 𝑭𝑭′ to figure 𝑭𝑭′′. Draw figures 𝑭𝑭′ and 𝑭𝑭′′, and then find the center 𝑶𝑶 and scale factor 𝒓𝒓 of the
dilation that takes 𝑭𝑭 to 𝑭𝑭′′.
4.
𝟏𝟏
maps figure 𝑭𝑭 to figure 𝑭𝑭′. A dilation with center 𝑶𝑶𝟐𝟐 and scale
𝟐𝟐
𝟑𝟑
𝟒𝟒
A figure 𝑻𝑻 is dilated from center 𝑶𝑶𝟏𝟏 with a scale factor 𝒓𝒓𝟏𝟏 =
center 𝑶𝑶𝟐𝟐 with a scale factor 𝒓𝒓𝟐𝟐 =
𝟑𝟑
to yield image 𝑻𝑻′, and figure 𝑻𝑻′ is then dilated from
𝟒𝟒
𝟒𝟒
to yield figure 𝑻𝑻′′. Explain why 𝑻𝑻 ≅ 𝑻𝑻′′.
𝟑𝟑
𝟑𝟑
For any distance, 𝒂𝒂, between two points in figure 𝑻𝑻, the distance between corresponding points in figure 𝑻𝑻′ is 𝒂𝒂.
𝟑𝟑
For the said distance between points in 𝑻𝑻′, 𝒂𝒂, the distance between corresponding points in figure 𝑻𝑻′′
is
𝟒𝟒 𝟑𝟑
� 𝒂𝒂� = 𝟏𝟏𝟏𝟏.
𝟑𝟑 𝟒𝟒
𝟒𝟒
𝟒𝟒
This implies that all distances between two points in figure 𝑻𝑻′′ are equal to the distances between
corresponding points in figure 𝑻𝑻. Furthermore, since dilations preserve angle measures, angles formed by any three
noncollinear points in figure 𝑻𝑻′′ are congruent to the angles formed by the corresponding three noncollinear points
in figure 𝑻𝑻. There is then a correspondence between 𝑻𝑻 and 𝑻𝑻′′ in which distance is preserved and angle measures
are preserved, implying that a sequence of rigid motions maps 𝑻𝑻 onto 𝑻𝑻′′; hence, a congruence exists between
figures 𝑻𝑻 and 𝑻𝑻′′.
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5.
A dilation with center 𝑶𝑶𝟏𝟏 and scale factor
𝟏𝟏
𝟐𝟐
maps figure 𝑯𝑯 to figure 𝑯𝑯′. A dilation with center 𝑶𝑶𝟐𝟐 and scale factor 𝟐𝟐
maps figure 𝑯𝑯′ to figure 𝑯𝑯′′. Draw figures 𝑯𝑯′ and 𝑯𝑯′′. Find a vector for a translation that maps 𝑯𝑯 to 𝑯𝑯′′.
Solution:
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6.
Figure 𝑾𝑾 is dilated from 𝑶𝑶𝟏𝟏 with a scale factor 𝒓𝒓𝟏𝟏 = 𝟐𝟐 to yield 𝑾𝑾′. Figure 𝑾𝑾′ is then dilated from center 𝑶𝑶𝟐𝟐 with a
scale factor 𝒓𝒓𝟐𝟐 =
𝟏𝟏
to yield 𝑾𝑾′′.
𝟒𝟒
a.
Construct the composition of dilations of figure 𝑾𝑾 described above.
b.
If you were to dilate figure 𝑾𝑾′′, what scale factor would be required to yield an image that is congruent to
figure 𝑾𝑾?
In a composition of dilations, for the resulting image to be congruent to the original pre-image, the product of
the scale factors of the dilations must be 𝟏𝟏.
𝒓𝒓𝟏𝟏 ⋅ 𝒓𝒓𝟐𝟐 ⋅ 𝒓𝒓𝟑𝟑
𝟏𝟏
𝟐𝟐 ⋅ ⋅ 𝒓𝒓𝟑𝟑
𝟒𝟒
𝟏𝟏
⋅ 𝒓𝒓
𝟐𝟐 𝟑𝟑
𝒓𝒓𝟑𝟑
= 𝟏𝟏
= 𝟏𝟏
= 𝟏𝟏
= 𝟐𝟐
The scale factor necessary to yield an image congruent to the original pre-image is 𝒓𝒓𝟑𝟑 = 𝟐𝟐.
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c.
7.
Locate the center of dilation that maps 𝑾𝑾′′ to 𝑾𝑾 using the scale factor that you identified in part (b).
Figures 𝑭𝑭𝟏𝟏 and 𝑭𝑭𝟐𝟐 in the diagram below are dilations of 𝑭𝑭 from centers 𝑶𝑶𝟏𝟏 and 𝑶𝑶𝟐𝟐 , respectively.
a.
Find 𝑭𝑭.
b.
If 𝑭𝑭𝟏𝟏 ≅ 𝑭𝑭𝟐𝟐 , what must be true of the scale factors 𝒓𝒓𝟏𝟏 and 𝒓𝒓𝟐𝟐 of each dilation?
The scale factors must be equal.
c.
Use direct measurement to determine each scale factor for 𝑫𝑫𝑶𝑶𝟏𝟏 ,𝒓𝒓𝟏𝟏 and 𝑫𝑫𝑶𝑶𝟐𝟐,𝒓𝒓𝟐𝟐 .
𝟏𝟏
𝟐𝟐
By direct measurement, the scale factor used for each dilation is 𝒓𝒓𝟏𝟏 = 𝒓𝒓𝟐𝟐 = 𝟐𝟐 .
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Note to the teacher: Parts of this next problem involve a great deal of mathematical reasoning and may not be suitable
for all students.
8.
Use a coordinate plane to complete each part below using 𝑼𝑼(𝟐𝟐, 𝟑𝟑), 𝑽𝑽(𝟔𝟔, 𝟔𝟔), and 𝑾𝑾(𝟔𝟔, −𝟏𝟏).
a.
Dilate △ 𝑼𝑼𝑼𝑼𝑼𝑼 from the origin with a scale factor 𝒓𝒓𝟏𝟏 = 𝟐𝟐. List the coordinates of image points 𝑼𝑼′, 𝑽𝑽′, and 𝑾𝑾′.
𝑼𝑼′(𝟒𝟒, 𝟔𝟔), 𝑽𝑽′(𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏), and 𝑾𝑾′(𝟏𝟏𝟏𝟏, −𝟐𝟐)
b.
𝟑𝟑
𝟒𝟒
Dilate △ 𝑼𝑼𝑼𝑼𝑼𝑼 from (𝟎𝟎, 𝟔𝟔) with a scale factor of 𝒓𝒓𝟐𝟐 = . List the coordinates of image points 𝑼𝑼′′, 𝑽𝑽′′, and 𝑾𝑾′′.
The center of this dilation is not the origin. The 𝒙𝒙-coordinate of the center is 𝟎𝟎, so the 𝒙𝒙-coordinates of the
image points can be calculated in the same manner as in part (a). However, the 𝒚𝒚-coordinates of the preimage must be considered as their distance from the 𝒚𝒚-coordinate of the center, 𝟔𝟔.
Point 𝑼𝑼 is 𝟑𝟑 units below the center of dilation, point 𝑽𝑽 is at the same vertical level as the center of dilation,
and point 𝑾𝑾 is 𝟕𝟕 units below the center of dilation.
𝟑𝟑
𝟒𝟒
𝒚𝒚𝑼𝑼′′ = 𝟔𝟔 + � (−𝟑𝟑)�
𝟗𝟗
𝟒𝟒
𝒚𝒚𝑼𝑼′′ = 𝟔𝟔 + �− �
𝒚𝒚𝑼𝑼′′ = 𝟑𝟑
𝟑𝟑
𝟒𝟒
𝟑𝟑
𝟒𝟒
𝒚𝒚𝑽𝑽′′ = 𝟔𝟔 + � (𝟎𝟎)�
𝒚𝒚𝑽𝑽′′ = 𝟔𝟔 + 
𝒚𝒚𝑽𝑽′′ = 𝟔𝟔
𝟏𝟏
𝟐𝟐
𝟑𝟑
𝟒𝟒
𝟏𝟏
𝟐𝟐
𝟑𝟑
𝟒𝟒
𝒚𝒚𝑾𝑾′′ = 𝟔𝟔 + � (−𝟕𝟕)�
𝒚𝒚𝑾𝑾′′ = 𝟔𝟔 + �−
𝒚𝒚𝑾𝑾′′ =
𝟏𝟏 𝟑𝟑
𝟐𝟐 𝟒𝟒
𝟑𝟑
𝟒𝟒
𝟐𝟐𝟐𝟐
�
𝟒𝟒
𝑼𝑼′′ �𝟏𝟏 , 𝟑𝟑 �, 𝑽𝑽′′(𝟒𝟒 , 𝟔𝟔), and 𝑾𝑾′′ �𝟒𝟒 , �
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c.
Find the scale factor, 𝒓𝒓𝟑𝟑 , from △ 𝑼𝑼′𝑽𝑽′𝑾𝑾′ to △ 𝑼𝑼′′𝑽𝑽′′𝑾𝑾′′.
△ 𝑼𝑼′𝑽𝑽′𝑾𝑾′ is the image of △ 𝑼𝑼𝑼𝑼𝑼𝑼 with a scale factor 𝒓𝒓𝟏𝟏 = 𝟐𝟐, so it follows that △ 𝑼𝑼𝑼𝑼𝑼𝑼 can be considered the
𝟏𝟏
𝟐𝟐
image of △ 𝑼𝑼′𝑽𝑽′𝑾𝑾′ with a scale factor of 𝒓𝒓𝟒𝟒 = . Therefore, △ 𝑼𝑼′′𝑽𝑽′′𝑾𝑾′′ can be considered the image of the
composition of dilations 𝑫𝑫(𝟎𝟎,𝟔𝟔),𝟑𝟑 �𝑫𝑫(𝟎𝟎,𝟎𝟎),𝟏𝟏 � of △ 𝑼𝑼′𝑽𝑽′𝑾𝑾′. This means that the scale factor 𝒓𝒓𝟑𝟑 = 𝒓𝒓𝟒𝟒 ⋅ 𝒓𝒓𝟐𝟐 .
𝟒𝟒
𝟐𝟐
𝒓𝒓𝟑𝟑 = 𝒓𝒓𝟒𝟒 ⋅ 𝒓𝒓𝟐𝟐
𝟏𝟏 𝟑𝟑
⋅
𝟐𝟐 𝟒𝟒
𝟑𝟑
𝒓𝒓𝟑𝟑 =
𝟖𝟖
𝒓𝒓𝟑𝟑 =
d.
Find the coordinates of the center of dilation that maps △ 𝑼𝑼′𝑽𝑽′𝑾𝑾′ to △ 𝑼𝑼′′𝑽𝑽′′𝑾𝑾′′.
The center of dilation 𝑶𝑶𝟑𝟑 must lie on the 𝒚𝒚-axis with centers (𝟎𝟎, 𝟎𝟎) and (𝟎𝟎, 𝟔𝟔). Therefore, the 𝒙𝒙-coordinate of
𝑶𝑶𝟑𝟑 is 𝟎𝟎. Using the graph, it appears that the 𝒚𝒚-coordinate of 𝑶𝑶𝟑𝟑 is a little more than 𝟐𝟐.
Considering the points 𝑽𝑽′ and 𝑽𝑽′′:
The center of dilation 𝑶𝑶𝟑𝟑 is (𝟎𝟎, 𝟐𝟐. 𝟒𝟒).
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𝟑𝟑
(𝟏𝟏𝟏𝟏 − 𝒚𝒚𝑶𝑶 ) = 𝟔𝟔 − 𝒚𝒚𝑶𝑶
𝟖𝟖
𝟗𝟗 𝟑𝟑
− 𝒚𝒚 = 𝟔𝟔 − 𝒚𝒚𝑶𝑶
𝟐𝟐 𝟖𝟖 𝑶𝑶
𝟓𝟓
𝟗𝟗
= 𝟔𝟔 − 𝒚𝒚𝑶𝑶
𝟖𝟖
𝟐𝟐
𝟓𝟓
𝟑𝟑
− = − 𝒚𝒚𝑶𝑶
𝟖𝟖
𝟐𝟐
𝟐𝟐𝟐𝟐
= 𝒚𝒚𝑶𝑶 = 𝟐𝟐. 𝟒𝟒
𝟏𝟏𝟏𝟏
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New York State Common Core
Mathematics Curriculum
GEOMETRY • MODULE 2
Topic C
Similarity and Dilations
G-SRT.A.2, G-SRT.A.3, G-SRT.B.5, G-MG.A.1
Focus Standards:
Instructional Days:
1Lesson
G-SRT.A.2
Given two figures, use the definition of similarity in terms of similarity transformations
to decide if they are similar; explain using similarity transformations the meaning of
similarity for triangles as the equality of all corresponding pairs of angles and the
proportionality of all corresponding pairs of sides.
G-SRT.A.3
Use the properties of similarity transformations to establish the AA criterion for two
triangles to be similar.
G-SRT.B.5
Use congruence and similarity criteria for triangles to solve problems and prove
relationships in geometric figures.
G-MG.A.1
Use geometric shapes, their measures, and their properties to describe objects (e.g.,
modeling a tree trunk or a human torso as a cylinder).★
9
Lesson 12:
What Are Similarity Transformations, and Why Do We Need Them? (P) 1
Lesson 13:
Properties of Similarity Transformations (P)
Lesson 14:
Similarity (P)
Lesson 15:
The Angle-Angle (AA) Criterion for Two Triangles to Be Similar (S)
Lesson 16:
Between-Figure and Within-Figure Ratios (P)
Lesson 17:
The Side-Angle-Side (SAS) and Side-Side-Side (SSS) Criteria for Two Triangles to Be Similar (E)
Lesson 18:
Similarity and the Angle Bisector Theorem (P)
Lesson 19:
Families of Parallel Lines and the Circumference of the Earth (S)
Lesson 20:
How Far Away Is the Moon? (S)
Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
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Topic C
M2
GEOMETRY
With an understanding of dilations, students are now ready to study similarity in Topic C. This is an
appropriate moment to pause and reflect on the change in how the study of similarity is conducted in this
curriculum versus traditional geometry curricula. It is common to open to a similarity unit in a traditional
textbook and read about polygons, chiefly triangles, which are of the same shape but different size. Some
may emphasize the proportional relationship between corresponding sides early in the unit. The point is that
grade-school mathematics has traditionally packaged the concept of similarity into a distilled version of the
bigger picture. The unpackaged view requires a more methodical journey to arrive at the concept of
similarity, including the use of transformations. It is in Topic C, after a foundation of scale drawings and
dilations, that we can discuss similarity.
Lesson 12 introduces the concept of a similarity transformation, which is needed to identify figures as being
similar. Just as with rigid motions and congruence, the lesson intentionally presents curvilinear examples to
emphasize that the use of similarity transformations allows us to compare both rectilinear and curvilinear
figures. Next, in Lesson 13, students apply similarity transformations to figures by construction. This is the
only lesson where students actually perform similarity transformations. The goals are to simply be able to
apply a similarity as well as observe how the properties of the individual transformations that compose each
similarity hold throughout construction. In Lesson 14, students observe the reflexive, symmetric, and
transitive properties of similarity. The scope of figures used in Lessons 15 through 18 narrows to triangles. In
these lessons, students discover and prove the AA, SSS, and SAS similarity criteria. Students use these criteria
and length relationships between similar figures and within figures to solve for unknown lengths in triangles
(G-SRT.A.3, G-SRT.B.5). Note that when students solve problems in Lesson 16, they are using geometric
shapes, their measures and properties to describe situations (e.g., similar triangles) is work related to the
modeling standard G-MG.A.1. Lessons 19 and 20 are modeling lessons (G-MG.A.1) that lead students
through the reasoning the ancient Greeks used to determine the circumference of the earth (Lesson 19) and
the distance from the earth to the moon (Lesson 20).
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Lesson 12: What Are Similarity Transformations, and Why
Do We Need Them?
Student Outcomes

Students define a similarity transformation as the composition of basic rigid motions and dilations. Students
define two figures to be similar if there is a similarity transformation that takes one to the other.

Students describe a similarity transformation applied to an arbitrary figure (i.e., not just triangles) and use
similarity to distinguish between figures that resemble each other versus those that are actually similar.
Lesson Notes
As noted earlier in this curriculum, congruence and similarity are presented differently here than in most previous
curricula. In the past, congruence criteria have first been defined for triangles (e.g., SSS or ASA), and then, by extension,
for polygonal figures, with similarity being treated in a like manner. However, the Common Core Standards approach
these concepts via transformations, allowing one to accommodate not only polygonal figures, but also curvilinear
figures, in one stroke.
Students begin Topic C with an understanding of what similarity transformations are and what it means for figures to be
similar. They should see how similarity transformations are like the rigid motions in their use to compare figures in the
plane. Unlike the work they did with similarity in Grade 8, students study similarity in the plane and not in the
coordinate plane. This is, of course, because we want students to fully realize what is referred to as the “abundance” of
transformations in the plane. It is not that transformations are limited in the coordinate system but rather that the
coordinate system simply encourages students to see certain transformations as more natural than others (e.g., a
translation parallel to the 𝑥𝑥-axis: (𝑥𝑥, 𝑦𝑦) ↦ (𝑥𝑥 + 𝑎𝑎, 𝑦𝑦)). Removing the coordinate system prevents this natural
gravitation toward certain transformations over others.
Classwork
Opening Exercise (3 minutes)
Opening Exercise
Observe Figures 1 and 2 and the images of the intermediate figures between them.
Figures 1 and 2 are called similar.
What observations can we make about Figures 1 and 2?
Answers will vary; students might say that the two figures look alike in terms of
shape but not size. Accept reasonable answers at this point to start the
conversation, and move on to filling out the chart below.
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Definition:
A
(or
) is a composition of a
finite number of dilations or basic rigid motions. The scale factor of a similarity transformation is
the product of the scale factors of the dilations in the composition. If there are no dilations in the
composition, the scale factor is defined to be 1.
similarity transformation, similarity
Scaffolding:
Depending on student ability,
consider using a fill-in-theblank approach to the
definitions listed here.
Definition:
Two figures in a plane are
other figure.
if there exists a similarity transformation taking one figure onto the
similar
Direct students to sketch possibilities for the Examples and Non-Examples boxes, and offer them the provided example
after they voice their ideas. Have them list characteristics of the Examples, and then provide them with the definitions
of similar and similarity transformation (see definitions above).
Definition
Characteristics
A similarity transformation (or similarity) is a
composition of a finite number of dilations or
basic rigid motions. The scale factor of a similarity
transformation is the product of the scale factors
of the dilations in the composition. If there are no
dilations in the composition, the scale factor is
defined to be 𝟏𝟏.
Similar figures should look the same, but one is a
different size, flipped, rotated, or translated
relative to the other.
Two figures in a plane are similar if there exists a
similarity transformation taking one figure onto
the other figure.
Examples
similar
Non-Examples
Discussion (10 minutes)

Consider what you know about congruence when thinking about similarity. One use of the rigid motions is to
establish whether two figures are identical in the plane. How did we use rigid motions to establish this?

If a series of rigid motions mapped one figure onto the other and the figures coincided, we could
conclude that they were congruent.
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
We can use similarity transformations in the same way. Consider Figure 1 below.
Figure 1

From our work on dilations, we can see that there is, in fact, a dilation that would map figure 𝐴𝐴 to 𝐴𝐴′. Note
that a similarity transformation does not have a minimum number of dilations or rigid motions (e.g., a single
reflection or a single dilation is a similarity transformation.)
Note that the similarity symbol “~” has not been mentioned in this lesson. If students remember it and have no trouble
with it, feel free to discuss it. The symbol is addressed in Lesson 14.

Now examine Figure 2. In Figure 2, the figure 𝐴𝐴′ was rotated 90° and is now labeled as 𝐴𝐴′′.
Figure 2

Would it be correct to say that 𝐴𝐴′′ is a dilation of 𝐴𝐴?


No. This is not a dilation because corresponding segments are neither parallel nor collinear.
Yet we saw in Figure 1 that it is possible to transform 𝐴𝐴 to 𝐴𝐴′, which we know to be congruent to 𝐴𝐴′′. So what
are the necessary steps to map 𝐴𝐴 to 𝐴𝐴′′?
Allow a moment for students to discuss this. Confirm that either both the composition of a dilation and rotation or the
composition of a dilation, rotation, and translation maps 𝐴𝐴 to 𝐴𝐴′′.

The series of steps needed to map 𝐴𝐴 to 𝐴𝐴′′, the dilation and rotation or rotation and dilation, can be thought of
as a composition of transformations, or more specifically, a similarity transformation. 𝐴𝐴′′ ≅ 𝑅𝑅𝐶𝐶,𝜃𝜃 �𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐴𝐴)�

If a similarity transformation maps one figure to another, we say the figures are similar.

Note this important distinction. We know that it is not enough to say, “If two figures look identical, they must
be congruent.” We know that they are congruent only if a series of rigid motions maps one figure to the other.
In the same way, it is not enough to say that two figures look like they have the same shape; we have to show
that a similarity transformation maps one figure to the other to be sure that the figures really do have the
same shape.
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
Recall also that a scale drawing of a figure is one whose corresponding lengths are proportional and whose
corresponding angles are equal in measurement. We know that a dilation produces a scale drawing.
Therefore, figures that are similar must be scale drawings. Why must this be true?

Any figure that maps onto another figure by similarity transformation 𝑇𝑇 either has a finite number of
dilations or does not have any dilations. If there are dilations involved, we have seen that dilations
result in figures with proportional, corresponding lengths and corresponding angles of equal
measurement. If there are no dilations, then the rigid motions that compose the similarity
transformation have a scale factor of 𝑟𝑟 = 1 by definition. Therefore, in either case, the two similar
figures are scale drawings of each other.
Note that we have not said that figures that are scale drawings must be similar. This is discussed in Lesson 14.

We denote a similarity transformation with 𝑇𝑇. The transformations that compose a similarity transformation
can be in any order; however, as a matter of convention, we usually begin a similarity transformation with the
dilation (as we did in Grade 8) and follow with rigid motions.
Note that this convention is apparent in problems where students must describe the series of transformations that map
one figure onto its similar image; the convention is adhered to so that in the first step the two figures become
congruent, and then students are left to determine the congruence transformation that map one to the other.


If 𝑇𝑇 is a similarity transformation, then 𝑇𝑇 is the composition of basic rigid motions and dilations. The scale
factor 𝑟𝑟 of 𝑇𝑇 is the product of the scale factors of the dilations in the transformation. With respect to the
above example, 𝑇𝑇(𝐴𝐴) = 𝑅𝑅𝐶𝐶,𝜃𝜃 �𝐷𝐷𝑂𝑂,𝑟𝑟 (𝐴𝐴)�.
If there is no dilation in the similarity transformation, it is a congruence. However, a congruence is simply a
more specific similarity transformation, which is why the definition allows for the composition of
transformations that need not include a dilation.
Example 1 (5 minutes)
Scaffolding:
Students identify the transformations that map one figure onto another. Remind students
that, as a matter of convention, any dilation in a similarity transformation is identified first.
Example 1
Figure 𝒁𝒁′ is similar to Figure 𝒁𝒁. Describe a transformation that maps Figure 𝒁𝒁 onto Figure 𝒁𝒁′.
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Curvilinear figures, such as that
shown in Example 1, may be
difficult for some students. Try
using the same similarity
transformations with simpler
figures, such as asymmetrical
letters like 𝐿𝐿 and 𝐹𝐹, and then
scaffold up to those involving
curves such as 𝑃𝑃 and 𝑄𝑄.
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




We are not looking for specific parameters (e.g., scale factor or degree of rotation of each transformation);
rather, we want to identify the series of transformations needed to map Figure 𝑍𝑍 to Figure 𝑍𝑍′.
Step 1: The dilation has a scale factor of 𝑟𝑟 < 1 since 𝑍𝑍′ is smaller than 𝑍𝑍.
Step 2: Notice that 𝑍𝑍′ is flipped from 𝑍𝑍1 . So, take a reflection of 𝑍𝑍1 to get 𝑍𝑍2
over a line 𝑙𝑙.
Step 3: Translate the plane such that a point of 𝑍𝑍2 maps to a corresponding
point in 𝑍𝑍′. Call the new figure 𝑍𝑍3 .
Step 4: Rotate until 𝑍𝑍3 coincides with 𝑍𝑍′.
Scaffolding:
Consider having more
general locations of centers of
rotation or dilation, lines of
reflection, and translation
vectors.
𝑙𝑙
Figure 𝒁𝒁 maps to Figure 𝒁𝒁′ by first dilating by a scale factor of 𝒓𝒓 until the corresponding lengths are equal in measurement
and then reflecting over line 𝒍𝒍 to ensure that both figures have the same orientation. Next, translate along a vector so
that one point of the image corresponds to 𝒁𝒁′ . Finally, a rotation around a center 𝑪𝑪 of degree 𝜽𝜽 orients Figure 𝒁𝒁 so that it
coincides with Figure 𝒁𝒁′.
Exercises 1–3 (8 minutes)
Exercises 1–3 allow students the opportunity to practice describing transformations that map an original figure onto its
corresponding transformed figure. If time is an issue, have students complete one exercise that seems appropriate, and
move on to Example 2.
Exercises 1–3
1.
Figure 1 is similar to Figure 2. Which transformations compose the similarity transformation that maps Figure 1
onto Figure 2?
First, dilate Figure 1 by a scale factor of 𝒓𝒓 > 𝟏𝟏 until the corresponding
lengths are equal in measurement, and then reflect over a line 𝒍𝒍 so that
Figure 1 coincides with Figure 2.
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2.
Figure 𝑺𝑺 is similar to Figure 𝑺𝑺′. Which transformations compose the similarity transformation that maps 𝑺𝑺 onto 𝑺𝑺′?
First, dilate 𝑺𝑺 by a scale factor of 𝒓𝒓 > 𝟏𝟏 until the corresponding segment
lengths are equal in measurement to those of 𝑺𝑺′. Then, 𝑺𝑺 must be rotated
around a center 𝑪𝑪 of degree 𝜽𝜽 so that 𝑺𝑺 coincides with 𝑺𝑺′.
3.
Figure 1 is similar to Figure 2. Which transformations compose the similarity transformation that maps Figure 1
onto Figure 2?
It is possible to only use two transformations: a rotation followed by a reflection. However, to do this, the correct
center must be found. The solution image reflects this approach, but students may say to first rotate Figure 1
around a center 𝑪𝑪 by 𝟗𝟗𝟗𝟗° in the clockwise direction. Then, reflect Figure 𝟏𝟏 over a vertical line 𝓵𝓵. Finally, translate
Figure 1 by a vector so that Figure 1 coincides with Figure 2.
Reemphasize to students that a similarity transformation does not need to have a dilation. Just as a square is a special
type of rectangle, but because the relationship does not work in reverse, so is a congruence transformation. Similarity
transformations generalize the notion of congruency.
Example 2 (5 minutes)
Remind students that the question asks them to take measurements in Example 2. If needed, further prompt them to
consider measurements of segments, not of any curved segment. This should alert them that there are few possible
measurements to make and that a relationship must exist between these measurements as well as what the question is
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Example 2
Show that no sequence of basic rigid motions and dilations takes the small figure to the large figure. Take measurements
as needed.
A similarity transformation that maps the small outer circle to the large outer circle would need to have scale factor of
about 𝟐𝟐. A similarity transformation that maps the small line segment 𝑨𝑨𝑨𝑨 to the large line segment 𝑪𝑪𝑪𝑪 would need to
have scale factor of about 𝟒𝟒. So, there is no similarity transformation that maps the small figure onto the large figure.
Exercises 4–5 (7 minutes)
Exercises 4–5
4.
Is there a sequence of dilations and basic rigid motions that takes the large figure to the small figure? Take
measurements as needed.
A similarity transformation that maps segment 𝑨𝑨𝑨𝑨 to
segment 𝑾𝑾𝑾𝑾 would need to have scale factor of
𝟐𝟐
about , but a similarity transformation that maps the
𝟑𝟑
small segment 𝑪𝑪𝑪𝑪 to segment 𝒀𝒀𝒀𝒀 would need to have
𝟏𝟏
scale factor of about . So, there is no similarity
𝟐𝟐
transformation that maps the large figure onto the
small figure.
5.
What purpose do transformations serve? Compare and contrast the application of rigid motions to the application
of similarity transformations.
We use all the transformations to compare figures in the plane. Rigid motions are distance preserving while
dilations, integral to similarity transformations, are not distance preserving. We use compositions of rigid motions
to determine whether two figures are congruent, and we use compositions of rigid motions and dilations, specifically
similarity transformations, to determine whether figures are similar.
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Closing (2 minutes)

What does it mean for two figures to be similar?


What are similarity transformations, and how can we use them?


We classify two figures as similar if there exists a similarity transformation that maps one figure onto
the other.
A similarity transformation is a composition of a finite number of dilations or rigid motions. Similarity
transformations precisely determine whether two figures have the same shape (i.e., two figures are
similar). If a similarity transformation does map one figure onto another, we know that one figure is a
scale drawing of the other.
How do congruence and similarity transformations compare to each other?

Both congruence and similarity transformations are a means of comparing figures in the plane. A
congruence transformation is also a similarity transformation, but a similarity transformation does not
need to be a congruence transformation.
Lesson Summary
Two figures are similar if there exists a similarity transformation that maps one figure onto the other.
A similarity transformation is a composition of a finite number of dilations or rigid motions.
Exit Ticket (5 minutes)
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Name
Date
Lesson 12: What Are Similarity Transformations, and Why Do We
Need Them?
Exit Ticket
1.
Figure A' is similar to Figure A. Which transformations compose the similarity transformation that maps Figure A
onto Figure A'?
Figure A'
Figure A
2.
Is there a sequence of dilations and basic rigid motions that takes the small figure to the large figure? Take
measurements as needed.
Figure A
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Figure B
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Exit Ticket Sample Solutions
1.
Figure A' is similar to Figure A. Which transformations compose the similarity transformation that maps Figure A
onto Figure A'?
Figure A'
Figure A
2.
One possible solution: We first take a dilation of Figure A with a scale factor of 𝒓𝒓 < 𝟏𝟏 and center 𝑶𝑶, the point where
the two line segments meet, until the corresponding lengths are equal to those in Figure A'. Next, take a rotation
(𝟏𝟏𝟏𝟏𝟏𝟏°) about 𝑶𝑶, and then, finally, take a reflection over a (vertical) line 𝓵𝓵.
Is there a sequence of dilations and basic rigid motions that takes the small figure to the large figure? Take
measurements as needed.
Figure A
Figure B
No similarity transformation exists because the circled corresponding distances and the corresponding distances
marked by the arrows on Figure B are not in the same ratio.
Problem Set Sample Solutions
1.
What is the relationship between scale drawings, dilations, and similar figures?
a.
How are scale drawings and dilations alike?
Scale drawings and dilated figures are alike in that all corresponding angles are congruent and all
corresponding distances are in the equivalent ratio, 𝒓𝒓, called the scale factor. A dilation of a figure produces a
scale drawing of that figure.
b.
How can scale drawings and dilations differ?
Dilations are a transformation of the plane in which all corresponding points from the image and pre-image
are mapped along rays that originate at the center of dilation. This is not a requirement for scale drawings.
c.
What is the relationship of similar figures to scale drawings and dilations?
Similar figures are scale drawings because they can be mapped together by a series of dilations and rigid
motions.
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2.
Given the diagram below, identify a similarity transformation, if one exists, that maps Figure A onto Figure B. If one
does not exist, explain why.
(Note to the teacher: The solution below is only one of many valid solutions to this problem.)
ℓ
𝟏𝟏
First, Figure A is dilated from center 𝑶𝑶 with a scale factor of . Next, the image is rotated −𝟗𝟗𝟗𝟗° about center 𝑶𝑶.
𝟑𝟑
Finally, the image is reflected over horizontal line 𝓵𝓵 onto Figure B.
3.
Teddy correctly identified a similarity transformation with at least one dilation that maps Figure 𝑰𝑰 onto Figure 𝑰𝑰𝑰𝑰.
Megan correctly identified a congruence transformation that maps Figure 𝑰𝑰 onto Figure 𝑰𝑰𝑰𝑰. What must be true
If Megan correctly identified a congruence transformation that maps Figure 𝑰𝑰 onto Figure 𝑰𝑰𝑰𝑰, then Figure 𝑰𝑰 and
Figure 𝑰𝑰𝑰𝑰 must be congruent. Therefore, Teddy’s similarity transformation must have either included a single
dilation with a scale factor of 𝟏𝟏 or must have included more than one dilation of which the product of all scale
factors was 𝟏𝟏 because it included at least one dilation.
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4.
Given the coordinate plane shown, identify a similarity transformation, if one exists, that maps 𝑿𝑿 onto 𝒀𝒀. If one
does not exist, explain why.
(Note to the teacher: The solution below is only one of many valid solutions to this problem.)
𝟏𝟏
First, reflect 𝑿𝑿 over line 𝒙𝒙 = 𝟏𝟏𝟏𝟏. Then, dilate the image from center (𝟏𝟏𝟏𝟏, 𝟏𝟏) with a scale factor of to obtain 𝒀𝒀.
𝟐𝟐
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5.
Given the diagram below, identify a similarity transformation, if one exists, that maps 𝑮𝑮 onto 𝑯𝑯. If one does not
A similarity transformation does not exist
that maps 𝑮𝑮 onto 𝑯𝑯 because the side
lengths of the figures are not all
proportional. Figure 𝑮𝑮 is a rectangle (not a
square), whereas Figure 𝑯𝑯 is a square.
6.
Given the coordinate plane shown, identify a similarity transformation, if one exists, that maps 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 onto
𝑨𝑨′′′𝑩𝑩′′′𝑪𝑪′′′𝑫𝑫′′′. If one does not exist, explain why.
(Note to the teacher: Students need to use a protractor to obtain the correct degree measure of rotation. The
solution below is only one of many valid solutions to this problem.)
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𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 can be mapped onto 𝑨𝑨′′′𝑩𝑩′′′𝑪𝑪′′′𝑫𝑫′′′ by first translating along the vector ���������⃗
𝑪𝑪𝑪𝑪′′′, then rotating about point 𝑪𝑪′′′
𝟏𝟏
by 𝟖𝟖𝟖𝟖°, and finally dilating from point 𝑪𝑪′′′ using a scale factor of .
𝟒𝟒
7.
The diagram below shows a dilation of the plane … or does it? Explain your answer.
The diagram does not show a dilation of the plane from point 𝑶𝑶, even though the corresponding points are collinear
with the center 𝑶𝑶. To be a dilation of the plane, a constant scale factor must be used for all points from the center of
dilation; however, the scale factor relating the distances from the center in the diagram range from 𝟐𝟐 to 𝟐𝟐. 𝟓𝟓.
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Lesson 13: Properties of Similarity Transformations
Student Outcomes

Students know the properties of a similarity transformation are determined by the transformations that
compose the similarity transformation.

Students are able to apply a similarity transformation to a figure by construction.
Lesson Notes
In Lesson 13, students apply similarity transformations to figures by construction. It is important to note here that
teachers should emphasize unhurried, methodical drawing and careful use of tools to students. Each exercise entails
many construction marks, and part of students’ success depends on their perseverance. This is the only lesson where
students actually construct what happens to a figure that undergoes a similarity transformation; students experience
this process once to witness how the points of a figure move about the plane instead of just hearing about it or
describing it.
Just as part of any lesson preparation, it is a good idea to do the examples to better anticipate where students might
struggle. Teachers should tailor the number of examples for their respective classes. Examples of varying difficulty have
been provided so that teachers have options to differentiate for their diverse classrooms.
Finally, space is available in the student books, but teachers may prefer to work outside of the books to maximize
available space. This can be done by photocopying the initial image onto blank paper. The initial image of each problem
is provided at the close of the lesson.
Classwork
Scaffolding:
Opening (10 minutes)

We have spent a good deal of time discussing the properties of
transformations. For example, we know that the property that
distinguishes dilations from rigid motions is that dilations do not preserve
distance, whereas translations, rotations, and reflections do preserve
distance.

Take a few moments with a partner to list the properties that all
transformations have in common.

These properties that are true for all dilations, reflections, rotations, and
translations (the transformations that comprise similarity transformations)
also hold true for similarity transformations in general. Title your list as
“Properties of Similarity Transformations.”
Allow students time to develop as complete a list as possible of the properties.
Develop a comprehensive class list. Consider having a premade poster with all the
properties listed and keeping each property covered. This can be done on a poster
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 To help organize students’
thinking, write a list of all
possible similarity
transformations that are
composed of exactly two
different transformations.
Consider why those that
contain a dilation do not
preserve distance, and keep all
other properties that are
consistent across each of the
similarity transformations in
the list as properties of
similarity transformations.
 Two examples of similarity
transformations are (1) a
translation and reflection and
(2) a reflection and dilation.
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with a strip of paper or on a projector or interactive white board. As students list the properties they recall, reveal that
property from the poster. After students have offered their lists, review any remaining properties that were not
mentioned. A few words are mentioned below in anticipation of the properties that students may not recall.
Properties of similarity transformations:
1.
Distinct points are mapped to distinct points.
•
2.
Each point 𝑃𝑃′ in the plane has a pre-image.
•
3.
This means that if 𝑃𝑃 ≠ 𝑄𝑄, then for a transformation 𝐺𝐺, 𝐺𝐺(𝑃𝑃) ≠ 𝐺𝐺(𝑄𝑄).
If 𝑃𝑃′ is a point in the plane, then 𝑃𝑃′ = 𝐺𝐺(𝑃𝑃) for some point 𝑃𝑃.
Scaffolding:
Post the complete list of
properties of similarity
transformations in a prominent
place in the classroom.
There is a scale factor 𝑟𝑟 for 𝐺𝐺 so that for any pair of points 𝑃𝑃 and 𝑄𝑄 with images 𝑃𝑃′ = 𝐺𝐺(𝑃𝑃) and 𝑄𝑄′ = 𝐺𝐺(𝑄𝑄), then
𝑃𝑃′ 𝑄𝑄′ = 𝑟𝑟𝑟𝑟𝑟𝑟.
•
The scale factor for a similarity transformation is the product of the scale factors. Remember, the scale factor
associated to any congruence transformation is 1. The scale factor of a similarity transformation is really the
product of the scale factors of all the transformations that compose the similarity; however, since we know the
scale factor of all rigid motions is 1, the scale factor of the similarity transformation is the product of the scale
factors of all the dilations in the similarity.
4.
A similarity transformation sends lines to lines, rays to rays, line segments to line segments, and parallel lines to
parallel lines.
5.
A similarity transformation sends angles to angles of equal measure.
6.
A similarity transformation maps a circle of radius 𝑅𝑅 to a circle of radius 𝑟𝑟𝑟𝑟, where 𝑟𝑟 is the scale factor of the
similarity transformation.
•
All of the properties are satisfied by a similarity transformation consisting of a single translation, reflection,
rotation, or dilation. If the similarity transformation consists of more than one such transformation, then the
properties still hold because they hold one step at a time.
•
For instance, if 𝐺𝐺 is the composition of three transformations 𝐺𝐺1 , 𝐺𝐺2 , 𝐺𝐺3 , where each of 𝐺𝐺1 , 𝐺𝐺2 , 𝐺𝐺3 is a
translation, reflection, rotation, or dilation, then 𝐺𝐺1 maps a pair of parallel lines to a second pair of parallel lines
that are then taken by 𝐺𝐺2 to another pair of parallel lines that are then taken by 𝐺𝐺3 to yet another pair of
parallel lines. The composition of 𝐺𝐺1 , 𝐺𝐺2 , 𝐺𝐺3 takes any pair of parallel lines to another pair of parallel lines.
•
We keep these properties in mind as we work on examples where multiple transformations comprise the
similarity transformation.
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Example 1 (10 minutes)
Students apply a similarity transformation to an initial figure and determine the image.
Review the steps to apply a reflection and rotation in Example 1.

Use a compass, protractor, and straightedge to determine the image of the
triangle.
Example 1
Similarity transformation 𝑮𝑮 consists of a rotation about the point 𝑷𝑷 by 𝟗𝟗𝟗𝟗°, followed by a dilation
centered at 𝑷𝑷 with scale factor of 𝒓𝒓 = 𝟐𝟐, and then followed by a reflection across line 𝓵𝓵. Find the
image of the triangle.

Scaffolding:
 Depending on student
ability and time, consider
limiting 𝐺𝐺 to the first two
transformations.
 Consider placing examples
on the coordinate plane,
on grid paper, and/or
using transparencies, patty
paper, or geometry
software.
Before getting to the actual construction process, draw a predictive sketch of the image of the triangle after
transformation. You should have three sketches.
Drawing a predictive sketch helps illuminate the path ahead before getting into the details of each construction. The
sketches also provide a time to reflect on how the properties are true for each transformation. For example, ask
students to select a property from their lists and describe where in their sketches they see it in the transformation. For
instance, a student might select property (2), which states that each point has a pre-image 𝑃𝑃 and can point to each preimage and image with each passing transformation.
ℓ
MP.1
ℓ
Complete this example with students. Remind them that the rotation requires all three geometry tools.
Note to the teacher: As an alternative strategy, consider using coordinate geometry by placing the image over an
appropriately scaled grid. Students gained familiarity with coordinate geometry as used with transformations of the
plane in Grade 8. The image on the right above has been provided for this alternative. Placement of the 𝑥𝑥- and 𝑦𝑦-axes
can be determined where convenient for this example. Given this flexibility, most students likely choose point 𝑃𝑃 to be
the origin of the coordinate plane since two of the given transformations are centered at 𝑃𝑃.

How will we rotate the triangle 90°?

To locate the vertices of the triangle’s image, draw rays through each vertex of the triangle originating
from 𝑃𝑃. Use each ray to form three 90° angles, and then use the compass to locate each new vertex on
the corresponding ray.
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Point out to students that in their constructions, the ray through 𝐵𝐵′ and 𝐶𝐶′ coincides and appears to be one ray, giving an
appearance of only two rays, not three.
ℓ
Next, allow students time to dilate △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′. If necessary, review the steps to create a dilation.
MP.1
ℓ

How will we reflect the triangle over the line?

Create the construction marks that determine the image of each vertex so that the line of reflection is
the perpendicular bisector of the segment that joins each vertex with its image. In other words, we
must create the construction marks so that the images of the vertices are located so that the line of
𝐵𝐵′′𝐵𝐵′′′, ��������
𝐶𝐶′′𝐶𝐶′′′.
reflection is the perpendicular bisector of ��������
𝐴𝐴′′𝐴𝐴′′′, ��������
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The steps to determine 𝐵𝐵′′′ are shown below.
Scaffolding:
 Diagrams with more than
two transformations of a
figure can become
cluttered very quickly.
Consider allowing students
to use different colored
pencils (or pens) to
complete each stage of
the similarity
transformation.
ℓ
 Also, to reduce clutter,
draw only construction
arcs as opposed to full
construction circles (as are
shown in the diagram).
MP.1


We have applied the outlined similarity transformation to △ 𝐴𝐴𝐴𝐴𝐴𝐴 and found its image, that is, the similar figure
△ 𝐴𝐴′′′𝐵𝐵′′′𝐶𝐶′′′.
Since 𝐺𝐺 comprises three individual transformations and each of the transformations satisfies the known
properties, we know that 𝐺𝐺 also satisfies the properties.
Example 2 (10 minutes)
Example 2 incorporates a translation, a reflection, and a dilation in the similarity transformation. Review the steps of
how to apply a translation in Example 2.
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Example 2
A similarity transformation 𝑮𝑮 applied to trapezoid 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 consists of a translation by vector �����⃗
𝑿𝑿𝑿𝑿, followed by a reflection
across line 𝓶𝓶, and then followed by a dilation centered at 𝑷𝑷 with a scale factor of 𝒓𝒓 = 𝟐𝟐. Recall that we can describe the
same sequence using the following notation: 𝑫𝑫𝑷𝑷,𝟐𝟐 �𝒓𝒓𝓶𝓶 �𝑻𝑻𝑿𝑿𝑿𝑿 (𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨)��. Find the image of 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨.
𝓂𝓂
Encourage students to draw a predictive sketch for each stage of the transformation before beginning the construction.

Describe the steps to apply the translation by vector 𝑋𝑋𝑋𝑋 to one point of the figure.

To apply the translation to 𝐴𝐴, construct 𝐶𝐶1 : center 𝐴𝐴, radius 𝑋𝑋𝑋𝑋. Then, construct 𝐶𝐶2 : center 𝑌𝑌, radius
𝑋𝑋𝑋𝑋.
𝐶𝐶2
𝐶𝐶1
Scaffolding:
Help students get started at
translating point 𝐴𝐴 by finding
the fourth vertex, 𝐴𝐴′, of
parallelogram 𝑋𝑋𝑋𝑋𝑋𝑋′𝑌𝑌. Finding
the images of the remaining
points 𝐵𝐵, 𝐶𝐶, and 𝐷𝐷 under the
translation can be found by
following similar processes.
𝓂𝓂
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The process for locating the image of 𝐵𝐵 under the translation is shown below:
𝓂𝓂
Allow students time to complete the rest of the example before reviewing it.
The following image shows the reflection of vertex 𝐶𝐶′:
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The following image shows the dilation of 𝐴𝐴′′𝐵𝐵′′𝐶𝐶′′𝐷𝐷′′:
Exercise 1 (8 minutes)
Allow students to work on Exercise 1 independently. Encourage students to draw a predictive sketch for each stage of
the transformation before beginning the construction.
Exercise 1
A similarity transformation for triangle 𝑫𝑫𝑫𝑫𝑫𝑫 is described by 𝒓𝒓𝓷𝓷 �𝑫𝑫
triangle 𝑫𝑫𝑫𝑫𝑫𝑫 under the similarity.
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𝟏𝟏 �𝑹𝑹𝑨𝑨,𝟗𝟗𝟗𝟗° (𝑫𝑫𝑫𝑫𝑫𝑫)��.
𝟐𝟐
𝑨𝑨,
Locate and label the image of
𝓃𝓃
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𝓃𝓃
Closing (2 minutes)

Why are the properties of a similarity transformation the same as those of both dilations and rigid motions?

The properties enjoyed by individual transformations are true for a similarity transformation, as each
transformation in a composition is done one transformation at a time.
Review the properties of similarity transformations:
1.
Distinct points are mapped to distinct points.
2.
Each point 𝑃𝑃′ in the plane has a pre-image.
3.
4.
There is a scale factor of 𝑟𝑟 for 𝐺𝐺 so that for any pair of points 𝑃𝑃 and 𝑄𝑄 with images 𝑃𝑃′ = 𝐺𝐺(𝑃𝑃) and
𝑄𝑄′ = 𝐺𝐺(𝑄𝑄), then 𝑃𝑃′ 𝑄𝑄′ = 𝑟𝑟𝑟𝑟𝑟𝑟.
A similarity transformation sends lines to lines, rays to rays, line segments to line segments, and parallel lines to
parallel lines.
5.
A similarity transformation sends angles to angles of equal measure.
6.
A similarity transformation maps a circle of radius 𝑅𝑅 to a circle of radius 𝑟𝑟𝑟𝑟, where 𝑟𝑟 is the scale factor of the
similarity transformation.
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Lesson Summary
Properties of similarity transformations:
1.
Distinct points are mapped to distinct points.
2.
Each point 𝑷𝑷′ in the plane has a pre-image.
3.
There is a scale factor of 𝒓𝒓 for 𝑮𝑮 so that for any pair of points 𝑷𝑷 and 𝑸𝑸 with images 𝑷𝑷′ = 𝑮𝑮(𝑷𝑷) and
𝑸𝑸′ = 𝑮𝑮(𝑸𝑸), then 𝑷𝑷′ 𝑸𝑸′ = 𝒓𝒓𝒓𝒓𝒓𝒓.
4.
A similarity transformation sends lines to lines, rays to rays, line segments to line segments, and parallel lines
to parallel lines.
5.
A similarity transformation sends angles to angles of equal measure.
6.
A similarity transformation maps a circle of radius 𝑹𝑹 to a circle of radius 𝒓𝒓𝒓𝒓, where 𝒓𝒓 is the scale factor of the
similarity transformation.
Exit Ticket (5 minutes)
Note to the teacher: The Exit Ticket contains a sequence of three transformations on the plane that may require more
than 5 minutes to complete. The second step in the sequence is a dilation, so students should be directed to complete
at least the first two transformations in the sequence to find 𝐴𝐴′′𝐶𝐶′′𝐷𝐷′′𝐸𝐸′′.
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Name
Date
Lesson 13: Properties of Similarity Transformations
Exit Ticket
A similarity transformation consists of a translation along the vector �����⃗
𝐹𝐹𝐹𝐹 , followed by a dilation from point 𝑃𝑃 with a scale
factor of 𝑟𝑟 = 2, and finally a reflection over line 𝑚𝑚. Use construction tools to find 𝐴𝐴′′′𝐶𝐶′′′𝐷𝐷′′′𝐸𝐸′′′.
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Exit Ticket Sample Solutions
����⃗, followed by a dilation from point 𝑷𝑷 with a scale
A similarity transformation consists of a translation along the vector 𝑭𝑭𝑭𝑭
factor of 𝒓𝒓 = 𝟐𝟐, and finally a reflection over line 𝒎𝒎. Use construction tools to find 𝑨𝑨′′′𝑪𝑪′′′𝑫𝑫′′′𝑬𝑬′′′.
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Problem Set Sample Solutions
1.
A similarity transformation consists of a reflection over line 𝓵𝓵, followed by a dilation from 𝑶𝑶 with a scale factor of
𝟑𝟑
𝟒𝟒
𝒓𝒓 = . Use construction tools to find △ 𝑮𝑮′′𝑯𝑯′′𝑰𝑰′′.
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2.
about 𝑶𝑶 of −𝟗𝟗𝟗𝟗°. Use construction tools to find kite 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′𝑫𝑫′′.
3.
For the Figure 𝒁𝒁, find the image of 𝒓𝒓𝓵𝓵 (𝑹𝑹𝑷𝑷,𝟗𝟗𝟗𝟗˚ �𝑫𝑫
𝟏𝟏 (𝒁𝒁)�.
𝟐𝟐
𝑷𝑷,
ℓ
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𝟏𝟏
𝟐𝟐
A similarity transformation consists of a dilation from point 𝑶𝑶 with a scale factor of 𝒓𝒓 = 𝟐𝟐 , followed by a rotation
ℓ
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4.
A similarity transformation consists of a translation along vector ������⃗
𝑼𝑼𝑼𝑼, followed by a rotation of 𝟔𝟔𝟔𝟔° about 𝑷𝑷, then
𝟏𝟏
𝟑𝟑
followed by a dilation from 𝑷𝑷 with a scale factor of 𝒓𝒓 = . Use construction tools to find △ 𝑿𝑿′′′𝒀𝒀′′′𝒁𝒁′′′.
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5.
Given the quarter-circular figure determined by points 𝑨𝑨, 𝑩𝑩, and 𝑪𝑪, a similarity transformation consists of a −𝟔𝟔𝟔𝟔°
𝟏𝟏
𝟐𝟐
rotation about point 𝑩𝑩, followed by a dilation from point 𝑶𝑶 with a scale factor of 𝒓𝒓 = . Find the image of the figure
determined by points 𝑨𝑨′′ , 𝑩𝑩′′ , and 𝑪𝑪′′.
Describe a different similarity transformation that would map quarter-circle 𝑨𝑨𝑨𝑨𝑨𝑨 to quarter-circle 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′.
𝟏𝟏
The quarter-circular region could have first been dilated from point 𝑶𝑶 using a scale factor of , followed by a
rotation of 𝟔𝟔𝟔𝟔° about point 𝑩𝑩′.
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𝟐𝟐
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6.
𝟏𝟏
A similarity transformation consists of a dilation from center 𝑶𝑶 with a scale factor of , followed by a rotation of 𝟔𝟔𝟔𝟔°
𝟐𝟐
about point 𝑶𝑶. Complete the similarity transformation on Figure 𝑻𝑻 to complete the drawing of Figure 𝑻𝑻′′.
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7.
Given Figure 𝑹𝑹 on the coordinate plane shown below, a similarity transformation consists of a dilation from (𝟎𝟎, 𝟔𝟔)
𝟏𝟏
with a scale factor of , followed by a reflection over line 𝒙𝒙 = −𝟏𝟏, and then followed by a vertical translation of 𝟓𝟓
𝟒𝟒
units down. Find the image of Figure 𝑹𝑹.
𝑅𝑅′
8.
𝑅𝑅′′
Given △ 𝑨𝑨𝑨𝑨𝑨𝑨, with vertices 𝑨𝑨(𝟐𝟐, −𝟕𝟕), 𝑩𝑩(−𝟐𝟐, −𝟏𝟏), and 𝑪𝑪(𝟑𝟑, −𝟒𝟒), locate and label the image of the triangle under
the similarity transformation 𝑫𝑫𝑩𝑩′,𝟏𝟏 �𝑹𝑹𝑨𝑨,𝟏𝟏𝟏𝟏𝟏𝟏° �𝒓𝒓𝒙𝒙=𝟐𝟐 (𝑨𝑨𝑨𝑨𝑨𝑨)��.
𝟐𝟐
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9.
In Problem 8, describe the relationship of 𝑨𝑨′′′ to �����
𝑨𝑨′′′ is the midpoint of �����
𝑨𝑨𝑨𝑨′. I know this because 𝑨𝑨 = 𝑨𝑨′ = 𝑨𝑨′′, and since the dilation at the end of the similarity
transformation was centered at 𝑩𝑩′, the image of 𝑨𝑨′′ must lie on the ray joining it to the center 𝑩𝑩′, which means that
𝑨𝑨′′′ lies on �����
𝑨𝑨𝑨𝑨′. Furthermore, by the definition of dilation, 𝑨𝑨′′′ 𝑩𝑩′ = 𝒓𝒓(𝑨𝑨′′ 𝑩𝑩′ ), and since the given scale factor was
𝟏𝟏
𝟏𝟏
𝑨𝑨𝑨𝑨′.
𝒓𝒓 = , it follows that 𝑨𝑨′′′ 𝑩𝑩′ = (𝑨𝑨′′ 𝑩𝑩′ ), so 𝑨𝑨′′′ must, therefore, be the midpoint of �����
𝟐𝟐
𝟐𝟐
10. Given 𝑶𝑶(−𝟖𝟖, 𝟑𝟑) and quadrilateral 𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩, with 𝑩𝑩(−𝟓𝟓, 𝟏𝟏), 𝑪𝑪(−𝟔𝟔, −𝟏𝟏), 𝑫𝑫(−𝟒𝟒, −𝟏𝟏), and 𝑬𝑬(−𝟒𝟒, 𝟐𝟐), what are the
coordinates of the vertices of the image of 𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩 under the similarity transformation 𝒓𝒓𝒙𝒙−𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 �𝑫𝑫𝑶𝑶,𝟑𝟑 (𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩)�?
The coordinates of the image of 𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩 are 𝑩𝑩′′(𝟏𝟏, 𝟑𝟑), 𝑪𝑪′′(−𝟐𝟐, 𝟗𝟗), 𝑫𝑫′′(𝟒𝟒, 𝟗𝟗), and 𝑬𝑬′′(𝟒𝟒, 𝟎𝟎).
11. Given triangle 𝑨𝑨𝑨𝑨𝑨𝑨 as shown on the diagram of the coordinate plane:
a.
Perform a translation so that vertex 𝑨𝑨 maps to the origin.
See diagram.
b.
𝟏𝟏
Next, dilate the image 𝑨𝑨′𝑩𝑩′𝑪𝑪′ from the origin using a scale factor of .
𝟑𝟑
See diagram.
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c.
Finally, translate the image 𝑨𝑨′′𝑩𝑩′′𝑪𝑪′′ so that the vertex 𝑨𝑨′′ maps to the original point 𝑨𝑨.
See diagram.
𝐀𝐀′ = 𝐀𝐀′′
𝐀𝐀 = 𝐀𝐀′′′
d.
Using transformations, describe how the resulting image 𝑨𝑨′′′𝑩𝑩′′′𝑪𝑪′′′ relates to the original figure 𝑨𝑨𝑨𝑨𝑨𝑨.
𝟏𝟏
The image 𝑨𝑨′′′𝑩𝑩′′′𝑪𝑪′′′ is a dilation of figure 𝑨𝑨𝑨𝑨𝑨𝑨 from point 𝑨𝑨 with a scale factor of .
𝟑𝟑
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12.
a.
In the coordinate plane, name the single transformation resulting from the composition of the two dilations
shown below:
(Hint: Try it!)
𝑫𝑫(𝟎𝟎,𝟎𝟎),𝟐𝟐 followed by 𝑫𝑫(𝟎𝟎,𝟒𝟒),𝟏𝟏
𝟐𝟐
The image can be obtained by a translation two units to the right (a vector that has half the distance as the
distance between the centers of dilation).
b.
In the coordinate plane, name the single transformation resulting from the composition of the two dilations
shown below:
(Hint: Try it!)
𝑫𝑫(𝟎𝟎,𝟎𝟎),𝟐𝟐 followed by 𝑫𝑫(𝟒𝟒,𝟒𝟒),𝟏𝟏
𝟐𝟐
The image can be obtained by a translation two units to the right (a vector that has half the distance as the
distance between the centers of dilation).
c.
Using the results from parts (a) and (b), describe what happens to the origin under both similarity
transformations.
The origin maps to the midpoint of a segment joining the centers used for each dilation.
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Example 1
ℓ
Example 2
𝓂𝓂
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Lesson 14: Similarity
Student Outcomes

Students understand that similarity is reflexive, symmetric, and transitive.

Students recognize that if two triangles are similar, there is a correspondence such that corresponding pairs of
angles have the same measure and corresponding sides are proportional. Conversely, they know that if there
is a correspondence satisfying these conditions, then there is a similarity transformation taking one triangle to
the other respecting the correspondence.
Lesson Notes
In Lesson 14, students delve more deeply into what it means for figures to be similar. Examples address the properties
of similarity and also focus on circles and, finally, triangles. Lessons 12–13 were intentionally not focused on triangles to
emphasize that instructors need to move away from equating similarity with triangles or strictly rectilinear figures and
rather to think of and teach similarity as a broader concept. With this idea covered in Lessons 12 and 13, attention is
shifted to triangles in the second half of this lesson in order to prepare students for triangle similarity criteria in Lessons
15 and 17.
Classwork
Opening (4 minutes)



As initially mentioned in Lesson 12, two figures in the plane are similar if there is a similarity transformation
that takes one to the other. If 𝐴𝐴 and 𝐵𝐵 are similar, we write 𝐴𝐴~𝐵𝐵 where “~” denotes similarity.
Which of the following properties do you believe to be true? Vote yes by raising your hand.
For a Figure 𝐴𝐴 in the plane, do you believe that 𝐴𝐴~𝐴𝐴?
After taking the vote, show students a simple figure such as the following triangle, and ask them to reconsider if the
figure is similar to itself.
MP.3
Allow students 30 seconds to justify their responses in a sentence, and then have them compare reasons with a neighbor
before sharing out and moving on to the next property.
Scaffolding:
For struggling students, place
the triangles in the coordinate
plane.

For two figures 𝐴𝐴 and 𝐵𝐵 in the plane, do you believe that if 𝐴𝐴~𝐵𝐵, then 𝐵𝐵~𝐴𝐴?
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MP.3
After taking the vote, show Figures 𝐴𝐴 and 𝐵𝐵, and ask them to reconsider the notion that if 𝐴𝐴 is similar to 𝐵𝐵, then is 𝐵𝐵
similar to 𝐴𝐴.
Allow students 30 seconds to justify their responses in a sentence, and then have them compare reasons with a neighbor
before sharing out and moving on to the next property.

MP.3
For Figures 𝐴𝐴, 𝐵𝐵, and 𝐶𝐶 in the plane, do you believe that if 𝐴𝐴~𝐵𝐵 and 𝐵𝐵~𝐶𝐶, then 𝐴𝐴~𝐶𝐶?
After taking the vote, show students Figures 𝐴𝐴, 𝐵𝐵, and 𝐶𝐶, and ask them to reconsider if 𝐴𝐴 is similar to 𝐵𝐵 and 𝐵𝐵 is similar
to 𝐶𝐶, then is 𝐴𝐴 similar to 𝐶𝐶.
Allow students 30 seconds to justify their responses in a sentence, and then have them compare reasons with a neighbor
before sharing out and moving on to the next property.
Announce that the properties are in fact true, and state them:




For each figure 𝐴𝐴 in the plane, 𝐴𝐴~𝐴𝐴. Similarity is reflexive.
If 𝐴𝐴 and 𝐵𝐵 are figures in the plane so that 𝐴𝐴~𝐵𝐵, then 𝐵𝐵~𝐴𝐴. Similarity is symmetric.
If 𝐴𝐴, 𝐵𝐵, and 𝐶𝐶 are figures in the plane such that 𝐴𝐴~𝐵𝐵 and 𝐵𝐵~𝐶𝐶, then 𝐴𝐴~𝐶𝐶. Similarity is transitive.
In Examples 1 and 2, we form informal arguments to prove why the conditions on similarity must be true.
Example 1 (4 minutes)
Present the question to the class, and then consider employing any discussion strategies commonly used for a brief
brainstorming session, whether it is a whole-group share out, a timed talk-and-turn session with a neighbor, or a Quick
Write. Allow about two minutes for whichever strategy is selected and the remaining two minutes for sharing out as a
whole group, demonstrating (as needed) any students’ suggestions on the board.
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Example 1
We said that for a figure 𝑨𝑨 in the plane, it must be true that 𝑨𝑨~𝑨𝑨. Describe why this must be true.

Remember, to show that for two figures to be similar, there must be a similarity transformation that maps one
to the other. Are there such transformations to show that 𝐴𝐴 maps to 𝐴𝐴?
Take multiple suggestions of transformations that map 𝐴𝐴 to 𝐴𝐴:


MP.3




There are several different transformations that map 𝐴𝐴 onto itself such as a rotation of 0° or a rotation of
360°.
A reflection of 𝐴𝐴 across a line and a reflection right back achieves the same result.
A translation with a vector of length 0 also maps 𝐴𝐴 to 𝐴𝐴.
A dilation with scale factor 1 maps 𝐴𝐴 to 𝐴𝐴, and any combination of these transformations also maps
𝐴𝐴 to 𝐴𝐴.
Therefore, 𝐴𝐴 must be similar to 𝐴𝐴 because there are many similarity transformations that map 𝐴𝐴 to 𝐴𝐴.
This condition is labeled as reflexive because every figure is similar to itself.
Example 2 (4 minutes)
Present the question to the class, and then consider employing any discussion strategies commonly used for a brief
brainstorming session, whether it is a whole-group share out, a timed talk-and-turn session with a neighbor, or a Quick
Write. Allow about two minutes for whichever strategy is selected and the remaining two minutes for sharing out as a
whole group, demonstrating (as needed) any students’ suggestions on the board.
Example 2
We said that for figures 𝑨𝑨 and 𝑩𝑩 in the plane so that 𝑨𝑨~𝑩𝑩, then it must be true that 𝑩𝑩~𝑨𝑨. Describe why this must be
true.
Now that students have completed Example 1, allow them time to discuss Example 2 among themselves.

This condition must be true because for any composition of transformations that maps 𝐴𝐴 to 𝐵𝐵, there is a
composition of transformations that can undo the first composition. For example, if a translation by vector �����⃑
𝑋𝑋𝑋𝑋
maps 𝐴𝐴 to 𝐵𝐵, then the vector �����⃗
𝑌𝑌𝑌𝑌 will undo the transformation and map 𝐵𝐵 to 𝐴𝐴.

A counterclockwise rotation of 90° can be undone by a clockwise rotation of 90°.

A reflection across a line can be undone by a reflection back across the same line.



1
2
A dilation by a scale factor of 𝑟𝑟 = 2 can be undone with a dilation by a scale factor of 𝑟𝑟 = .
Therefore, it must be true that if a figure 𝐴𝐴 is similar to a figure 𝐵𝐵 in the plane (i.e., if there is a similarity
transformation that maps 𝐴𝐴 to 𝐵𝐵), then there must also be a composition of transformations that can undo
that similarity transformation and map 𝐵𝐵 back to 𝐴𝐴.
This condition is labeled as symmetric because of its likeness to the symmetric property of equality where if
one number is equal to another number, then they both must have the same value (if 𝑎𝑎 = 𝑏𝑏, then 𝑏𝑏 = 𝑎𝑎).
We leave the third condition, that similarity is transitive, for the Problem Set.
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Example 3 (10 minutes)
In Example 3, students must show that any circle is similar to any other circle. Note the term similar is used here, unlike
in Lesson 8 where students proved the dilation theorem for circles. Encourage students to first discuss the question with
a partner.
Example 3
Based on the definition of similar, how would you show that any two circles are similar?
Based on their discussions, provide students with the following cases to help them along:

Consider the different cases you must address in showing that any two circles
are similar to each other:
a.
Circles with different centers but radii of equal length

MP.3
b.
Circles with the same center but radii of different lengths

c.
If two circles have different centers but have radii of equal length, then
the circles are congruent, and a translation along a vector that brings
one center to the other will map one circle onto the other.
If two circles have the same center, but one circle has radius 𝑅𝑅 and the
other has radius 𝑅𝑅′, then a dilation about the center with a scale factor of
𝑟𝑟 =
𝑅𝑅′
maps one circle onto the other so that 𝑟𝑟𝑟𝑟 = 𝑅𝑅′.
𝑅𝑅
Circles with different centers and radii of different lengths

If two circles have different centers and radii of different lengths, then
the composition of a dilation described in (b) and the translation
described in (a) maps one onto the other.
Scaffolding:
learners, have students
discuss what cases must
be considered in order to
show similarity between
circles.
 Consider assigning cases
(a), (b), and (c) to small
groups. Share out results
after allowing for group
discussion. Case (a) is
likely the easiest to
consider since the circles
are congruent by a
translation and, therefore,
also similar.
Students may notice that two circles with different centers and different radii length may alternatively be mapped onto
each other by a single dilation as shown in the following image.
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Discussion (3 minutes)
This Discussion leads to the proof of the converse of the theorem on similar triangles in Example 4.
What do we mean when we say that two triangles are similar?

Two triangles △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝐷𝐷𝐷𝐷𝐷𝐷 are similar if there is a similarity transformation that maps △ 𝐴𝐴𝐴𝐴𝐴𝐴 to
△ 𝐷𝐷𝐷𝐷𝐷𝐷.

When we write △ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐷𝐷𝐷𝐷𝐷𝐷, we mean, additionally, that the similarity transformation takes 𝐴𝐴 to 𝐷𝐷, 𝐵𝐵 to
𝐸𝐸, and 𝐶𝐶 to 𝐹𝐹. So, when we see △ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐷𝐷𝐷𝐷𝐷𝐷, we know that a correspondence exists such that the
corresponding angles are equal in measurement and the corresponding lengths of sides are proportional (i.e.,

𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐷𝐷, 𝑚𝑚∠𝐵𝐵 = 𝑚𝑚∠𝐸𝐸, 𝑚𝑚∠𝐶𝐶 = 𝑚𝑚∠𝐹𝐹, and
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
=
𝐸𝐸𝐸𝐸
𝐵𝐵𝐵𝐵
=
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
).
THEOREM ON SIMILAR TRIANGLES: If △ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐷𝐷𝐷𝐷𝐷𝐷, then 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐷𝐷, 𝑚𝑚∠𝐵𝐵 = 𝑚𝑚∠𝐸𝐸, 𝑚𝑚∠𝐶𝐶 = 𝑚𝑚∠𝐹𝐹, and
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
=
𝐸𝐸𝐸𝐸
𝐵𝐵𝐵𝐵
=
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
.
Example 4 (7 minutes)
Example 4 asks students whether the converse of the theorem on similar triangles is true. Students begin with the fact
that a correspondence exists between two triangles or that the corresponding lengths are proportional and
corresponding angles are equal in measurement. Students must argue whether this makes the triangles similar.
The argument may not be a stretch for students to make since they have worked through the original argument. Allow
students a few minutes to attempt the informal proof before reviewing it with them.
Example 4
Suppose △ 𝑨𝑨𝑨𝑨𝑨𝑨 ↔ △ 𝑫𝑫𝑫𝑫𝑫𝑫 and that, under this correspondence, corresponding angles are equal and corresponding sides
are proportional. Does this guarantee that △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑫𝑫𝑫𝑫𝑫𝑫 are similar?

We have already shown that if two figures (e.g., triangles) are similar, then corresponding angles are of equal
measurement and corresponding sides are proportional in length.

This question is asking whether the converse of the theorem is true. We know that a correspondence exists
between △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝐷𝐷𝐷𝐷𝐷𝐷. What does the correspondence imply?



𝐴𝐴𝐴𝐴
=
𝐸𝐸𝐸𝐸
𝐵𝐵𝐵𝐵
=
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
.
Since the side lengths are proportional under the correspondence, then 𝑟𝑟 =
𝐷𝐷𝐷𝐷 𝐸𝐸𝐸𝐸 𝐷𝐷𝐷𝐷
=
= .
𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴
The dilation that takes △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐷𝐷𝐷𝐷𝐷𝐷 can have any point for a center. The dilation maps 𝐴𝐴 to 𝐴𝐴′ , 𝐵𝐵 to 𝐵𝐵′,
and 𝐶𝐶 to 𝐶𝐶′ so that △ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′.
How would you describe the length 𝐴𝐴′ 𝐵𝐵′ ?


𝐷𝐷𝐷𝐷
We will show there is a similarity transformation taking △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐷𝐷𝐷𝐷𝐷𝐷 that starts with a dilation. How can
we use what we know about the correspondence to express the scale factor 𝑟𝑟?


The correspondence implies that 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐷𝐷, 𝑚𝑚∠𝐵𝐵 = 𝑚𝑚∠𝐸𝐸, 𝑚𝑚∠𝐶𝐶 = 𝑚𝑚∠𝐹𝐹, and
𝐴𝐴′ 𝐵𝐵′ = 𝑟𝑟𝑟𝑟𝑟𝑟
We take this one step further and say that 𝐴𝐴′ 𝐵𝐵′ = 𝑟𝑟𝑟𝑟𝑟𝑟 =
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴 = 𝐷𝐷𝐷𝐷. Similarly, 𝐵𝐵′ 𝐶𝐶 ′ = 𝐸𝐸𝐸𝐸 and 𝐴𝐴′ 𝐶𝐶 ′ = 𝐷𝐷𝐷𝐷.
𝐴𝐴𝐴𝐴
So, all the sides and all the angles of △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ and △ 𝐷𝐷𝐷𝐷𝐷𝐷 match up, and the triangles are congruent.
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

Since they are congruent, a sequence of basic rigid motions takes △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ to △ 𝐷𝐷𝐷𝐷𝐷𝐷.
So, a dilation takes △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′, and a congruence transformation takes △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ to △ 𝐷𝐷𝐷𝐷𝐷𝐷, and we
conclude that △ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐷𝐷𝐷𝐷𝐷𝐷.
Example 5 (7 minutes)
The intent of Example 5 is to highlight how an efficient sequence of transformations can be found to map one figure
onto its similar image. There are many sequences that get the job done, but the goal here is to show the least number of
needed transformations to map one triangle onto the other.
Scaffolding:
Example 5
a.
In the diagram below, △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′. Describe a similarity transformation that
maps △ 𝑨𝑨𝑨𝑨𝑨𝑨 to △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′.
Example 5 is directed at more
proficiency with similarity
transformations. The question
poses an interesting
consideration but is not a
requirement. In general, time
and focus in this lesson should
be placed on the content
preceding Example 5.
Students most likely describe some sequence that involves a dilation, a reflection, a translation, and a rotation. Direct
them to part (b).
b.
Joel says the sequence must require a dilation and three rigid motions, but Sharon is sure there is a similarity
transformation composed of just a dilation and two rigid motions. Who is right?
Allow students to struggle with the question before revealing that it is, in fact, possible to describe a similarity composed
of a dilation and just two rigid motions shown in the following sequence:

Step 1: Dilate by a scale factor of 𝑟𝑟 and center 𝑂𝑂 so that (1) the resulting △ 𝐴𝐴𝐴𝐴𝐴𝐴 is congruent to
△ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ and (2) one pair of corresponding vertices coincide.
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
Step 2: Reflect the dilated triangle across ������
𝐵𝐵′𝐶𝐶′.

Step 3: Rotate the reflected triangle until it coincides with △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′.

We have found a similarity transformation that maps △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ with just one dilation and two rigid
Closing (1 minute)

What does it mean for similarity to be reflexive? Symmetric?
Students develop an informal argument for why similarity is transitive in the Problem Set.


Similarity is reflexive because a figure is similar to itself. Similarity is symmetric because once a
similarity transformation is determined to take a figure to another, there are inverse transformations
that can take the figure back to the original.
We have seen in our earlier work that if two figures (e.g., two triangles) are similar, then there exists a
correspondence such that corresponding lengths are proportional and corresponding angles are equal in
measurement. What is the converse of this statement? Is it true?
Lesson Summary
Similarity is reflexive because a figure is similar to itself.
Similarity is symmetric because once a similarity transformation is determined to take a figure to another, there are
inverse transformations that can take the figure back to the original.
Exit Ticket (5 minutes)
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Name
Date
Lesson 14: Similarity
Exit Ticket
1.
In the diagram, △ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐷𝐷𝐷𝐷𝐷𝐷 by the dilation with center 𝑂𝑂 and a scale factor of 𝑟𝑟. Explain why △ 𝐷𝐷𝐷𝐷𝐷𝐷~ △ 𝐴𝐴𝐴𝐴𝐴𝐴.
2.
���� and �𝑇𝑇𝑇𝑇
��� are parallel. Is circle 𝐶𝐶𝐶𝐶,𝐶𝐶𝐶𝐶 similar to circle 𝐶𝐶𝑇𝑇,𝑇𝑇𝑇𝑇 ? Explain.
3.
Two triangles, △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝐷𝐷𝐷𝐷𝐷𝐷, are in the plane so that 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐷𝐷, 𝑚𝑚∠𝐵𝐵 = 𝑚𝑚∠𝐸𝐸, 𝑚𝑚∠𝐶𝐶 = 𝑚𝑚∠𝐹𝐹, and
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
=
𝐸𝐸𝐸𝐸
𝐵𝐵𝐵𝐵
=
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
. Summarize the argument that proves that the triangles must be similar.
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Exit Ticket Sample Solutions
1.
In the diagram, △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑫𝑫𝑫𝑫𝑫𝑫 by the dilation with center 𝑶𝑶 and a scale factor of 𝒓𝒓. Explain why △ 𝑫𝑫𝑬𝑬𝑬𝑬~ △ 𝑨𝑨𝑨𝑨𝑨𝑨.
We know that △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫 by a dilation
with center 𝑶𝑶 and a scale factor of 𝒓𝒓. A
dilation with the same center 𝑶𝑶 but a scale
𝟏𝟏
factor of maps △ 𝑫𝑫𝑫𝑫𝑫𝑫 onto △ 𝑨𝑨𝑨𝑨𝑨𝑨; this
𝒓𝒓
means △ 𝑫𝑫𝑫𝑫𝑫𝑫~ △ 𝑨𝑨𝑨𝑨𝑨𝑨.
2.
𝑪𝑪𝑪𝑪 and ����
𝑻𝑻𝑻𝑻 are parallel. Is circle 𝑪𝑪,𝑪𝑪𝑪𝑪 similar to circle 𝑪𝑪𝑻𝑻,𝑻𝑻𝑻𝑻? Explain.
Yes, the circles are similar because a dilation with center 𝑶𝑶 and
a scale factor of 𝒓𝒓 exists that maps 𝑪𝑪,𝑪𝑪𝑪𝑪 onto 𝑪𝑪𝑻𝑻,𝑻𝑻𝑻𝑻.
3.
Two triangles, △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑫𝑫𝑫𝑫𝑫𝑫, are in the plane so that 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑫𝑫, 𝒎𝒎∠𝑩𝑩 = 𝒎𝒎∠𝑬𝑬, 𝒎𝒎∠𝑪𝑪 = 𝒎𝒎∠𝑭𝑭, and
𝑫𝑫𝑫𝑫
𝑨𝑨𝑨𝑨
=
𝑬𝑬𝑬𝑬
𝑩𝑩𝑩𝑩
=
𝑫𝑫𝑫𝑫
𝑨𝑨𝑨𝑨
. Summarize the argument that proves that the triangles must be similar.
A dilation exists such that the lengths of one triangle can be made equal to the lengths of the other triangle. Once
the triangles have lengths and angles of equal measurement, or are congruent, then a sequence of rigid motions
maps one triangle to the other. Therefore, a similarity transformation exists that maps one triangle onto the other,
and the triangles must be similar.
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Problem Set Sample Solutions
1.
If you are given any two congruent triangles, describe a sequence of basic rigid motions that takes one to the other.
Translate one triangle to the other by a vector �����⃗
𝑿𝑿𝑿𝑿 so that the triangles coincide at a vertex.
Case 1: If both triangles are of the same orientation, simply rotate about the common vertex until the triangles
coincide.
Case 2: If the triangles are of opposite orientations, reflect the one triangle over one of its two sides that include the
common vertex, and then rotate around the common vertex until the triangles coincide.
2.
If you are given two similar triangles that are not congruent triangles, describe a sequence of dilations and basic
rigid motions that takes one to the other.
Dilate one triangle with a center 𝑶𝑶 such that when the lengths of its sides are equal to the corresponding lengths of
the other triangle, one pair of corresponding vertices coincide. Then, follow one of the sequences described in Case 1
or Case 2 of Problem 1.
Students may have other similarity transformations that map one triangle to the other consisting of translations,
reflections, rotations, and dilations. For example, they might dilate one triangle until the corresponding lengths are
equal, then translate one triangle to the other by a vector �����⃗
𝑿𝑿𝑿𝑿 so that the triangles coincide at a vertex, and then
follow one of the sequences described in Case 1 or Case 2 of Problem 1.
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3.
Given two line segments, ����
𝑨𝑨𝑨𝑨 and ����
𝑪𝑪𝑪𝑪, of different lengths, answer the following questions:
���� to ����
a.
It is always possible to find a similarity transformation that maps 𝑨𝑨𝑨𝑨
𝑪𝑪𝑪𝑪 sending 𝑨𝑨 to 𝑪𝑪 and 𝑩𝑩 to 𝑫𝑫.
Describe one such similarity transformation.
���� with corresponding points 𝑨𝑨 and 𝑪𝑪 (and likewise 𝑩𝑩 and 𝑫𝑫) on the same
Rotate ����
𝑨𝑨𝑩𝑩 so that it is parallel to 𝑪𝑪𝑪𝑪
�����⃗ so that the midpoint of 𝑨𝑨𝑨𝑨
����
side of each line segment. Then, translate the image of ����
𝑨𝑨𝑨𝑨 by a vector 𝑿𝑿𝑿𝑿
𝑪𝑪𝑪𝑪. Finally, dilate ����
coincides with the midpoint of ����
𝑨𝑨𝑨𝑨 until the two segments are equal in length.
b.
���� and ����
If you are given that 𝑨𝑨𝑨𝑨
𝑪𝑪𝑪𝑪 are not parallel, are not congruent, do not share any points, and do not lie in
����? Which
���� to 𝑪𝑪𝑪𝑪
the same line, what is the fewest number of transformations needed in a sequence to map 𝑨𝑨𝑨𝑨
transformations make this work?
������ and ����
Rotate ����
𝑨𝑨𝑨𝑨 to ������
𝑨𝑨′𝑩𝑩′ so that 𝑨𝑨′𝑩𝑩′
𝑪𝑪𝑪𝑪 are parallel and oriented in the same direction; then, use the fact that
����.
there is a dilation that takes ������
𝑨𝑨′𝑩𝑩′ to 𝑪𝑪𝑪𝑪
c.
If you performed a similarity transformation that instead takes 𝑨𝑨 to 𝑫𝑫 and 𝑩𝑩 to 𝑪𝑪, either describe what
mistake was made in the similarity transformation, or describe what additional transformation is needed to
fix the error so that 𝑨𝑨 maps to 𝑪𝑪 and 𝑩𝑩 maps to 𝑫𝑫.
The rotation in the similarity transformation was not sufficient to orient the directed line segments in the
same direction, resulting in mismatched corresponding points. This error could be fixed by changing the
rotation such that the desired corresponding endpoints lie on the same end of each segment.
If the desired endpoints do not coincide after a similarity transformation, rotate the line segment about its
midpoint by 𝟏𝟏𝟏𝟏𝟏𝟏°.
4.
We claim that similarity is transitive (i.e., if 𝑨𝑨, 𝑩𝑩, and 𝑪𝑪 are figures in the plane such that 𝑨𝑨~𝑩𝑩 and 𝑩𝑩~𝑪𝑪, then 𝑨𝑨~𝑪𝑪).
Describe why this must be true.
If similarity transformation 𝑻𝑻𝟏𝟏 maps 𝑨𝑨 to 𝑩𝑩 and similarity transformation 𝑻𝑻𝟐𝟐 maps 𝑩𝑩 to 𝑪𝑪, then the composition of
basic rigid motions and dilations that takes 𝑨𝑨 to 𝑩𝑩 together with the composition of basic rigid motions and dilations
that takes 𝑩𝑩 to 𝑪𝑪 shows that 𝑨𝑨~𝑪𝑪.
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Lesson 14
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GEOMETRY
5.
6.
Given two line segments, ����
𝑨𝑨𝑨𝑨 and ����
𝑪𝑪𝑪𝑪, of different lengths, we have seen that it is always possible to find a similarity
����, sending 𝑨𝑨 to 𝑪𝑪 and 𝑩𝑩 to 𝑫𝑫 with one rotation and one dilation. Can you do this
transformation that maps ����
𝑨𝑨𝑨𝑨 to 𝑪𝑪𝑪𝑪
with one reflection and one dilation?
������ and ����
���� to ������
Yes. Reflect 𝑨𝑨𝑨𝑨
𝑨𝑨′𝑩𝑩′so that 𝑨𝑨′𝑩𝑩′
𝑪𝑪𝑪𝑪 are parallel, and then use the fact that there is a dilation that takes
������
����
𝑨𝑨′𝑩𝑩′ to 𝑪𝑪𝑪𝑪.
Given two triangles, △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑫𝑫𝑫𝑫𝑫𝑫, is it always possible to rotate △ 𝑨𝑨𝑨𝑨𝑨𝑨 so that the sides of △ 𝑨𝑨𝑨𝑨𝑨𝑨 are parallel
����)?
to the corresponding sides in △ 𝑫𝑫𝑫𝑫𝑫𝑫 (e.g., ����
𝑨𝑨𝑨𝑨 ∥ 𝑫𝑫𝑫𝑫
No, it is not always possible. Sometimes a
reflection is necessary. If you consider the
diagram to the right, △ 𝑫𝑫𝑫𝑫𝑫𝑫 can be rotated in
various ways such that ����
𝑫𝑫𝑫𝑫 either coincides with
����, where corresponding points
or is parallel to 𝑨𝑨𝑨𝑨
𝑨𝑨 and 𝑫𝑫 are located at the same end of the
segment. However, in each case, 𝑭𝑭 lies on one
side of ����
𝑫𝑫𝑫𝑫 but is opposite the side on which 𝑪𝑪 lies
with regard to ����
𝑨𝑨𝑨𝑨. This means that a reflection
is necessary to reorient the third vertex of the
triangle.
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Lesson 15: The Angle-Angle (AA) Criterion for Two Triangles
to Be Similar
Student Outcomes

Students prove the angle-angle criterion for two triangles to be similar and use it to solve triangle problems.
Lesson Notes
In Lesson 4, students learned that the triangle side splitter theorem is an important theorem because it is the central
ingredient in proving the AA criterion for similar triangles. In this lesson, this statement is substantiated by using the
theorem (in the form of the dilation theorem) to prove the AA criterion. The AA criterion is arguably one of the most
useful theorems for recognizing and proving that two triangles are similar. However, students may not need to accept
that this statement is true without justification: They get plenty of opportunities in the remaining lessons (and modules)
to see how useful the AA criterion is.
Classwork
Exercises 1–5 (10 minutes)
Exercises
1.
MP.5
Draw two triangles of different sizes with two pairs of equal angles. Then, measure the lengths of the corresponding
sides to verify that the ratio of their lengths is proportional. Use a ruler, compass, or protractor, as necessary.
Students’ work will vary. Verify that students have drawn a pair of triangles with two pairs of equal angles and
have shown via direct measurement that the ratios of corresponding side lengths are proportional.
2.
Are the triangles you drew in Exercise 1 similar? Explain.
Yes, the triangles are similar. The converse of the theorem on similar triangles states that when we have two
triangles △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′ with corresponding angles that are equal and corresponding side lengths that are
proportional, then the triangles are similar.
MP.3
3.
Why is it that you only need to construct triangles where two pairs of angles are equal but not three?
If we are given the measure of two angles of a triangle, then we also know the third measure because of the triangle
sum theorem. All three angles must add to 𝟏𝟏𝟏𝟏𝟏𝟏°, so showing two pairs of angles are equal in measure is just like
showing all three pairs of angles are equal in measure.
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4.
Why were the ratios of the corresponding sides proportional?
Since the triangles are similar, we know that there exists a similarity transformation that maps one triangle onto
another. Then, corresponding sides of similar triangles must be in proportion because of what we know about
similarity, dilation, and scale factor. Specifically, the length of a dilated segment is equal to the length of the
original segment multiplied by the scale factor. For example,
𝑨𝑨′ 𝑩𝑩′
𝑨𝑨𝑨𝑨
5.
= 𝒓𝒓 and 𝑨𝑨′ 𝑩𝑩′ = 𝒓𝒓 ⋅ 𝑨𝑨𝑨𝑨.
Do you think that what you observed will be true when you construct a pair of triangles with two pairs of equal
angles? Explain.
Accept any reasonable explanation. Use students’ responses as a springboard for the Opening discussion and the
presentation of the AA criterion for similarity.
Opening (4 minutes)
Debrief the work that students completed in Exercises 1–5 by having them share their responses to Exercise 5. Then,
continue the discussion with the points below and the presentation of the theorem.

Based on our understanding of similarity transformations, we know that we can show two figures in the plane
are similar by describing a sequence of dilations and rigid motions that would map one figure onto another.

Since a similarity implies the properties observed in Exercises 1–5 about corresponding side lengths and angle
measures, will it be necessary to show that all 6 conditions (3 sides and 3 angles) are met before concluding
that triangles are similar?
Provide time for students to discuss this question in small groups and to make conjectures about the answers. Consider
having students share their conjectures with the class.

Instead of having to check all 6 conditions, it would be nice to simplify our work by checking just two or three
of the conditions. Our work in Exercises 1–5 leads us to our next theorem:
THEOREM: Two triangles with two pairs of equal corresponding angles are similar. (This is known as the AA criterion for
similarity.)
Exercise 6 (4 minutes)
This exercise is optional and can be used if students require more time to explore whether two pairs of equal
corresponding angles can produce similar triangles.
6.
MP.5
Draw another two triangles of different sizes with two pairs of equal angles. Then, measure the lengths of the
corresponding sides to verify that the ratio of their lengths is proportional. Use a ruler, compass, or protractor, as
necessary.
Students’ work will vary. Verify that they have drawn a pair of triangles with two pairs of equal angles and have
shown via direct measurement that the ratios of corresponding side lengths are proportional.
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Discussion (9 minutes)

To prove the AA criterion, we need to show that two triangles with two pairs of
equal corresponding angles are in fact similar. To do so, we apply our
knowledge of both congruence and dilation.

Recall the ASA criterion for congruent triangles. If two triangles have two pairs
of equal angles and an included side equal in length, then the triangles are
congruent. The proof of the AA criterion for similarity is related to the ASA
criterion for congruence. Can you think of how they are related?
Scaffolding:
Prior to discussing the
relationship between ASA
criterion for congruence and
AA criterion for similarity, it
may be necessary for students
to review the congruence
criterion learned in Module 1.
Provide students time to discuss the relationship between ASA and AA in small groups.


The ASA criterion for congruence requires the included side to be equal in length between the two
figures. Since AA is for similarity, we would not expect the lengths to be equal in measure but more
likely proportional to the scale factor. Both ASA and AA criteria require two pairs of equal angles.
Given two triangles, △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝐷𝐷𝐷𝐷𝐷𝐷, where 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐷𝐷 and 𝑚𝑚∠𝐵𝐵 = 𝑚𝑚∠𝐸𝐸, show that △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐷𝐷𝐷𝐷𝐷𝐷.
Scaffolding:
 Consider using cardboard
cutouts (reproducible
available at the end of the
lesson) of the triangles as
manipulatives to make the
discussion of the proof less
abstract. Cutouts can be
given to small groups of
students or used only by
the teacher.

MP.2
If we can show that △ 𝐷𝐷𝐷𝐷𝐷𝐷 is congruent to a triangle that is a dilated version of
△ 𝐴𝐴𝐴𝐴𝐴𝐴, then we can describe the similarity transformation to prove that
△ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐷𝐷𝐷𝐷𝐷𝐷. To do so, what scale factor should we choose?


Since
We should let 𝑟𝑟 =
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
𝐷𝐷𝐷𝐷
.
𝐴𝐴𝐴𝐴
= 𝑟𝑟, we dilate triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴 from center 𝐴𝐴 by a scale factor of 𝑟𝑟 to
produce △ 𝐴𝐴𝐴𝐴′𝐶𝐶′ as shown below.
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to make their own cutouts.
Have students create
triangles with the same
two angles (e.g., 50° and
70°) and then compare
their triangles with their
neighbors’ triangles.
The Angle-Angle (AA) Criterion for Two Triangles to Be Similar
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
Have we constructed a triangle so that △ 𝐴𝐴𝐵𝐵′ 𝐶𝐶 ′ ≅ △ 𝐷𝐷𝐷𝐷𝐷𝐷? Explain. (Hint: Use ASA for congruence.)
Provide students time to discuss this question in small groups.

Proof using ASA for congruence:

Angle: 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐷𝐷. Given

Side: 𝐴𝐴𝐵𝐵′ = 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐷𝐷𝐷𝐷. The first equality is true because △ 𝐴𝐴𝐴𝐴′𝐶𝐶′ is a
dilation of △ 𝐴𝐴𝐴𝐴𝐴𝐴 by scale factor 𝑟𝑟. The second equality is true because
𝑟𝑟 is defined by 𝑟𝑟 =
MP.5


Scaffolding:
𝐷𝐷𝐷𝐷
.
𝐴𝐴𝐴𝐴
𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴 = 𝐷𝐷𝐷𝐷
𝐴𝐴𝐴𝐴
Angle: 𝑚𝑚∠𝐴𝐴𝐵𝐵′ 𝐶𝐶 ′ = 𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑚𝑚∠𝐸𝐸. By the dilation theorem,
⃖�������⃗
𝐵𝐵′ 𝐶𝐶 ′ ∥ ⃖����⃗
𝐵𝐵𝐵𝐵 . Therefore, 𝑚𝑚∠𝐴𝐴𝐵𝐵′ 𝐶𝐶 ′ = 𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 because corresponding
angles of parallel lines are equal in measure. Finally,
𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑚𝑚∠𝐷𝐷𝐷𝐷𝐷𝐷 is given.
𝐴𝐴𝐵𝐵′ = 𝑟𝑟𝑟𝑟𝑟𝑟 =
Ask a direct question: Why is
△ 𝐴𝐴𝐵𝐵′ 𝐶𝐶 ′ ≅ △ 𝐷𝐷𝐷𝐷𝐷𝐷? Then,
instruct students to prove the
triangles are congruent using
ASA. If necessary, ask students to
explain why 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐷𝐷, why
𝐴𝐴𝐵𝐵′ = 𝐷𝐷𝐷𝐷, and why
𝑚𝑚∠𝐴𝐴𝐵𝐵′ 𝐶𝐶 ′ = 𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑚𝑚∠𝐸𝐸.
Therefore, △ 𝐴𝐴𝐵𝐵′ 𝐶𝐶 ′ ≅ △ 𝐷𝐷𝐸𝐸𝐹𝐹 by ASA.

Now we have the following diagram:

Therefore, there is a composition of basic rigid motions that takes △ 𝐴𝐴𝐵𝐵′ 𝐶𝐶 ′ to △ 𝐷𝐷𝐷𝐷𝐷𝐷. Thus, a dilation with
scale factor 𝑟𝑟 composed with basic rigid motions takes △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐷𝐷𝐷𝐷𝐷𝐷. Since a similarity transformation
exists that maps △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐷𝐷𝐷𝐷𝐷𝐷, then △ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐷𝐷𝐷𝐷𝐷𝐷.
Exercises 7–10 (9 minutes)
In Exercises 7–10, students practice using the AA criterion to determine if two triangles are similar and then determine
unknown side lengths and/or angle measures of the triangles.
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7.
Are the triangles shown below similar? Explain. If the triangles are similar, identify any missing angle and sidelength measures.
28
Yes, the triangles are similar because they have two pairs of equal corresponding angles. By the AA criterion, they
must be similar. The angle measures and side lengths are shown below.
8.
Are the triangles shown below similar? Explain. If the triangles are similar, identify any missing angle and sidelength measures.
The triangles are not similar because they have just one pair of corresponding
equal angles. By the triangle sum theorem, 𝒎𝒎∠𝑪𝑪 = 𝟔𝟔𝟔𝟔° and 𝒎𝒎∠𝑫𝑫 = 𝟔𝟔𝟔𝟔°. Since
similar triangles must have equal corresponding angles, we can conclude that the
triangles shown are not similar.
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9.
The triangles shown below are similar. Use what you know about similar triangles to find the missing side lengths 𝒙𝒙
and 𝒚𝒚.
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏. 𝟓𝟓
=
𝒙𝒙
𝟒𝟒
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟔𝟔𝟔𝟔
𝒙𝒙 = 𝟓𝟓. 𝟓𝟓
𝒚𝒚
𝟏𝟏𝟏𝟏
=
𝟑𝟑. 𝟏𝟏𝟏𝟏
𝟒𝟒
𝟑𝟑𝟑𝟑. 𝟔𝟔𝟔𝟔 = 𝟒𝟒𝟒𝟒
𝟗𝟗. 𝟒𝟒𝟒𝟒 = 𝒚𝒚
Side length 𝒙𝒙 is 𝟓𝟓. 𝟓𝟓 units, and side
length 𝒚𝒚 is 𝟗𝟗. 𝟒𝟒𝟒𝟒 units.
10. The triangles shown below are similar. Write an explanation to a student, Claudia, of how to find the lengths of 𝒙𝒙
and 𝒚𝒚.
Claudia,
We are given that the triangles are similar; therefore, we know that they have equal corresponding angles and
corresponding sides that are equal in ratio. For that reason, we can write
corresponding sides of the two triangles. We can solve for 𝒚𝒚 as follows:
𝟑𝟑
=
𝟐𝟐+ 𝒚𝒚
𝟗𝟗
𝟑𝟑
𝟗𝟗
=
, which represents two pairs of
𝟐𝟐+𝒚𝒚
𝟔𝟔 + 𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏
𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏
𝒚𝒚 = 𝟒𝟒
We can solve for 𝒙𝒙 in a similar manner. We begin by writing two pairs of corresponding sides as equal ratios,
making sure that one of the ratios contains the length 𝒙𝒙.
𝟑𝟑
𝒙𝒙
=
𝟏𝟏𝟏𝟏 𝟗𝟗
𝟗𝟗𝟗𝟗 = 𝟑𝟑𝟑𝟑
𝒙𝒙 = 𝟒𝟒
Therefore, side length 𝒙𝒙 is 𝟒𝟒 units, and side length 𝒚𝒚 is 𝟒𝟒 units.
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Closing (4 minutes)

Explain what the AA criterion means.


It means that when two pairs of corresponding angles of two triangles are equal, the triangles are
similar.
Why is it enough to check only two conditions, two pairs of corresponding angles, as opposed to all six
conditions (3 angles and 3 sides), to conclude that a pair of triangles are similar?

We know that for two triangles, when two pairs of corresponding angles are equal and the included
corresponding sides are equal in length, the triangles are congruent. By the triangle sum theorem, we
can actually state that all three pairs of corresponding angles of the triangles are equal. Since a unique
triangle is formed by two fixed angles and a fixed included side length, the other two sides are also
fixed by the construction, meeting all six criteria. In the case of similarity, given two pairs of equal
angles, we would expect the lengths of the corresponding included sides to be equal in ratio to the scale
factor, again meeting all six conditions. For this reason, we can conclude that two triangles are similar
by verifying that two pairs of corresponding angles are equal.
Exit Ticket (5 minutes)
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Name
Date
Lesson 15: The Angle-Angle (AA) Criterion for Two Triangles to Be
Similar
Exit Ticket
1.
Given the diagram to the right, ����
𝑈𝑈𝑈𝑈 ⊥ �����
𝑉𝑉𝑉𝑉 , and �����
𝑊𝑊𝑊𝑊 ⊥ ����
𝑈𝑈𝑈𝑈 .
Show that △ 𝑈𝑈𝑈𝑈𝑈𝑈~ △ 𝑊𝑊𝑊𝑊𝑊𝑊.
2.
Given the diagram to the right and ����
𝐷𝐷𝐷𝐷 ∥ ����
𝐾𝐾𝐾𝐾, find 𝐹𝐹𝐹𝐹
and 𝐹𝐹𝐹𝐹.
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Exit Ticket Sample Solutions
1.
�� ⊥ 𝑽𝑽𝑽𝑽
����, and ����
����.
Given the diagram to the right, ��
𝑼𝑼𝑼𝑼
𝑾𝑾𝑾𝑾 ⊥ 𝑼𝑼𝑼𝑼
Show that △ 𝑼𝑼𝑼𝑼𝑼𝑼~ △ 𝑾𝑾𝑾𝑾𝑾𝑾.
By the definition of perpendicular lines, ∠𝑾𝑾𝑾𝑾𝑾𝑾 and ∠𝑼𝑼𝑼𝑼𝑼𝑼
are right angles, and all right angles are congruent, so
∠𝑾𝑾𝑾𝑾𝑾𝑾 ≅ ∠𝑼𝑼𝑼𝑼𝑼𝑼. Both △ 𝑼𝑼𝑼𝑼𝑼𝑼 and △ 𝑾𝑾𝑾𝑾𝑾𝑾 share ∠𝑽𝑽, and
by the reflexive property ∠𝑽𝑽 ≅ ∠𝑽𝑽. Therefore, by the 𝑨𝑨𝑨𝑨
criterion for proving similar triangles, △ 𝑼𝑼𝑼𝑼𝑼𝑼~ △ 𝑾𝑾𝑾𝑾𝑾𝑾.
2.
���� ∥ ���
Given the diagram to the right and 𝑫𝑫𝑫𝑫
𝑲𝑲𝑲𝑲,
find 𝑭𝑭𝑭𝑭 and 𝑭𝑭𝑭𝑭.
����, it follows
By alternate interior ∠’s, ����
𝑫𝑫𝑫𝑫 ∥ 𝑲𝑲𝑲𝑲
that ∠𝑲𝑲 ≅ ∠𝑬𝑬 (and by a similar argument,
∠𝑫𝑫 ≅ ∠𝑳𝑳). ∠𝑫𝑫𝑫𝑫𝑫𝑫 and ∠𝑲𝑲𝑲𝑲𝑲𝑲 are vertically
opposite angles and therefore congruent. By
AA criterion for proving similar triangles,
△ 𝑫𝑫𝑫𝑫𝑫𝑫~ △ 𝑳𝑳𝑳𝑳𝑳𝑳. Therefore,
𝑫𝑫𝑫𝑫 𝑬𝑬𝑬𝑬 𝑫𝑫𝑫𝑫
=
=
.
𝑳𝑳𝑳𝑳 𝑲𝑲𝑲𝑲 𝑳𝑳𝑳𝑳
𝑫𝑫𝑫𝑫 𝑬𝑬𝑬𝑬
=
𝑳𝑳𝑳𝑳 𝑲𝑲𝑲𝑲
𝟏𝟏𝟏𝟏 𝑬𝑬𝑬𝑬
=
𝟔𝟔
𝟒𝟒
𝟔𝟔𝟔𝟔𝟔𝟔 = 𝟔𝟔𝟔𝟔
𝑬𝑬𝑬𝑬 = 𝟏𝟏𝟏𝟏
𝑫𝑫𝑫𝑫 𝑫𝑫𝑫𝑫
=
𝑳𝑳𝑳𝑳 𝑳𝑳𝑳𝑳
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
=
𝟔𝟔
𝑳𝑳𝑳𝑳
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟕𝟕𝟕𝟕
𝑳𝑳𝑳𝑳 = 𝟒𝟒
𝟒𝟒
𝟓𝟓
𝑳𝑳𝑳𝑳 = 𝟒𝟒. 𝟖𝟖
Using the relationship of similar triangles, 𝑬𝑬𝑬𝑬 = 𝟏𝟏𝟏𝟏, and 𝑳𝑳𝑳𝑳 = 𝟒𝟒. 𝟖𝟖.
Problem Set Sample Solutions
1.
In the figure to the right, △ 𝑳𝑳𝑳𝑳𝑳𝑳~ △ 𝑴𝑴𝑴𝑴𝑴𝑴.
a.
Classify △ 𝑳𝑳𝑳𝑳𝑳𝑳 based on what you know about similar triangles, and justify your reasoning.
By the given similarity statement, 𝑴𝑴 and 𝑷𝑷 are corresponding vertices; therefore, the angles at 𝑴𝑴 and 𝑷𝑷 must
be congruent. This means that △ 𝑳𝑳𝑳𝑳𝑳𝑳 is an isosceles triangle by base ∠'s converse.
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b.
If 𝒎𝒎∠𝑷𝑷 = 𝟐𝟐𝟐𝟐°, find the remaining angles in the diagram.
𝒎𝒎∠𝑴𝑴 = 𝟐𝟐𝟐𝟐°, 𝒎𝒎∠𝑴𝑴𝑴𝑴𝑴𝑴 = 𝟐𝟐𝟐𝟐°, 𝒎𝒎∠𝑴𝑴𝑴𝑴𝑴𝑴 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝒎𝒎∠𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝒎𝒎∠𝑴𝑴𝑴𝑴𝑴𝑴 = 𝟏𝟏𝟏𝟏𝟏𝟏°, and 𝒎𝒎∠𝑳𝑳𝑳𝑳𝑳𝑳 = 𝟒𝟒𝟒𝟒°.
Triangle 𝑴𝑴𝑴𝑴𝑴𝑴 is also isosceles.
2.
In the diagram below, △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑨𝑨𝑨𝑨𝑨𝑨. Determine whether the following statements must be true from the given
information, and explain why.
a.
△ 𝑪𝑪𝑪𝑪𝑪𝑪~ △ 𝑫𝑫𝑫𝑫𝑫𝑫
This statement is true because
corresponding vertices are the
same as in the given similarity
statement but are listed in a
different order.
b.
△ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑪𝑪𝑪𝑪𝑪𝑪
There is no information given to draw this conclusion.
c.
△ 𝑩𝑩𝑩𝑩𝑩𝑩~ △ 𝑨𝑨𝑨𝑨𝑨𝑨
There is no information given to draw this conclusion.
d.
△ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑨𝑨𝑨𝑨𝑨𝑨
This statement is true because corresponding vertices are the same as in the given similarity statement but
are listed in a different order.
3.
����. Prove that
In the diagram below, 𝑫𝑫 is the midpoint of ����
𝑨𝑨𝑨𝑨, 𝑭𝑭 is the midpoint of ����
𝑩𝑩𝑩𝑩, and 𝑬𝑬 is the midpoint of 𝑨𝑨𝑨𝑨
△ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑭𝑭𝑭𝑭𝑭𝑭.
Using the triangle side splitter theorem, since 𝑫𝑫, 𝑭𝑭, and 𝑬𝑬 are all midpoints of the sides of △ 𝑨𝑨𝑨𝑨𝑨𝑨, the sides are cut
����, ����
����, and ����
���� ∥ 𝑨𝑨𝑨𝑨
����. This provides multiple pairs of parallel lines with
proportionally; therefore, 𝑫𝑫𝑫𝑫
𝑫𝑫𝑫𝑫 ∥ 𝑩𝑩𝑩𝑩
𝑬𝑬𝑬𝑬 ∥ 𝑨𝑨𝑨𝑨
parallel transversals.
����, and ∠𝑫𝑫𝑫𝑫𝑫𝑫 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 by alternate interior ∠'s, ����
����, so by
∠𝑨𝑨 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 by corresponding ∠'s, ����
𝑫𝑫𝑫𝑫 ∥ 𝑨𝑨𝑨𝑨
𝑬𝑬𝑬𝑬 ∥ 𝑨𝑨𝑨𝑨
transitivity, ∠𝑨𝑨 ≅ ∠𝑫𝑫𝑫𝑫𝑫𝑫.
����, and ∠𝑬𝑬𝑬𝑬𝑬𝑬 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 by alternate interior ∠'s, ����
����, so by
∠𝑪𝑪 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 by corresponding ∠'s, ����
𝑫𝑫𝑫𝑫 ∥ 𝑨𝑨𝑨𝑨
𝑫𝑫𝑫𝑫 ∥ 𝑩𝑩𝑩𝑩
transitivity, ∠𝑪𝑪 ≅ ∠𝑬𝑬𝑬𝑬𝑬𝑬.
△ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑭𝑭𝑭𝑭𝑭𝑭 by the AA criterion for proving similar triangles.
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4.
Use the diagram below to answer each part.
a.
���� ∥ 𝑬𝑬𝑬𝑬
����, ����
����, and ����
����, prove that the triangles are similar.
If 𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨 ∥ 𝑬𝑬𝑬𝑬
𝑪𝑪𝑪𝑪 ∥ 𝑫𝑫𝑫𝑫
By extending all sides of both triangles, there are several pairs of parallel lines cut by parallel transversals.
Using corresponding angles within parallel lines and transitivity, △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑬𝑬𝑬𝑬𝑬𝑬 by the AA criterion for
proving similar triangles.
b.
The triangles are not congruent. Find the dilation that takes one to the other.
Extend lines joining corresponding vertices to find their intersection 𝑶𝑶, which is the center of dilation.
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5.
����, prove that △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫.
Given trapezoid 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨, and ����
𝑨𝑨𝑨𝑨 ∥ 𝑬𝑬𝑬𝑬
���� (by the same argument,
From the given information, ∠𝑬𝑬𝑬𝑬𝑬𝑬 ≅ ∠𝑫𝑫𝑫𝑫𝑫𝑫 by alternate interior ∠'s, ����
𝑨𝑨𝑨𝑨 ∥ 𝑬𝑬𝑬𝑬
∠𝑫𝑫𝑫𝑫𝑫𝑫 ≅ ∠𝑨𝑨𝑨𝑨𝑨𝑨). Furthermore, ∠𝑬𝑬𝑬𝑬𝑬𝑬 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 because vertical angles are congruent. Therefore, △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑫𝑫𝑫𝑫𝑫𝑫
by the AA criterion for proving similar triangles.
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Cutouts to use for in-class discussion:
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Lesson 16
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GEOMETRY
Lesson 16: Between-Figure and Within-Figure Ratios
Student Outcomes

Students indirectly solve for measurements involving right triangles using scale factors, ratios between similar
figures, and ratios within similar figures.

Students use trigonometric ratios to solve applied problems.
Lesson Notes
By now, students are very familiar with how to use a scale factor with similar figures to determine unknown lengths of
figures. This lesson demonstrates how the values of the ratios of corresponding sides between figures can be rewritten,
equivalently, as ratios of corresponding lengths within figures. This work foreshadows ratios related to trigonometry.
Though sine, cosine, and tangent are not formally defined in this lesson, students are essentially using the ratios as a
premise for formal treatment of trigonometric ratios in Topic E.
Classwork
Opening Exercise (2 minutes)
Opening Exercise
At a certain time of day, a 𝟏𝟏𝟏𝟏 𝐦𝐦 flagpole casts an 𝟖𝟖 𝐦𝐦 shadow. Write an equation that would allow you to find the
height, 𝒉𝒉, of the tree that uses the length, 𝒔𝒔, of
𝟏𝟏𝟏𝟏 𝒉𝒉
=
𝒔𝒔
𝟖𝟖
𝟑𝟑
𝒔𝒔 = 𝒉𝒉
𝟐𝟐
OR
𝟏𝟏𝟏𝟏 𝟖𝟖
=
𝒔𝒔
𝒉𝒉
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟖𝟖𝟖𝟖
𝟑𝟑
𝒔𝒔 = 𝒉𝒉
𝟐𝟐
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Example 1 (14 minutes)
The discussion following this exercise highlights length relationships within figures. Many students rely on ratios of
lengths between figures to solve for unknown measurements of sides rather than use ratios of lengths within a figure.
This example shows students that comparing ratios between and within figures is mathematically equivalent.
Example 1
Given △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑨𝑨′𝑩𝑩′𝑪𝑪′, find the missing side lengths.


����
(Take a poll.) How many people used the ratio of lengths 4: 12 or 12: 4 (i.e., the corresponding lengths of 𝐴𝐴𝐴𝐴
������?
���� and 𝐴𝐴′𝐵𝐵′
and �����
𝐴𝐴′𝐶𝐶′) to determine the measurements of 𝐵𝐵𝐵𝐵
Are there any other length relationships we could use to set up an equation and solve for the missing side
lengths?
Provide students time to think about and try different equations that would lead to the same answer.

MP.7


𝐴𝐴′ 𝐵𝐵′ = 15.
���� , we can use the ratios 𝐴𝐴′ 𝐶𝐶 ′ : 𝐵𝐵′𝐶𝐶′ and 𝐴𝐴𝐴𝐴: 𝐵𝐵𝐵𝐵. Equating the values of the
To find the length of 𝐵𝐵𝐵𝐵
ratios, we get the following:
4
12
=
𝐵𝐵𝐵𝐵
6
𝐵𝐵𝐵𝐵 = 2.
Why is it possible to use this length relationship, 𝐴𝐴𝐴𝐴: 𝐴𝐴𝐴𝐴, to solve for a missing side length?


In addition to the ratio of corresponding lengths 𝐴𝐴𝐴𝐴: 𝐴𝐴′𝐶𝐶′, we can use a ratio of lengths within one of
the triangles. For example, we could use the ratio of 𝐴𝐴𝐴𝐴: 𝐴𝐴𝐴𝐴 to find the length of ������
𝐴𝐴′𝐵𝐵′. Equating the
values of the ratios, we get the following:
4
12
=
5 𝐴𝐴′𝐵𝐵′
Take student responses, and then offer the following explanation:
There are three methods that we can use to determine the missing side lengths.
Method 1 (Scale Factor): Find the scale factor, and use it to compute for the desired side lengths.
Since we know 𝐴𝐴𝐴𝐴 = 4 and the corresponding side 𝐴𝐴′ 𝐶𝐶 ′ = 12, the scale factor of 𝑟𝑟 satisfies 4𝑟𝑟 = 12. So,
𝑟𝑟 = 3, 𝐴𝐴′ 𝐵𝐵′ = 𝑟𝑟𝑟𝑟𝑟𝑟 = 3 ⋅ 5 = 15, and 𝐵𝐵𝐵𝐵 =
𝐵𝐵′𝐶𝐶′ 6
= = 2.
𝑟𝑟
3
Method 2 (Ratios Between-Figures): Equate the values of the ratios of the corresponding sides.
𝐴𝐴′𝐵𝐵′
𝐴𝐴𝐴𝐴
=
𝐴𝐴′𝐶𝐶 ′
𝐴𝐴𝐴𝐴
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implies
𝐴𝐴′ 𝐵𝐵′
5
=
12
4
and 𝐴𝐴′ 𝐵𝐵′ = 15.
𝐵𝐵𝐵𝐵
𝐵𝐵′ 𝐶𝐶 ′
=
𝐴𝐴𝐴𝐴
𝐴𝐴′ 𝐶𝐶 ′
implies
𝐵𝐵𝐵𝐵
6
=
4
12
and 𝐵𝐵𝐵𝐵 = 2.
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Method 3 (Ratios Within-Figures): Equate the values of the ratios within each triangle.
𝐴𝐴′ 𝐵𝐵′

𝐴𝐴′𝐶𝐶 ′
=
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
implies
𝐴𝐴′𝐵𝐵′
12
=
5
4
and 𝐴𝐴′ 𝐵𝐵′ = 15.
𝐵𝐵𝐵𝐵
𝐴𝐴𝐴𝐴
=
𝐵𝐵′ 𝐶𝐶 ′
𝐴𝐴′ 𝐶𝐶 ′
implies
𝐵𝐵𝐵𝐵
4
=
6
12
and 𝐵𝐵𝐵𝐵 = 2.
Why does Method 3 work? That is, why can we use values of ratios within each triangle to find the missing
side lengths?
Provide students time to explain why Method 3 works, and then offer the following explanation:

If △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′, then it is true that the ratio between any two side lengths of the first triangle is equal to
the ratio between the corresponding side lengths of the second triangle; for example, 𝐴𝐴𝐴𝐴: 𝐵𝐵𝐵𝐵 = 𝐴𝐴′𝐵𝐵′: 𝐵𝐵′𝐶𝐶′.
This is because if 𝑟𝑟 is the scale factor, then 𝐴𝐴𝐴𝐴: 𝐵𝐵𝐵𝐵 = 𝑟𝑟𝑟𝑟𝑟𝑟: 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐴𝐴′ 𝐵𝐵′ : 𝐵𝐵′𝐶𝐶′.
Revisit the Opening Exercise. Ask students to explain which method they used to write their equations for the height of
the tree. Answers may vary, but students likely have used Method 2 because that is what is most familiar to them.
Example 2 (7 minutes)
In this example, students use what they know about similar figures to indirectly measure the height of a building. If
possible, use an image from the school or community to personalize the example.

Suppose you want to know the height of a building. Would it make sense to climb to the roof and use a tape
measure? Possibly. Suppose you want to know the distance between the earth and the moon. Would a tape
measure work in that situation? Not likely. For now, we will take indirect measurements of trees and
buildings, but we will soon learn how the Greeks measured the distance from the earth to the moon.
Example 2
In the diagram above, a large flagpole stands outside of an office building. Marquis realizes that when he looks up from
the ground 𝟔𝟔𝟔𝟔 𝐦𝐦 away from the flagpole, the top of the flagpole and the top of the building line up. If the flagpole is
𝟑𝟑𝟑𝟑 𝐦𝐦 tall and Marquis is 𝟏𝟏𝟕𝟕𝟎𝟎 𝐦𝐦 from the building, how tall is the building?
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a.
Are the triangles in the diagram similar? Explain.
Yes, the triangle formed by Marquis, the ground, and the flagpole is similar to the triangle formed by
Marquis, the ground, and the building. They are similar by the AA criterion. Each of the triangles has a
common angle with Marquis at the vertex, and each triangle has a right angle, where the flagpole and the
building meet the ground.
b.
Determine the height of the building using what you know about scale factors.
The scale factor is
c.
𝟏𝟏𝟏𝟏𝟏𝟏
𝟔𝟔𝟔𝟔
=
𝟏𝟏𝟏𝟏
𝟔𝟔
�. Then, the height of the building is 𝟑𝟑𝟑𝟑(𝟐𝟐. 𝟖𝟖𝟑𝟑�) = 𝟗𝟗𝟗𝟗. 𝟏𝟏𝟔𝟔� 𝐦𝐦.
= 𝟐𝟐. 𝟖𝟖𝟑𝟑
Determine the height of the building using ratios between similar figures.
𝟑𝟑𝟑𝟑
𝟔𝟔𝟔𝟔
=
𝒉𝒉
𝟏𝟏𝟏𝟏𝟏𝟏
𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 = 𝟔𝟔𝟔𝟔𝟔𝟔
�
𝒉𝒉 = 𝟗𝟗𝟗𝟗. 𝟏𝟏𝟔𝟔
d.
Determine the height of the building using ratios within similar figures.
𝟑𝟑𝟑𝟑
𝒉𝒉
=
𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏
𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 = 𝟔𝟔𝟔𝟔𝟔𝟔
�
𝒉𝒉 = 𝟗𝟗𝟗𝟗. 𝟏𝟏𝟔𝟔
Example 3 (12 minutes)
Work through the following example with students. The goal is for them to recognize how the value of the ratio within a
figure can be used to determine an unknown side length of a similar figure. Make clear to students that more than one
ratio (from parts (a)–(c)) can be used to determine the requested unknown lengths in parts (d)–(f).
Example 3
The following right triangles are similar. We will determine the unknown side lengths by using
ratios within the first triangle. For each of the triangles below, we define the base as the
horizontal length of the triangle and the height as the vertical length.
Scaffolding:
 It may benefit some
groups of students to
show only the 3– 4– 5
triangle as they complete
parts (a)–(c) of the
example.
consider why the triangle
is called a 3– 4– 5 triangle
and determine the
significance of these
numbers to any triangle
that is similar to it.
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a.
b.
c.
Write and find the value of the ratio that compares the height to the hypotenuse of the leftmost triangle.
𝟒𝟒
= 𝟎𝟎. 𝟖𝟖
𝟓𝟓
Write and find the value of the ratio that compares the base to the hypotenuse of the leftmost triangle.
𝟑𝟑
= 𝟎𝟎. 𝟔𝟔
𝟓𝟓
Write and find the value of the ratio that compares the height to the base of the leftmost triangle.
𝟒𝟒
�
= 𝟏𝟏. 𝟑𝟑
𝟑𝟑
d.
Use the triangle with lengths 𝟑𝟑– 𝟒𝟒– 𝟓𝟓 and triangle 𝑨𝑨 to answer the following questions:
i.
Which ratio can be used to determine the height of triangle 𝑨𝑨?
The ratio that compares the height to the base can be used to determine the height of triangle 𝑨𝑨.
ii.
Which ratio can be used to determine the hypotenuse of triangle 𝑨𝑨?
The ratio that compares the base to the hypotenuse can be used to determine the hypotenuse of
triangle 𝑨𝑨.
iii.
Find the unknown lengths of triangle 𝑨𝑨.
Let 𝒉𝒉 represent the height of the triangle; then
𝒉𝒉
�
= 𝟏𝟏. 𝟑𝟑
𝟏𝟏. 𝟓𝟓
𝒉𝒉 = 𝟐𝟐.
Let 𝒄𝒄 represent the length of the hypotenuse; then
𝟏𝟏. 𝟓𝟓
= 𝟎𝟎. 𝟔𝟔
𝒄𝒄
𝒄𝒄 = 𝟐𝟐. 𝟓𝟓.
e.
Use the triangle with lengths 𝟑𝟑– 𝟒𝟒– 𝟓𝟓 and triangle 𝑩𝑩 to answer the following questions:
i.
Which ratio can be used to determine the base of triangle 𝑩𝑩?
The ratio that compares the height to the base can be used to determine the base of triangle 𝑩𝑩.
ii.
Which ratio can be used to determine the hypotenuse of triangle 𝑩𝑩?
The ratio that compares the height to the hypotenuse can be used to determine the hypotenuse of
triangle 𝑩𝑩.
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iii.
Find the unknown lengths of triangle 𝑩𝑩.
Let 𝒃𝒃 represent the length of the base; then
𝟓𝟓
𝟏𝟏
= 𝟏𝟏
𝒃𝒃
𝟑𝟑
𝒃𝒃 = 𝟑𝟑. 𝟕𝟕𝟕𝟕.
Let 𝒄𝒄 represent the length of the hypotenuse; then
𝟓𝟓
= 𝟎𝟎. 𝟖𝟖
𝒄𝒄
𝒄𝒄 = 𝟔𝟔. 𝟐𝟐𝟐𝟐.
f.
Use the triangle with lengths 𝟑𝟑– 𝟒𝟒– 𝟓𝟓 and triangle 𝑪𝑪 to answer the following questions:
i.
Which ratio can be used to determine the height of triangle 𝑪𝑪?
The ratio that compares the height to the hypotenuse can be used to determine the height of triangle 𝑪𝑪.
ii.
Which ratio can be used to determine the base of triangle 𝑪𝑪?
The ratio that compares the base to the hypotenuse can be used to determine the base of triangle 𝑪𝑪.
iii.
Find the unknown lengths of triangle 𝑪𝑪.
Let 𝒉𝒉 represent the height of the triangle; then
𝒉𝒉
= 𝟎𝟎. 𝟖𝟖
𝟖𝟖
𝒉𝒉 = 𝟔𝟔. 𝟒𝟒.
Let 𝒃𝒃 represent the length of the base; then
𝒃𝒃
= 𝟎𝟎. 𝟔𝟔
𝟖𝟖
𝒃𝒃 = 𝟒𝟒. 𝟖𝟖.
g.
Explain the relationship of the ratio of the corresponding sides within a figure to the ratio of corresponding
sides within a similar figure.
Corresponding lengths of similar figures have proportional lengths, but the ratio of two lengths within a
figure is equal to the corresponding ratio of two lengths in a similar figure.
h.
How does the relationship you noted in part (g) allow you to determine the length of an unknown side of a
triangle?
This relationship allows us to find the length of an unknown side of a triangle. We can write the ratio of
corresponding sides within the figure that contains the unknown side length. This ratio will be equal to the
value of the ratio of corresponding sides where the lengths are known. Since the ratios are equal, the
unknown length can be found by solving a simple equation.
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Closing (5 minutes)
To close the lesson, the following three questions may be asked separately. Or let students complete a Quick Write for
all three prompts, and then discuss the three questions all at one time.

What does it mean to use between-figure ratios of corresponding sides of similar triangles?


What does it mean to use within-figure ratios of corresponding sides of similar triangles?


Between-figure ratios are those where each ratio compares corresponding side lengths between two
similar figures. That is, one number of the ratio comes from one triangle while the other number in the
ratio comes from a different, similar triangle.
Within-figure ratios are those where each ratio is composed of side lengths from one figure of the
similar figures. That is, one ratio contains numbers that represent the side lengths from one triangle,
and a second ratio contains numbers that represent the side lengths from a second, similar triangle.
How can within-figure ratios be used to find unknown side lengths of similar triangles?

If a triangle is similar to another, then the within-figure ratios can be used to find the unknown lengths
of the other triangle. For example, if the ratio that compares the height to the hypotenuse is known for
one triangle and that ratio is equal to 0.6, then a ratio that compares the height to the hypotenuse of a
similar triangle will also be equal to 0.6.
Exit Ticket (5 minutes)
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Name
Date
Lesson 16: Between-Figure and Within-Figure Ratios
Exit Ticket
Dennis needs to fix a leaky roof on his house but does not own a ladder. He thinks that a 25 ft. ladder will be long
enough to reach the roof, but he needs to be sure before he spends the money to buy one. He chooses a point 𝑃𝑃 on the
ground where he can visually align the roof of his car with the edge of the house roof. Help Dennis determine if a 25 ft.
ladder will be long enough for him to safely reach his roof.
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Exit Ticket Sample Solutions
Dennis needs to fix a leaky roof on his house but does not own a ladder. He thinks that a 𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟. ladder will be long
enough to reach the roof, but he needs to be sure before he spends the money to buy one. He chooses a point 𝑷𝑷 on the
ground where he can visually align the roof of his car with the edge of the house roof. Help Dennis determine if a 𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟.
ladder will be long enough for him to safely reach his roof.
The height of the edge of the roof from the ground is
unknown, so let 𝒙𝒙 represent the distance from the ground
to the edge of the roof. The nested triangles are similar
triangles by the AA criterion for similar triangles because
they share ∠𝑷𝑷, and the height of the car and the distance
from the ground to the edge of the house roof are both
measured perpendicular to the ground. Therefore, the
following is true:
(𝟖𝟖. 𝟓𝟓 + 𝟐𝟐𝟐𝟐 )
𝟖𝟖. 𝟓𝟓
=
𝒙𝒙
𝟒𝟒. 𝟐𝟐𝟐𝟐
𝟖𝟖. 𝟓𝟓𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖
𝒙𝒙 = 𝟏𝟏𝟏𝟏. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 …
𝒙𝒙 ≈ 𝟏𝟏𝟏𝟏. 𝟕𝟕𝟕𝟕.
The distance from the ground to the edge of the roof is 𝟏𝟏𝟏𝟏. 𝟕𝟕𝟕𝟕 𝐟𝐟𝐟𝐟. (or 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟. 𝟗𝟗 𝐢𝐢𝐢𝐢.) The 𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟. ladder is clearly long enough
for Dennis to safely reach his roof.
Problem Set Sample Solutions
1.
△ 𝑫𝑫𝑫𝑫𝑫𝑫 ~ △ 𝑨𝑨𝑨𝑨𝑨𝑨 All side length measurements are in centimeters. Use between-figure ratios and/or within-figure
ratios to determine the unknown side lengths.
Using the given similarity statement, ∠𝑫𝑫 corresponds with ∠𝑨𝑨, and
����,
∠𝑪𝑪 corresponds with ∠𝑭𝑭, so it follows that ����
𝑨𝑨𝑨𝑨 corresponds with 𝑫𝑫𝑫𝑫
���� with ����
𝑫𝑫𝑫𝑫, and ����
𝑩𝑩𝑩𝑩 with ����
𝑬𝑬𝑬𝑬.
𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨 𝑫𝑫𝑫𝑫
=
𝑩𝑩𝑩𝑩 𝑬𝑬𝑬𝑬
𝟐𝟐. 𝟓𝟓 𝟑𝟑. 𝟕𝟕𝟕𝟕
=
𝑬𝑬𝑬𝑬
𝟏𝟏
𝟐𝟐. 𝟓𝟓(𝑬𝑬𝑬𝑬) = 𝟑𝟑. 𝟕𝟕𝟕𝟕
𝑬𝑬𝑬𝑬 = 𝟏𝟏. 𝟓𝟓
2.
𝑨𝑨𝑨𝑨 𝑨𝑨𝑨𝑨
=
𝑫𝑫𝑫𝑫 𝑫𝑫𝑫𝑫
𝑨𝑨𝑨𝑨
𝟐𝟐. 𝟓𝟓
=
𝟑𝟑. 𝟕𝟕𝟕𝟕 𝟒𝟒. 𝟓𝟓
𝟑𝟑. 𝟕𝟕𝟕𝟕(𝑨𝑨𝑨𝑨) = 𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐
𝑨𝑨𝑨𝑨 = 𝟑𝟑
Given △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑿𝑿𝑿𝑿𝑿𝑿, answer the following questions:
a.
b.
Write and find the value of the ratio that compares
the height ����
𝑨𝑨𝑨𝑨 to the hypotenuse of △ 𝑨𝑨𝑨𝑨𝑨𝑨.
𝟓𝟓
𝟏𝟏𝟏𝟏
Write and find the value of the ratio that compares
���� to the hypotenuse of △ 𝑨𝑨𝑨𝑨𝑨𝑨.
the base 𝑨𝑨𝑨𝑨
𝟏𝟏𝟐𝟐
𝟏𝟏𝟏𝟏
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c.
d.
Write and find the value of the ratio that compares the height 𝑨𝑨𝑨𝑨 to the base 𝑨𝑨𝑨𝑨 of △ 𝑨𝑨𝑨𝑨𝑨𝑨.
𝟓𝟓
𝟏𝟏𝟏𝟏
Use within-figure ratios to find the corresponding height of △ 𝑿𝑿𝑿𝑿𝑿𝑿.
𝟓𝟓
𝑿𝑿𝑿𝑿
=
𝟏𝟏𝟏𝟏 𝟒𝟒. 𝟓𝟓
𝟕𝟕
𝑿𝑿𝑿𝑿 = 𝟏𝟏
𝟖𝟖
e.
Use within-figure ratios to find the hypotenuse of △ 𝑿𝑿𝑿𝑿𝑿𝑿.
𝟏𝟏𝟏𝟏 𝟒𝟒. 𝟓𝟓
=
𝟏𝟏𝟏𝟏 𝒀𝒀𝒀𝒀
𝟕𝟕
𝒀𝒀𝒀𝒀 = 𝟒𝟒
𝟖𝟖
3.
Right triangles 𝑨𝑨, 𝑩𝑩, 𝑪𝑪, and 𝑫𝑫 are similar. Determine the unknown side lengths of each triangle by using ratios of
side lengths within triangle 𝑨𝑨.
C
A
a.
b.
c.
B
D
Write and find the value of the ratio that compares the height to the hypotenuse of triangle 𝑨𝑨.
√𝟑𝟑
≈ 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖
𝟐𝟐
Write and find the value of the ratio that compares the base to the hypotenuse of triangle 𝑨𝑨.
𝟏𝟏
= 𝟎𝟎. 𝟓𝟓
𝟐𝟐
Write and find the value of the ratio that compares the height to the base of triangle 𝑨𝑨.
√𝟑𝟑
= √𝟑𝟑 ≈ 𝟏𝟏. 𝟕𝟕𝟕𝟕
𝟏𝟏
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d.
Which ratio can be used to determine the height of triangle 𝑩𝑩? Find the height of triangle 𝑩𝑩.
The ratio that compares the height to the hypotenuse of triangle 𝑨𝑨 can be used to determine the height of
triangle 𝑩𝑩. Let the unknown height of triangle 𝑩𝑩 be 𝒎𝒎.
√𝟑𝟑 𝒎𝒎
=
𝟐𝟐
𝟏𝟏
𝒎𝒎 =
e.
√𝟑𝟑
𝟐𝟐
Which ratio can be used to determine the base of triangle 𝑩𝑩? Find the base of triangle 𝑩𝑩.
The ratio that compares the base to the hypotenuse of triangle 𝑨𝑨 can be used to determine the base of
triangle 𝑩𝑩. Let the unknown base of triangle 𝑩𝑩 be 𝒏𝒏.
f.
Find the unknown lengths of triangle 𝑪𝑪.
𝟏𝟏 𝒏𝒏
=
𝟐𝟐 𝟏𝟏
𝟏𝟏
𝒏𝒏 =
𝟐𝟐
The base of triangle 𝑪𝑪 is √𝟑𝟑.
The height of triangle 𝑪𝑪 is 𝟑𝟑.
g.
Find the unknown lengths of triangle 𝑫𝑫.
The base of triangle 𝑫𝑫 is
√𝟑𝟑
.
𝟑𝟑
The hypotenuse of triangle 𝑫𝑫 is
h.
𝟐𝟐√𝟑𝟑
𝟑𝟑
.
Triangle 𝑬𝑬 is also similar to triangles 𝑨𝑨, 𝑩𝑩, 𝑪𝑪, and 𝑫𝑫. Find the lengths of the missing sides in terms of 𝒙𝒙.
The base of triangle 𝑬𝑬 is 𝒙𝒙.
The height of triangle 𝑬𝑬 is 𝒙𝒙√𝟑𝟑.
The hypotenuse of triangle 𝑬𝑬 is 𝟐𝟐𝟐𝟐.
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4.
Brian is photographing the Washington Monument and wonders how tall the monument is. Brian places his
𝟓𝟓 𝐟𝐟𝐟𝐟. camera tripod approximately 𝟏𝟏𝟏𝟏𝟏𝟏 𝐲𝐲𝐲𝐲. from the base of the monument. Lying on the ground, he visually aligns
the top of his tripod with the top of the monument and marks his location on the ground approximately
𝟐𝟐 𝐟𝐟𝐟𝐟. 𝟗𝟗 𝐢𝐢𝐢𝐢. from the center of his tripod. Use Brian’s measurements to approximate the height of the Washington
Monument.
Brian’s location on the ground is approximately
𝟑𝟑𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕𝟕 𝐟𝐟𝐟𝐟. from the base of the monument. His visual
line forms two similar right triangles with the height of
the monument and the height of his camera tripod.
𝒉𝒉
𝟓𝟓
=
𝟐𝟐. 𝟕𝟕𝟕𝟕 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕𝟕
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟕𝟕𝟕𝟕 = 𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕
𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓 ≈ 𝒉𝒉
According to Brian’s measurements, the height of the
Washington Monument is approximately 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓 𝐟𝐟𝐟𝐟.
Students may want to check the accuracy of this
problem. The actual height of the Washington
Monument is 𝟓𝟓𝟓𝟓𝟓𝟓 𝐟𝐟𝐟𝐟.
5.
Catarina’s boat has come untied and floated away on the lake. She is standing atop a cliff that is 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. above the
water in a lake. If she stands 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟. from the edge of the cliff, she can visually align the top of the cliff with the
water at the back of her boat. Her eye level is 𝟓𝟓
Catarina’s boat?
𝟏𝟏
𝐟𝐟𝐟𝐟. above the ground. Approximately how far out from the cliff is
𝟐𝟐
Catarina’s height and the height of the cliff both are measured perpendicularly to the ground (and water), so both
triangles formed by her visual path are right triangles. Assuming that the ground level on the cliff and the water are
parallel, Catarina’s visual path forms the same angle with the cliff as it does with the surface of the water
(corresponding ∠’s, parallel lines). So, the right triangles are similar triangles by the AA criterion for showing
triangle similarity, which means that the ratios within triangles will be equal. Let 𝒅𝒅 represent the distance from the
boat to the cliff:
𝟓𝟓. 𝟓𝟓 𝟑𝟑𝟑𝟑
=
𝒅𝒅
𝟏𝟏𝟏𝟏
𝟓𝟓. 𝟓𝟓𝒅𝒅 = 𝟑𝟑𝟑𝟑𝟑𝟑
𝒅𝒅 ≈ 𝟔𝟔𝟔𝟔. 𝟔𝟔
Catarina’s boat is approximately 𝟔𝟔𝟔𝟔. 𝟔𝟔 𝐟𝐟𝐟𝐟. away from the cliff.
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6.
Given the diagram below and △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑿𝑿𝑿𝑿𝑿𝑿, find the unknown lengths 𝒙𝒙, 𝟐𝟐𝟐𝟐, and 𝟑𝟑𝟑𝟑.
The triangles are given as similar, so the values of the ratios of the sides within each triangle must be equal.
𝒙𝒙
𝟐𝟐
=
𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑
The sides of the larger triangle are unknown; however, the lengths include the same factor 𝒙𝒙, which is clearly
𝟏𝟏
nonzero, so the ratio of the sides must then be . Similarly, the sides of the smaller triangle have the same factor 𝟐𝟐,
𝟏𝟏
so the value of the ratio can be rewritten as .
𝒙𝒙
𝟑𝟑
𝟏𝟏 𝟏𝟏
=
𝒙𝒙 𝟑𝟑
𝒙𝒙 = 𝟑𝟑
Since the value of 𝒙𝒙 is 𝟑𝟑, it follows that 𝒀𝒀𝒀𝒀 = 𝟑𝟑, 𝑨𝑨𝑨𝑨 = 𝟔𝟔, and 𝑿𝑿𝑿𝑿 = 𝟗𝟗.
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Lesson 17: The Side-Angle-Side (SAS) and Side-Side-Side
(SSS) Criteria for Two Triangles to Be Similar
Student Outcomes

Students prove the side-angle-side criterion for two triangles to be similar and use it to solve triangle
problems.

Students prove the side-side-side criterion for two triangles to be similar and use it to solve triangle problems.
Lesson Notes
At this point, students know that two triangles can be considered similar if they have two pairs of corresponding equal
angles. In this lesson, students learn the other conditions that can be used to deem two triangles similar, specifically the
SAS and SSS criteria.
Classwork
Opening Exercise (3 minutes)
Opening Exercise
a.
b.
c.
MP.3
Choose three lengths that represent the sides of a triangle. Draw the triangle with your chosen lengths using
construction tools.
Answers will vary. Sample response: 𝟔𝟔 𝐜𝐜𝐜𝐜, 𝟕𝟕 𝐜𝐜𝐜𝐜, and 𝟖𝟖 𝐜𝐜𝐜𝐜.
Multiply each length in your original triangle by 𝟐𝟐 to get three corresponding lengths of sides for a second
triangle. Draw your second triangle using construction tools.
Answers will be twice the lengths given in part (a). Sample response: 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜, 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜, and 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜.
The triangles appear to be similar. Their corresponding sides are given as having lengths in the ratio 𝟐𝟐: 𝟏𝟏, and
the corresponding angles appear to be equal in measure. (This can be verified using either a protractor or
patty paper.)
d.
Do you think that the triangles can be shown to be similar without knowing the angle measures?

In Lesson 16, we discovered that if two triangles have two pairs of angles with equal measures, it is not
necessary to check all 6 conditions (3 sides and 3 angles) to show that the two triangles are similar. We called
it the AA criterion. In this lesson, we will use other combinations of conditions to conclude that two triangles
are similar.
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Exploratory Challenge 1/Exercises 1–2 (8 minutes)
This challenge gives students the opportunity to discover the SAS criterion for similar triangles. The Discussion that
follows solidifies this fact. Consider splitting the class into two groups so that one group works on Exercise 1 and the
other group completes Exercise 2. Then, have students share their work with the whole class.
Exploratory Challenge 1/Exercises 1–2
1.
Examine the figure, and answer the questions to determine whether or not the triangles shown are similar.
Scaffolding:
students how many
triangles they see in Figure
1, and then have them
identify each of the
triangles.
a.
What information is given about the triangles in Figure 1?
We are given that ∠𝑨𝑨 is common to both △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑩𝑩′ 𝑪𝑪′ . We are also given
information about some of the side lengths.
b.
How can the information provided be used to determine whether △ 𝑨𝑨𝑨𝑨𝑨𝑨 is similar to
△ 𝑨𝑨𝑩𝑩′ 𝑪𝑪′ ?
 Have students create
triangles that meet specific
criteria. Given a starting
triangle with lengths
3, 6, 7, generate similar
triangles by multiplying
the sides of the “starting”
triangle by a single factor,
such as 1, 1.5, 2, 3, and so
on.
We know that similar triangles will have ratios of corresponding sides that are proportional; therefore, we
can use the side lengths to check for proportionality.
c.
Compare the corresponding side lengths of △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑩𝑩′ 𝑪𝑪′. What do you notice?
𝟒𝟒
𝟑𝟑
=
𝟏𝟏𝟏𝟏 𝟗𝟗. 𝟕𝟕𝟕𝟕
𝟑𝟑𝟑𝟑 = 𝟑𝟑𝟑𝟑
The cross-products are equal; therefore, the side lengths are proportional.
d.
Based on your work in parts (a)–(c), draw a conclusion about the relationship between △ 𝑨𝑨𝑨𝑨𝑨𝑨 and
△ 𝑨𝑨𝑩𝑩′ 𝑪𝑪′ . Explain your reasoning.
The triangles are similar. By the triangle side splitter theorem, I know that when the sides of a triangle are
′ 𝑪𝑪′ . Then, I can conclude that the triangles are similar because they have two
������
cut proportionally, then ����
𝑩𝑩𝑩𝑩 ∥ 𝑩𝑩
pairs of corresponding angles that are equal.
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2.
Examine the figure, and answer the questions to determine whether or not the triangles shown are similar.
Scaffolding:
students how many triangles
they see in Figure 2; then, have
them identify each of the
triangles.
a.
What information is given about the triangles in Figure 2?
We are given that ∠𝑷𝑷 is common to both △ 𝑷𝑷𝑷𝑷𝑷𝑷 and △ 𝑷𝑷𝑷𝑷′𝑹𝑹′. We are also given information about some of
the side lengths.
b.
How can the information provided be used to determine whether △ 𝑷𝑷𝑷𝑷𝑷𝑷 is similar to △ 𝑷𝑷𝑷𝑷′𝑹𝑹′?
We know that similar triangles will have ratios of corresponding sides that are proportional; therefore, we
can use the side lengths to check for proportionality.
c.
Compare the corresponding side lengths of △ 𝑷𝑷𝑷𝑷𝑷𝑷 and △ 𝑷𝑷𝑷𝑷′𝑹𝑹′. What do you notice?
The side lengths are not proportional.
d.
𝟑𝟑
𝟐𝟐
≠
𝟏𝟏𝟏𝟏 𝟗𝟗
𝟐𝟐𝟐𝟐 ≠ 𝟐𝟐𝟐𝟐
Based on your work in parts (a)–(c), draw a conclusion about the relationship between △ 𝑷𝑷𝑷𝑷𝑷𝑷 and △ 𝑷𝑷𝑷𝑷′𝑹𝑹′.
The triangles are not similar. The side lengths are not proportional, which is what I would expect in similar
triangles. I know that the triangles have one common angle, but I cannot determine from the information
given whether there is another pair of equal angles. Therefore, I conclude that the triangles are not similar.
Discussion (5 minutes)
Have students reference their work in part (d) of Exercises 1 and 2 while leading the discussion below.

Which figure contained triangles that are similar? How do you know?


Figure 1 has the similar triangles. I know because the triangles have side lengths that are proportional.
Since the side lengths are split proportionally by segment 𝐵𝐵𝐵𝐵, then 𝐵𝐵𝐵𝐵 ∥ 𝐵𝐵′ 𝐶𝐶 ′ , and the triangles are
similar by the AA criterion. The same cannot be said about Figure 2.
The triangles in Figure 1 are similar. Take note of the information that is given about the triangles in the
diagrams, not what you are able to deduce about the angles.

The diagram shows information related to the side lengths and the angle between the sides.
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Once the information contained in the above two bullets is clear to students, explain the side-angle-side criterion for
triangle similarity below.
THEOREM: The side-angle-side criterion for two triangles to be similar is as follows:
Given two triangles 𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐴𝐴′𝐵𝐵′𝐶𝐶′ so that
𝐴𝐴′𝐵𝐵′
𝐴𝐴𝐴𝐴
=
𝐴𝐴′𝐶𝐶 ′
𝐴𝐴𝐴𝐴
and 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐴𝐴′ , the triangles are similar;
△ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐴𝐴′ 𝐵𝐵′ 𝐶𝐶 ′ . That is, two triangles are similar if they have one pair of corresponding angles that are
congruent and the sides adjacent to that angle are proportional.


The proof of this theorem is simply to take any dilation with a scale factor of 𝑟𝑟 =
𝐴𝐴′ 𝐵𝐵′
𝐴𝐴𝐴𝐴
=
𝐴𝐴′𝐶𝐶 ′
𝐴𝐴𝐴𝐴
. This dilation
maps △ 𝐴𝐴𝐴𝐴𝐴𝐴 to a triangle that is congruent to △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′ by the side-angle-side congruence criterion.
We refer to the angle between the two sides as the included angle. Or we can say the sides are adjacent to the
given angle. When the side lengths adjacent to the angle are in proportion, then we can conclude that the
triangles are similar by the side-angle-side criterion.
The question below requires students to apply their new knowledge of the SAS criterion to the triangles in Figure 2.
Students should have concluded that they were not similar in part (d) of Exercise 2. The question below pushes students
to apply the SAS and AA theorems related to triangle similarity to show definitively that the triangles in Figure 2 are not
similar.

Does Figure 2 have side lengths that are proportional? What can you conclude about the triangles in Figure 2?


No, the side lengths are not proportional, so we cannot use the SAS criterion for similarity to say that
the triangles are similar. If the side lengths were proportional, we could conclude that the lines
𝑃𝑃′ 𝑄𝑄′ are parallel, but the side lengths are not proportional; therefore, the lines
containing ����
𝑃𝑃𝑃𝑃 and ������
𝑃𝑃′ 𝑄𝑄′ are not parallel. This then means that the corresponding angles are not equal,
containing ����
𝑃𝑃𝑃𝑃 and ������
and the AA criterion cannot be used to say that the triangles are similar.
We can use the SAS criterion to determine if a pair of triangles are similar. Since the side lengths in Figure 2
were not proportional, we can conclude that the triangles are not similar.
Exploratory Challenge 2/Exercises 3–4 (8 minutes)
This challenge gives students the opportunity to discover the SSS criterion for similar triangles. The Discussion that
follows solidifies this fact. Consider splitting the class into two groups so that one group works on Exercise 3 and the
other group completes Exercise 4. Then, have students share their work with the whole class.
Exploratory Challenge 2/Exercises 3–4
3.
Examine the figure, and answer the questions to determine whether or not the triangles shown are similar.
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a.
What information is given about the triangles in Figure 3?
We are only given information related to the side lengths of the triangles.
b.
How can the information provided be used to determine whether △ 𝑨𝑨𝑨𝑨𝑨𝑨 is similar to △ 𝑨𝑨′𝑩𝑩′ 𝑪𝑪′ ?
We know that similar triangles will have ratios of corresponding sides that are proportional; therefore, we
can use the side lengths to check for proportionality.
c.
Compare the corresponding side lengths of △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨′𝑩𝑩′ 𝑪𝑪′ . What do you notice?
The side lengths are proportional.
d.
𝟏𝟏. 𝟒𝟒𝟒𝟒 𝟑𝟑. 𝟓𝟓 𝟐𝟐. 𝟕𝟕 𝟏𝟏
=
=
=
𝟐𝟐. 𝟖𝟖𝟖𝟖
𝟕𝟕
𝟓𝟓. 𝟒𝟒 𝟐𝟐
Based on your work in parts (a)–(c), make a conjecture about the relationship between △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨′𝑩𝑩′ 𝑪𝑪′ .
I think that the triangles are similar. The side lengths are proportional, which is what I would expect in similar
triangles.
4.
Examine the figure, and answer the questions to determine whether or not the triangles shown are similar.
a.
What information is given about the triangles in Figure 4?
We are only given information related to the side lengths of the triangles.
b.
How can the information provided be used to determine whether △ 𝑨𝑨𝑨𝑨𝑨𝑨 is similar to △ 𝑨𝑨′𝑩𝑩′ 𝑪𝑪′ ?
We know that similar triangles will have ratios of corresponding sides that are proportional; therefore, we
can use the side lengths to check for proportionality.
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c.
Compare the corresponding side lengths of △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨′𝑩𝑩′ 𝑪𝑪′ . What do you notice?
The side lengths are not proportional.
d.
𝟐𝟐. 𝟏𝟏
𝟖𝟖. 𝟐𝟐𝟐𝟐 𝟕𝟕. 𝟐𝟐𝟐𝟐
≠
≠
𝟏𝟏. 𝟒𝟒𝟒𝟒 𝟔𝟔. 𝟎𝟎𝟎𝟎 𝟓𝟓. 𝟐𝟐𝟐𝟐
Based on your work in parts (a)–(c), make a conjecture about the relationship between △ 𝑨𝑨𝑨𝑨𝑨𝑨 and
△ 𝑨𝑨′𝑩𝑩′ 𝑪𝑪′ . Explain your reasoning.
I think that the triangles are not similar. I would expect the side lengths to be proportional if the triangles
were similar.
Discussion (5 minutes)

Which figure contains triangles that are similar? What made you think they are similar?
MP.8


The triangles in Figure 3 are similar. Take note of the information given about the triangles in the diagrams.


Figure 3 has the similar triangles. I think Figure 3 has the similar triangles because all three pairs of
corresponding sides are in proportion. That is not the case for Figure 4.
The diagram shows information related only to the side lengths of each of the triangles.
THEOREM: The side-side-side criterion for two triangles to be similar is as follows.
Given two triangles 𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐴𝐴′𝐵𝐵′𝐶𝐶′ so that
′

′
′
𝐴𝐴′𝐵𝐵′
𝐴𝐴𝐴𝐴
=
𝐵𝐵′ 𝐶𝐶 ′
𝐵𝐵𝐵𝐵
=
𝐴𝐴′𝐶𝐶 ′
𝐴𝐴𝐴𝐴
, the triangles are similar;
△ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐴𝐴 𝐵𝐵 𝐶𝐶 . In other words, two triangles are similar if their corresponding sides are proportional.
What would the scale factor, 𝑟𝑟, need to be to show that these triangles are similar? Explain.

A scale factor of 𝑟𝑟 =
𝐴𝐴′ 𝐵𝐵′
𝐴𝐴𝐴𝐴
or 𝑟𝑟 =
𝐵𝐵′ 𝐶𝐶 ′
𝐵𝐵𝐵𝐵
or 𝑟𝑟 =
𝐴𝐴′ 𝐶𝐶 ′
𝐴𝐴𝐴𝐴
would show that these triangles are similar.
Dilating by one of those scale factors guarantees that the corresponding sides of the triangles will be
proportional.

Then, a dilation by scale factor 𝑟𝑟 =
𝐴𝐴′𝐵𝐵′
𝐴𝐴𝐴𝐴
=
𝐵𝐵′ 𝐶𝐶 ′
𝐵𝐵𝐵𝐵
𝐴𝐴′𝐵𝐵′𝐶𝐶′ by the side-side-side congruence criterion.
=
𝐴𝐴′ 𝐶𝐶 ′
𝐴𝐴𝐴𝐴
maps △ 𝐴𝐴𝐴𝐴𝐴𝐴 to a triangle that is congruent to △

When all three pairs of corresponding sides are in proportion, we can conclude that the triangles are similar by
the side-side-side criterion.

Does Figure 4 have side lengths that are proportional? What can you conclude about the triangles in Figure 4?


No, the side lengths are not proportional; therefore, the triangles are not similar.
We can use the SSS criterion to determine if two triangles are similar. Since the side lengths in Figure 4 are not
proportional, we can conclude that the triangles are not similar.
Exercises 5–10 (8 minutes)
Students identify whether or not the triangles are similar. For pairs of triangles that are similar, students identify the
criterion used: AA, SAS, or SSS.
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Exercises 5–10
5.
Are the triangles shown below similar? Explain. If the triangles are similar, write the similarity statement.
𝟏𝟏
𝟔𝟔. 𝟒𝟒
𝟓𝟓. 𝟖𝟖𝟖𝟖
≠
≠
𝟎𝟎. 𝟔𝟔𝟔𝟔 𝟐𝟐. 𝟒𝟒𝟒𝟒 𝟐𝟐. 𝟏𝟏𝟏𝟏
There is no information about the angle measures, so we
cannot use AA or SAS to conclude the triangles are similar.
Since the side lengths are not proportional, we cannot use SSS
to conclude the triangles are similar. Therefore, the triangles
shown are not similar.
6.
Are the triangles shown below similar? Explain. If the triangles are similar, write the similarity statement.
𝟑𝟑. 𝟏𝟏𝟏𝟏
𝟑𝟑
=
𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓 𝟏𝟏. 𝟓𝟓
Yes, the triangles shown are similar. △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫 by SAS because 𝒎𝒎∠𝑩𝑩 = 𝒎𝒎∠𝑬𝑬, and the adjacent sides are
proportional.
7.
Are the triangles shown below similar? Explain. If the triangles are similar, write the similarity statement.
Yes, the triangles shown are similar. △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑨𝑨𝑨𝑨𝑨𝑨 by AA
because 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 = 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨, and both triangles share ∠𝑨𝑨.
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8.
Are the triangles shown below similar? Explain. If the triangles are similar, write the similarity statement.
𝟓𝟓 𝟑𝟑
≠
𝟑𝟑 𝟏𝟏
There is no information about the angle measures other
than the right angle, so we cannot use AA to conclude the
triangles are similar. We only have information about two
of the three side lengths for each triangle, so we cannot use
SSS to conclude they are similar. If the triangles are similar,
we would have to use the SAS criterion, and since the side
lengths are not proportional, the triangles shown are not
similar. (Note that students could also utilize the
Pythagorean theorem to determine the lengths of the
hypotenuses and then use SSS similarity criterion to answer
the question.)
9.
Are the triangles shown below similar? Explain. If the triangles are similar, write the similarity statement.
𝟑𝟑. 𝟔𝟔 𝟕𝟕. 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏
=
=
𝟐𝟐. 𝟕𝟕 𝟓𝟓. 𝟒𝟒𝟒𝟒 𝟕𝟕. 𝟓𝟓
Yes, the triangles are similar. △ 𝑷𝑷𝑷𝑷𝑷𝑷 ~ △ 𝑿𝑿𝑿𝑿𝑿𝑿 by SSS.
10. Are the triangles shown below similar? Explain. If the triangles are similar, write the similarity statement.
𝟐𝟐. 𝟐𝟐𝟐𝟐 𝟑𝟑. 𝟏𝟏𝟏𝟏
=
𝟒𝟒. 𝟒𝟒𝟒𝟒 𝟔𝟔. 𝟑𝟑𝟑𝟑
Yes, the triangles are similar. △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑫𝑫𝑪𝑪𝑪𝑪 by SAS because 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 = 𝒎𝒎∠𝑫𝑫𝑫𝑫𝑫𝑫 (vertical angles are congruent),
and the sides adjacent to those angles are proportional.
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Closing (3 minutes)
Ask the following three questions to informally assess students’ understanding of the similarity criterion for triangles:

Given only information about the angles of a pair of triangles, how can you determine if the given triangles are
similar?


Given only information about one pair of angles for two triangles, how can you determine if the given triangles
are similar?


The AA criterion can be used to determine if two triangles are similar. The triangles must have two
pairs of corresponding angles that are equal in measure.
The SAS criterion can be used to determine if two triangles are similar. The triangles must have one
pair of corresponding angles that are equal in measure, and the ratios of the corresponding adjacent
sides must be in proportion.
Given no information about the angles of a pair of triangles, how can you determine if the given triangles are
similar?

The SSS criterion can be used to determine if two triangles are similar. The triangles must have three
pairs of corresponding side lengths in proportion.
Exit Ticket (5 minutes)
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Name
Date
Lesson 17: The Side-Angle-Side (SAS) and Side-Side-Side (SSS)
Criteria for Two Triangles to Be Similar
Exit Ticket
1.
Given △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝑀𝑀𝑀𝑀𝑀𝑀 in the diagram below and ∠𝐵𝐵 ≅ ∠𝐿𝐿, determine if the triangles are similar. If so, write a
similarity statement, and state the criterion used to support your claim.
2.
Given △ 𝐷𝐷𝐷𝐷𝐷𝐷 and △ 𝐸𝐸𝐸𝐸𝐸𝐸 in the diagram below, determine if the triangles are similar. If so, write a similarity
statement, and state the criterion used to support your claim.
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Exit Ticket Sample Solutions
1.
Given △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑴𝑴𝑴𝑴𝑴𝑴 in the diagram below, determine if the triangles are similar. If so, write a similarity
statement, and state the criterion used to support your claim.
In comparing the ratios of sides between figures, I found that
𝑨𝑨𝑨𝑨
𝑴𝑴𝑴𝑴
=
𝑪𝑪𝑪𝑪
𝑵𝑵𝑵𝑵
because the cross-products of the proportion
𝟖𝟖
𝟑𝟑
are both 𝟑𝟑𝟑𝟑. We are given that ∠𝑳𝑳 ≅ ∠𝑩𝑩. Therefore,
△ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑴𝑴𝑴𝑴𝑴𝑴 by the SAS criterion for proving similar
triangles.
2.
=
𝟐𝟐
𝟑𝟑
𝟑𝟑
𝟒𝟒
𝟒𝟒
𝟏𝟏𝟏𝟏
Given △ 𝑫𝑫𝑫𝑫𝑫𝑫 and △ 𝑬𝑬𝑬𝑬𝑬𝑬 in the diagram below, determine if the triangles are similar. If so, write a similarity
statement, and state the criterion used to support your claim.
By comparison, if the triangles are in fact similar,
then the longest sides of each triangle will
correspond, and likewise the shortest sides will
correspond. The corresponding sides from each
triangle are proportional since
𝟓𝟓
𝟏𝟏
𝟑𝟑
𝟖𝟖
=
𝟒𝟒
𝟏𝟏
𝟐𝟐
𝟐𝟐
=
𝟔𝟔
𝟐𝟐
𝟓𝟓
𝟒𝟒
=
𝟖𝟖
𝟓𝟓
, so
𝑫𝑫𝑫𝑫
𝑬𝑬𝑬𝑬
=
𝑬𝑬𝑬𝑬
𝑮𝑮𝑮𝑮
=
𝑫𝑫𝑫𝑫
𝑬𝑬𝑬𝑬
.
Therefore, by the SSS criterion for showing
triangle similarity, △ 𝑫𝑫𝑫𝑫𝑫𝑫 ~ △ 𝑬𝑬𝑬𝑬𝑬𝑬.
Problem Set Sample Solutions
1.
For parts (a) through (d) below, state which of the three triangles, if any, are similar and why.
a.
Triangles 𝑩𝑩 and 𝑪𝑪 are similar because they share three pairs of corresponding sides that are in the same ratio.
𝟔𝟔
𝟖𝟖
𝟏𝟏
𝟒𝟒
=
=
=
𝟖𝟖 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐
Triangles 𝑨𝑨 and 𝑩𝑩 are not similar because the ratios of their corresponding sides are not in the same ratio.
𝟔𝟔
𝟒𝟒
≠
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
Further, if triangle 𝑨𝑨 is not similar to triangle 𝑩𝑩, then triangle 𝑨𝑨 is not similar to triangle 𝑪𝑪.
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Triangles to be Similar
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b.
Triangles 𝑨𝑨 and 𝑩𝑩 are not similar because their corresponding sides are not all in the same ratio. Two pairs of
corresponding sides are proportional, but the third pair of corresponding sides is not.
𝟔𝟔
𝟓𝟓
𝟑𝟑
≠
=
𝟔𝟔 𝟕𝟕. 𝟓𝟓 𝟗𝟗
Triangles 𝑩𝑩 and 𝑪𝑪 are not similar because their corresponding sides are not in the same ratio.
𝟕𝟕
𝟖𝟖
𝟔𝟔
≠
≠
𝟔𝟔 𝟕𝟕. 𝟓𝟓 𝟗𝟗
Triangles 𝑨𝑨 and 𝑪𝑪 are not similar because their corresponding sides are not in the same ratio.
𝟑𝟑 𝟓𝟓 𝟔𝟔
≠ ≠
𝟔𝟔 𝟕𝟕 𝟖𝟖
c.
Triangles 𝑩𝑩 and 𝑫𝑫 are the only similar triangles because they have the same angle measures. Using the angle
sum of a triangle, both triangles 𝑩𝑩 and 𝑫𝑫 have angles of 𝟕𝟕𝟕𝟕°, 𝟔𝟔𝟔𝟔°, and 𝟒𝟒𝟒𝟒°.
d.
Triangles 𝑨𝑨 and 𝑩𝑩 are similar because they have two pairs of corresponding sides that are in the same ratio,
and their included angles are equal measures. Triangle 𝑪𝑪 cannot be shown to be similar because, even
though it has two sides that are the same length as two sides of triangle 𝑨𝑨, the 𝟕𝟕𝟕𝟕° angle in triangle 𝑪𝑪 is not
the included angle and, therefore, does not correspond to the 𝟕𝟕𝟕𝟕° angle in triangle 𝑨𝑨.
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2.
For each given pair of triangles, determine if the triangles are similar or not, and provide your reasoning. If the
triangles are similar, write a similarity statement relating the triangles.
a.
The triangles are similar because, using the angle sum of a triangle, each triangle has angle measures of 𝟓𝟓𝟓𝟓°,
𝟔𝟔𝟔𝟔°, and 𝟕𝟕𝟕𝟕°. Therefore, △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑻𝑻𝑻𝑻𝑻𝑻.
b.
The triangles are not similar, because the ratios of corresponding sides are not all in proportion.
𝑨𝑨𝑨𝑨
𝑫𝑫𝑫𝑫
=
𝑩𝑩𝑩𝑩
𝑬𝑬𝑬𝑬
= 𝟐𝟐; however,
𝑨𝑨𝑨𝑨
𝑫𝑫𝑫𝑫
=
𝟕𝟕
𝟒𝟒
≠ 𝟐𝟐.
c.
△ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑭𝑭𝑭𝑭𝑭𝑭 by the SSS criterion for showing similar triangles because the ratio of all pairs of
corresponding sides
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𝑨𝑨𝑨𝑨
𝑭𝑭𝑭𝑭
=
𝑨𝑨𝑨𝑨
𝑭𝑭𝑭𝑭
=
𝑩𝑩𝑩𝑩
𝑫𝑫𝑫𝑫
= 𝟐𝟐.
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d.
𝑨𝑨𝑨𝑨
𝑫𝑫𝑫𝑫
=
𝑩𝑩𝑩𝑩
𝑫𝑫𝑫𝑫
𝟒𝟒
= , and included angles 𝑩𝑩 and 𝑫𝑫 are both 𝟑𝟑𝟑𝟑° and, therefore, congruent, so △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑬𝑬𝑬𝑬𝑬𝑬 by
𝟓𝟓
the SAS criterion for showing similar triangles.
3.
For each pair of similar triangles below, determine the unknown lengths of the sides labeled with letters.
a.
The ratios of corresponding sides must be equal, so
𝟓𝟓
𝒏𝒏
=
𝟑𝟑
𝟒𝟒
𝟑𝟑
𝟗𝟗
𝟖𝟖
𝒎𝒎 = 𝟑𝟑.
, giving 𝒏𝒏 = 𝟏𝟏𝟏𝟏 𝟏𝟏𝟐𝟐.
Likewise,
𝒎𝒎
𝟏𝟏
𝟕𝟕
𝟐𝟐
=
𝟑𝟑
𝟒𝟒
𝟑𝟑
𝟗𝟗
𝟖𝟖
, giving
b.
The ratios of corresponding sides must
be equal, so
Likewise,
4.
𝟖𝟖
𝟔𝟔
𝟖𝟖
𝟔𝟔
=
=
𝟕𝟕
𝒕𝒕
𝒔𝒔
𝟔𝟔
𝟑𝟑 , giving
𝟒𝟒
𝒔𝒔 = 𝟗𝟗.
𝟏𝟏
𝟒𝟒
, giving 𝒕𝒕 = 𝟓𝟓 .
���� intersect at 𝑬𝑬 and ����
����, show that △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫.
���� and 𝑩𝑩𝑩𝑩
Given that 𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨 ∥ 𝑪𝑪𝑪𝑪
∠𝑨𝑨𝑨𝑨𝑨𝑨 and ∠𝑫𝑫𝑫𝑫𝑫𝑫 are vertically opposite angles and,
therefore, congruent. ∠𝑨𝑨 ≅ ∠𝑫𝑫 by alt. int. ∠'s,
����. Therefore, △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫 by the AA
����
𝑨𝑨𝑨𝑨 ∥ 𝑪𝑪𝑪𝑪
criterion for showing similar triangles.
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5.
Given 𝑩𝑩𝑩𝑩 = 𝟏𝟏𝟏𝟏, 𝑬𝑬𝑬𝑬 = 𝟏𝟏𝟏𝟏, 𝑩𝑩𝑩𝑩 = 𝟕𝟕, and 𝑫𝑫𝑫𝑫 = 𝟕𝟕, show that △ 𝑩𝑩𝑩𝑩𝑩𝑩 ~ △ 𝑩𝑩𝑩𝑩𝑩𝑩.
Both triangles share angle 𝑩𝑩, and by the reflexive
property, ∠𝑩𝑩 ≅ ∠𝑩𝑩. 𝑩𝑩𝑩𝑩 = 𝑩𝑩𝑩𝑩 + 𝑬𝑬𝑬𝑬, so 𝑩𝑩𝑩𝑩 = 𝟐𝟐𝟐𝟐,
and 𝑩𝑩𝑩𝑩 = 𝑩𝑩𝑩𝑩 + 𝑫𝑫𝑫𝑫, so 𝑩𝑩𝑩𝑩 = 𝟏𝟏𝟏𝟏. The ratios of
corresponding sides
𝑩𝑩𝑩𝑩
=
𝑩𝑩𝑩𝑩
𝟏𝟏
= . Therefore,
𝟐𝟐
△ 𝑩𝑩𝑩𝑩𝑩𝑩 ~ △ 𝑩𝑩𝑩𝑩𝑩𝑩 by the SAS criterion for triangle
similarity.
6.
Given the diagram below, 𝑿𝑿 is on ����
𝑹𝑹𝑹𝑹 and 𝒀𝒀 is on ����
𝑹𝑹𝑹𝑹, 𝑿𝑿𝑿𝑿 = 𝟐𝟐, 𝑿𝑿𝑿𝑿 = 𝟔𝟔, 𝑺𝑺𝑺𝑺 = 𝟗𝟗, and 𝒀𝒀𝒀𝒀 = 𝟒𝟒.
a.
Show that △ 𝑹𝑹𝑹𝑹𝑹𝑹 ~ △ 𝑹𝑹𝑹𝑹𝑹𝑹.
The diagram shows ∠𝑹𝑹𝑹𝑹𝑹𝑹 ≅ ∠𝑹𝑹𝑹𝑹𝑹𝑹. Both triangle 𝑹𝑹𝑹𝑹𝑹𝑹 and 𝑹𝑹𝑹𝑹𝑹𝑹 share angle 𝑹𝑹, and by the reflexive
property, ∠𝑹𝑹 ≅ ∠𝑹𝑹, so △ 𝑹𝑹𝑹𝑹𝑹𝑹 ~ △ 𝑹𝑹𝑹𝑹𝑹𝑹 by the AA criterion show triangle similarity.
b.
Find 𝑹𝑹𝑹𝑹 and 𝑹𝑹𝑹𝑹.
Since the triangles are similar, their corresponding sides must be in the same ratio.
𝑹𝑹𝑹𝑹
=
𝑹𝑹𝑹𝑹
=
𝑿𝑿𝑿𝑿 𝟔𝟔
=
𝑺𝑺𝑺𝑺 𝟗𝟗
𝑹𝑹𝑹𝑹 = 𝑹𝑹𝑹𝑹 + 𝑿𝑿𝑿𝑿, so 𝑹𝑹𝑹𝑹 = 𝑹𝑹𝑹𝑹 + 𝟐𝟐 and 𝑹𝑹𝑹𝑹 = 𝑹𝑹𝑹𝑹 + 𝒀𝒀𝒀𝒀, so 𝑹𝑹𝑹𝑹 = 𝑹𝑹𝑹𝑹 + 𝟒𝟒.
𝑹𝑹𝑹𝑹
𝟔𝟔
=
𝑹𝑹𝑹𝑹 + 𝟐𝟐 𝟗𝟗
𝟗𝟗(𝑹𝑹𝑹𝑹) = 𝟔𝟔(𝑹𝑹𝑹𝑹) + 𝟏𝟏𝟏𝟏
𝑹𝑹𝑹𝑹 = 𝟒𝟒
7.
𝑹𝑹𝑹𝑹
𝟔𝟔
=
𝑹𝑹𝑹𝑹 + 𝟒𝟒 𝟗𝟗
𝟗𝟗(𝑹𝑹𝑹𝑹) = 𝟔𝟔(𝑹𝑹𝑹𝑹) + 𝟐𝟐𝟐𝟐
𝑹𝑹𝑹𝑹 = 𝟖𝟖
One triangle has a 𝟏𝟏𝟏𝟏𝟏𝟏° angle, and a second triangle has a 𝟔𝟔𝟔𝟔° angle. Is it possible that the two triangles are
similar? Explain why or why not.
No The triangles cannot be similar because in the first triangle, the sum of the remaining angles is 𝟔𝟔𝟔𝟔°, which means
that it is not possible for the triangle to have a 𝟔𝟔𝟔𝟔° angle. For the triangles to be similar, both triangles would have
to have angles measuring 𝟏𝟏𝟏𝟏𝟏𝟏° and 𝟔𝟔𝟔𝟔°, but this is impossible due to the angle sum of a triangle.
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8.
A right triangle has a leg that is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜, and another right triangle has a leg that is 𝟔𝟔 𝐜𝐜𝐜𝐜. Can you tell whether the
two triangles are similar? If so, explain why. If not, what other information would be needed to show they are
similar?
The two triangles may or may not be similar. There is not enough information to make this claim. If the second leg
of the first triangle is twice the length of the second leg of the first triangle, then the triangles are similar by SAS
criterion for showing similar triangles.
9.
Given the diagram below, 𝑱𝑱𝑱𝑱 = 𝟕𝟕. 𝟓𝟓, 𝑯𝑯𝑯𝑯 = 𝟔𝟔, and 𝑲𝑲𝑲𝑲 = 𝟗𝟗, is there a pair of similar triangles? If so, write a
similarity statement, and explain why. If not, explain your reasoning.
△ 𝑳𝑳𝑳𝑳𝑳𝑳 ~ △ 𝑯𝑯𝑯𝑯𝑯𝑯 by the SAS criterion for showing
triangle similarity. Both triangles share ∠𝑲𝑲, and by the
reflexive property, ∠𝑲𝑲 ≅ ∠𝑲𝑲. Furthermore,
𝑳𝑳𝑳𝑳
𝑱𝑱𝑱𝑱
=
𝑯𝑯𝑯𝑯
𝑳𝑳𝑳𝑳
=
𝟐𝟐
𝟑𝟑
, giving two pairs of corresponding sides
in the same ratio and included angles of the same size.
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Lesson 18: Similarity and the Angle Bisector Theorem
Student Outcomes

Students state, understand, and prove the angle bisector theorem.

Students use the angle bisector theorem to solve problems.
Classwork
Opening Exercise (5 minutes)
The Opening Exercise should activate students’ prior knowledge acquired in Module 1 that is helpful in proving the angle
bisector theorem.
Scaffolding:
Opening Exercise
a.
What is an angle bisector?
The bisector of an angle is a ray in the interior of the angle such that the two adjacent
angles formed by it have equal measures.
b.
 If necessary, provide
visuals to accompany
these questions.
 For part (a):
Describe the angle relationships formed when parallel lines are cut by a transversal.
When parallel lines are cut by a transversal, the corresponding, alternate interior, and
alternate exterior angles are all congruent; the same-side interior angles are
supplementary.
c.
 For part (b):
What are the properties of an isosceles triangle?
An isosceles triangle has at least two congruent sides, and the angles opposite to the
congruent sides (i.e., the base angles) are also congruent.
Discussion (20 minutes)
Prior to proving the angle bisector theorem, students observe the length relationships of
the sides of a triangle when one of the angles of the triangle has been bisected.

 For part (c):
In this lesson, we investigate the length relationships of the sides of a triangle
when one angle of the triangle has been bisected.
Provide students time to look for relationships between the side lengths. This requires
trial and error on the part of students and may take several minutes.
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Discussion
���� at point 𝑫𝑫. Does the angle bisector create any
In the diagram below, the angle bisector of ∠𝑨𝑨 in △ 𝑨𝑨𝑨𝑨𝑨𝑨 meets side 𝑩𝑩𝑩𝑩
observable relationships with respect to the side lengths of the triangle?
MP.7
Acknowledge any relationships students may find, but highlight the relationship 𝑩𝑩𝑩𝑩: 𝑪𝑪𝑪𝑪 = 𝑩𝑩𝑩𝑩: 𝑪𝑪𝑪𝑪. Then, continue the
discussion below that proves this relationship.

The following theorem generalizes our observation:
THEOREM: The angle bisector theorem: In △ 𝐴𝐴𝐴𝐴𝐴𝐴, if the angle bisector of ∠𝐴𝐴 meets side ����
𝐵𝐵𝐵𝐵 at point 𝐷𝐷, then
𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶.
In words, the bisector of an angle of a triangle splits the opposite side into segments that have the same ratio
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
Our goal now is to prove this relationship for all triangles. We begin by constructing a line through vertex 𝐶𝐶
that is parallel to side ����
𝐴𝐴𝐴𝐴. Let 𝐸𝐸 be the point where this parallel line meets the angle bisector, as shown.
Scaffolding:
In place of the formal proof,
students may construct angle
bisectors for a series of
triangles and take
measurements to verify the
relationship inductively.
If we can show that △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐸𝐸𝐸𝐸𝐸𝐸, then we can use what we know about similar triangles to prove the
relationship 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶.
Consider asking students why it is that if they can show △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐸𝐸𝐸𝐸𝐸𝐸, they are closer to the goal of showing
𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶. Students should respond that similar triangles have proportional length relationships. The triangles
𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐸𝐸𝐸𝐸𝐸𝐸, if shown similar, would give 𝐵𝐵𝐵𝐵: 𝐵𝐵𝐵𝐵 = 𝐶𝐶𝐶𝐶: 𝐶𝐶𝐶𝐶, three out of the four lengths needed in the ratio
𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶.

How can we show that △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐸𝐸𝐸𝐸𝐸𝐸?
Provide students time to discuss how to show that the triangles are similar. Elicit student responses; then, continue with
the discussion below.

It is true that △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐸𝐸𝐸𝐸𝐸𝐸 by the AA criterion for similarity. Vertical angles 𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐸𝐸𝐸𝐸𝐸𝐸 are congruent
and, therefore, equal. ∠𝐷𝐷𝐷𝐷𝐷𝐷 and ∠𝐵𝐵𝐵𝐵𝐵𝐵 are congruent and equal because they are alternate interior angles of
parallel lines cut by a transversal. (Show the diagram below.)

Since the triangles are similar, we know that 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶. This is very close to what we are trying to
show: 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶. What must we do now to prove the theorem?

We have to show that 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶.
Once students have identified what needs to be done, that is, show that 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶, provide them time to discuss how to
show it. The prompts below can be used to guide students’ thinking.
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
We need to show that 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶. Notice that the segments 𝐶𝐶𝐶𝐶 and 𝐶𝐶𝐶𝐶 are two sides of △ 𝐴𝐴𝐴𝐴𝐴𝐴. How might
that be useful?


How can we show that △ 𝐴𝐴𝐴𝐴𝐴𝐴 is an isosceles triangle?


If we could show that △ 𝐴𝐴𝐴𝐴𝐴𝐴 is an isosceles triangle, then we would know that 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶.
We were first given that angle 𝐴𝐴 was bisected by ����
𝐴𝐴𝐴𝐴 , which means that ∠𝐵𝐵𝐵𝐵𝐵𝐵 ≅ ∠𝐶𝐶𝐶𝐶𝐶𝐶. Then, by
alternate interior ∠’s, ����
𝐶𝐶𝐶𝐶 ∥ ����
𝐴𝐴𝐴𝐴 , it follows that ∠𝐶𝐶𝐶𝐶𝐶𝐶 = ∠𝐶𝐶𝐶𝐶𝐶𝐶. We can use the converse of the base
angles of isosceles triangle theorem (i.e., base ∠’s converse). Since ∠𝐶𝐶𝐶𝐶𝐶𝐶 = ∠𝐶𝐶𝐶𝐶𝐶𝐶, then △ 𝐴𝐴𝐴𝐴𝐴𝐴 must
be an isosceles triangle.
Now that we know △ 𝐴𝐴𝐴𝐴𝐴𝐴 is isosceles, we can conclude that 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶 and finish the proof of the angle
bisector theorem. All we must do now is substitute 𝐶𝐶𝐶𝐶 for 𝐶𝐶𝐶𝐶 in 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶. Therefore,
𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶 = 𝐵𝐵𝐵𝐵: 𝐶𝐶𝐶𝐶, and the theorem is proved.
Consider asking students to restate what was just proved and summarize the steps of the proof. Students should
respond that the bisector of an angle of a triangle splits the opposite side into segments that have the same ratio as the
Exercises 1–4 (10 minutes)
Students complete Exercises 1–4 independently.
Exercises 1–4
1.
The sides of a triangle are 𝟖𝟖, 𝟏𝟏𝟏𝟏, and 𝟏𝟏𝟏𝟏. An angle bisector
meets the side of length 𝟏𝟏𝟏𝟏. Find the lengths 𝒙𝒙 and 𝒚𝒚. Explain
𝒚𝒚 𝟏𝟏𝟏𝟏
=
𝒙𝒙
𝟖𝟖
𝒙𝒙 = 𝟏𝟏𝟏𝟏 − 𝟗𝟗
𝒙𝒙 = 𝟔𝟔
𝒙𝒙 = 𝟏𝟏𝟏𝟏 − 𝒚𝒚
𝟖𝟖𝟖𝟖 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝟖𝟖𝟖𝟖 = 𝟏𝟏𝟏𝟏(𝟏𝟏𝟏𝟏 − 𝒚𝒚)
𝟖𝟖𝟖𝟖 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝒚𝒚 = 𝟗𝟗
The length 𝒙𝒙 is 𝟔𝟔, and the length 𝒚𝒚 is 𝟗𝟗.
Since I know that ∠𝑨𝑨 is bisected, I applied what I knew about the angle bisector theorem to determine the lengths 𝒙𝒙
and 𝒚𝒚. Specifically, the angle bisector cuts the side that is opposite the bisected angle so that 𝒚𝒚: 𝒙𝒙 = 𝟏𝟏𝟏𝟏: 𝟖𝟖. I set up
an equation using the values of the ratios, which could be solved once I rewrote one of the variables 𝒙𝒙 or 𝒚𝒚. I
rewrote 𝒙𝒙 as 𝟏𝟏𝟏𝟏 − 𝒚𝒚 and then solved for 𝒚𝒚. Once I had a value for 𝒚𝒚, I could replace it in the equation 𝒙𝒙 = 𝟏𝟏𝟏𝟏 − 𝒚𝒚 to
determine the value of 𝒙𝒙.
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2.
The sides of a triangle are 𝟖𝟖, 𝟏𝟏𝟏𝟏, and 𝟏𝟏𝟏𝟏. An angle bisector meets the side of length 𝟏𝟏𝟏𝟏. Find the lengths 𝒙𝒙 and 𝒚𝒚.
𝒚𝒚 = 𝟏𝟏𝟏𝟏 − 𝒙𝒙
𝒚𝒚 = 𝟏𝟏𝟏𝟏 − 𝟒𝟒
𝒙𝒙
𝟖𝟖
=
𝟏𝟏𝟏𝟏 − 𝒙𝒙 𝟏𝟏𝟏𝟏
𝒚𝒚 = 𝟕𝟕
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟖𝟖(𝟏𝟏𝟏𝟏 − 𝒙𝒙)
𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐
𝟒𝟒
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟗𝟗𝟗𝟗 − 𝟖𝟖𝟖𝟖
𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟗𝟗𝟗𝟗
The length of 𝒙𝒙 is 𝟒𝟒
3.
𝒙𝒙 =
𝟗𝟗𝟗𝟗
𝟒𝟒
= 𝟒𝟒
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝟒𝟒
𝟏𝟏𝟏𝟏
, and the length of 𝒚𝒚 is 𝟕𝟕 .
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
The sides of a triangle are 𝟖𝟖, 𝟏𝟏𝟏𝟏, and 𝟏𝟏𝟏𝟏. An angle bisector meets the side of length 𝟖𝟖. Find the lengths 𝒙𝒙 and 𝒚𝒚.
𝒚𝒚 𝟏𝟏𝟏𝟏
=
𝒙𝒙 𝟏𝟏𝟏𝟏
𝒚𝒚 = 𝟖𝟖 − 𝒙𝒙
𝒚𝒚 = 𝟖𝟖 − 𝟑𝟑
𝒚𝒚 = 𝟒𝟒
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝟒𝟒
𝟗𝟗
𝟓𝟓
𝟗𝟗
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏(𝟖𝟖 − 𝒙𝒙)
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟗𝟗𝟗𝟗 − 𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐𝟕𝟕𝟕𝟕 = 𝟗𝟗𝟗𝟗
𝟓𝟓
𝟗𝟗
𝒙𝒙 =
𝟗𝟗𝟗𝟗
𝟏𝟏𝟏𝟏
𝟓𝟓
= 𝟑𝟑
= 𝟑𝟑
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝟗𝟗
𝟒𝟒
𝟗𝟗
The length of 𝒙𝒙 is 𝟑𝟑 , and the length of 𝒚𝒚 is 𝟒𝟒 .
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4.
The angle bisector of an angle splits the opposite side of a triangle into lengths 𝟓𝟓 and 𝟔𝟔. The perimeter of the
triangle is 𝟑𝟑𝟑𝟑. Find the lengths of the other two sides.
Let 𝒛𝒛 be the scale factor of a similarity. By the angle bisector theorem, the side of the triangle adjacent to the
segment of length 𝟓𝟓 has length of 𝟓𝟓𝟓𝟓, and the side of the triangle adjacent to the segment of length 𝟔𝟔 has length of
𝟔𝟔𝟔𝟔. The sum of the sides is equal to the perimeter.
𝟓𝟓 + 𝟔𝟔 + 𝟓𝟓𝟓𝟓 + 𝟔𝟔𝟔𝟔 = 𝟑𝟑𝟑𝟑
𝟏𝟏𝟏𝟏 + 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟐𝟐𝟐𝟐
𝒛𝒛 = 𝟐𝟐
𝟓𝟓(𝟐𝟐) = 𝟏𝟏𝟏𝟏, and 𝟔𝟔(𝟐𝟐) = 𝟏𝟏𝟏𝟏 The lengths of the other two sides are 𝟏𝟏𝟏𝟏 and 𝟏𝟏𝟏𝟏.
Closing (5 minutes)

Explain the angle bisector theorem in your own words.

Explain how knowing that one of the angles of a triangle has been bisected allows you to determine unknown
side lengths of a triangle.
Exit Ticket (5 minutes)
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Name
Date
Lesson 18: Similarity and the Angle Bisector Theorem
Exit Ticket
1.
The sides of a triangle have lengths of 12, 16, and 21. An angle bisector meets the side of length 21. Find the
lengths 𝑥𝑥 and 𝑦𝑦.
2.
1
1
The perimeter of △ 𝑈𝑈𝑈𝑈𝑈𝑈 is 22 . ������⃗
𝑊𝑊𝑊𝑊 bisects ∠𝑈𝑈𝑈𝑈𝑈𝑈, 𝑈𝑈𝑈𝑈 = 2, and 𝑉𝑉𝑉𝑉 = 2 . Find 𝑈𝑈𝑈𝑈 and 𝑉𝑉𝑉𝑉.
2
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Exit Ticket Sample Solutions
1.
The sides of a triangle have lengths of 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, and 𝟐𝟐𝟐𝟐. An angle bisector meets the side of length 𝟐𝟐𝟐𝟐. Find the
lengths 𝒙𝒙 and 𝒚𝒚.
By the angle bisector theorem,
𝟐𝟐𝟐𝟐 − 𝒙𝒙 𝟏𝟏𝟏𝟏
=
𝒙𝒙
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏(𝟐𝟐𝟐𝟐 − 𝒙𝒙) = 𝟏𝟏𝟏𝟏𝟏𝟏
𝒚𝒚
=
𝒙𝒙
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
, and 𝒚𝒚 = 𝟐𝟐𝟐𝟐 − 𝒙𝒙, so
𝒚𝒚 = 𝟐𝟐𝟐𝟐 − 𝒙𝒙
𝒚𝒚 = 𝟐𝟐𝟐𝟐 − 𝟗𝟗
𝒚𝒚 = 𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟐𝟐𝟐𝟐𝟐𝟐
𝟗𝟗 = 𝒙𝒙
2.
𝟏𝟏
𝟏𝟏
The perimeter of △ 𝑼𝑼𝑼𝑼𝑼𝑼 is 𝟐𝟐𝟐𝟐 . �������⃗
𝑾𝑾𝑾𝑾 bisects ∠𝑼𝑼𝑼𝑼𝑼𝑼, 𝑼𝑼𝑼𝑼 = 𝟐𝟐, and 𝑽𝑽𝑽𝑽 = 𝟐𝟐 . Find 𝑼𝑼𝑼𝑼 and 𝑽𝑽𝑽𝑽.
By the angle bisector theorem,
𝟐𝟐
𝟐𝟐
.𝟓𝟓
=
𝑼𝑼𝑼𝑼
𝑽𝑽𝑽𝑽
𝟐𝟐
, so 𝑼𝑼𝑼𝑼 = 𝟐𝟐𝟐𝟐 and 𝑽𝑽𝑽𝑽 = 𝟐𝟐. 𝟓𝟓𝟓𝟓 for
𝟏𝟏
𝟐𝟐
some positive number 𝒙𝒙. The perimeter of the triangle is 𝟐𝟐𝟐𝟐 , so
𝟐𝟐 + 𝟐𝟐. 𝟓𝟓 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐. 𝟓𝟓𝟓𝟓 = 𝟐𝟐𝟐𝟐. 𝟓𝟓
𝟒𝟒. 𝟓𝟓 + 𝟒𝟒. 𝟓𝟓𝟓𝟓 = 𝟐𝟐𝟐𝟐. 𝟓𝟓
𝟒𝟒. 𝟓𝟓𝟓𝟓 = 𝟏𝟏𝟏𝟏
𝑼𝑼𝑼𝑼 = 𝟐𝟐𝟐𝟐 = 𝟐𝟐(𝟒𝟒) = 𝟖𝟖
𝒙𝒙 = 𝟒𝟒
𝑽𝑽𝑽𝑽 = 𝟐𝟐. 𝟓𝟓𝟓𝟓 = 𝟐𝟐. 𝟓𝟓(𝟒𝟒) = 𝟏𝟏𝟏𝟏
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Problem Set Sample Solutions
1.
𝒙𝒙 and 𝒚𝒚.
Using the angle bisector theorem,
𝒙𝒙
𝟓𝟓
=
𝟏𝟏
𝟔𝟔 − 𝒙𝒙 𝟖𝟖
𝟐𝟐
𝟏𝟏
𝟓𝟓 �𝟔𝟔 − 𝒙𝒙� = 𝟖𝟖𝟖𝟖
𝟐𝟐
𝟏𝟏
𝟑𝟑𝟑𝟑 − 𝟓𝟓𝟓𝟓 = 𝟖𝟖𝟖𝟖
𝟐𝟐
𝟏𝟏
𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐
𝟏𝟏
𝒙𝒙 = 𝟐𝟐
𝟐𝟐
2.
𝟏𝟏
𝟐𝟐
𝟏𝟏
𝟐𝟐
The sides of a triangle have lengths of 𝟓𝟓, 𝟖𝟖, and 𝟔𝟔 . An angle bisector meets the side of length 𝟔𝟔 . Find the lengths
𝒙𝒙
𝟓𝟓
𝟏𝟏
𝟐𝟐
= , and 𝒚𝒚 = 𝟔𝟔 − 𝒙𝒙, so
𝟖𝟖
𝒚𝒚
𝟏𝟏
𝒚𝒚 = 𝟔𝟔 − 𝒙𝒙
𝟐𝟐
𝟏𝟏
𝟏𝟏
𝒚𝒚 = 𝟔𝟔 − 𝟐𝟐
𝟐𝟐
𝟐𝟐
𝒚𝒚 = 𝟒𝟒
𝟏𝟏
𝟐𝟐
𝟏𝟏
𝟐𝟐
The sides of a triangle are 𝟏𝟏𝟏𝟏 , 𝟏𝟏𝟏𝟏 , and 𝟗𝟗. An angle bisector meets the side of length 𝟗𝟗. Find the lengths 𝒙𝒙 and 𝒚𝒚.
By the angle bisector theorem,
𝒙𝒙
𝟏𝟏𝟏𝟏. 𝟓𝟓
=
𝟗𝟗 − 𝒙𝒙 𝟏𝟏𝟏𝟏. 𝟓𝟓
𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 = 𝟏𝟏𝟏𝟏. 𝟓𝟓(𝟗𝟗 − 𝒙𝒙)
𝒙𝒙
𝒚𝒚
𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓 − 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓
𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓
=
𝟏𝟏
𝟐𝟐
𝟏𝟏, and
𝟏𝟏𝟏𝟏
𝟐𝟐
𝟏𝟏𝟏𝟏
𝒚𝒚 = 𝟗𝟗 − 𝒙𝒙, so
𝒚𝒚 = 𝟗𝟗 − 𝒙𝒙
𝒚𝒚 = 𝟗𝟗 − 𝟓𝟓. 𝟓𝟓
𝒚𝒚 = 𝟑𝟑. 𝟓𝟓
𝒙𝒙 = 𝟓𝟓. 𝟓𝟓
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Lesson 18
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
3.
𝟏𝟏
𝟔𝟔
In the diagram of triangle 𝑫𝑫𝑫𝑫𝑫𝑫 below, ����
𝑫𝑫𝑫𝑫 is an angle bisector, 𝑫𝑫𝑫𝑫 = 𝟖𝟖, 𝑫𝑫𝑫𝑫 = 𝟔𝟔, and 𝑬𝑬𝑬𝑬 = 𝟖𝟖 . Find 𝑭𝑭𝑭𝑭 and 𝑬𝑬𝑬𝑬.
𝟏𝟏
𝟔𝟔
���� is the angle bisector of angle 𝑫𝑫, 𝑭𝑭𝑭𝑭: 𝑮𝑮𝑮𝑮 = 𝑭𝑭𝑭𝑭: 𝑬𝑬𝑬𝑬 by the angle bisector theorem. If 𝑬𝑬𝑬𝑬 = 𝟖𝟖 , then
Since 𝑫𝑫𝑫𝑫
𝟏𝟏
𝟔𝟔
𝑭𝑭𝑭𝑭 = 𝟖𝟖 − 𝑮𝑮𝑮𝑮.
𝟏𝟏
�𝟖𝟖 − 𝑮𝑮𝑮𝑮� 𝟔𝟔
𝟔𝟔
=
𝑮𝑮𝑮𝑮
𝟖𝟖
𝟔𝟔
𝟏𝟏
𝟖𝟖 − 𝑮𝑮𝑮𝑮 = ∙ 𝑮𝑮𝑮𝑮
𝟖𝟖
𝟔𝟔
𝟏𝟏 𝟏𝟏𝟏𝟏
∙ 𝑮𝑮𝑮𝑮
𝟖𝟖 =
𝟖𝟖
𝟔𝟔
𝟒𝟒𝟒𝟒 𝟖𝟖
∙
= 𝑮𝑮𝑮𝑮
𝟔𝟔 𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐
= 𝑮𝑮𝑮𝑮
𝟔𝟔
𝟐𝟐
𝑮𝑮𝑮𝑮 = 𝟒𝟒
𝟑𝟑
4.
𝑭𝑭𝑭𝑭 = 𝑮𝑮𝑮𝑮 + 𝑭𝑭𝑭𝑭
𝟏𝟏
𝟐𝟐
= 𝟒𝟒 + 𝑭𝑭𝑭𝑭
𝟔𝟔
𝟑𝟑
𝟏𝟏
𝑭𝑭𝑭𝑭 = 𝟑𝟑
𝟐𝟐
𝟖𝟖
Given the diagram below and ∠𝑩𝑩𝑩𝑩𝑩𝑩 ≅ ∠𝑫𝑫𝑫𝑫𝑫𝑫, show that 𝑩𝑩𝑩𝑩: 𝑩𝑩𝑩𝑩 = 𝑪𝑪𝑪𝑪: 𝑪𝑪𝑪𝑪.
Using the given information, ����
𝑨𝑨𝑨𝑨 is the angle bisector of angle 𝑨𝑨. By the angle bisector theorem,
𝑩𝑩𝑩𝑩: 𝑪𝑪𝑪𝑪 = 𝑩𝑩𝑩𝑩: 𝑪𝑪𝑪𝑪, so
𝑩𝑩𝑩𝑩 𝑩𝑩𝑩𝑩
=
𝑪𝑪𝑪𝑪 𝑪𝑪𝑪𝑪
𝑩𝑩𝑩𝑩 ∙ 𝑪𝑪𝑪𝑪 = 𝑪𝑪𝑪𝑪 ∙ 𝑩𝑩𝑩𝑩
𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪
=
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Lesson 18
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M2
GEOMETRY
5.
The perimeter of triangle 𝑳𝑳𝑳𝑳𝑳𝑳 is 𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜. �����
𝑵𝑵𝑵𝑵 is the angle bisector of angle 𝑵𝑵, 𝑳𝑳𝑳𝑳 = 𝟑𝟑 𝐜𝐜𝐜𝐜, and 𝑿𝑿𝑿𝑿 = 𝟓𝟓 𝐜𝐜𝐜𝐜. Find
𝑳𝑳𝑳𝑳 and 𝑴𝑴𝑴𝑴.
����� is an angle bisector of angle 𝑵𝑵, by the angle bisector theorem, 𝑿𝑿𝑿𝑿: 𝑿𝑿𝑿𝑿 = 𝑳𝑳𝑳𝑳: 𝑴𝑴𝑴𝑴; thus, 𝑳𝑳𝑳𝑳: 𝑴𝑴𝑴𝑴 = 𝟑𝟑: 𝟓𝟓.
Since 𝑵𝑵𝑵𝑵
Therefore, 𝑳𝑳𝑳𝑳 = 𝟑𝟑𝟑𝟑 and 𝑴𝑴𝑴𝑴 = 𝟓𝟓𝟓𝟓 for some positive number 𝒙𝒙. The perimeter of the triangle is 𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜, so
𝑿𝑿𝑿𝑿 + 𝑿𝑿𝑿𝑿 + 𝑴𝑴𝑴𝑴 + 𝑳𝑳𝑳𝑳 = 𝟑𝟑𝟑𝟑
𝟑𝟑 + 𝟓𝟓 + 𝟑𝟑𝒙𝒙 + 𝟓𝟓𝟓𝟓 = 𝟑𝟑𝟑𝟑
𝟖𝟖 + 𝟖𝟖𝟖𝟖 = 𝟑𝟑𝟑𝟑
𝟖𝟖𝟖𝟖 = 𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑 = 𝟑𝟑(𝟑𝟑) = 𝟗𝟗 and 𝟓𝟓𝟓𝟓 = 𝟓𝟓(𝟑𝟑) = 𝟏𝟏𝟏𝟏
𝒙𝒙 = 𝟑𝟑
𝑳𝑳𝑳𝑳 = 𝟗𝟗 𝐜𝐜𝐜𝐜 and 𝑴𝑴𝑴𝑴 = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
6.
Given 𝑪𝑪𝑪𝑪 = 𝟑𝟑, 𝑫𝑫𝑫𝑫 = 𝟒𝟒, 𝑩𝑩𝑩𝑩 = 𝟒𝟒, 𝑭𝑭𝑭𝑭 = 𝟓𝟓, 𝑨𝑨𝑨𝑨 = 𝟔𝟔, and ∠𝑪𝑪𝑪𝑪𝑪𝑪 ≅ ∠𝑫𝑫𝑫𝑫𝑫𝑫 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩 ≅ ∠𝑭𝑭𝑭𝑭𝑭𝑭, find the perimeter of
���� is the angle bisector of angle 𝑪𝑪𝑪𝑪𝑪𝑪, so by the
𝑨𝑨𝑨𝑨
angle bisector theorem, 𝑪𝑪𝑪𝑪: 𝑩𝑩𝑩𝑩 = 𝑪𝑪𝑪𝑪: 𝑩𝑩𝑩𝑩.
𝟑𝟑 𝑪𝑪𝑪𝑪
=
𝟔𝟔
𝟒𝟒
𝟒𝟒 ∙ 𝑪𝑪𝑪𝑪 = 𝟏𝟏𝟏𝟏
𝑪𝑪𝑪𝑪 = 𝟒𝟒. 𝟓𝟓
����
𝑨𝑨𝑨𝑨 is the angle bisector of angle 𝑩𝑩𝑩𝑩𝑩𝑩, so by the
angle bisector theorem, 𝑩𝑩𝑩𝑩: 𝑬𝑬𝑬𝑬 = 𝑩𝑩𝑩𝑩: 𝑬𝑬𝑬𝑬.
𝟔𝟔
𝟒𝟒
=
𝟓𝟓 𝑬𝑬𝑬𝑬
𝟒𝟒 ∙ 𝑬𝑬𝑬𝑬 = 𝟑𝟑𝟑𝟑
𝑬𝑬𝑬𝑬 = 𝟕𝟕. 𝟓𝟓
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 = 𝑪𝑪𝑪𝑪 + 𝑫𝑫𝑫𝑫 + 𝑩𝑩𝑩𝑩 + 𝑭𝑭𝑭𝑭 + 𝑬𝑬𝑬𝑬 + 𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 = 𝟑𝟑 + 𝟒𝟒 + 𝟒𝟒 + 𝟓𝟓 + 𝟕𝟕. 𝟓𝟓 + 𝟒𝟒. 𝟓𝟓
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 = 𝟐𝟐𝟐𝟐
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Lesson 18
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
7.
���� at 𝑫𝑫 such that 𝑪𝑪𝑪𝑪: 𝑩𝑩𝑩𝑩 = 𝑪𝑪𝑪𝑪: 𝑩𝑩𝑩𝑩, show that ∠𝑪𝑪𝑪𝑪𝑪𝑪 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩. Explain how this proof relates to the
If ����
𝑨𝑨𝑨𝑨 meets 𝑩𝑩𝑩𝑩
angle bisector theorem.
This is a proof of the converse to the angle bisector theorem.
∠𝑪𝑪𝑪𝑪𝑪𝑪 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩
����
Alt. Int. ∠’s ����
𝑪𝑪𝑪𝑪 ∥ 𝑨𝑨𝑨𝑨
∆𝑪𝑪𝑪𝑪𝑪𝑪 ~∆𝑩𝑩𝑩𝑩𝑩𝑩
AA ~
Vert. ∠’s
∠𝑨𝑨𝑨𝑨𝑨𝑨 ≅ ∠𝑬𝑬𝑬𝑬𝑬𝑬
𝑪𝑪𝑪𝑪
𝑫𝑫𝑫𝑫
=
𝑪𝑪𝑪𝑪
𝑩𝑩𝑩𝑩
𝑪𝑪𝑪𝑪
implies then that
∠𝑪𝑪𝑪𝑪𝑪𝑪 ≅ ∠𝑪𝑪𝑪𝑪𝑪𝑪
Base ∠’s
Given
𝑪𝑪𝑪𝑪
Corr. sides of ~ △’s
𝑩𝑩𝑩𝑩
=
𝑩𝑩𝑩𝑩
𝑪𝑪𝑪𝑪 = 𝑪𝑪𝑪𝑪, so △ 𝑨𝑨𝑨𝑨𝑨𝑨 is isosceles.
Substitution
∠𝑪𝑪𝑪𝑪𝑪𝑪 ≅ ∠𝑩𝑩𝑩𝑩𝑩𝑩
8.
In the diagram below, ����
𝑬𝑬𝑬𝑬 ≅ �����
𝑫𝑫𝑫𝑫, ����
𝑩𝑩𝑩𝑩 bisects ∠𝑨𝑨𝑨𝑨𝑨𝑨, 𝑨𝑨𝑨𝑨 = 𝟒𝟒, and 𝑫𝑫𝑫𝑫 = 𝟖𝟖. Prove that △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑪𝑪𝑪𝑪𝑪𝑪.
Using given information, 𝑬𝑬𝑬𝑬 = 𝑫𝑫𝑫𝑫, so it follows that 𝑬𝑬𝑬𝑬 = 𝟐𝟐𝟐𝟐𝟐𝟐. Since ����
𝑩𝑩𝑩𝑩 bisects ∠𝑨𝑨𝑨𝑨𝑨𝑨, ∠𝑨𝑨𝑨𝑨𝑨𝑨 ≅ ∠𝑪𝑪𝑪𝑪𝑪𝑪. Also,
by the angle bisector theorem,
Since
𝑬𝑬𝑬𝑬
𝑫𝑫𝑫𝑫
=
𝑪𝑪𝑪𝑪
𝑨𝑨𝑨𝑨
𝟐𝟐
GEO-M2-TE-1.3.0-07.2015
=
𝑩𝑩𝑩𝑩
, which means
𝑩𝑩𝑩𝑩
=
𝟖𝟖
𝟒𝟒
𝟐𝟐
= .
𝟏𝟏
= , and ∠𝑨𝑨𝑨𝑨𝑨𝑨 ≅ ∠𝑪𝑪𝑪𝑪𝑪𝑪, △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑪𝑪𝑪𝑪𝑪𝑪 by the SAS criterion for showing similar triangles.
𝟏𝟏
Lesson 18:
𝑪𝑪𝑪𝑪
𝑨𝑨𝑨𝑨
Similarity and the Angle Bisector Theorem
282
Lesson 19
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Lesson 19: Families of Parallel Lines and the Circumference
of the Earth
Student Outcomes

Students understand that parallel lines cut transversals into proportional segments. They use ratios between
corresponding line segments in different transversals and ratios within line segments on the same transversal.

Students understand Eratosthenes’ method for measuring the earth and solve related problems.
Lesson Notes
Students revisit their study of side splitters and the triangle side-splitter theorem to understand how parallel lines cut
transversals into proportional segments. The theorem is a natural consequence of side splitters; allow students the
opportunity to make as many connections of their own while guiding them forward. The second half of the lesson is
teacher-led and describes Eratosthenes’ calculation of the earth’s circumference. The segment lays the foundation for
Lesson 20, where students study another application of geometry by the ancient Greeks.
Classwork
Opening (7 minutes)

Consider △ 𝑂𝑂𝑂𝑂𝑂𝑂 below with side splitter 𝐶𝐶𝐶𝐶.

Recall that we say line segment 𝐶𝐶𝐶𝐶 splits sides ����
𝑂𝑂𝑂𝑂 and ����
𝑂𝑂𝑂𝑂 proportionally if
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
.
Lesson 19:
GEO-M2-TE-1.3.0-07.2015
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
or equivalently if
Families of Parallel Lines and the Circumference of the Earth
283
Lesson 19
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY

Using 𝑥𝑥, 𝑦𝑦, 𝑥𝑥 ′ , 𝑦𝑦′ as the lengths of the indicated segments, how can we rewrite
possible.

MP.7

𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
is the same as
𝑥𝑥+𝑦𝑦
𝑥𝑥
=
𝑥𝑥 ′ +𝑦𝑦 ′
𝑥𝑥 ′
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
=
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
? Simplify as much as
.
𝑥𝑥 + 𝑦𝑦 𝑥𝑥 ′ + 𝑦𝑦 ′
=
𝑥𝑥 ′
𝑥𝑥
𝑦𝑦 ′
𝑦𝑦
1+ = 1+ ′
𝑥𝑥
𝑥𝑥
𝑦𝑦 𝑦𝑦 ′
=
𝑥𝑥 𝑥𝑥 ′
Thus, another way to say that segment 𝐶𝐶𝐶𝐶 splits the sides proportionally is to say that the ratios 𝑥𝑥: 𝑦𝑦 and 𝑥𝑥 ′ : 𝑦𝑦′
are equal.
As scaffolding, consider reviewing a parallel example with numerical values:
Opening Exercise (4 minutes)
Suggest students use the numerical example from the Opening to help them with the Opening Exercise.
Opening Exercise
Show 𝒙𝒙: 𝒚𝒚 = 𝒙𝒙′ : 𝒚𝒚′ is equivalent to 𝒙𝒙: 𝒙𝒙′ = 𝒚𝒚: 𝒚𝒚′ .
𝒙𝒙 𝒙𝒙′
=
𝒚𝒚 𝒚𝒚′
𝒙𝒙𝒙𝒙′ = 𝒙𝒙′𝒚𝒚
𝒙𝒙𝒙𝒙′
𝒙𝒙′ 𝒚𝒚
=
𝒙𝒙′𝒚𝒚′ 𝒙𝒙′𝒚𝒚′
𝒙𝒙
𝒚𝒚
=
𝒙𝒙′ 𝒚𝒚′
Discussion (10 minutes)
Lead students through a discussion to prove that parallel lines cut transversals into proportional segments.

We will use our understanding of side splitters to prove the following theorem.
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Lesson 19
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY

THEOREM: Parallel lines cut transversals into proportional segments. If parallel lines are intersected by two
transversals, then the ratios of the segments determined along each transversal between the parallel lines are
equal.

Draw three parallel lines that are cut by two transversals. Label the following lengths of the line segments as
𝑥𝑥, 𝑦𝑦, 𝑥𝑥′, and 𝑦𝑦′.

To prove the theorem, we must show that 𝑥𝑥: 𝑦𝑦 = 𝑥𝑥 ′ : 𝑦𝑦 ′ . Why would this be enough to show that the ratios of
the segments along each transversal between the parallel lines are equal?

This would be enough because the relationship 𝑥𝑥: 𝑦𝑦 = 𝑥𝑥 ′ : 𝑦𝑦 ′ implies that 𝑥𝑥: 𝑥𝑥′ = 𝑦𝑦: 𝑦𝑦 ′ .

Draw a segment so that two triangles are formed between the parallel lines and between the transversals.

Label each portion of the segment separated by a pair of parallel lines as 𝑎𝑎 and 𝑏𝑏.

Are there any conclusions we can draw based on the diagram?


We can apply the triangle side-splitter theorem twice to see that 𝑥𝑥: 𝑦𝑦 = 𝑎𝑎: 𝑏𝑏 and 𝑎𝑎: 𝑏𝑏 = 𝑥𝑥 ′ : 𝑦𝑦′.
So, 𝑥𝑥: 𝑦𝑦 = 𝑎𝑎: 𝑏𝑏, and 𝑥𝑥 ′ : 𝑦𝑦 ′ = 𝑎𝑎: 𝑏𝑏. Thus, 𝑥𝑥: 𝑦𝑦 = 𝑥𝑥 ′ : 𝑦𝑦 ′ .

Therefore, we have proved the theorem: Parallel lines cut transversals into proportional segments.

Notice that the two equations 𝑥𝑥: 𝑦𝑦 = 𝑥𝑥 ′ : 𝑦𝑦′ and 𝑥𝑥: 𝑥𝑥 ′ = 𝑦𝑦: 𝑦𝑦′ are equivalent as described above.
Exercises 1–2 (4 minutes)
Students apply their understanding that parallel lines cut transversals into proportional segments to determine the
unknown length in each problem.
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Exercises 1–2
Lines that appear to be parallel are in fact parallel.
1.
𝒙𝒙 = 𝟒𝟒
2.
𝒙𝒙 = 𝟏𝟏. 𝟓𝟓
Discussion (13 minutes)

The word geometry is Greek for geos (earth) and metron (measure). A Greek named Eratosthenes, who lived
over 2,200 years ago, used geometry to measure the circumference of the earth. The Greeks knew the earth
was a sphere. They also knew the sun was so far away that rays from the sun (as they met the earth) were, for
all practical purposes, parallel to each other. Using these two facts, here is how Eratosthenes calculated the
circumference of the earth.

Eratosthenes lived in Egypt. He calculated the circumference of the earth without ever leaving Egypt. Every
summer solstice in the city of Syene, the sun was directly overhead at noon. Eratosthenes knew this because
at noon on that day alone, the sun would reflect directly (perpendicularly) off the bottom of a deep well.

On the same day at noon in Alexandria, a city north of Syene, the angle between the perpendicular to the
ground and the rays of the sun was about 7.2°. We do not have a record of how Eratosthenes found this
measurement, but here is one possible explanation:

Imagine a pole perpendicular to the ground (in Alexandria), as well as its shadow. With a tool such as a
protractor, the shadow can be used to determine the measurement of the angle between the ray and the pole,
regardless of the height of the pole.
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
You might argue that we cannot theorize such a method because we do not know the height of the pole.
However, because of the angle the rays make with the ground and the 90° angle of the pole with the ground,
the triangle formed by the ray, the pole, and the shadow are all similar triangles, regardless of the height of the
pole. Why must the triangles all be similar?

The situation satisfies the AA criterion; therefore, the triangles must be similar.
Remind students that the discussion points here are similar to the work done in Lesson 16 on indirect measurement.

Therefore, if the pole had a height of 10 meters, the shadow had a length of 1.26 meters, or if the pole had a
height of 1 meter, the shadow had a length of 0.126 meters.

This measurement of the angle between the sun’s rays and the pole was instrumental to the calculation of the
circumference. Eratosthenes used it to calculate the angle between the two cities from the center of the
earth.

You might think it necessary to go to the center of the earth to determine this measurement, but it is not.
Eratosthenes extrapolated both the sun’s rays and the ray perpendicular to the ground in Alexandria. Notice
that the perpendicular at Alexandria acts as a transversal to the parallel rays of the sun, namely, the sun ray
that passes through Syene and the center of the earth, and the sun ray that forms the triangle with the top of
the pole and the shadow of the pole. Using the alternate interior angles determined by the transversal (the
extrapolated pole) that intersects the parallel lines (extrapolated sun rays), Eratosthenes found the angle
between the two cities to be 7.2°.

How can this measurement be critical to finding the entire circumference?

We can divide 7.2° into 360°, which gives us a fraction of the circumference, or how many times we
have to multiply the distance between Alexandria to Syene to get the whole circumference.
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


Eratosthenes divided 360° by 7.2°, which yielded 50. So, the distance from Syene to Alexandria is
1
50
of the
circumference of the earth. The only thing that is missing is that distance between Syene and Alexandria,
which was known to be about 5,000 stades; the stade was a Greek unit of measurement, and
So, Eratosthenes’ estimate was about 50 ⋅ 5,000 ⋅ 600 feet, or about 28,400 miles. A modern-day estimate
for the circumference of the earth at the equator is about 24,900 miles.
It is remarkable that around 240 B.C.E., basic geometry helped determine a very close approximation of the
circumference of the earth.
Closing (2 minutes)

In your own words, explain how parallel lines cut transversals into proportional segments.

What are some assumptions that Eratosthenes must have made as part of his calculation?

The rays of the sun are all parallel.

The earth is perfectly spherical.
Share the following link to a video on Eratosthenes and his calculation of the earth’s circumference.
http://www.youtube.com/watch?v=wnElDaV4esg&feature=youtu.be (The same video was used in Module 1, Lesson 11.)
Exit Ticket (5 minutes)
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Name
Date
Lesson 19: Families of Parallel Lines and the Circumference of the
Earth
Exit Ticket
1.
Given the diagram shown, ����
𝐴𝐴𝐴𝐴 ∥ ����
𝐵𝐵𝐵𝐵 ∥ ���
𝐶𝐶𝐶𝐶 , 𝐴𝐴𝐴𝐴 = 6.5 cm, 𝐺𝐺𝐺𝐺 = 7.5 cm, and 𝐻𝐻𝐻𝐻 = 18 cm. Find 𝐵𝐵𝐵𝐵.
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2.
Martin the Martian lives on Planet Mart. Martin wants to know the circumference of Planet Mart, but it is too large
to measure directly. He uses the same method as Eratosthenes by measuring the angle of the sun’s rays in two
locations. The sun shines on a flagpole in Martinsburg, but there is no shadow. At the same time, the sun shines on
a flagpole in Martville, and a shadow forms a 10° angle with the pole. The distance from Martville to Martinsburg is
294 miles. What is the circumference of Planet Mart?
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Exit Ticket Sample Solutions
1.
Given the diagram shown, ���
𝑨𝑨𝑨𝑨
���, 𝑨𝑨𝑨𝑨 = 𝟔𝟔. 𝟓𝟓 𝐜𝐜𝐜𝐜, 𝑮𝑮𝑮𝑮 = 𝟕𝟕. 𝟓𝟓 𝐜𝐜𝐜𝐜, and 𝑯𝑯𝑯𝑯 = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜. Find 𝑩𝑩𝑩𝑩.
���� ∥ 𝑪𝑪𝑪𝑪
∥ 𝑩𝑩𝑩𝑩
Parallel lines cut transversals proportionally; therefore,
𝑩𝑩𝑩𝑩
𝑨𝑨𝑨𝑨
=
𝑯𝑯𝑯𝑯
, and thus,
𝑮𝑮𝑮𝑮
𝑯𝑯𝑯𝑯
(𝑨𝑨𝑨𝑨)
𝑮𝑮𝑮𝑮
𝟏𝟏𝟏𝟏
(𝟔𝟔. 𝟓𝟓)
𝒙𝒙 =
𝟕𝟕. 𝟓𝟓
𝑩𝑩𝑩𝑩 =
The length of ����
𝑩𝑩𝑩𝑩 is 𝟏𝟏𝟏𝟏. 𝟔𝟔 𝐜𝐜𝐜𝐜.
2.
𝒙𝒙 = 𝟏𝟏𝟏𝟏. 𝟔𝟔.
Martin the Martian lives on Planet Mart. Martin wants to know the circumference of Planet Mart, but it is too large
to measure directly. He uses the same method as Eratosthenes by measuring the angle of the sun’s rays in two
locations. The sun shines on a flagpole in Martinsburg, but there is no shadow. At the same time, the sun shines on
a flagpole in Martville, and a shadow forms a 𝟏𝟏𝟏𝟏° angle with the pole. The distance from Martville to Martinsburg is
𝟐𝟐𝟐𝟐𝟐𝟐 miles. What is the circumference of Planet Mart?
The distance from Martinsburg to Martville makes up only 𝟏𝟏𝟏𝟏° of the total rotation about the planet. There are
𝟑𝟑𝟑𝟑𝟑𝟑° in the complete circumference of the planet, and 𝟑𝟑𝟑𝟑 ⋅ 𝟏𝟏𝟏𝟏° = 𝟑𝟑𝟑𝟑𝟑𝟑°, so 𝟑𝟑𝟑𝟑 ⋅ 𝟐𝟐𝟐𝟐𝟐𝟐 miles = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 miles.
The circumference of planet Mart is 𝟏𝟏𝟏𝟏, 𝟓𝟓𝟓𝟓𝟓𝟓 miles.
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Problem Set Sample Solutions
1.
��� ∥ 𝑳𝑳𝑳𝑳
���� ∥ 𝑸𝑸𝑸𝑸
����, and ����
���� ∥ 𝑫𝑫𝑫𝑫
���� ∥ 𝑪𝑪𝑪𝑪
����. Use the additional information given in
Given the diagram shown, ����
𝑨𝑨𝑨𝑨 ∥ 𝑮𝑮𝑮𝑮
𝑨𝑨𝑨𝑨 ∥ 𝑩𝑩𝑩𝑩
each part below to answer the questions:
a.
If 𝑮𝑮𝑮𝑮 = 𝟒𝟒, what is 𝑯𝑯𝑯𝑯?
𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮 forms a parallelogram since opposite sides are parallel, and opposite sides of a parallelogram are
equal in length; therefore, 𝑯𝑯𝑯𝑯 = 𝑮𝑮𝑮𝑮 = 𝟒𝟒.
b.
If 𝑮𝑮𝑮𝑮 = 𝟒𝟒, 𝑳𝑳𝑳𝑳 = 𝟗𝟗, and 𝑿𝑿𝑿𝑿 = 𝟓𝟓, what is 𝒀𝒀𝒀𝒀?
Parallel lines cut transversals proportionally; therefore, it is true that
𝟗𝟗
(𝟓𝟓)
𝟒𝟒
𝟏𝟏
𝒀𝒀𝒀𝒀 = 𝟏𝟏𝟏𝟏
𝟒𝟒
𝒀𝒀𝒀𝒀 =
c.
𝑮𝑮𝑮𝑮
𝑳𝑳𝑳𝑳
=
𝑿𝑿𝑿𝑿
𝒀𝒀𝒀𝒀
, and likewise 𝒀𝒀𝒀𝒀 =
𝑳𝑳𝑳𝑳
(𝑿𝑿𝑿𝑿).
𝑮𝑮𝑮𝑮
Using information from part (b), if 𝑪𝑪𝑪𝑪 = 𝟏𝟏𝟏𝟏, what is 𝑾𝑾𝑾𝑾?
By the same argument used in part (a), 𝑰𝑰𝑰𝑰 = 𝑮𝑮𝑮𝑮 = 𝟒𝟒. Parallel lines cut transversals proportionally; therefore,
it is true that
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𝑪𝑪𝑪𝑪
𝑰𝑰𝑰𝑰
=
𝑾𝑾𝑾𝑾
𝑿𝑿𝑿𝑿
, and likewise 𝑾𝑾𝑾𝑾 =
𝑪𝑪𝑪𝑪
(𝑿𝑿𝑿𝑿).
𝑰𝑰𝑰𝑰
𝟏𝟏𝟏𝟏
(𝟓𝟓)
𝟒𝟒
𝟏𝟏
𝑾𝑾𝑾𝑾 = 𝟐𝟐𝟐𝟐
𝟐𝟐
𝑾𝑾𝑾𝑾 =
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2.
Use your knowledge about families of parallel lines to find the coordinates of point 𝑷𝑷 on the coordinate plane
below.
The given lines on the coordinate plane are parallel because they have the same slope 𝒎𝒎 = 𝟑𝟑. First, draw a
horizontal transversal through points (𝟖𝟖, −𝟐𝟐) and (𝟏𝟏𝟏𝟏, −𝟐𝟐) and a second transversal through points (𝟏𝟏𝟏𝟏, 𝟒𝟒) and
(𝟏𝟏𝟏𝟏, −𝟒𝟒). The transversals intersect at (𝟏𝟏𝟏𝟏, −𝟐𝟐). Parallel lines cut transversals proportionally, so using horizontal
and vertical distances,
Point 𝑷𝑷 is
𝟐𝟐
𝟑𝟑
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𝟐𝟐
=
𝟐𝟐
𝒙𝒙
, where 𝒙𝒙 represents the distance from point (𝟏𝟏𝟏𝟏, −𝟐𝟐) to 𝑷𝑷.
𝟐𝟐
(𝟐𝟐)
𝟔𝟔
𝟒𝟒 𝟐𝟐
𝒙𝒙 = =
𝟔𝟔 𝟑𝟑
𝒙𝒙 =
𝟐𝟐
𝟑𝟑
𝟐𝟐
𝟑𝟑
unit more than 𝟏𝟏𝟏𝟏, or 𝟏𝟏𝟏𝟏 , so the coordinates of point 𝑷𝑷 are �𝟏𝟏𝟏𝟏 , −𝟐𝟐�.
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3.
𝟏𝟏
𝟐𝟐
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 and 𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 are both trapezoids with bases ����
𝑨𝑨𝑨𝑨, ����
𝑭𝑭𝑭𝑭, and ����
𝑪𝑪𝑪𝑪. The perimeter of trapezoid 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 is 𝟐𝟐𝟐𝟐 . If the
𝟓𝟓
𝟖𝟖
ratio of 𝑨𝑨𝑨𝑨: 𝑭𝑭𝑭𝑭 is 𝟏𝟏: 𝟑𝟑, 𝑨𝑨𝑨𝑨 = 𝟕𝟕, and 𝑬𝑬𝑬𝑬 = 𝟓𝟓 , find 𝑨𝑨𝑨𝑨, 𝑭𝑭𝑭𝑭, and 𝑩𝑩𝑩𝑩.
The bases of a trapezoid are parallel, and since ����
𝑪𝑪𝑪𝑪 serves as a base for both trapezoids, it follows that ����
𝑨𝑨𝑨𝑨, ����
𝑭𝑭𝑭𝑭, and
���� are parallel line segments. Parallel lines cut transversals proportionally, so it must be true that
𝑪𝑪𝑪𝑪
By the given information,
𝑨𝑨𝑨𝑨
𝑭𝑭𝑭𝑭
𝟏𝟏
𝑩𝑩𝑩𝑩 𝟏𝟏
=
𝟓𝟓 𝟑𝟑
𝟓𝟓
𝟖𝟖
𝟏𝟏 𝟓𝟓
𝑩𝑩𝑩𝑩 = �𝟓𝟓 �
𝟑𝟑 𝟖𝟖
𝟕𝟕
𝟏𝟏𝟏𝟏
= 𝟏𝟏
𝑩𝑩𝑩𝑩 =
𝟖𝟖
𝟖𝟖
𝟕𝟕
𝟖𝟖
𝑨𝑨𝑨𝑨
𝑭𝑭𝑭𝑭
=
𝟓𝟓
𝟖𝟖
𝑩𝑩𝑩𝑩
𝑬𝑬𝑬𝑬
𝟏𝟏
= .
𝟑𝟑
𝟏𝟏
𝟐𝟐
= , it follows that 𝑭𝑭𝑭𝑭 = 𝟑𝟑𝟑𝟑𝟑𝟑. Also, 𝑩𝑩𝑩𝑩 = 𝑩𝑩𝑩𝑩 + 𝑬𝑬𝑬𝑬, so 𝑩𝑩𝑩𝑩 = 𝟏𝟏 + 𝟓𝟓 = 𝟕𝟕 .
𝟑𝟑
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = 𝑨𝑨𝑨𝑨 + 𝑩𝑩𝑩𝑩 + 𝑫𝑫𝑫𝑫 + 𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨 + 𝑭𝑭𝑭𝑭 = 𝑨𝑨𝑨𝑨, and, by the given ratio, 𝑭𝑭𝑭𝑭 = 𝟑𝟑𝟑𝟑𝟑𝟑, so 𝑨𝑨𝑨𝑨 = 𝑨𝑨𝑨𝑨 + 𝟑𝟑𝟑𝟑𝟑𝟑 = 𝟒𝟒𝟒𝟒𝟒𝟒.
𝟏𝟏
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = 𝟕𝟕 + 𝟕𝟕 + 𝟒𝟒 + 𝟒𝟒𝟒𝟒𝟒𝟒
𝟐𝟐
𝟏𝟏
𝟏𝟏
𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟏𝟏 + 𝟒𝟒𝟒𝟒𝟒𝟒
𝟐𝟐
𝟐𝟐
𝟔𝟔 = 𝟒𝟒𝟒𝟒𝟒𝟒
By substituting
𝟑𝟑
𝟐𝟐
for 𝑨𝑨𝑨𝑨,
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𝟑𝟑
= 𝑨𝑨𝑨𝑨
𝟐𝟐
𝟑𝟑
𝑭𝑭𝑭𝑭 = 𝟑𝟑 � �
𝟐𝟐
𝟗𝟗
𝟏𝟏
𝑭𝑭𝑭𝑭 = = 𝟒𝟒 .
𝟐𝟐
𝟐𝟐
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4.
Given the diagram and the ratio of 𝒂𝒂: 𝒃𝒃 is 𝟑𝟑: 𝟐𝟐, answer each question below:
a.
Write an equation for 𝒂𝒂𝒏𝒏 in terms of 𝒃𝒃𝒏𝒏 .
𝒂𝒂𝒏𝒏 =
b.
Write an equation for 𝒃𝒃𝒏𝒏 in terms of 𝒂𝒂𝒏𝒏 .
𝒃𝒃𝒏𝒏 =
c.
𝟐𝟐
𝒂𝒂
𝟑𝟑 𝒏𝒏
Use one of your equations to find 𝒃𝒃𝟏𝟏 in terms of 𝒂𝒂 if 𝒂𝒂𝟏𝟏 = 𝟏𝟏. 𝟐𝟐(𝒂𝒂).
𝒃𝒃𝒏𝒏 =
𝒃𝒃𝟏𝟏 =
𝒃𝒃𝟏𝟏 =
d.
𝟑𝟑
𝒃𝒃
𝟐𝟐 𝒏𝒏
𝟐𝟐
𝒂𝒂
𝟑𝟑 𝒏𝒏
𝟐𝟐 𝟏𝟏𝟏𝟏
� 𝒂𝒂�
𝟑𝟑 𝟏𝟏𝟏𝟏
𝟒𝟒
𝒂𝒂
𝟓𝟓
What is the relationship between 𝒃𝒃𝟏𝟏 and 𝒃𝒃?
𝟑𝟑
𝟐𝟐
𝟒𝟒 𝟑𝟑
𝟓𝟓 𝟐𝟐
𝟔𝟔
𝟓𝟓
Using the equation from parts (a) and (c), 𝒂𝒂 = 𝒃𝒃, so 𝒃𝒃𝟏𝟏 = � 𝒃𝒃�; thus, 𝒃𝒃𝟏𝟏 = 𝒃𝒃.
e.
What constant, 𝒄𝒄, relates 𝒃𝒃𝟏𝟏 and 𝒃𝒃? Is this surprising? Why or why not?
The constant relating 𝒃𝒃𝟏𝟏 and 𝒃𝒃 is the same constant relating 𝒂𝒂𝟏𝟏 to 𝒂𝒂, 𝒄𝒄 =
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𝟏𝟏𝟏𝟏
= 𝟏𝟏. 𝟐𝟐.
𝟏𝟏𝟏𝟏
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f.
Using the formula 𝒂𝒂𝒏𝒏 = 𝒄𝒄 ⋅ 𝒂𝒂𝒏𝒏−𝟏𝟏, find 𝒂𝒂𝟑𝟑 in terms of 𝒂𝒂.
𝒂𝒂𝟏𝟏 =
𝒂𝒂𝟐𝟐 =
𝒂𝒂𝟐𝟐 =
𝒂𝒂𝟐𝟐 =
g.
5.
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
� 𝒂𝒂�
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
𝒂𝒂𝟑𝟑 =
𝟑𝟑𝟑𝟑
𝒂𝒂
𝟐𝟐𝟐𝟐
𝒂𝒂𝟑𝟑 =
𝟏𝟏𝟏𝟏𝟏𝟏
𝒂𝒂
𝟏𝟏𝟏𝟏𝟏𝟏
𝒂𝒂𝟑𝟑 =
Using the formula 𝒃𝒃𝒏𝒏 = 𝒄𝒄 ⋅ 𝒃𝒃𝒏𝒏−𝟏𝟏, find 𝒃𝒃𝟑𝟑 in terms of 𝒃𝒃.
𝒃𝒃𝟑𝟑 =
h.
𝟏𝟏𝟏𝟏
(𝒂𝒂)
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑
� 𝒂𝒂�
𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐
𝟒𝟒𝟒𝟒𝟒𝟒
𝒂𝒂
𝟐𝟐𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐
𝒂𝒂
𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐𝟐𝟐
𝒃𝒃
𝟏𝟏𝟏𝟏𝟏𝟏
Use your answers from parts (f) and (g) to calculate the value of the ratio of 𝒂𝒂𝟑𝟑 : 𝒃𝒃𝟑𝟑 .
𝟐𝟐𝟐𝟐𝟐𝟐
𝒂𝒂𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏 𝒂𝒂 𝒂𝒂 𝟑𝟑
=
= =
𝒃𝒃𝟑𝟑 𝟐𝟐𝟐𝟐𝟐𝟐 𝒃𝒃 𝒃𝒃 𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏
Julius wants to try to estimate the circumference of the earth based on measurements made near his home. He
cannot find a location near his home where the sun is straight overhead. Will he be able to calculate the
circumference of the earth? If so, explain and draw a diagram to support your claim.
Note to the teacher: This problem is very open-ended, requires critical thinking, and may not be suitable for all
students. The problem may be scaffolded by providing a diagram with possible measurements that Julius made
based on the description in the student solution below.
Possible solution: If Julius can find two locations such that those locations and their shadows lie in the same straight
path, then the difference of the shadows’ angles can be used as part of the 𝟑𝟑𝟑𝟑𝟑𝟑° in the earth’s circumference. The
distance between those two locations corresponds with that difference of angles.
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Lesson 20: How Far Away Is the Moon?
Student Outcomes

Students understand how the Greeks measured the distance from the earth to the moon and solve related
problems.
Lesson Notes
In Lesson 20, students learn how to approximate the distance of the moon from the earth. Around the same time that
Eratosthenes approximated the circumference of the earth, Greek astronomer Aristarchus found a way to determine the
distance to the moon. Less of the history is provided in the lesson, as a more comprehensive look at eclipses had to be
included alongside the central calculations. The objective of the lesson is to provide an understandable method of how
the distance was calculated; assumptions are folded in to expedite the complete story, but teachers are encouraged to
bring to light details that are left in the notes as they see fit for students.
Classwork
Opening Exercise (4 minutes)
Opening Exercise
What is a solar eclipse? What is a lunar eclipse?
A solar eclipse occurs when the moon passes between the earth and the sun, and a lunar eclipse occurs when the earth
passes between the moon and the sun.

In fact, we should imagine a solar eclipse occurring when the moon passes between the earth and the sun, and
the earth, sun, and moon lie on a straight line, and similarly so for a lunar eclipse.
Discussion (30 minutes)
Lead students through a conversation regarding the details of solar and lunar eclipses.

A total solar eclipse lasts only a few minutes because the sun and moon appear to be the same size.

How would appearances change if the moon were closer to the earth?


The moon would appear larger, and the eclipse would last longer. What if the moon were farther away from
the earth? Would we experience a total solar eclipse?


The moon would appear larger.
The moon would appear smaller, and it would not be possible for a total solar eclipse to occur, because
the moon would appear as a dark dot blocking only part of the sun.
Sketch a diagram of a solar eclipse.
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Students’ knowledge on eclipses varies. This is an opportunity for students to share what they know regarding eclipses.
Allow a minute of discussion, and then guide them through a basic description of the moon’s shadow and how it is
conical (in 3D view), but on paper in a profile view, the shadow appears as an isosceles triangle whose base coincides
with the diameter of the earth. Discuss what makes the shadows of celestial bodies similar. Describe the two parts of
the moon’s shadow, the umbra and penumbra.
Note that the distances are not drawn to scale in the following image.
Discussion
Solar Eclipse
3D view:

The umbra is the portion of the shadow where all sunlight is blocked, while the penumbra is the part of the
shadow where light is only partially blocked. For the purposes of our discussion today, we will be simplifying
the situation and considering only the umbra.

What is remarkable about the full shadow caused by the eclipse? That is, what is remarkable about the umbra
and the portion of the moon that is dark? Consider the relationship between the 3D and 2D image of it.


If a cone represents the portion of the moon that is dark as well as the umbra, then the part that is
entirely dark in the 2D image is an isosceles triangle.
We assume that shadows from the moon and the earth are all similar isosceles triangles.
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This is, in fact, not the case for all planetary objects, as the shadow formed depends on how far away the light source is
from the celestial body and the size of the planet. For the sake of simplification as well as approximation, this
assumption is made. Then, shadows created by the moon and the earth will have the following relationship:
Earth
Moon
𝑎𝑎 = 𝑏𝑏

You can imagine simulating a solar eclipse using a marble that is one inch in diameter. If you hold it one arm
length away, it will easily block (more than block) the sun from one eye. Do not try this; you will damage your
eye!

To make the marble just barely block the sun, it must be about 9 feet (i.e., 108 inches) away from your eye. So
the cone of shadow behind the marble tapers to a point, which is where your eye is, and at this point, the
marble just blocks out the sun. This means that the ratio of the length of the shadow of the marble to the
diameter of the marble is about 108: 1.



In fact, by experimenting with different-size spheres, we find that this ratio
holds true regardless of the size of the sphere (or circular object) as long as the
sphere is at the point where it just blocks the sun from our vantage point. In
other words, whether you are using a marble, a tennis ball, or a basketball to
model the eclipse, the distance the sphere must be held from the eye is 108
times the diameter of the sphere.
We conclude that this ratio also holds for the moon and the earth during a solar
eclipse. Since the moon and sun appear to be the same size in the sky (the
moon just blocks the sun in a solar eclipse), we can conclude that the distance
from the earth to the moon must also be roughly 108 times the diameter of the
moon because the earth is at the tip of the moon’s shadow.
Scaffolding:
Emphasize the altitude:base
ratio of lengths by citing other
everyday objects, such as a
tennis ball; for example, a
tennis ball just blocks out the
sun if the tennis ball is viewed
from 108 tennis ball diameters
away.
Let us now consider what is happening in a lunar eclipse. A lunar eclipse occurs when the moon passes behind
the earth. Once again, the earth, sun, and moon lie on a straight line, but this time the earth is between the
moon and the sun.

Consequently, during a lunar eclipse, the moon is still faintly visible from the earth because of light reflected
off the earth.

Sketch a diagram of a lunar eclipse.
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Lunar Eclipse
3D view:

Based on what we know about the ratio of distances for an object to just block the sun, we can conclude that
the moon must be within 108 earth diameters; if it were not within that distance, it would not pass through
the earth’s shadow, and the earth could not block the sun out completely.

Studying total lunar eclipses was critical to finding the distance to the moon. Other types of eclipses exist, but
they involve the penumbra. The following measurement required a total eclipse, or one involving the umbra.

By carefully examining the shadow of the earth during a total lunar eclipse, it was determined that the width,
or more specifically, the diameter of the cross section of the earth’s conical shadow at the distance of the
1
2
moon, is about 2 moon diameters.
Share the following animation of a lunar eclipse: http://galileoandeinstein.physics.virginia.edu/lectures/eclipse3.htm
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
With this measurement, a diagram like the following was constructed, where the shadow of the moon was
reversed along the shadow of the earth:

What length and angle relationships can we label in this diagram?
Pose the following questions one at a time, and allow students to discuss and mark their diagrams.

Based on what is known about the shadows of the moon and the earth, what do we know about the measures
of ∠𝐴𝐴𝐴𝐴𝐴𝐴 and ∠𝐵𝐵𝐵𝐵𝐵𝐵?


What is the relationship between 𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴?


The angle measures are equal since we are assuming the shadows can be modeled by similar isosceles
triangles.
Since the distance an object must be from the eye to block out the sun is 108 times the object’s
diameter, the length of 𝐴𝐴𝐴𝐴 must be 108 times that of 𝐴𝐴𝐴𝐴.
What is the relationship between 𝐴𝐴𝐴𝐴 and 𝐹𝐹𝐹𝐹?

We know that the diameter of the cross section of the earth’s conical shadow at the distance of the
1
2
moon is about 2 moon diameters, so the length of 𝐹𝐹𝐹𝐹 is 2.5 times the length of 𝐴𝐴𝐴𝐴.

Reversing the moon’s shadow completes parallelogram 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. How can we be sure that 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 is a
parallelogram?
Allow students time to discuss why this must be true, using the relationships determined in the diagram. Share out ideas
before explaining further.


���� must be parallel to ����
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵 by construction. The diameter of the earth is parallel to the segment formed by the
moon’s diameter and the diameter of the shadow at the distance of the moon.
We know that 𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵 since the shadows are similar triangles. This means that ����
𝐴𝐴𝐴𝐴 is parallel to
����
𝐷𝐷𝐷𝐷 (alternate interior angles).
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



Since we know that 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 is a parallelogram, then 𝐴𝐴𝐴𝐴 = 𝐵𝐵𝐵𝐵, or 𝐵𝐵𝐵𝐵 = 3.5 units.
From our work in Lesson 19, we know the Greeks had the circumference of the earth at this time, which means
they also had the earth’s diameter. Then, it was known that 𝐵𝐵𝐵𝐵 ≈ 8000 miles. With this information, we
conclude:
3.5 units ≈ 8000
1 unit ≈ 2300.
We already know that the distance from the moon to the earth is 108 times the diameter of the moon. If the
moon has a diameter of approximately 2,300 miles, then the distance from the moon to the earth is roughly
(2300 × 108) miles, or 248,000 miles.
With some careful observation and measurement (that of a lunar eclipse) and basic geometry, Aristarchus was
able to determine a fairly accurate measure of the distance from the earth to the moon.
Example (5 minutes)
Example
a.
If the circumference of the earth is about 𝟐𝟐𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎 miles, what is the earth’s diameter in miles?
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝝅𝝅
b.
Using part (a), what is the moon’s diameter in miles?
𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝟕𝟕
c.
⋅
𝝅𝝅
The moon’s diameter is approximately 𝟐𝟐, 𝟑𝟑𝟑𝟑𝟑𝟑 miles.
≈ 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
How far away is the moon in miles?
𝟏𝟏𝟏𝟏𝟏𝟏 ⋅

The earth’s diameter is approximately 𝟖𝟖, 𝟎𝟎𝟎𝟎𝟎𝟎 miles.
≈ 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
⋅
≈ 𝟐𝟐𝟐𝟐𝟐𝟐 𝟎𝟎𝟎𝟎𝟎𝟎
𝟕𝟕
𝝅𝝅
The moon is approximately 𝟐𝟐𝟐𝟐𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎 miles from the earth.
Modern-day calculations show that the distance from the earth to the moon varies between 225,622 miles
and 252,088 miles. The variation is because the orbit is actually elliptical versus circular.
Closing (1 minute)

With some methodical observations and the use of geometry, Aristarchus was able to make a remarkable
approximation of the distance from the earth to the moon using similarity.
Exit Ticket (5 minutes)
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Name
Date
Lesson 20: How Far Away Is the Moon?
Exit Ticket
1.
1
On Planet A, a -inch diameter ball must be held at a height of 72 inches to just block the sun. If a moon orbiting
4
Planet A just blocks the sun during an eclipse, approximately how many moon diameters is the moon from the
planet?
2.
1
Planet A has a circumference of 93,480 miles. Its moon has a diameter that is approximated to be that of Planet
A. Find the approximate distance of the moon from Planet A.
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Exit Ticket Sample Solutions
1.
𝟏𝟏
On Planet A, a -inch diameter ball must be held at a height of 𝟕𝟕𝟕𝟕 inches to just block the sun. If a moon orbiting
𝟒𝟒
Planet A just blocks the sun during an eclipse, approximately how many moon diameters is the moon from the
planet?
The ratio of the diameter of the ball to the specified height is
𝟏𝟏
𝟒𝟒
𝟕𝟕𝟕𝟕
=
𝟏𝟏
. The moon’s distance in order to just block
𝟐𝟐𝟐𝟐𝟐𝟐
the sun would be proportional since the shadows formed are similar triangles, so the moon would orbit
approximately 𝟐𝟐𝟐𝟐𝟐𝟐 moon diameters from Planet A.
2.
𝟏𝟏
Planet A has a circumference of 𝟗𝟗𝟗𝟗, 𝟒𝟒𝟒𝟒𝟒𝟒 miles. Its moon has a diameter that is approximately that of Planet A.
𝟖𝟖
Find the approximate distance of the moon from Planet A.
To find the diameter of Planet A:
𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝝅𝝅
The diameter of Planet A is approximately 𝟐𝟐𝟐𝟐, 𝟕𝟕𝟕𝟕𝟕𝟕 miles.
= 𝒅𝒅𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 𝐀𝐀
To find the diameter of the moon:
𝒅𝒅𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 =
𝟏𝟏
𝒅𝒅
𝟖𝟖 𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏 𝐀𝐀
𝒅𝒅𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 =
𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
=
𝟖𝟖𝟖𝟖
𝝅𝝅
𝒅𝒅𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 =
𝟏𝟏 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
�
�
𝟖𝟖
𝝅𝝅
The diameter of the moon is approximately 𝟑𝟑, 𝟕𝟕𝟕𝟕𝟕𝟕 miles.
To find the distance of the moon from Planet A:
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐧𝐧𝐜𝐜𝐞𝐞𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 = 𝟐𝟐𝟐𝟐𝟐𝟐(𝒅𝒅𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 )
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
�
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐞𝐞𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 = 𝟐𝟐𝟐𝟐𝟐𝟐 �
𝝅𝝅
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐞𝐞𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 =
𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐𝟐𝟐
𝝅𝝅
The distance from Planet A to its moon is approximately 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎, 𝟐𝟐𝟐𝟐𝟐𝟐 miles.
Problem Set Sample Solutions
1.
If the sun and the moon do not have the same diameter, explain how the sun’s light can be covered by the moon
during a solar eclipse.
The farther away an object is from the viewer, the smaller that object appears. The moon is closer to the earth than
the sun, and it casts a shadow where it blocks some of the light from the sun. The sun is much farther away from the
earth than the moon, and because of the distance, it appears much smaller in size.
2.
What would a lunar eclipse look like when viewed from the moon?
The sun would be completely blocked out by the earth for a time because the earth casts an umbra that spans a
greater distance than the diameter of the moon, meaning the moon would be passing through the earth’s shadow.
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3.
Suppose you live on a planet with a moon, where during a solar eclipse, the moon appears to be half the diameter of
the sun.
a.
Draw a diagram of how the moon would look against the sun during a solar eclipse.
Sample response:
b.
A 𝟏𝟏-inch diameter marble held 𝟏𝟏𝟏𝟏𝟏𝟏 inches away on the planet barely blocks the sun. How many moon
diameters away is the moon from the planet? Draw and label a diagram to support your answer.
If the diameter of the moon appears to be half the diameter of the sun as viewed from the planet, then the
moon will not cause a total eclipse of the sun. In the diagram, 𝑷𝑷𝑷𝑷 is the diameter of the moon, and 𝑻𝑻𝑻𝑻 is the
diameter of the sun as seen from the planet at point 𝑰𝑰.
The diameter of the moon is represented by distance 𝑷𝑷𝑷𝑷. For me to view a total eclipse, where the sun is just
blocked by the moon, the moon would have to be twice as wide, so 𝑻𝑻𝑻𝑻 = 𝟐𝟐(𝑷𝑷𝑷𝑷). △ 𝑻𝑻𝑻𝑻𝑻𝑻 and △ 𝒀𝒀𝒀𝒀′𝑿𝑿 are
both isosceles triangles, and their vertex angles are the same, so the triangles are similar by the SAS criterion.
If the triangles are similar, then their altitudes are in the same ratio as their bases. This means
substituting values,
the planet.
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𝑰𝑰𝑰𝑰
𝟐𝟐𝟐𝟐𝟐𝟐
=
𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏
𝑰𝑰𝑰𝑰
𝑻𝑻𝑻𝑻
=
𝑰𝑰′ 𝑨𝑨
𝒀𝒀𝒀𝒀
. By
, so 𝑰𝑰𝑰𝑰 = 𝟐𝟐𝟐𝟐𝟐𝟐(𝑷𝑷𝑷𝑷). The moon is approximately 𝟐𝟐𝟐𝟐𝟐𝟐 moon diameters from
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c.
𝟑𝟑
If the diameter of the moon is approximately of the diameter of the planet and the circumference of the
𝟓𝟓
planet is 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 miles, approximately how far is the moon from the planet?
The diameter of the planet:
𝒅𝒅𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 =
𝒅𝒅𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 =
𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎
𝝅𝝅
𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎
𝝅𝝅
The diameter of the planet is approximately 𝟓𝟓𝟓𝟓, 𝟖𝟖𝟖𝟖𝟖𝟖 miles.
The diameter of the moon:
𝟑𝟑
𝟓𝟓
𝒅𝒅𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 = 𝒅𝒅𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩
𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎
�
𝟓𝟓
𝝅𝝅
𝒅𝒅𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 = �
𝒅𝒅𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 =
𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎
𝝅𝝅
The diameter of the moon is approximately 𝟑𝟑𝟑𝟑, 𝟑𝟑𝟑𝟑𝟑𝟑 miles.
The distance of the moon from the planet:
𝑰𝑰𝑰𝑰 = 𝟐𝟐𝟐𝟐𝟐𝟐(𝑷𝑷𝑷𝑷)
𝑰𝑰𝑰𝑰 = 𝟐𝟐𝟐𝟐𝟐𝟐 �
𝑰𝑰𝑰𝑰 =
𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎
�
𝝅𝝅
𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟎𝟎𝟎𝟎𝟎𝟎
𝝅𝝅
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The planet’s moon is approximately 𝟕𝟕, 𝟎𝟎𝟎𝟎𝟎𝟎, 𝟒𝟒𝟒𝟒𝟒𝟒 miles from the
planet.
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Name
Date
1. The coordinates of △ 𝐴𝐴𝐴𝐴𝐴𝐴 are shown on the coordinate plane below. △ 𝐴𝐴𝐴𝐴𝐴𝐴 is dilated from the origin by
scale factor 𝑟𝑟 = 2.
a.
Identify the coordinates of the dilated △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′.
b.
Is △ 𝐴𝐴′ 𝐵𝐵′ 𝐶𝐶 ′ ~ △ 𝐴𝐴𝐴𝐴𝐴𝐴? Explain.
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�����⃗ to point 𝑃𝑃. Line ℓ passes through 𝑃𝑃 and
2. Points 𝐴𝐴, 𝐵𝐵, and 𝐶𝐶 are not collinear, forming ∠𝐵𝐵𝐵𝐵𝐵𝐵. Extend 𝐴𝐴𝐴𝐴
�����⃗ at point 𝑄𝑄.
is parallel to segment 𝐵𝐵𝐵𝐵. It meets 𝐴𝐴𝐴𝐴
a.
Draw a diagram to represent the situation described.
b.
���� ?
Is ����
𝑃𝑃𝑃𝑃 longer or shorter than 𝐵𝐵𝐵𝐵
c.
Prove that △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐴𝐴𝐴𝐴𝐴𝐴.
d.
What other pairs of segments in this figure have the same ratio of lengths that ����
𝑃𝑃𝑃𝑃 has to ����
𝐵𝐵𝐵𝐵 ?
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3. There is a triangular floor space △ 𝐴𝐴𝐴𝐴𝐴𝐴 in a restaurant. Currently, a square portion 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is covered with
tile. The owner wants to remove the existing tile and then tile the largest square possible within △ 𝐴𝐴𝐴𝐴𝐴𝐴,
���� .
keeping one edge of the square on 𝐴𝐴𝐴𝐴
a.
Describe a construction that uses a dilation with center 𝐴𝐴 that can be used to determine the
maximum square 𝐷𝐷′𝐸𝐸′𝐹𝐹′𝐺𝐺′ within △ 𝐴𝐴𝐴𝐴𝐴𝐴 with one edge on ����
𝐴𝐴𝐴𝐴 .
b.
����� in terms of the distances ����
�����?
���� to 𝐹𝐹′𝐺𝐺′
What is the scale factor of 𝐹𝐹𝐹𝐹
𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴′
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c.
The owner uses the construction in part (a) to mark off where the square would be located. He
measures 𝐴𝐴𝐴𝐴 to be 15 feet and 𝐸𝐸𝐸𝐸′ to be 5 feet. If the original square is 144 square feet, how many
square feet of tile does he need for 𝐷𝐷′𝐸𝐸′𝐹𝐹′𝐺𝐺′?
4. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 is a parallelogram, with the vertices listed counterclockwise around the figure. Points 𝑀𝑀, 𝑁𝑁, 𝑂𝑂,
����, 𝐵𝐵𝐵𝐵
���� , 𝐶𝐶𝐶𝐶
����, and 𝐷𝐷𝐷𝐷
����, respectively. The segments 𝑀𝑀𝑀𝑀 and 𝑁𝑁𝑁𝑁 cut the
and 𝑃𝑃 are the midpoints of sides 𝐴𝐴𝐴𝐴
parallelogram into four smaller parallelograms, with the point 𝑊𝑊 in the center of 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 as a common
vertex.
a.
Exhibit a sequence of similarity transformations that takes △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐶𝐶𝐶𝐶𝐶𝐶. Be specific in
describing the parameter of each transformation; for example, if describing a reflection, state the
line of reflection.
b.
Given the correspondence in △ 𝐴𝐴𝐴𝐴𝐴𝐴 similar to △ 𝐶𝐶𝐶𝐶𝐶𝐶, list all corresponding pairs of angles and
corresponding pairs of sides. What is the ratio of the corresponding pairs of angles? What is the
ratio of the corresponding pairs of sides?
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5. Given two triangles, △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝐷𝐷𝐷𝐷𝐷𝐷, 𝑚𝑚∠𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑚𝑚∠𝐹𝐹𝐹𝐹𝐹𝐹, and 𝑚𝑚∠𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑚𝑚∠𝐹𝐹𝐹𝐹𝐹𝐹. Points 𝐴𝐴, 𝐵𝐵, 𝐷𝐷,
and 𝐸𝐸 lie on line 𝑙𝑙 as shown. Describe a sequence of rigid motions and/or dilations to show that
△ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐷𝐷𝐷𝐷𝐷𝐷, and sketch an image of the triangles after each transformation.
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����.
6. △ 𝐽𝐽𝐽𝐽𝐽𝐽 is a right triangle; ����
𝑁𝑁𝑁𝑁 ⊥ ����
𝐾𝐾𝐾𝐾, ����
𝑁𝑁𝑁𝑁 ⊥ ���
𝐽𝐽𝐽𝐽, �����
𝑀𝑀𝑀𝑀 ∥ 𝑂𝑂𝑂𝑂
a.
List all sets of similar triangles. Explain how you know.
b.
Select any two similar triangles, and show why they are similar.
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7.
a.
The line 𝑃𝑃𝑃𝑃 contains point 𝑂𝑂. What happens to ⃖����⃗
𝑃𝑃𝑃𝑃 with a dilation about 𝑂𝑂 and scale factor of
b.
The line 𝑃𝑃𝑃𝑃 does not contain point 𝑂𝑂. What happens to ⃖����⃗
𝑃𝑃𝑃𝑃 with a dilation about 𝑂𝑂 and scale factor
of 𝑟𝑟 = 2?
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8. Use the diagram below to answer the following questions.
a.
State the pair of similar triangles. Which similarity criterion guarantees their similarity?
b.
Calculate 𝐷𝐷𝐷𝐷 to the hundredths place.
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9. In △ 𝐴𝐴𝐴𝐴𝐴𝐴, 𝑚𝑚∠𝐴𝐴 is 40°, 𝑚𝑚∠𝐵𝐵 is 60°, and 𝑚𝑚∠𝐶𝐶 is 80°. The triangle is dilated by a factor of 2 about point 𝑃𝑃
to form △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′. It is also dilated by a factor of 3 about point 𝑄𝑄 to form △ 𝐴𝐴′′𝐵𝐵′′𝐶𝐶′′. What is the
measure of the angle formed by line 𝐴𝐴′𝐵𝐵′ and line 𝐵𝐵′′𝐶𝐶′′? Explain how you know.
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10. In the diagram below, |𝐴𝐴𝐴𝐴| = |𝐶𝐶𝐶𝐶| = |𝐸𝐸𝐸𝐸|, and ∠𝐵𝐵𝐵𝐵𝐵𝐵, ∠𝐷𝐷𝐷𝐷𝐷𝐷, and ∠𝐹𝐹𝐹𝐹𝐹𝐹 are right. The two lines meet
at a point to the right. Are the triangles similar? Why or why not?
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11. The side lengths of the following right triangle are 16, 30, and 34. An altitude of a right triangle from the
right angle splits the hypotenuse into line segments of length 𝑥𝑥 and 𝑦𝑦.
a.
What is the relationship between the large triangle and the two sub-triangles? Why?
b.
Solve for ℎ, 𝑥𝑥, and 𝑦𝑦.
c.
Extension: Find an expression that gives ℎ in terms of 𝑥𝑥 and 𝑦𝑦.
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12. The sentence below, as shown, is being printed on a large banner for a birthday party. The height of the
banner is 18 inches. There must be a minimum 1-inch margin on all sides of the banner. Use the
dimensions in the image below to answer each question.
a.
Describe a reasonable figure in the plane to model the printed image.
b.
Find the scale factor that maximizes the size of the characters within the given constraints.
c.
What is the total length of the banner based on your answer to part (a)?
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A Progression Toward Mastery
Assessment
1
a–b
G-SRT.A.2
2
a–d
G-SRT.B.5
3
a–c
G-SRT.A.1
STEP 1
Missing or
and little evidence
of reasoning or
application of
mathematics to
solve the problem.
STEP 2
Missing or
but evidence of
some reasoning or
application of
mathematics to
solve the problem.
STEP 3
with some
evidence of
reasoning or
application of
mathematics to
solve the problem,
OR an incorrect
substantial
evidence of solid
reasoning or
application of
mathematics to
solve the problem.
STEP 4
supported by
substantial
evidence of solid
reasoning or
application of
mathematics to
solve the problem.
Student provides at
least two incorrect
coordinates of part (a),
and student does not
show clear
understanding of why
△ 𝐴𝐴′𝐵𝐵′𝐶𝐶′~ △ 𝐴𝐴𝐴𝐴𝐴𝐴 for
part (b).
Student provides only
partially correct and
parts (a) and (b).
(a) correctly, but part
(b) lacks clear
explanation regarding
the corresponding
angles and length
measurements.
OR
(b) correctly, and part
(a) has errors in the
coordinates of the
dilated vertices.
parts (a) and (b)
correctly.
Student provides a
correct response for
part (b), but student
shows insufficient
understanding or
parts (a), (c), and (d).
Student incorrectly
Student incorrectly
incorrect diagram in
part (a) or insufficient
evidence provided for
(c)).
Student correctly
Student provides a
response for part (a)
that is missing two or
more construction
steps, and student
parts (b) and (c).
Student provides a
response for part (a)
that is missing two or
more construction
steps, and student
(b) or part (c).
Student provides a
response for part (a)
that is missing two or
more construction steps
parts (b) and (c).
OR
Student correctly
(b) or part (c).
Student correctly
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4
a–b
G-SRT.A.2
Student provides an
incomplete or
otherwise inaccurate
description of a
similarity
transformation for part
(a).
Student provides an
incomplete list of
correspondences for
part (b) and makes
errors in the ratios of
pairs of angles or pairs
of sides.
Student provides an
incomplete or
otherwise inaccurate
description of a
similarity
transformation for part
(a).
Student provides an
incomplete list of
correspondences or
makes errors in the
ratios of pairs of angles
or pairs of sides in part
(b).
Student provides an
incomplete or
otherwise inaccurate
description of a
similarity
transformation for part
(a) but correctly
OR
Student provides a
(a) but provides an
incomplete list of
correspondences or
makes errors in the
ratios of pairs of angles
or pairs of sides in part
(b).
Student correctly
and (b).
5
G-SRT.A.3
Student describes an
incoherent sequence of
transformations or
provides no description
and does not explain
parameters to show
that one triangle maps
to the other.
Student describes a
sequence of
transformations that
maps one triangle to
the other but does not
provide detail regarding
the parameters (e.g., a
reflection is cited, but
no detail regarding the
line of reflection is
mentioned).
Student describes a
sequence of
transformations that
maps one triangle to
the other but does not
provide detail regarding
the parameters of one
of the transformations,
transformation and its
respective parameters
are needed to complete
a correct sequence of
transformations.
Student clearly
describes a sequence of
appropriate
transformations and
provides the
appropriate parameters
for each
transformation.
6
a–b
Student provides a
response that is missing
three or more similar
triangles out of both
sets.
Student lists all but two
of the similar triangles
out of both sets.
Student lists all but one
of the similar triangles
out of both sets.
Student correctly lists
all the similar triangles
in each set.
Student does not show
an understanding of the
properties of dilations
in either part (a) or part
(b).
Student correctly
(b).
Student correctly
does not provide
justification as to why
the line maps to itself in
part (a).
Student correctly
(b).
G-SRT.B.5
7
a–b
G-SRT.B.5
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8
a–b
G-SRT.B.5
Student provides
incorrect similarity
criteria for part (a) and
makes more than one
conceptual or
computational error in
part (b).
Student provides
incorrect similarity
criteria for part (a) and
makes one conceptual
or computational error
in part (b).
Student correctly
makes one conceptual
or computational error
in part (b).
OR
Student provides
incorrect similarity
criteria for part (a) but
(b).
Student correctly
(b).
9
G-SRT.A.1
Student shows little or
no understanding of
why the two dilations
lead to an angle of 60°
formed by 𝐴𝐴′𝐵𝐵′ and
𝐵𝐵′′𝐶𝐶′′.
Student includes an
attempted diagram but
is missing multiple
conclusive elements or
has an error in the
verbal justification.
Student includes an
attempted diagram but
is missing one
conclusive element
(e.g., missing one of the
dilations or an incorrect
dilation) or has an error
in the verbal
justification.
Student includes an
accurate diagram and
provides a complete
and correct justification.
10
G-SRT.B.4
Student shows little or
no understanding of
why the triangles are
not similar.
Student includes a
justification that
demonstrates why the
triangles are not similar
but has two conceptual
errors.
Student includes a
justification that
demonstrates why the
triangles are not similar
but has one conceptual
error.
Student includes an
accurate justification
that demonstrates why
the triangles are not
similar.
11
a–c
Student does not
provide fully correct
(b), and (c).
Student provides a fully
of the three parts.
Student provides fully
two of the three parts.
Student provides
parts (a), (b), and (c)
and clearly explains
each.
Student does not
provide fully correct
(b), and (c).
Student provides a fully
of the three parts.
Student correctly
computes scaled
dimensions for the
printed image but fails
to consider the required
margins.
Student provides
parts (a), (b), and (c)
and clearly explains
each.
G-SRT.B.5
12
a–c
G-MG.A.1
G-MG.A.3
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Name
Date
1. The coordinates of △ 𝐴𝐴𝐴𝐴𝐴𝐴 are shown on the coordinate plane below. △ 𝐴𝐴𝐴𝐴𝐴𝐴 is dilated from the origin by
scale factor 𝑟𝑟 = 2.
a.
Identify the coordinates of the dilated △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′.
Point A = (3, 2), then A' = (2×(3), 2×(2)) = (6, 4)
Point B = (0, –1), then B' = (2×(0), 2×(–1)) = (0, –2)
Point 𝐶𝐶 = (-3, 1), then C' = (2×(–3), 2×(1)) = (–6, 2)
b.
Is △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′~ △ 𝐴𝐴𝐴𝐴𝐴𝐴? Explain.
Yes. The side lengths of △A'B'C' are each two times the length of the sides of △ABC, and
corresponding sides are proportional in length. Also, the corresponding angles are equal
in measurement because dilations preserve the measurements of angles.
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�����⃗ to point 𝑃𝑃. Line ℓ passes through 𝑃𝑃 and
2. Points 𝐴𝐴, 𝐵𝐵, and 𝐶𝐶 are not collinear, forming ∠𝐵𝐵𝐵𝐵𝐵𝐵. Extend 𝐴𝐴𝐴𝐴
is parallel to segment 𝐵𝐵𝐵𝐵. It meets ray 𝐴𝐴𝐴𝐴 at point 𝑄𝑄.
a.
Draw a diagram to represent the situation described.
ℓ
b.
���� ?
Is ����
𝑃𝑃𝑃𝑃 longer or shorter than 𝐵𝐵𝐵𝐵
����� is longer than �����
PQ
BC.
c.
Prove that △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐴𝐴𝐴𝐴𝐴𝐴.
�����
PQ ∥ �����
BC. Since corresponding angles are equal in measure, then m∠ABC = m∠APQ.
Additionally, m∠A = m∠A. Then △ABC ~ △APQ by AA similarity criterion.
d.
What other pairs of segments in this figure have the same ratio of lengths that ����
𝑃𝑃𝑃𝑃 has to ����
𝐵𝐵𝐵𝐵 ?
PA:BA, QA:CA
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3. There is a triangular floor space △ 𝐴𝐴𝐴𝐴𝐴𝐴 in a restaurant. Currently, a square portion 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is covered with
tile. The owner wants to remove the existing tile and then tile the largest square possible within △ 𝐴𝐴𝐴𝐴𝐴𝐴,
keeping one edge of the square on 𝐴𝐴𝐴𝐴.
a.
Describe a construction that uses a dilation with center 𝐴𝐴 that can be used to determine the
maximum square 𝐷𝐷′𝐸𝐸′𝐹𝐹′𝐺𝐺′ within △ 𝐴𝐴𝐴𝐴𝐴𝐴 with one edge on ����
𝐴𝐴𝐴𝐴 .
1.
Use A as a center of dilation.
2.
Draw ������⃗
AF through BC.
3.
4.
5.
6.
Label the intersection of ������⃗
AF and BC as F'.
�������⃗ parallel to EF, where E' is the intersection of AB and the parallel line.
Construct E'F'
Construct ��������⃗
F'G' parallel to FG, where G' is the intersection of AC and the parallel line.
Construct ���������⃗
E'D′ parallel to ED, where D' is the intersection of AC and the parallel
line.
7.
b.
Connect D', E', F', G'.
What is the scale factor of 𝐹𝐹𝐹𝐹 to 𝐹𝐹′𝐺𝐺′ in terms of the distances 𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴′?
AF'
AF
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c.
The owner uses the construction in part (a) to mark off where the square would be located. He
measures 𝐴𝐴𝐴𝐴 to be 15 feet and 𝐸𝐸𝐸𝐸′ to be 5 feet. If the original square is 144 square feet, how many
square feet of tile does he need for 𝐷𝐷′𝐸𝐸′𝐹𝐹′𝐺𝐺′?
The distance AE' = 20 ft, so the scale factor from DEFG to D'E'F'G' is
20
15
4
= .
3
The areas of similar figures are related by the square of the scale factor; therefore,
4 2
Area(D'E'F'G') = � � Area(DEFG)
3
16
(144)
Area(D'E'F'G') =
9
Area(D'E'F'G') = 256
The owner needs 256 square feet of tile for D'E'F'G'.
4. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 is a parallelogram, with the vertices listed counterclockwise around the figure. Points 𝑀𝑀, 𝑁𝑁, 𝑂𝑂 and
����, 𝐵𝐵𝐵𝐵
���� , ����
����, respectively. The segments 𝑀𝑀𝑀𝑀 and 𝑁𝑁𝑁𝑁 cut the
𝑃𝑃 are the midpoints of sides 𝐴𝐴𝐴𝐴
𝐶𝐶𝐶𝐶, and 𝐷𝐷𝐷𝐷
parallelogram into four smaller parallelograms, with the point 𝑊𝑊 in the center of 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 as a common
vertex.
a.
Exhibit a sequence of similarity transformations that takes △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐶𝐶𝐶𝐶𝐶𝐶. Be specific in
describing the parameter of each transformation; for example, if describing a reflection, state the
line of reflection.
Answers will vary (e.g., 180˚ rotation about W; dilation with center of dilation C).
b.
Given the correspondence in △ 𝐴𝐴𝑀𝑀𝑀𝑀 similar to △ 𝐶𝐶𝐶𝐶𝐶𝐶, list all corresponding pairs of angles and
corresponding pairs of sides. What is the ratio of the corresponding pairs of angles? What is the
ratio of the corresponding pairs of sides?
∠MAW corresponds to ∠DCA; ∠AMW corresponds to ∠CDA; ∠AWM corresponds to
�����; ������
�����.
�����; ������
AM corresponds to CD
MW corresponds to DA
WA corresponds to AC
The ratio of corresponding pairs of angles is 1: 1.
The ratio of corresponding pairs of sides is 1: 2.
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5. Given two triangles, 𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐷𝐷𝐷𝐷𝐷𝐷, 𝑚𝑚∠𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑚𝑚∠𝐹𝐹𝐹𝐹𝐹𝐹, and 𝑚𝑚∠𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑚𝑚∠𝐹𝐹𝐹𝐹𝐹𝐹. Points 𝐴𝐴, 𝐵𝐵, 𝐷𝐷, and 𝐸𝐸
lie on line 𝑙𝑙 as shown. Describe a sequence of rigid motions and/or dilations to show that
△ 𝐴𝐴𝐴𝐴𝐴𝐴~ △ 𝐷𝐷𝐷𝐷𝐷𝐷, and sketch an image of the triangles after each transformation.
����. The reflection takes E to B and D to a
Reflect △DEF over the perpendicular bisector of BE
point on line AB. Since angle measures are preserved in rigid motions, F must map to a
point on �����
BC.
•
By the hypothesis, m∠A = m∠D; therefore, ������
D'F' ∥ �����
AC since corresponding angles are
equal in measure.
•
Since dilations map a segment to a parallel line segment, dilate △D'E'F' about E' and
by scale factor r =
•
By the dilation theorem, F' goes to C.
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and that sends D' to A.
BD'
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����.
6. △ 𝐽𝐽𝐽𝐽𝐽𝐽 is a right triangle; ����
𝑁𝑁𝑁𝑁 ⊥ ����
𝐾𝐾𝐾𝐾, ����
𝑁𝑁𝑁𝑁 ⊥ ���
𝐽𝐽𝐽𝐽, �����
𝑀𝑀𝑀𝑀 ∥ 𝑂𝑂𝑂𝑂
a.
List all sets of similar triangles. Explain how you know.
Set 1



Set 2



△MNO
△PON
△OPK
△JON
△JKL
△NPL
The triangles are similar because of the AA criterion.
b.
Select any two similar triangles, and show why they are similar.
Possible response:
△MNO~△PON by the AA criterion.
∠K is a right angle since △JKL is a right triangle.
����� ⊥ ����
∠MON is a right angle since NO
JK.
����� ∥ �����
m∠NMO = m∠POK since MN
OP, and ����
JK is a transversal that
intersects �����
MN and �����
OP; corresponding ∠’s are equal in measure.
Therefore, by the AA criterion, △MNO~△PON.
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7.
a.
The line 𝑃𝑃𝑃𝑃 contains point 𝑂𝑂. What happens to ⃖����⃗
𝑃𝑃𝑃𝑃 with a dilation about 𝑂𝑂 and scale factor of
•
Since the points P and Q are collinear with the center O, then by definition of a
dilation, both P' and Q' will also be collinear with the center O.
•
b.
The line PQ maps to itself.
⃖����⃗ with a dilation about 𝑂𝑂 and scale factor
The line ⃖����⃗
𝑃𝑃𝑃𝑃 does not contain point 𝑂𝑂. What happens to 𝑃𝑃𝑃𝑃
of 𝑟𝑟 = 2?
The line PQ maps to a parallel line P'Q'.
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8. Use the diagram below to answer the following questions.
a.
State the pair of similar triangles. Which similarity criterion guarantees their similarity?
△DEF ~ △FGH
SAS
b.
Calculate 𝐷𝐷𝐷𝐷 to the hundredths place.
DE
DF
=
FG
FH
DE
17
=
6
8
8DE = 102
DE ≈ 12.75
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9. In △ 𝐴𝐴𝐴𝐴𝐴𝐴, 𝑚𝑚∠𝐴𝐴 is 40°, 𝑚𝑚∠𝐵𝐵 is 60°, and 𝑚𝑚∠𝐶𝐶 is 80°. The triangle is dilated by a factor of 2 about point 𝑃𝑃
to form △ 𝐴𝐴′𝐵𝐵′𝐶𝐶′. It is also dilated by a factor of 3 about point 𝑄𝑄 to form △ 𝐴𝐴′′𝐵𝐵′′𝐶𝐶′′. What is the
measure of the angle formed by line 𝐴𝐴′𝐵𝐵′ and line 𝐵𝐵′′𝐶𝐶′′? Explain how you know.
60˚. ������
B'C' and ������
A'B' meet to form a 60˚ angle. Since dilations map segments to parallel
�������. Then the angle formed by lines A'B' and B''C'' is a corresponding angle
segments, ������
B'C'||B''C''
to ∠B and has a measure of 60˚.
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10. In the diagram below, 𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐶𝐶 = 𝐸𝐸𝐸𝐸, and ∠𝐵𝐵𝐵𝐵𝐵𝐵, ∠𝐷𝐷𝐷𝐷𝐷𝐷, and ∠𝐹𝐹𝐹𝐹𝐹𝐹 are right. The two lines meet at a
point to the right. Are the triangles similar? Why or why not?
The triangles are not similar.
If they were similar, then there would have to be a similarity transformation taking A to C,
B to D, and C to E. But then the dilation factor for this transformation would have to be
1, since AC = CE.
����� is the image of �����
But since the dilation factor is 1 and CD
AB, then it must be true that
AB = CD. Additionally, we know that �����
AB is parallel to �����
CD, and since ∠BAC and ∠DCE are
right, this implies ACDB is a rectangle. This implies that �����
BD is parallel to �����
AC, but this is
contrary to the given.
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11. The side lengths of the following right triangle are 16, 30, and 34. An altitude of a right triangle from the
right angle splits the hypotenuse into line segments of length 𝑥𝑥 and 𝑦𝑦.
a.
What is the relationship between the large triangle and the two sub-triangles? Why?
An altitude drawn from the vertex of the right angle of a right triangle to the
hypotenuse divides the right triangle into two sub-triangles that are similar to the
original triangle by the AA criterion.
b.
Solve for ℎ, 𝑥𝑥, and 𝑦𝑦.
h
16
=
30
34
240
h =
17
c.
y
30
=
30
34
450
y =
17
x
16
=
16
34
128
x =
17
Extension. Find an expression that gives ℎ in terms of 𝑥𝑥 and 𝑦𝑦.
The large triangle is similar to both sub-triangles, and both sub-triangles are, therefore,
similar. Then
h
x
=
y
h
h
2
= xy
h = �xy.
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12. The sentence below, as shown, is being printed on a large banner for a birthday party. The height of the
banner is 18 inches. There must be a minimum 1-inch margin on all sides of the banner. Use the
dimensions in the image below to answer each question.
a.
Describe a reasonable figure in the plane to model the printed image.
The sentence can be modeled as a rectangle with dimensions 4.78" × 0.44".
b.
Find the scale factor that maximizes the size of the characters within the given constraints.
The scaled height of the rectangle cannot exceed 16" to allow for 1" margins above and
below on the banner.
16 = k(0.44), where k represents the scale factor of the banner;
16
400
4
k=
=
= 36
0.44
11
11
c.
What is the total length of the banner based on your answer to part (a)?
Using the scale factor from part (a),
400
� ⋅(4.78)
11
9
y = 173
11
y = �
The total length of the image in the banner is 173
9
inches; however, the banner must
11
have a minimum 1" margin on all sides of the image, so the banner must be at least 2"
longer. The total length of the banner must be at least 175
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GEOMETRY • MODULE 2
Topic D
Applying Similarity to Right Triangles
G-SRT.B.4
Focus Standard:
G-SRT.B.4
Instructional Days:
4
Prove theorems about triangles. Theorems include: a line parallel to one side of a
triangle divides the other two proportionally, and conversely; the Pythagorean Theorem
proved using triangle similarity.
Lesson 21:
Special Relationships Within Right Triangles—Dividing into Two Similar Sub-Triangles (P) 1
Lesson 22:
Multiplying and Dividing Expressions with Radicals (P)
Lesson 23:
Lesson 24:
Proving the Pythagorean Theorem Using Similarity (E)
In Topic D, students use their understanding of similarity and focus on right triangles as a lead up to
trigonometry. In Lesson 21, students use the AA criterion to show how an altitude drawn from the vertex of
the right angle of a right triangle to the hypotenuse creates two right triangles similar to the original right
triangle. Students examine how the ratios within the three similar right triangles can be used to find
unknown side lengths. Work with lengths in right triangles lends itself to expressions with radicals. In
Lessons 22 and 23, students learn to rationalize fractions with radical expressions in the denominator and
also to simplify, add, and subtract radical expressions. In the final lesson of Topic D, students use the
relationships created by an altitude to the hypotenuse of a right triangle to prove the Pythagorean theorem.
1Lesson
Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
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Lesson 21: Special Relationships Within Right Triangles—
Dividing into Two Similar Sub-Triangles
Student Outcomes

Students understand that the altitude of a right triangle from the vertex of the right angle to the hypotenuse
divides the triangle into two similar right triangles that are also similar to the original right triangle.

Students complete a table of ratios for the corresponding sides of the similar triangles that are the result of
dividing a right triangle into two similar sub-triangles.
Lesson Notes
This lesson serves as a foundational piece for understanding trigonometric ratios related to right triangles. The goal of
the lesson is to show students how ratios within figures can be used to find lengths of sides in another triangle when
those triangles are known to be similar.
Classwork
Opening Exercise (5 minutes)
Opening Exercise
Use the diagram below to complete parts (a)–(c).
a.
Are the triangles shown above similar? Explain.
Yes, the triangles are similar by the AA criterion. Both triangles have a right angle, and 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑿𝑿 and
𝒎𝒎∠𝑪𝑪 = 𝒎𝒎∠𝒁𝒁.
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b.
Determine the unknown lengths of the triangles.
����.
Let 𝒙𝒙 represent the length of 𝒀𝒀𝒀𝒀
Let 𝒚𝒚 be the length of the hypotenuse of △ 𝑨𝑨𝑨𝑨𝑨𝑨.
Let 𝒛𝒛 be the length of the hypotenuse of △ 𝑿𝑿𝑿𝑿𝑿𝑿.
c.
Explain how you found the lengths in part (a).
𝟑𝟑
𝒙𝒙
=
𝟐𝟐 𝟏𝟏. 𝟓𝟓
𝟐𝟐𝟐𝟐 = 𝟒𝟒. 𝟓𝟓
𝒙𝒙 = 𝟐𝟐. 𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐 + 𝟑𝟑𝟐𝟐 = 𝒚𝒚𝟐𝟐
𝟒𝟒 + 𝟗𝟗 = 𝒚𝒚𝟐𝟐
𝟏𝟏𝟏𝟏 = 𝒚𝒚𝟐𝟐
√𝟏𝟏𝟏𝟏 = 𝒚𝒚
𝟏𝟏. 𝟓𝟓𝟐𝟐 + 𝟐𝟐. 𝟐𝟐𝟓𝟓𝟐𝟐 = 𝒛𝒛𝟐𝟐
𝟐𝟐. 𝟐𝟐𝟐𝟐 + 𝟓𝟓. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 = 𝒛𝒛𝟐𝟐
𝟕𝟕. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 = 𝒛𝒛𝟐𝟐
√𝟕𝟕. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 = 𝒛𝒛
Since the triangles are similar, I used the values of the ratios of the corresponding side lengths to determine
����. To determine the lengths of ����
the length of 𝒀𝒀𝒀𝒀
𝑨𝑨𝑨𝑨 and ����
𝑿𝑿𝑿𝑿, I used the Pythagorean theorem.
Example 1 (15 minutes)
In Example 1, students learn that when a perpendicular is drawn from the right angle to
the hypotenuse of a right triangle, the triangle is divided into two sub-triangles. Further,
students show that all three of the triangles, the original one and the two formed by the
perpendicular, are similar.
Example 1
Recall that an altitude of a triangle is a perpendicular line segment from a vertex to the line
����� is the altitude from vertex 𝑩𝑩 to the line
determined by the opposite side. In △ 𝑨𝑨𝑨𝑨𝑨𝑨 below, 𝑩𝑩𝑩𝑩
����.
containing 𝑨𝑨𝑨𝑨
a.
How many triangles do you see in the figure?
There are three triangles in the figure.
b.
Identify the three triangles by name.
Note that there are many ways to name the three triangles. Ensure that the names
students give show corresponding angles.
Scaffolding:
 A good hands-on visual
that can be used here
requires a 3 × 5 notecard.
Have students draw the
diagonal and then draw
the perpendicular line
from 𝐵𝐵 to side 𝐴𝐴𝐴𝐴.
 Make sure students label
all of the parts to match
the triangles before they
make the cuts. Next, have
students cut out the three
triangles. Students then
have a notecard version of
the three triangles shown
and are better able to see
the relationships among
them.
△ 𝑨𝑨𝑨𝑨𝑨𝑨, △ 𝑨𝑨𝑨𝑨𝑨𝑨, and △ 𝑩𝑩𝑩𝑩𝑩𝑩.
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We want to consider the altitude of a right triangle from the right angle to the hypotenuse. The altitude of a right triangle
splits the triangle into two right triangles, each of which shares a common acute angle with the original triangle. In
△ 𝑨𝑨𝑨𝑨𝑨𝑨, the altitude �����
𝑩𝑩𝑩𝑩 divides the right triangle into two sub-triangles, △ 𝑩𝑩𝑩𝑩𝑩𝑩 and △ 𝑨𝑨𝑨𝑨𝑨𝑨.
c.
Is △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑩𝑩𝑩𝑩𝑩𝑩? Is △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑨𝑨𝑨𝑨𝑨𝑨? Explain.
△ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑩𝑩𝑩𝑩𝑩𝑩 are similar by the AA criterion. Each has a right angle, and each shares ∠𝑪𝑪. △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △
𝑨𝑨𝑨𝑨𝑨𝑨 are similar because each has a right angle, and each shares ∠𝑨𝑨, so, again, these triangles are similar by
the AA criterion.
d.
Is △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫? Explain.
△ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑫𝑫𝑫𝑫𝑫𝑫 are not similar, because their corresponding angles, under the given correspondence of
vertices, do not have equal measure.
e.
Since △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑩𝑩𝑩𝑩𝑩𝑩 and △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑨𝑨𝑨𝑨𝑨𝑨, can we conclude that △ 𝑩𝑩𝑩𝑩𝑩𝑩 ~ △ 𝑨𝑨𝑨𝑨𝑨𝑨? Explain.
Since similarity is transitive, △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑩𝑩𝑩𝑩𝑩𝑩 and △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑨𝑨𝑨𝑨𝑨𝑨 implies that
△ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑩𝑩𝑩𝑩𝑩𝑩 ~ △ 𝑨𝑨𝑨𝑨𝑨𝑨.
f.
Identify the altitude drawn in △ 𝑬𝑬𝑬𝑬𝑬𝑬.
�����
𝑮𝑮𝑮𝑮 is the altitude from vertex 𝑮𝑮 to the line
containing ����
𝑬𝑬𝑬𝑬.
g.
As before, the altitude divides the triangle into two sub-triangles, resulting in a total of three triangles
including the given triangle. Identify them by name so that the corresponding angles match up.
△ 𝑬𝑬𝑬𝑬𝑬𝑬, △ 𝑮𝑮𝑮𝑮𝑮𝑮, and △ 𝑬𝑬𝑬𝑬𝑬𝑬
h.
Does the altitude divide △ 𝑬𝑬𝑬𝑬𝑬𝑬 into two similar sub-triangles as the altitude did with △ 𝑨𝑨𝑨𝑨𝑨𝑨?
Yes.
Allow students time to investigate whether the triangles are similar. Students should conclude that
△ 𝐸𝐸𝐸𝐸𝐸𝐸~ △ 𝐺𝐺𝐺𝐺𝐺𝐺~ △ 𝐸𝐸𝐸𝐸𝐸𝐸 using the same reasoning as before, that is, the AA criterion and the fact that similarity is
transitive.
The fact that the altitude drawn from the right angle of a right triangle divides the triangle into two similar sub-triangles,
which are also similar to the original triangle, allows us to determine the unknown lengths of right triangles.
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Example 2 (15 minutes)
In this example, students use ratios within figures to determine unknown side lengths of triangles.
Example 2
Consider the right triangle △ 𝑨𝑨𝑨𝑨𝑨𝑨 below.
Scaffolding:
���� as 𝒚𝒚, and �����
����� from vertex 𝑩𝑩 to the line containing ����
Draw the altitude 𝑩𝑩𝑩𝑩
𝑨𝑨𝑨𝑨. Label ����
𝑨𝑨𝑨𝑨 as 𝒙𝒙, 𝑫𝑫𝑫𝑫
𝑩𝑩𝑩𝑩
as 𝒛𝒛.
It may be helpful for students
to use the cutouts from
Example 1. Have students label
the side lengths on the front
and back of the notecard
according to this diagram and
then use the cutouts to
complete the table of ratios.
Find the values of 𝒙𝒙, 𝒚𝒚, and 𝒛𝒛.
MP.1
Provide students time to find the values of 𝑥𝑥, 𝑦𝑦, and 𝑧𝑧. Allow students to use any reasonable strategy to complete the
task. The suggested time allotment for this part of the example is 5 minutes. Next, have students briefly share their
solutions and explanations for finding the lengths 𝑥𝑥 = 1
12
1
8
, 𝑦𝑦 = 11 , and 𝑧𝑧 = 4 . For example, students may first
13
13
13
use what they know about similar triangles and corresponding side lengths having equal ratios to determine 𝑥𝑥, then use
the equation 𝑥𝑥 + 𝑦𝑦 = 13 to determine the value of 𝑦𝑦, and finally use the Pythagorean theorem to determine the length
of 𝑧𝑧.
Now we will look at a different strategy for determining the lengths of 𝒙𝒙, 𝒚𝒚, and 𝒛𝒛. The strategy requires that we
complete a table of ratios that compares different parts of each triangle.
Students may struggle with the initial task of finding the values of 𝑥𝑥, 𝑦𝑦, and 𝑧𝑧. Encourage them by letting them know
that they have all the necessary tools to find these values. When transitioning to the use of ratios to find values, explain
any method that yields acceptable answers. We are simply looking to add another tool to the toolbox of strategies that
apply in this situation.
Provide students a moment to complete the ratios related to △ 𝐴𝐴𝐴𝐴𝐴𝐴.
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Make a table of ratios for each triangle that relates the sides listed in the column headers.
△ 𝑨𝑨𝑨𝑨𝑨𝑨
shorter leg: hypotenuse
longer leg: hypotenuse
shorter leg: longer leg
𝟓𝟓: 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏: 𝟏𝟏𝟏𝟏
𝟓𝟓: 𝟏𝟏𝟏𝟏
MP.2
& Ensure that students have written the correct ratios before moving on to complete the table of ratios for △ 𝐴𝐴𝐴𝐴𝐴𝐴 and
MP.7 △ 𝐶𝐶𝐶𝐶𝐶𝐶.
shorter leg: hypotenuse
△ 𝑨𝑨𝑨𝑨𝑨𝑨
△ 𝑪𝑪𝑪𝑪𝑪𝑪
longer leg: hypotenuse
𝒙𝒙: 𝟓𝟓
shorter leg: longer leg
𝒛𝒛: 𝟓𝟓
𝒛𝒛: 𝟏𝟏𝟏𝟏
𝒚𝒚: 𝟏𝟏𝟏𝟏
𝒙𝒙: 𝒛𝒛
𝒛𝒛: 𝒚𝒚
Our work in Example 1 showed us that △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑪𝑪𝑪𝑪𝑪𝑪. Since the triangles are similar, the ratios of their
corresponding sides are equal. For example, we can find the length of 𝒙𝒙 by equating the values of shorter leg:
hypotenuse ratios of △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑨𝑨𝑨𝑨.
𝟓𝟓
𝒙𝒙
=
𝟓𝟓 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝒙𝒙 = 𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏
𝒙𝒙 =
= 𝟏𝟏
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
Why can we use these ratios to determine the length of 𝒙𝒙?
We can use these ratios because the triangles are similar. Similar triangles have ratios of corresponding sides that are
equal. We also know that we can use ratios between figures or within figures. The ratios used were within-figure ratios.
Which ratios can we use to determine the length of 𝒚𝒚?
To determine the value of 𝒚𝒚, we can equate the values longer leg: hypotenuse ratios for △ 𝑪𝑪𝑪𝑪𝑪𝑪 and △ 𝑨𝑨𝑨𝑨𝑨𝑨.
𝟏𝟏𝟏𝟏
𝒚𝒚
=
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏
𝒚𝒚 =
= 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
Use ratios to determine the length of 𝒛𝒛.
Students have several options of ratios to determine the length of 𝑧𝑧. As students work, identify those students using
different ratios, and ask them to share their work with the class.
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To determine the value of 𝒛𝒛, we can equate the values of longer leg: hypotenuse ratios for △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑨𝑨𝑨𝑨:
𝒛𝒛 𝟏𝟏𝟏𝟏
=
𝟓𝟓 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟔𝟔𝟔𝟔
𝟖𝟖
𝟔𝟔𝟔𝟔
𝒛𝒛 =
= 𝟒𝟒
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
To determine the value of 𝒛𝒛, we can equate the values of shorter leg: hypotenuse ratios for △ 𝑪𝑪𝑪𝑪𝑪𝑪 and △ 𝑨𝑨𝑨𝑨𝑨𝑨:
𝟓𝟓
𝒛𝒛
=
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟔𝟔𝟔𝟔
𝟖𝟖
𝟔𝟔𝟔𝟔
𝒛𝒛 =
= 𝟒𝟒
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
To determine the value of 𝒛𝒛, we can equate the values of shorter leg: longer leg ratios for △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑨𝑨𝑨𝑨:
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏 = 𝟓𝟓
𝒛𝒛
𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐
(𝟏𝟏𝟏𝟏)
𝟓𝟓𝟓𝟓 =
𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏
𝒛𝒛 =
� �
𝟏𝟏𝟏𝟏 𝟓𝟓
𝟖𝟖
𝟔𝟔𝟔𝟔
𝒛𝒛 =
= 𝟒𝟒
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
To determine the value of 𝒛𝒛, we can equate the values of shorter leg: longer leg ratios for △ 𝑪𝑪𝑪𝑪𝑪𝑪 and △ 𝑨𝑨𝑨𝑨𝑨𝑨:
𝟓𝟓
𝒛𝒛
=
𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 =
(𝟓𝟓)
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓
𝒛𝒛 =
� �
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
𝟖𝟖
𝟔𝟔𝟔𝟔
𝒛𝒛 =
= 𝟒𝟒
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
Since corresponding ratios within similar triangles are equal, we can solve for any unknown side length by equating the
values of the corresponding ratios. In the coming lessons, we will learn about more useful ratios for determining
unknown side lengths of right triangles.
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Closing (5 minutes)
Ask students the following questions. Students may respond in writing, to a partner, or to the whole class.

What is an altitude, and what happens when an altitude is drawn from the right angle of a right triangle?


What is the relationship between the original right triangle and the two similar sub-triangles?


An altitude is the perpendicular line segment from a vertex of a triangle to the line containing the
opposite side. When an altitude is drawn from the right angle of a right triangle, then the triangle is
divided into two similar sub-triangles.
By the AA criterion and transitive property, we can show that all three triangles are similar.
Explain how to use the ratios of the similar right triangles to determine the unknown lengths of a triangle.
Note that we have used shorter leg and longer leg in the lesson and would expect students to do the same in responding
to this prompt. It may be valuable to point out to students that an isosceles right triangle would not have a shorter or
longer leg.

Ratios of side lengths can be written using “shorter leg,” “longer leg,” and hypotenuse. The ratios of
corresponding sides of similar triangles are equivalent and can be used to find unknown lengths of a
triangle.
Exit Ticket (5 minutes)
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Name
Date
Lesson 21: Special Relationships Within Right Triangles—Dividing
into Two Similar Sub-Triangles
Exit Ticket
Given △ 𝑅𝑅𝑅𝑅𝑅𝑅, with altitude ����
𝑆𝑆𝑆𝑆 drawn to its hypotenuse, 𝑆𝑆𝑆𝑆 = 15, 𝑅𝑅𝑅𝑅 = 36, and 𝑅𝑅𝑅𝑅 = 39, answer the questions below.
1.
2.
Complete the similarity statement relating the three triangles in the diagram.
△ 𝑅𝑅𝑅𝑅𝑅𝑅 ~ △
Complete the table of ratios specified below.
shorter leg: hypotenuse
~△
longer leg: hypotenuse
shorter leg: longer leg
△ 𝑹𝑹𝑹𝑹𝑹𝑹
△ 𝑹𝑹𝑹𝑹𝑹𝑹
△ 𝑺𝑺𝑺𝑺𝑺𝑺
3.
Use the values of the ratios you calculated to find the length of ����
𝑆𝑆𝑆𝑆.
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Exit Ticket Sample Solutions
Given △ 𝑹𝑹𝑹𝑹𝑹𝑹, with altitude ����
𝑺𝑺𝑺𝑺 drawn to its hypotenuse, 𝑺𝑺𝑺𝑺 = 𝟏𝟏𝟏𝟏, 𝑹𝑹𝑹𝑹 = 𝟑𝟑𝟑𝟑, and 𝑹𝑹𝑹𝑹 = 𝟑𝟑𝟑𝟑, answer the questions below.
1.
Complete the similarity statement relating the three triangles in the diagram.
△ 𝑹𝑹𝑹𝑹𝑹𝑹 ~ △
𝑹𝑹𝑹𝑹𝑹𝑹
~△
𝑺𝑺𝑺𝑺𝑺𝑺
Using the right angles and shared angles, the triangles are similar by AA criterion. The transitive property may also
be used.
2.
Complete the table of ratios specified below.
△ 𝑹𝑹𝑹𝑹𝑹𝑹
△ 𝑹𝑹𝑹𝑹𝑹𝑹
△ 𝑺𝑺𝑺𝑺𝑺𝑺
3.
shorter leg: hypotenuse
longer leg: hypotenuse
shorter leg: longer leg
𝟏𝟏𝟏𝟏
𝟑𝟑𝟑𝟑
𝟑𝟑𝟑𝟑
𝟑𝟑𝟑𝟑
𝟏𝟏𝟏𝟏
𝟑𝟑𝟑𝟑
𝑺𝑺𝑺𝑺
𝟑𝟑𝟑𝟑
𝑹𝑹𝑹𝑹
𝟑𝟑𝟑𝟑
𝑻𝑻𝑻𝑻
𝟏𝟏𝟏𝟏
𝑺𝑺𝑺𝑺
𝟏𝟏𝟏𝟏
𝑺𝑺𝑺𝑺
𝑹𝑹𝑹𝑹
𝑻𝑻𝑻𝑻
𝑺𝑺𝑺𝑺
����.
Use the values of the ratios you calculated to find the length of 𝑺𝑺𝑺𝑺
𝟏𝟏𝟏𝟏 𝑺𝑺𝑺𝑺
=
𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑
𝟓𝟓𝟓𝟓𝟓𝟓 = 𝟑𝟑𝟑𝟑(𝑺𝑺𝑺𝑺)
𝟓𝟓𝟓𝟓𝟓𝟓
= 𝑺𝑺𝑺𝑺
𝟑𝟑𝟑𝟑
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
= 𝑺𝑺𝑺𝑺
𝟏𝟏𝟏𝟏
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Problem Set Sample Solutions
1.
Use similar triangles to find the length of the altitudes labeled with variables in each triangle below.
a.
△ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫 by AA criterion, so corresponding sides are
proportional.
𝒙𝒙 𝟗𝟗
=
𝟒𝟒 𝒙𝒙
𝒙𝒙𝟐𝟐 = 𝟑𝟑𝟑𝟑
𝒙𝒙 = √𝟑𝟑𝟑𝟑 = 𝟔𝟔
b.
△ 𝑮𝑮𝑮𝑮𝑮𝑮 ~ △ 𝑮𝑮𝑮𝑮𝑮𝑮 by AA criterion, so corresponding sides are proportional.
𝒚𝒚 𝟏𝟏𝟏𝟏
=
𝒚𝒚
𝟒𝟒
𝒚𝒚𝟐𝟐 = 𝟔𝟔𝟔𝟔
𝒚𝒚 = √𝟔𝟔𝟔𝟔 = 𝟖𝟖
c.
△ 𝑳𝑳𝑳𝑳𝑳𝑳 ~ △ 𝑱𝑱𝑱𝑱𝑱𝑱 by AA criterion,
so corresponding sides are
proportional:
𝟒𝟒
𝒛𝒛
=
𝟐𝟐𝟐𝟐 𝒛𝒛
𝒛𝒛𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝒛𝒛 = √𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏
d.
Describe the pattern that you see in your calculations for parts (a) through (c).
For each of the given right triangles, the length of the altitude drawn to its hypotenuse is equal to the square
root of the product of the lengths of the pieces of the hypotenuse that it cuts.
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2.
���� drawn to the hypotenuse, find the lengths of 𝑬𝑬𝑬𝑬
�����, 𝑭𝑭𝑭𝑭
����, and �����
Given right triangle 𝑬𝑬𝑬𝑬𝑬𝑬 with altitude 𝑭𝑭𝑭𝑭
𝑮𝑮𝑮𝑮.
The altitude drawn from 𝑭𝑭 to 𝑯𝑯 cuts triangle 𝑬𝑬𝑬𝑬𝑬𝑬 into two similar sub-triangles, providing the following
correspondence:
△ 𝑬𝑬𝑬𝑬𝑬𝑬 ~ △ 𝑬𝑬𝑬𝑬𝑬𝑬~ △ 𝑭𝑭𝑭𝑭𝑭𝑭
Using the ratio shorter leg: hypotenuse for the similar triangles:
𝟏𝟏𝟏𝟏 𝑯𝑯𝑯𝑯
=
𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟐𝟐𝟐𝟐(𝑯𝑯𝑯𝑯)
𝟏𝟏𝟏𝟏𝟏𝟏
= 𝑯𝑯𝑯𝑯
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏
𝟑𝟑
𝟗𝟗
= 𝟗𝟗 = 𝑯𝑯𝑯𝑯
𝟐𝟐𝟐𝟐
𝟓𝟓
𝟏𝟏𝟏𝟏 𝑯𝑯𝑯𝑯
=
𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟐𝟐𝟐𝟐(𝑯𝑯𝑯𝑯)
𝟏𝟏𝟏𝟏𝟏𝟏
= 𝑯𝑯𝑯𝑯
𝟐𝟐𝟐𝟐
𝟒𝟒
𝟏𝟏
𝟕𝟕
= 𝟕𝟕 = 𝑯𝑯𝑯𝑯
𝟐𝟐𝟐𝟐
𝟓𝟓
𝑬𝑬𝑬𝑬 + 𝑮𝑮𝑮𝑮 = 𝑬𝑬𝑬𝑬
𝟏𝟏
𝟕𝟕 + 𝑮𝑮𝑮𝑮 = 𝟐𝟐𝟐𝟐
𝟓𝟓
𝑮𝑮𝑮𝑮 = 𝟏𝟏𝟏𝟏
3.
𝟒𝟒
𝟓𝟓
Given triangle 𝑰𝑰𝑰𝑰𝑰𝑰 with altitude ���
𝑱𝑱𝑱𝑱, 𝑱𝑱𝑱𝑱 = 𝟑𝟑𝟑𝟑, and 𝑰𝑰𝑰𝑰 = 𝟐𝟐𝟐𝟐, find 𝑰𝑰𝑰𝑰, 𝑱𝑱𝑱𝑱, 𝑳𝑳𝑳𝑳, and 𝑰𝑰𝑰𝑰.
��� cuts △ 𝑰𝑰𝑰𝑰𝑰𝑰 into two similar sub-triangles such that △ 𝑰𝑰𝑰𝑰𝑰𝑰 ~ △ 𝑱𝑱𝑱𝑱𝑱𝑱~ △ 𝑰𝑰𝑰𝑰𝑰𝑰.
Altitude 𝑱𝑱𝑱𝑱
By the Pythagorean theorem:
𝟐𝟐𝟒𝟒𝟐𝟐 + 𝟑𝟑𝟐𝟐𝟐𝟐 = 𝑰𝑰𝑱𝑱𝟐𝟐
𝟓𝟓𝟓𝟓𝟓𝟓 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝑰𝑰𝑱𝑱𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝑰𝑰𝑱𝑱𝟐𝟐
Using the ratio shorter leg: longer leg:
𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒
=
𝟑𝟑𝟑𝟑 𝑱𝑱𝑱𝑱
𝟐𝟐𝟐𝟐(𝑱𝑱𝑱𝑱) = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝑱𝑱𝑱𝑱 =
𝟐𝟐𝟐𝟐
𝟏𝟏
𝑱𝑱𝑱𝑱 = 𝟓𝟓𝟓𝟓
𝟑𝟑
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√𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝑰𝑰𝑰𝑰
𝟒𝟒𝟒𝟒 = 𝑰𝑰𝑰𝑰
Using the ratio shorter leg: hypotenuse:
𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒
=
𝟒𝟒𝟒𝟒 𝑰𝑰𝑰𝑰
𝟐𝟐𝟐𝟐(𝑰𝑰𝑰𝑰) = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝑰𝑰𝑰𝑰 =
𝟐𝟐𝟐𝟐
𝟐𝟐
𝑰𝑰𝑰𝑰 = 𝟔𝟔𝟔𝟔
𝟑𝟑
𝑰𝑰𝑰𝑰 + 𝑳𝑳𝑳𝑳 = 𝑰𝑰𝑰𝑰
𝟐𝟐
𝟑𝟑
𝟐𝟐
𝑳𝑳𝑳𝑳 = 𝟒𝟒𝟒𝟒
𝟑𝟑
𝟐𝟐𝟐𝟐 + 𝑳𝑳𝑳𝑳 = 𝟔𝟔𝟔𝟔
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GEOMETRY
4.
���� to its hypotenuse, 𝑻𝑻𝑻𝑻 = 𝟏𝟏
Given right triangle 𝑹𝑹𝑹𝑹𝑹𝑹 with altitude 𝑹𝑹𝑹𝑹
sides of △ 𝑹𝑹𝑹𝑹𝑹𝑹.
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏
, and 𝑹𝑹𝑹𝑹 = 𝟔𝟔 , find the lengths of the
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
���� cuts △ 𝑹𝑹𝑹𝑹𝑹𝑹 into similar sub-triangles: △ 𝑼𝑼𝑼𝑼𝑼𝑼 ~ △ 𝑼𝑼𝑼𝑼𝑼𝑼.
Altitude 𝑹𝑹𝑹𝑹
Using the Pythagorean theorem:
Using the ratio shorter leg: hypotenuse:
𝑻𝑻𝑼𝑼𝟐𝟐 + 𝑹𝑹𝑼𝑼𝟐𝟐 = 𝑻𝑻𝑹𝑹𝟐𝟐
𝟒𝟒𝟒𝟒
𝟐𝟐𝟐𝟐 = 𝟕𝟕
𝟕𝟕
𝑺𝑺𝑺𝑺
𝟐𝟐
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏 𝟐𝟐
�𝟏𝟏 � + �𝟔𝟔 � = 𝑻𝑻𝑹𝑹𝟐𝟐
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝟒𝟒𝟒𝟒
(𝑺𝑺𝑺𝑺) = 𝟒𝟒𝟒𝟒
𝟐𝟐𝟐𝟐
𝟒𝟒𝟒𝟒 𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐
� � +�
� = 𝑻𝑻𝑹𝑹𝟐𝟐
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
+
= 𝑻𝑻𝑹𝑹𝟐𝟐
𝟔𝟔𝟔𝟔𝟔𝟔
𝟔𝟔𝟔𝟔𝟔𝟔
𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
= 𝑻𝑻𝑹𝑹𝟐𝟐
𝟔𝟔𝟔𝟔𝟔𝟔
𝑺𝑺𝑺𝑺 = 𝟐𝟐𝟐𝟐
𝟒𝟒𝟒𝟒 = 𝑻𝑻𝑹𝑹𝟐𝟐
√𝟒𝟒𝟒𝟒 = 𝑻𝑻𝑻𝑻
Using the Pythagorean theorem:
𝟕𝟕 = 𝑻𝑻𝑻𝑻
𝑹𝑹𝑺𝑺𝟐𝟐 + 𝑹𝑹𝑻𝑻𝟐𝟐 = 𝑺𝑺𝑻𝑻𝟐𝟐
𝑹𝑹𝑺𝑺𝟐𝟐 + 𝟕𝟕𝟐𝟐 = 𝟐𝟐𝟓𝟓𝟐𝟐
𝑹𝑹𝑺𝑺𝟐𝟐 + 𝟒𝟒𝟒𝟒 = 𝟔𝟔𝟔𝟔𝟔𝟔
𝑹𝑹𝑺𝑺𝟐𝟐 = 𝟓𝟓𝟓𝟓𝟓𝟓
𝑹𝑹𝑹𝑹 = √𝟓𝟓𝟓𝟓𝟓𝟓 = 𝟐𝟐𝟐𝟐
Lesson 21:
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Sub-Triangles
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Lesson 21
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M2
GEOMETRY
Note to the teacher: The next problem involves radical values that students have used previously; however, rationalizing
is a focus of Lessons 23 and 24, so the solutions provided in this problem involve nonrationalized values.
5.
Given right triangle 𝑨𝑨𝑨𝑨𝑪𝑪 with altitude ����
𝑪𝑪𝑪𝑪, find 𝑨𝑨𝑨𝑨, 𝑩𝑩𝑩𝑩, 𝑨𝑨𝑨𝑨, and 𝑫𝑫𝑫𝑫.
Using the Pythagorean theorem:
𝟐𝟐
𝟐𝟐
�𝟐𝟐√𝟓𝟓� + �√𝟕𝟕� = 𝑨𝑨𝑩𝑩𝟐𝟐
𝟐𝟐𝟐𝟐 + 𝟕𝟕 = 𝑨𝑨𝑩𝑩𝟐𝟐
𝟐𝟐𝟐𝟐 = 𝑨𝑨𝑩𝑩𝟐𝟐
√𝟐𝟐𝟐𝟐 = 𝑨𝑨𝑨𝑨
𝟑𝟑√𝟑𝟑 = 𝑨𝑨𝑨𝑨
An altitude from the right angle in a right triangle to the hypotenuse cuts the triangle into two similar right
triangles: △ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑨𝑨𝑨𝑨𝑨𝑨~ △ 𝑪𝑪𝑪𝑪𝑪𝑪.
Using the ratio shorter leg: hypotenuse:
√𝟕𝟕
𝟑𝟑√𝟑𝟑
=
𝑫𝑫𝑫𝑫
𝟐𝟐√𝟓𝟓
𝟐𝟐√𝟑𝟑𝟑𝟑 = 𝑫𝑫𝑫𝑫�𝟑𝟑√𝟑𝟑�
𝟐𝟐√𝟑𝟑𝟑𝟑
𝟑𝟑√𝟑𝟑
= 𝑫𝑫𝑫𝑫
√𝟕𝟕
𝟑𝟑√𝟑𝟑
𝟕𝟕
=
𝑫𝑫𝑫𝑫
√𝟕𝟕
𝟕𝟕 = 𝑫𝑫𝑫𝑫�𝟑𝟑√𝟑𝟑�
𝟑𝟑√𝟑𝟑
= 𝑫𝑫𝑫𝑫
Using the Pythagorean theorem:
𝑨𝑨𝑫𝑫𝟐𝟐 + 𝑫𝑫𝑪𝑪𝟐𝟐 = 𝑨𝑨𝑪𝑪𝟐𝟐
𝟐𝟐
𝟐𝟐
𝟐𝟐√𝟑𝟑𝟑𝟑
𝑨𝑨𝑫𝑫𝟐𝟐 + �
� = �𝟐𝟐√𝟓𝟓�
𝟑𝟑√𝟑𝟑
𝟒𝟒 ⋅ 𝟑𝟑𝟑𝟑
= (𝟒𝟒 ⋅ 𝟓𝟓)
𝟗𝟗 ⋅ 𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏
= 𝟐𝟐𝟐𝟐
𝑨𝑨𝑫𝑫𝟐𝟐 +
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝑨𝑨𝑫𝑫𝟐𝟐 = 𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐
𝑨𝑨𝑫𝑫𝟐𝟐 +
𝑨𝑨𝑨𝑨 = �𝟏𝟏𝟏𝟏
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𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
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Lesson 21
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M2
GEOMETRY
6.
���� is an altitude of
Right triangle 𝑫𝑫𝑫𝑫𝑫𝑫 is inscribed in a circle with radius 𝑨𝑨𝑨𝑨 = 𝟓𝟓. ����
𝑫𝑫𝑫𝑫 is a diameter of the circle, 𝑬𝑬𝑬𝑬
△ 𝑫𝑫𝑫𝑫𝑫𝑫, and 𝑫𝑫𝑫𝑫 = 𝟔𝟔. Find the lengths 𝒙𝒙 and 𝒚𝒚.
The radius of the circle is 𝟓𝟓, and
𝑫𝑫𝑫𝑫 = 𝟐𝟐(𝟓𝟓) = 𝟏𝟏𝟏𝟏.
By the Pythagorean theorem:
𝟔𝟔𝟐𝟐 + 𝑬𝑬𝑪𝑪𝟐𝟐 = 𝟏𝟏𝟎𝟎𝟐𝟐
𝟑𝟑𝟑𝟑 + 𝑬𝑬𝑬𝑬𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝑬𝑬𝑬𝑬𝟐𝟐 = 𝟔𝟔𝟔𝟔
𝑬𝑬𝑬𝑬 = 𝟖𝟖
(𝑬𝑬𝑬𝑬 = −𝟖𝟖 is also a solution; however,
since 𝑬𝑬𝑬𝑬 represents a distance, its value
must be positive, so the solution
𝑬𝑬𝑬𝑬 = −𝟖𝟖 is disregarded.)
We showed that an altitude from the
right angle of a right triangle to the
hypotenuse cuts the triangle into two
similar sub-triangles, so
△ 𝑫𝑫𝑫𝑫𝑫𝑫 ~ △ 𝑫𝑫𝑫𝑫𝑫𝑫 ~ △ 𝑬𝑬𝑬𝑬𝑬𝑬.
𝒚𝒚
𝟔𝟔
=
𝟏𝟏
𝟏𝟏
𝟏𝟏
𝟏𝟏
𝟔𝟔
𝟔𝟔
Using the ratio shorter leg: hypotenuse for similar right triangles:
7.
𝟔𝟔
𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝟑𝟑.
=
𝒙𝒙
𝟖𝟖
𝟒𝟒𝟒𝟒 = 𝟏𝟏𝟏𝟏𝟏𝟏
= 𝒚𝒚
𝟒𝟒. 𝟖𝟖 = 𝒙𝒙
In right triangle 𝑨𝑨𝑨𝑨𝑨𝑨, 𝑨𝑨𝑨𝑨 = 𝟓𝟓𝟓𝟓, and altitude 𝑫𝑫𝑫𝑫 = 𝟏𝟏𝟏𝟏. Find the lengths of ����
𝑩𝑩𝑩𝑩 and ����
𝑨𝑨𝑨𝑨.
Let length 𝑩𝑩𝑩𝑩 = 𝒙𝒙. Then, 𝑨𝑨𝑨𝑨 = 𝟓𝟓𝟓𝟓 − 𝒙𝒙.
Using the pattern from Problem 1,
𝟏𝟏𝟒𝟒𝟐𝟐 = 𝒙𝒙(𝟓𝟓𝟓𝟓 − 𝒙𝒙)
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟓𝟓𝟓𝟓𝟓𝟓 − 𝒙𝒙𝟐𝟐
𝒙𝒙𝟐𝟐 − 𝟓𝟓𝟓𝟓𝟓𝟓 + 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟎𝟎
(𝒙𝒙 − 𝟒𝟒𝟒𝟒)(𝒙𝒙 − 𝟒𝟒) = 𝟎𝟎
𝒙𝒙 = 𝟒𝟒𝟒𝟒 or 𝒙𝒙 = 𝟒𝟒.
Using the solutions from the equation and the given information, either 𝑩𝑩𝑩𝑩 = 𝟒𝟒 and 𝑨𝑨𝑨𝑨 = 𝟒𝟒𝟒𝟒, or 𝑩𝑩𝑩𝑩 = 𝟒𝟒𝟒𝟒 and
𝑨𝑨𝑨𝑨 = 𝟒𝟒.
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M2
Lesson 22
NYS COMMON CORE MATHEMATICS CURRICULUM
GEOMETRY
Lesson 22: Multiplying and Dividing Expressions with
Student Outcomes

Students multiply and divide expressions that contain radicals to simplify their answers.

Students rationalize the denominator of a number expressed as a fraction.
Lesson Notes
Exercises 1–5 and the Discussion are meant to remind students of what they learned about roots in Grade 8 and Algebra
I. In Grade 8, students learned the notation related to square roots and understood that the square root symbol
automatically denotes the positive root (Grade 8, Module 7). In Algebra I, students used both the positive and negative
roots of a number to find the location of the roots of a quadratic function. Because of the upcoming work with special
triangles in this module, this lesson reviews what students learned about roots in Grade 8 Module 7 Lesson 4. For
example, students need to understand that
1 √2
√2 2
when they are writing the trigonometric ratios of right triangles. To
achieve this understanding, students must learn how to rationalize the denominator of numbers expressed as fractions.
It is also important for students to get a sense of the value of a number. When a radical is in the denominator or is not
simplified, it is more challenging to estimate its value, for example, √3750 compared to 25√6.
For students who are struggling with the concepts of multiplying and dividing expressions with radicals, it may be
necessary to divide the lesson so that multiplication is the focus one day and division the next. This lesson is a steppingstone, as it moves students toward an understanding of how to rewrite expressions involving radical and rational
exponents using the properties of exponents (N.RN.A.2), which are not mastered until Algebra II.
The lesson focuses on simplifying expressions and solving equations that contain terms with roots. By the end of the
lesson, students should understand that one reason the denominator of a number expressed as a fraction is rationalized
is to better estimate the value of the number. For example, students can more accurately estimate the value of
when written as
3√3
3
or simply √3. Further, putting numbers in this form allows students to more easily recognize when
numbers can be combined. For example, if adding √3 and
until
1
√3
is rewritten as
√3
3
. Then, the sum of √3 and
√3
3
is
1
Lesson 22:
, students may not recognize that they can be combined
√3
4√3
3
an expected standard form such as a rationalized expression.
GEO-M2-TE-1.3.0-07.2015
3
√3
.
As a teacher, it is easier to check answers when there is
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Classwork
Exercises 1–5 (8 minutes)
The first three exercises review square roots that are perfect squares. The last two exercises require students to
compare the value of two radical expressions and make a conjecture about their relationship. These last two exercises
exemplify what is studied in this lesson. Students may need to be reminded that the square root symbol automatically
denotes the positive root of the number.
Scaffolding:
Exercises 1–5
Simplify as much as possible.
1.
√𝟏𝟏𝟕𝟕𝟐𝟐 =
�𝟏𝟏𝟕𝟕𝟐𝟐 = 𝟏𝟏𝟏𝟏
2.
√𝟓𝟓𝟏𝟏𝟏𝟏 =
�𝟓𝟓𝟏𝟏𝟏𝟏 = �𝟓𝟓𝟐𝟐 × �𝟓𝟓𝟐𝟐 × �𝟓𝟓𝟐𝟐 × �𝟓𝟓𝟐𝟐 × �𝟓𝟓𝟐𝟐
√𝟒𝟒𝒙𝒙𝟒𝟒 =
�𝟒𝟒𝒙𝒙𝟒𝟒 = √𝟒𝟒 × �𝒙𝒙𝟐𝟐 × �𝒙𝒙𝟐𝟐
3.
4.
= 𝟓𝟓 × 𝟓𝟓 × 𝟓𝟓 × 𝟓𝟓 × 𝟓𝟓
= 𝟓𝟓𝟓𝟓
= 𝟐𝟐 × 𝒙𝒙 × 𝒙𝒙
a.
 Consider doing a fluency
activity that allows
students to learn their
perfect squares up to 30.
This may include choral
recitation.
 English language learners
may benefit from choral
practice with the word
= 𝟐𝟐|𝒙𝒙|𝟐𝟐
Complete parts (a) through (c).
 Some students may need
to review the perfect
squares. A reproducible
sheet for squares of
numbers 1–30 is provided
at the end of the lesson.
Compare the value of √𝟑𝟑𝟑𝟑 to the value of √𝟗𝟗 × √𝟒𝟒.
The value of the two expressions is equal. The square root of 𝟑𝟑𝟑𝟑 is 𝟔𝟔, and the product of the square roots of 𝟗𝟗
and 𝟒𝟒 is also 𝟔𝟔.
b.
Make a conjecture about the validity of the following statement: For nonnegative real numbers 𝒂𝒂 and 𝒃𝒃,
√𝒂𝒂𝒂𝒂 = √𝒂𝒂 ∙ √𝒃𝒃. Explain.
Answers will vary. Students should say that the statement √𝒂𝒂𝒂𝒂 = √𝒂𝒂 ∙ √𝒃𝒃 is valid because of the problem
that was just completed: √𝟑𝟑𝟑𝟑 = √𝟗𝟗 × √𝟒𝟒 = 𝟔𝟔.
c.
Does your conjecture hold true for 𝒂𝒂 = −𝟒𝟒 and 𝒃𝒃 = −𝟗𝟗?
No. The conjecture is not true when the numbers are negative because we cannot take the square root of a
negative number. �(−𝟒𝟒)(−𝟗𝟗) = √𝟑𝟑𝟑𝟑 = 𝟔𝟔, but we cannot calculate √−𝟒𝟒 × √−𝟗𝟗 in order to compare.
5.
Complete parts (a) through (c).
a.
𝟏𝟏𝟏𝟏𝟏𝟏
Compare the value of �
𝟐𝟐𝟐𝟐
to the value of
√𝟏𝟏𝟏𝟏𝟏𝟏
√𝟐𝟐𝟐𝟐
.
The value of the two expressions is equal. The fraction
𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐
simplifies to 𝟒𝟒, and the square root of 𝟒𝟒 is 𝟐𝟐. The
square root of 𝟏𝟏𝟏𝟏𝟏𝟏 divided by the square root of 𝟐𝟐𝟐𝟐 is equal to
Lesson 22:
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𝟏𝟏𝟏𝟏
𝟓𝟓
, which is equal to 𝟐𝟐.
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b.
Make a conjecture about the validity of the following statement: For nonnegative real numbers 𝒂𝒂 and 𝒃𝒃,
𝒂𝒂
𝒃𝒃
when 𝒃𝒃 ≠ 𝟎𝟎, � =
√𝒂𝒂
. Explain.
�𝒃𝒃
𝒂𝒂
𝒃𝒃
Answers will vary. Students should say that the statement � =
𝟏𝟏𝟏𝟏𝟏𝟏 �𝟏𝟏𝟏𝟏𝟏𝟏
just completed: �
=
= 𝟐𝟐.
𝟐𝟐𝟐𝟐
c.
�𝟐𝟐𝟐𝟐
√𝒂𝒂
�𝒃𝒃
is valid because of the problem that was
Does your conjecture hold true for 𝒂𝒂 = −𝟏𝟏𝟏𝟏𝟏𝟏 and 𝒃𝒃 = −𝟐𝟐𝟐𝟐?
No. The conjecture is not true when the numbers are negative because we cannot take the square root of a
negative number.
−𝟏𝟏𝟏𝟏𝟏𝟏
�
−𝟐𝟐𝟐𝟐
=
√𝟒𝟒 = 𝟐𝟐, but we cannot calculate
√−𝟏𝟏𝟏𝟏𝟏𝟏
√−𝟐𝟐𝟐𝟐
in order to compare.
Discussion (8 minutes)
Debrief Exercises 1–5 by reminding students of the definition for square root, the facts given by the definition, and the
rules associated with square roots for positive radicands. Whenever possible, elicit the facts and definitions from
students based on their work in Exercises 1–5. Within this Discussion is the important distinction between a square root
of a number and the square root of a number. A square root of a number may be negative; however, the square root of
a number always refers to the principle square root or the positive root of the number.

Definition of the square root: If 𝑥𝑥 ≥ 0, then √𝑥𝑥 is the nonnegative number 𝑝𝑝 so that 𝑝𝑝2 = 𝑥𝑥. This definition
gives us four facts. The definition should not be confused with finding a square root of a number. For
example, −2 is a square root of 4, but the square root of 4, that is, √4, is 2.
Consider asking students to give an example of each fact using concrete numbers. Sample responses are included below
each fact.
Fact 1: √𝑎𝑎2 = 𝑎𝑎 if 𝑎𝑎 ≥ 0

√122 = 12, for any positive squared number in the radicand
Fact 2: √𝑎𝑎2 = −𝑎𝑎 if 𝑎𝑎 < 0
This may require additional explanation because students see the answer as “negative 𝑎𝑎,” as opposed to the opposite of
𝑎𝑎. For this fact, it is assumed that 𝑎𝑎 is a negative number; therefore, −𝑎𝑎 is a positive number. It is similar to how
students think about the absolute value of a number 𝑎𝑎: |𝑎𝑎| = 𝑎𝑎 if 𝑎𝑎 > 0, but |𝑎𝑎| = −𝑎𝑎 if 𝑎𝑎 < 0. Simply put, the minus
sign is just telling students they need to take the opposite of the negative number 𝑎𝑎 to produce the desired result, that
is, −(−𝑎𝑎) = 𝑎𝑎.

�(−5)2 = 5, for any negative squared number in the radicand
Fact 3: √𝑎𝑎2 = |𝑎𝑎| for all real numbers 𝑎𝑎

√132 = |13|, and �(−13)2 = | − 13|
Fact 4: √𝑎𝑎2𝑛𝑛 = �(𝑎𝑎𝑛𝑛 )2 = 𝑎𝑎𝑛𝑛 when 𝑎𝑎𝑛𝑛 is nonnegative

√716 = �(78 )2 = 78
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Consider asking students which of the first five exercises used Rule 1.

When 𝑎𝑎 ≥ 0, 𝑏𝑏 ≥ 0, and 𝑐𝑐 ≥ 0, then the following rules can be applied to square roots:
Rule 1: √𝑎𝑎𝑎𝑎 = √𝑎𝑎 ∙ √𝑏𝑏
A consequence of Rule 1 and the associative property gives us the following:
Rule 2:
𝑎𝑎
�𝑏𝑏
=
√𝑎𝑎
√𝑏𝑏
when 𝑏𝑏 ≠ 0
√𝑎𝑎𝑎𝑎𝑎𝑎 = �𝑎𝑎(𝑏𝑏𝑏𝑏) = √𝑎𝑎 ⋅ √𝑏𝑏𝑏𝑏 = √𝑎𝑎 ∙ √𝑏𝑏 ∙ √𝑐𝑐
We want to show that √𝑎𝑎𝑎𝑎 = √𝑎𝑎 ∙ √𝑏𝑏 for 𝑎𝑎 ≥ 0 and 𝑏𝑏 ≥ 0. To do so, we use the definition of square root.
Consider allowing time for students to discuss with partners how they can prove Rule 1. Shown below are three proofs
of Rule 1. Share one or all with the class.
The proof of Rule 1: Let 𝑝𝑝 be the nonnegative number so that 𝑝𝑝2 = 𝑎𝑎, and let 𝑞𝑞 be the nonnegative number so
that 𝑞𝑞 2 = 𝑏𝑏. Then, √𝑎𝑎𝑎𝑎 = 𝑝𝑝𝑝𝑝 because 𝑝𝑝𝑝𝑝 is nonnegative (it is the product of two nonnegative numbers), and
(𝑝𝑝𝑝𝑝)2 = 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝑝𝑝2 𝑞𝑞 2 = 𝑎𝑎𝑎𝑎. Then, by definition, √𝑎𝑎𝑎𝑎 = 𝑝𝑝𝑝𝑝 = √𝑎𝑎 ⋅ √𝑏𝑏. Since both sides equal 𝑝𝑝𝑝𝑝, the
equation is true.
The proof of Rule 1: Let 𝐶𝐶 = √𝑎𝑎 ⋅ �𝑏𝑏, and let 𝐷𝐷 = √𝑎𝑎𝑎𝑎. We need to show that 𝐶𝐶 = 𝐷𝐷. Given positive
numbers 𝐶𝐶, 𝐷𝐷, and exponent 2, if we can show that 𝐶𝐶 2 = 𝐷𝐷2 , then we know that 𝐶𝐶 = 𝐷𝐷, and Rule 1 is proved.
Consider asking students why it is true that if they can show that 𝐶𝐶 2 = 𝐷𝐷2 , then they know that 𝐶𝐶 = 𝐷𝐷. Students should
refer to what they know about the definition of exponents. That is, since 𝐶𝐶 2 = 𝐶𝐶 × 𝐶𝐶 and 𝐷𝐷2 = 𝐷𝐷 × 𝐷𝐷 and
𝐶𝐶 × 𝐶𝐶 = 𝐷𝐷 × 𝐷𝐷, then 𝐶𝐶 must be the same number as 𝐷𝐷.

With that goal in mind, we take each of 𝐶𝐶 = √𝑎𝑎√𝑏𝑏 and 𝐷𝐷 = √𝑎𝑎𝑎𝑎, and by the multiplication property of
equality, we raise both sides of each equation to a power of 2.
2
𝐶𝐶 2 = �√𝑎𝑎 ⋅ √𝑏𝑏�
= �√𝑎𝑎 ⋅ √𝑏𝑏� ⋅ �√𝑎𝑎 ⋅ √𝑏𝑏�
= √𝑎𝑎 ⋅ √𝑎𝑎 ⋅ √𝑏𝑏 ⋅ √𝑏𝑏
= 𝑎𝑎𝑎𝑎
𝐷𝐷2 = �√𝑎𝑎𝑎𝑎�
2
= √𝑎𝑎𝑎𝑎 ⋅ √𝑎𝑎𝑎𝑎
= 𝑎𝑎𝑎𝑎
Since 𝐶𝐶 2 = 𝐷𝐷2 implies 𝐶𝐶 = 𝐷𝐷, then √𝑎𝑎𝑎𝑎 = √𝑎𝑎√𝑏𝑏.
The proof of Rule 1: Let 𝐶𝐶, 𝐷𝐷 > 0. If 𝐶𝐶 2 = 𝐷𝐷2 , then 𝐶𝐶 = 𝐷𝐷. Assume 𝐶𝐶 2 = 𝐷𝐷2 ; then, 𝐶𝐶 2 − 𝐷𝐷2 = 0. By
factoring the difference of squares, we have (𝐶𝐶 + 𝐷𝐷)(𝐶𝐶 − 𝐷𝐷) = 0. Since both 𝐶𝐶 and 𝐷𝐷 are positive, then
𝐶𝐶 + 𝐷𝐷 > 0, which means that 𝐶𝐶 − 𝐷𝐷 must be equal to zero because of the zero product property. Since
𝐶𝐶 − 𝐷𝐷 = 0, then 𝐶𝐶 = 𝐷𝐷.
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Example 1 (4 minutes)

We can use Rule 1 to rationalize the denominators of fractional expressions. One reason we do this is so that
we can better estimate the value of a number. For example, if we know that √2 ≈ 1.414, what is the value of
1
√2

? Isn’t it is easier to determine the value of
1
√2
and
√2
2
are equivalent.
Another reason to rationalize the denominators of fractional expressions is because putting numbers in this
form allows us to more easily recognize when numbers can be combined. For example, if you have to add √3
and

2
? The fractional expressions
Notice that the first expression has the irrational number √2 as its denominator, and the second expression
has the rational number 2 as its denominator. What we will learn today is how to rationalize the denominator
of a fractional expression using Rule 1.
and

√2
1
, you may not recognize that they can be combined until
√3
√3
3
is
4√3
3
.
1
√3
is rewritten as
√3
3
. Then, the sum of √3
We want to express numbers in their simplest radical form. An expression is in its simplest radical form when
the radicand (the expression under the radical sign) has no factor that can be raised to a power greater than or
equal to the index (either 2 or 3), and there is no radical in the denominator.
Using Rule 1 for square roots, we can simplify expressions that contain square
roots by writing the factors of the number under the square root sign as
products of perfect squares, if possible. For example, to simplify √75, we
consider all of the factors of 75, focusing on those factors that are perfect
squares. Which factors should we use?

Then,
Scaffolding:
Consider showing multiple
simple examples. For example:
√28 = √4 ⋅ √7
= 2√7
We should use 25 and 3 because 25 is a perfect square.
√75 = √25 ∙ √3
= 5√3
√45 = √9 ⋅ √5
= 3√5
√32 = √16 ⋅ √2
= 4√2
Example 2 (2 minutes)
In Example 2, we first use Rule 2 to rewrite a number as a rational expression and then use Rule 1 to rationalize a
denominator, that is, rewrite the denominator as an integer. We have not yet proved this rule, because it is an exercise
in the Problem Set. Consider mentioning this fact to students.

Rules 1 and 2 for square roots are used to rationalize denominators of fractional expressions.
Consider asking students what it means to rationalize the denominator. Students should understand that rationalizing
the denominator means expressing it as an integer.
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
3
3
Consider the expression � . By Rule 2, � =
5
5
with a rational number for the denominator.
√3
√5
×
√5
√5
=
=
=
√3√5
√5√5
√15
√25
√3
√5
. We want to write an expression that is equivalent to
√3
√5
By multiplication rule for fractional expressions
By Rule 1
√15
.
5
Example 3 (3 minutes)

Demarcus found the scale factor of a dilation to be
√2
2
1
√2
.
When he compared his answer to Yesenia’s, which was
, he told her that one of them must have made a mistake.
Show work, and provide an explanation to
Demarcus and Yesenia that proves they are both correct.

Student work:
1
MP.3
√2
×
√2
√2
=
=
=
1√2
√2√2
√2
√4
√2
2
By multiplication rule for fractional expressions
By Rule 1
By definition of square root
If Demarcus were to rationalize the denominator of his answer, he would see that it is equal to
Yesenia’s answer. Therefore, they are both correct.
Example 4 (5 minutes)

Assume 𝑥𝑥 > 0. Rationalize the denominator of
𝑥𝑥
√𝑥𝑥 3
Provide time for students to work independently or in pairs. Use the question below if necessary. Seek out students
who multiplied by different factors to produce an equivalent fractional expression to simplify this problem. For example,
some students may have multiplied by
√𝑥𝑥 3
√𝑥𝑥 3
, while others may have used
√𝑥𝑥
√𝑥𝑥
or some other fractional expression that
would produce an exponent of 𝑥𝑥 with an even number, which can be simplified. Have students share their work and

We need to multiply √𝑥𝑥 3 by a number so that it becomes a perfect square. What should we multiply by?
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Students may say to multiply by √𝑥𝑥 3 because that is what was done in the two previous examples. If so, finish the
problem that way, and then show that we can multiply by √𝑥𝑥 and get the same answer. Ask students why both methods
work. They should mention equivalent expressions and the role that the number
equivalent expression.

Student work:
𝑥𝑥
√𝑥𝑥 3
×
√𝑥𝑥 3
√𝑥𝑥 3
=
=
𝑥𝑥
𝑥𝑥√𝑥𝑥 3
√𝑥𝑥 3
√𝑥𝑥 3 √𝑥𝑥 3
𝑥𝑥√𝑥𝑥 3
√𝑥𝑥 6
𝑥𝑥𝑥𝑥 √𝑥𝑥
=
𝑥𝑥 3
2
𝑥𝑥 √𝑥𝑥
= 3
𝑥𝑥
√𝑥𝑥
=
𝑥𝑥
×
√𝑥𝑥 3
√𝑥𝑥 3
or
√𝑥𝑥
√𝑥𝑥
plays in producing the
𝑥𝑥 √𝑥𝑥
√𝑥𝑥
=
√𝑥𝑥 √𝑥𝑥 3 √𝑥𝑥
𝑥𝑥 √𝑥𝑥
=
√𝑥𝑥 4
𝑥𝑥 √𝑥𝑥
= 2
𝑥𝑥
√𝑥𝑥
=
𝑥𝑥
Exercises 6–17 (7 minutes)
Have students work through all of the exercises in this set, or select problems for students to complete based on their
level. Students who are struggling should complete Exercises 6–10. Students who are on level should complete
Exercises 9–13. Students who are accelerated should complete Exercises 13–16. All students should attempt to
complete Exercise 17.
Exercises 6–17
Simplify each expression as much as possible, and rationalize denominators when applicable.
6.
√𝟕𝟕𝟕𝟕 =
7.
√𝟕𝟕𝟕𝟕 = √𝟑𝟑𝟑𝟑 ⋅ √𝟐𝟐
= 𝟔𝟔√𝟐𝟐
𝟏𝟏𝟏𝟏
=
𝟐𝟐𝟐𝟐
�
𝟏𝟏𝟏𝟏 √𝟏𝟏𝟏𝟏
� =
𝟐𝟐𝟐𝟐 √𝟐𝟐𝟐𝟐
=
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𝟓𝟓
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8.
√𝟑𝟑𝟑𝟑𝟑𝟑 =
9.
√𝟑𝟑𝟑𝟑𝟑𝟑 = √𝟏𝟏𝟏𝟏√𝟐𝟐𝟐𝟐
𝟏𝟏
�
=
𝟑𝟑
𝟏𝟏 √𝟏𝟏
� =
𝟑𝟑 √𝟑𝟑
= 𝟒𝟒√𝟐𝟐𝟐𝟐
=
=
=
10. √𝟓𝟓𝟓𝟓𝒙𝒙𝟐𝟐 =
11.
�𝟓𝟓𝟓𝟓𝒙𝒙𝟐𝟐 = √𝟗𝟗√𝟔𝟔�𝒙𝒙𝟐𝟐
�
𝟒𝟒
𝒙𝒙𝟒𝟒
𝟓𝟓
𝟓𝟓
√𝒙𝒙𝟕𝟕
=
=
=
=
𝟓𝟓
𝟓𝟓√𝒙𝒙
√𝒙𝒙𝟖𝟖
×
√𝒙𝒙
√𝒙𝒙
𝟓𝟓√𝒙𝒙
𝒙𝒙𝟒𝟒
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=
𝟒𝟒𝟒𝟒
=
�𝟔𝟔𝟔𝟔𝒙𝒙𝟐𝟐
=
√𝟔𝟔𝟔𝟔𝒙𝒙𝟐𝟐
15.
√𝒙𝒙𝟕𝟕
√𝟑𝟑
√𝟑𝟑
𝟑𝟑
𝟒𝟒𝟒𝟒
𝟐𝟐
= 𝟐𝟐
𝒙𝒙
�𝒙𝒙𝟕𝟕
√𝟗𝟗
√𝟑𝟑
= √𝟐𝟐
𝟒𝟒
√𝟒𝟒
� 𝟒𝟒 =
𝒙𝒙
√𝒙𝒙𝟒𝟒
14.
√𝟑𝟑
×
𝟑𝟑𝟑𝟑
=�
𝟏𝟏𝟏𝟏
√𝟏𝟏𝟏𝟏
13.
=
√𝟏𝟏𝟏𝟏
√𝟑𝟑𝟑𝟑
= 𝟑𝟑|𝒙𝒙|√𝟔𝟔
12.
√𝟑𝟑𝟑𝟑
√𝟏𝟏
√𝟑𝟑
𝒙𝒙𝟓𝟓
�
𝟐𝟐
=
𝟒𝟒𝟒𝟒
𝟖𝟖𝟖𝟖
𝟏𝟏
𝟐𝟐
=
𝒙𝒙𝟓𝟓 √𝒙𝒙𝟓𝟓
� =
𝟐𝟐
√𝟐𝟐
=
=
𝒙𝒙𝟐𝟐 √𝒙𝒙
√𝟐𝟐
𝟐𝟐
×
𝒙𝒙 √𝟐𝟐𝟐𝟐
𝟐𝟐
√𝟐𝟐
√𝟐𝟐
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16.
√𝟏𝟏𝟏𝟏𝟏𝟏
𝟑𝟑�𝒙𝒙𝟓𝟓
√𝟏𝟏𝟏𝟏𝟏𝟏
𝟑𝟑√𝒙𝒙𝟓𝟓
=
=
=
=
=
=
√𝟗𝟗√𝟐𝟐𝟐𝟐
𝟑𝟑𝒙𝒙𝟐𝟐 √𝒙𝒙
𝟑𝟑√𝟐𝟐𝟐𝟐
𝟑𝟑𝒙𝒙𝟐𝟐 √𝒙𝒙
√𝟐𝟐𝟐𝟐 √𝒙𝒙
×
𝒙𝒙𝟐𝟐 √𝒙𝒙 √𝒙𝒙
𝒙𝒙√𝟐𝟐
𝒙𝒙𝟑𝟑
√𝟐𝟐
𝒙𝒙𝟐𝟐
17. The captain of a ship recorded the ship’s coordinates, then sailed north and then west, and then recorded the new
coordinates. The coordinates were used to calculate the distance they traveled, √𝟓𝟓𝟓𝟓𝟓𝟓 𝐤𝐤𝐤𝐤. When the captain asked
how far they traveled, the navigator said, “About 𝟐𝟐𝟐𝟐 𝐤𝐤𝐤𝐤.” Is the navigator correct? Under what conditions might
he need to be more precise in his answer?
Sample student responses:
The number √𝟓𝟓𝟓𝟓𝟓𝟓 is close to the perfect square √𝟓𝟓𝟓𝟓𝟓𝟓. The perfect square √𝟓𝟓𝟓𝟓𝟓𝟓 = 𝟐𝟐𝟐𝟐; therefore, the navigator is
correct in his estimate of the distance traveled.
When the number √𝟓𝟓𝟓𝟓𝟓𝟓 is simplified, the result is 𝟏𝟏𝟏𝟏√𝟐𝟐. The number 𝟓𝟓𝟓𝟓𝟓𝟓 has factors of 𝟐𝟐𝟐𝟐𝟐𝟐 and 𝟐𝟐. Then:
√𝟓𝟓𝟓𝟓𝟓𝟓 = √𝟐𝟐𝟐𝟐𝟐𝟐√𝟐𝟐
= 𝟏𝟏𝟏𝟏√𝟐𝟐
= 𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 …
≈ 𝟐𝟐𝟐𝟐
Yes, the navigator is correct in his estimate of the distance traveled.
A more precise answer may be needed if the captain were looking for a particular location, such as the location of a
shipwreck or buried treasure.
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Closing (4 minutes)
Ask the following questions. Have students respond in writing, to a partner, or to the whole class.

What are some of the basic facts and rules related to square roots?


The basic facts about square roots show us how to simplify a square root when it is a perfect square.
For example, √52 = 5, �(−5)2 = 5, �(−5)2 = | − 5|, and √512 = �(56 )2 = 56 . The rules allow us to
simplify square roots. Rule 1 shows that we can rewrite radicands as factors and simplify the factors, if
possible. Rule 2 shows us that the square root of a fractional expression can be expressed as the square
root of the numerator divided by the square root of a denominator.
What does it mean to rationalize the denominator of a fractional expression? Why might we want to do it?

Rationalizing a denominator means that the fractional expression must be expressed with a rational
number in the denominator. We might want to rationalize the denominator of a fractional expression
to better estimate the value of the number. Another reason is to verify whether two numbers are equal
or can be combined.
Exit Ticket (4 minutes)
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Name
Date
Lesson 22: Multiplying and Dividing Expressions with Radicals
Exit Ticket
Write each expression in its simplest radical form.
1.
√243 =
2.
�
3.
Teja missed class today. Explain to her how to write the length of the hypotenuse in simplest radical form.
7
5
=
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GEOMETRY
Exit Ticket Sample Solutions
Write each expression in its simplest radical form.
1.
2.
√𝟐𝟐𝟐𝟐𝟐𝟐 =
𝟕𝟕
�
𝟓𝟓
√𝟐𝟐𝟐𝟐𝟐𝟐 = √𝟖𝟖𝟖𝟖 ∙ √𝟑𝟑
= 𝟗𝟗√𝟑𝟑
𝟕𝟕 √𝟕𝟕 √𝟓𝟓
� =
×
𝟓𝟓 √𝟓𝟓 √𝟓𝟓
=
=
=
=
3.
√𝟕𝟕√𝟓𝟓
√𝟓𝟓√𝟓𝟓
√𝟑𝟑𝟑𝟑
√𝟐𝟐𝟐𝟐
√𝟑𝟑𝟑𝟑
𝟓𝟓
Teja missed class today. Explain to her how to write the length of the hypotenuse in simplest radical form.
Use the Pythagorean theorem to determine the length of the hypotenuse, 𝒄𝒄:
𝟓𝟓𝟐𝟐 + 𝟏𝟏𝟑𝟑𝟐𝟐 = 𝒄𝒄𝟐𝟐
𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝒄𝒄𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝒄𝒄𝟐𝟐
√𝟏𝟏𝟏𝟏𝟏𝟏 = 𝒄𝒄
To simplify the square root, rewrite the radicand as a product of its factors. The goal is to find a
factor that is a perfect square and can then be simplified. There are no perfect square factors of
the radicand; therefore, the length of the hypotenuse in simplest radical form is √𝟏𝟏𝟏𝟏𝟏𝟏.
Problem Set Sample Solutions
Express each number in its simplest radical form.
1.
√𝟔𝟔 ⋅ √𝟔𝟔𝟔𝟔 =
2.
√𝟔𝟔 ⋅ √𝟔𝟔𝟔𝟔 = √𝟔𝟔 ⋅ √𝟔𝟔 ⋅ √𝟏𝟏𝟏𝟏
√𝟏𝟏𝟏𝟏𝟏𝟏 = √𝟗𝟗 ∙ √𝟒𝟒 ∙ √𝟑𝟑
= 𝟔𝟔√𝟏𝟏𝟏𝟏
3.
√𝟏𝟏𝟏𝟏𝟏𝟏 =
= 𝟑𝟑 ∙ 𝟐𝟐√𝟑𝟑
= 𝟔𝟔√𝟑𝟑
Pablo found the length of the hypotenuse of a right triangle to be √𝟒𝟒𝟒𝟒. Can the length be simplified? Explain.
√𝟒𝟒𝟒𝟒 = √𝟗𝟗√𝟓𝟓
= 𝟑𝟑√𝟓𝟓
Yes, the length can be simplified because the number 𝟒𝟒𝟒𝟒 has a factor that is a perfect square.
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4.
√𝟏𝟏𝟏𝟏𝒙𝒙𝟒𝟒 =
�𝟏𝟏𝟏𝟏𝒙𝒙𝟒𝟒 = √𝟒𝟒√𝟑𝟑�𝒙𝒙𝟒𝟒
= 𝟐𝟐𝒙𝒙𝟐𝟐 √𝟑𝟑
5.
Sarahi found the distance between two points on a coordinate plane to be √𝟕𝟕𝟕𝟕. Can this answer be simplified?
Explain.
The number 𝟕𝟕𝟕𝟕 can be factored, but none of the factors are perfect squares, which are necessary to simplify.
Therefore, √𝟕𝟕𝟕𝟕 cannot be simplified.
6.
√𝟏𝟏𝟏𝟏𝒙𝒙𝟑𝟑 =
�𝟏𝟏𝟏𝟏𝒙𝒙𝟑𝟑 = √𝟏𝟏𝟏𝟏�𝒙𝒙𝟐𝟐 √𝒙𝒙
= 𝟒𝟒𝟒𝟒√𝒙𝒙
7.
√𝟐𝟐𝟐𝟐
√𝟑𝟑
√𝟐𝟐𝟐𝟐
√𝟑𝟑
=
𝟐𝟐𝟐𝟐
=�
𝟑𝟑
= √𝟗𝟗
= 𝟑𝟑
8.
√𝟑𝟑
. Show and explain that their answers are equivalent.
𝟑𝟑
𝟏𝟏
√𝟑𝟑
=
=
𝟏𝟏
√𝟑𝟑
√𝟑𝟑
𝟑𝟑
×
𝟏𝟏
, and Joffrey says the
√𝟑𝟑
√𝟑𝟑
√𝟑𝟑
If Nazem were to rationalize the denominator in his answer, he would see that it is equal to Joffrey’s answer.
9.
𝟓𝟓
�
𝟖𝟖
=
𝟓𝟓 √𝟓𝟓 √𝟐𝟐
� =
×
𝟖𝟖 √𝟖𝟖 √𝟐𝟐
=
=
√𝟏𝟏𝟏𝟏
√𝟏𝟏𝟏𝟏
√𝟏𝟏𝟏𝟏
𝟒𝟒
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10. Determine the area of a square with side length 𝟐𝟐√𝟕𝟕 𝐢𝐢𝐢𝐢.
𝟐𝟐
𝑨𝑨 = �𝟐𝟐√𝟕𝟕�
𝟐𝟐
= 𝟐𝟐𝟐𝟐 �√𝟕𝟕�
= 𝟒𝟒(𝟕𝟕)
= 𝟐𝟐𝟐𝟐
The area of the square is 𝟐𝟐𝟐𝟐 𝐢𝐢𝐧𝐧𝟐𝟐 .
11. Determine the exact area of the shaded region shown below.
Let 𝒓𝒓 be the length of the radius.
By special triangles or the Pythagorean theorem, 𝒓𝒓 = 𝟓𝟓√𝟐𝟐.
The area of the rectangle containing the shaded region is
𝑨𝑨 = 𝟐𝟐�𝟓𝟓√𝟐𝟐��𝟓𝟓√𝟐𝟐� = 𝟐𝟐(𝟐𝟐𝟐𝟐)(𝟐𝟐) = 𝟏𝟏𝟏𝟏𝟏𝟏.
The sum of the two quarter circles in the rectangular region is
𝟏𝟏
𝟐𝟐
𝝅𝝅�𝟓𝟓√𝟐𝟐�
𝟐𝟐
𝟏𝟏
= 𝝅𝝅(𝟐𝟐𝟐𝟐)(𝟐𝟐)
𝟐𝟐
= 𝟐𝟐𝟐𝟐𝟐𝟐.
𝑨𝑨 =
The area of the shaded region is 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟐𝟐𝟐𝟐𝟐𝟐 square units.
12. Determine the exact area of the shaded region shown to the right.
The radius of each quarter circle is
𝟏𝟏
𝟐𝟐
𝟏𝟏
�√𝟐𝟐𝟐𝟐� = �𝟐𝟐√𝟓𝟓� = √𝟓𝟓.
𝟐𝟐
𝟐𝟐
The sum of the area of the four circular regions is 𝑨𝑨 = 𝝅𝝅�√𝟓𝟓� = 𝟓𝟓𝟓𝟓.
𝟐𝟐
The area of the square is 𝑨𝑨 = �√𝟐𝟐𝟐𝟐� = 𝟐𝟐𝟐𝟐.
The area of the shaded region is 𝟐𝟐𝟐𝟐 − 𝟓𝟓𝟓𝟓 square units.
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13. Calculate the area of the triangle to the right.
𝑨𝑨 =
=
=
𝟐𝟐
𝟏𝟏
�√𝟏𝟏𝟏𝟏� � �
𝟐𝟐
√𝟓𝟓
𝟏𝟏 𝟐𝟐√𝟏𝟏𝟏𝟏
�
�
𝟐𝟐 √𝟓𝟓
√𝟏𝟏𝟏𝟏
√𝟓𝟓
𝟏𝟏𝟏𝟏
=�
𝟓𝟓
= √𝟐𝟐
The area of the triangle is √𝟐𝟐 square units.
14.
�𝟐𝟐𝒙𝒙𝟑𝟑 ⋅√𝟖𝟖𝒙𝒙
�𝒙𝒙𝟑𝟑
=
√𝟐𝟐𝒙𝒙𝟑𝟑 ⋅ √𝟖𝟖𝟖𝟖
√𝒙𝒙𝟑𝟑
=
=
=
=
=
√𝟏𝟏𝟏𝟏𝒙𝒙𝟒𝟒
√𝒙𝒙𝟑𝟑
𝟒𝟒𝒙𝒙𝟐𝟐
𝒙𝒙√𝒙𝒙
𝟒𝟒𝟒𝟒
√𝒙𝒙
𝟒𝟒𝟒𝟒
√𝒙𝒙
×
√𝒙𝒙
√𝒙𝒙
𝟒𝟒𝟒𝟒√𝒙𝒙
𝒙𝒙
= 𝟒𝟒√𝒙𝒙
15. Prove Rule 2 for square roots:
𝒂𝒂
√𝒂𝒂
�𝒃𝒃 = √𝒃𝒃
(𝒂𝒂 ≥ 𝟎𝟎, 𝒃𝒃 > 𝟎𝟎)
Let 𝒑𝒑 be the nonnegative number so that 𝒑𝒑𝟐𝟐 = 𝒂𝒂, and let 𝒒𝒒 be the nonnegative number so that 𝒒𝒒𝟐𝟐 = 𝒃𝒃. Then,
𝒂𝒂
𝒑𝒑𝟐𝟐
� = � 𝟐𝟐
𝒒𝒒
𝒃𝒃
𝒑𝒑 𝟐𝟐
= �� �
𝒒𝒒
=
=
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𝒑𝒑
𝒒𝒒
√𝒂𝒂
√𝒃𝒃
By substitution
By the laws of exponents for integers
By definition of square root
By substitution
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Perfect Squares of Numbers 1–30
12 = 1
162 = 256
32 = 9
182 = 324
22 = 4
172 = 289
42 = 16
192 = 361
52 = 25
202 = 400
62 = 36
212 = 441
72 = 49
222 = 484
82 = 64
232 = 529
92 = 81
242 = 576
102 = 100
252 = 625
112 = 121
262 = 676
122 = 144
272 = 729
132 = 169
282 = 784
142 = 196
292 = 841
152 = 225
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GEOMETRY
Lesson 23: Adding and Subtracting Expressions with
Student Outcomes

Students use the distributive property to simplify expressions that contain radicals.
Lesson Notes
In this lesson, students add and subtract expressions with radicals, continuing their work from the previous lesson.
Again, the overarching goal is for students to rewrite expressions involving radical and rational exponents using the
properties of exponents (N.RN.A.2), which is mastered in Algebra II. Practice adding and subtracting expressions in a
geometric setting is provided to achieve this goal.
Classwork
Exercises 1–5 (8 minutes)
The first three exercises are designed to informally assess students’ ability to simplify square root expressions, a skill that
is necessary for the topic of this lesson. The last two exercises provide a springboard for discussing how to add
expressions that contain radicals. Encourage students to discuss their conjectures with a partner. The Discussion that
follows debriefs Exercises 1–5.
Exercises 1–5
Simplify each expression as much as possible.
1.
2.
√𝟑𝟑𝟑𝟑 =
√𝟑𝟑𝟑𝟑 = √𝟏𝟏𝟏𝟏√𝟐𝟐
3.
= 𝟒𝟒√𝟐𝟐
√𝟑𝟑𝟑𝟑𝟑𝟑 =
√𝟒𝟒𝟒𝟒 =
√𝟒𝟒𝟒𝟒 = √𝟗𝟗√𝟓𝟓
= 𝟑𝟑√𝟓𝟓
√𝟑𝟑𝟑𝟑𝟑𝟑 = √𝟏𝟏𝟏𝟏𝟏𝟏√𝟑𝟑
= 𝟏𝟏𝟏𝟏√𝟑𝟑
4.
The triangle shown below has a perimeter of 𝟔𝟔. 𝟓𝟓√𝟐𝟐 units. Make a conjecture about how this answer was reached.
It appears that when all three sides of the triangle were added, the numbers that
preceded the square roots were the only numbers that were added:
𝟑𝟑 + 𝟐𝟐 + 𝟏𝟏. 𝟓𝟓 = 𝟔𝟔. 𝟓𝟓. The √𝟐𝟐 shown as part of each length remained √𝟐𝟐.
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5.
The sides of a triangle are 𝟒𝟒√𝟑𝟑, √𝟏𝟏𝟏𝟏, and √𝟕𝟕𝟕𝟕. Make a conjecture about how to determine the perimeter of this
triangle.
Answers will vary. The goal is for students to realize that √𝟏𝟏𝟏𝟏 and √𝟕𝟕𝟕𝟕 can be rewritten so that each has a factor of
√𝟑𝟑, which then strongly resembles Exercise 4. By rewriting each side length as a multiple of √𝟑𝟑, we get
𝟒𝟒√𝟑𝟑 + 𝟐𝟐√𝟑𝟑 + 𝟓𝟓√𝟑𝟑 = 𝟏𝟏𝟏𝟏√𝟑𝟑.
Some students may answer incorrectly by adding 𝟑𝟑 + 𝟏𝟏𝟏𝟏 + 𝟕𝟕𝟕𝟕. Show that this is incorrect using a simpler example.
√𝟗𝟗 + √𝟏𝟏𝟏𝟏 ≠ √𝟐𝟐𝟐𝟐
Discussion (6 minutes)
Scaffolding:
Ensure that students are correctly simplifying the expressions in Exercises 1–3 because it is
a skill that is required to add and subtract the expressions in this lesson. Then, continue
with the discussion that follows.

Share your conjecture for Exercise 4. How can we explain that
1.5√2 + 2√2 + 3√2 = 6.5√2?
Select students to share their conjectures. The expectation is that students recognize that
the rational parts of each length were added.

Does this remind you of anything else you have done before? Give an example.
MP.7


The distributive property is true for all real numbers. Is √2 a real number?



It is reminiscent of combining like terms using the distributive property.
For example, 1.5𝑥𝑥 + 2𝑥𝑥 + 3𝑥𝑥 = (1.5 + 2 + 3)𝑥𝑥 = 6.5𝑥𝑥.
It may be necessary to show
students that √2 is a number
between 1 and 2 on the
number line. If they consider a
unit square on the number line
with a diagonal, 𝑠𝑠, as shown,
then they can use a compass
with a radius equal in length to
the diagonal and center at 0 to
show that the length of the
diagonal of the square (√2) is a
number.
Yes, √2 is a real number. It is irrational, but it is a real number.
For this reason, we can apply the distributive property to radical expressions.
Share your conjecture for Exercise 5. How might we add 4√3, √12, and √75?
Now that students know that they can apply the distributive property to radical
expressions, they may need another minute or two to evaluate the conjectures they
developed while working on Exercise 5. Select students to share their conjectures. The
expectation is that students apply their knowledge from the previous lesson to determine
a strategy for finding the perimeter of the triangle. This strategy should include simplifying each expression and
combining those with the same irrational factor. Now would be a good time to point out that the �𝑝𝑝 for any prime
number 𝑝𝑝 is an irrational number. This explains why many of the simplifications tend to have radicands that are either
prime or a product of prime numbers.
Exercise 6 (10 minutes)
To complete Exercise 6, students need to circle the expressions on the list that they think can be simplified. As students
complete the task, pair them so that they can compare their lists and discuss any discrepancies. Have students who are
ready for a challenge generate their own examples, possibly including expressions that contain variables.
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Exercise 6
Circle the expressions that can be simplified using the distributive property. Be prepared to explain your choices.
𝟖𝟖. 𝟑𝟑√𝟐𝟐 + 𝟕𝟕. 𝟗𝟗√𝟐𝟐
√𝟏𝟏𝟏𝟏 − √𝟔𝟔
−𝟏𝟏𝟏𝟏√𝟓𝟓 + √𝟒𝟒𝟒𝟒
𝟏𝟏𝟏𝟏√𝟕𝟕 − 𝟔𝟔√𝟕𝟕 + 𝟑𝟑√𝟐𝟐
𝟏𝟏𝟏𝟏√𝟐𝟐 + 𝟐𝟐√𝟖𝟖
𝟒𝟒 + √𝟏𝟏𝟏𝟏
√𝟕𝟕 + 𝟐𝟐√𝟏𝟏𝟏𝟏
√𝟏𝟏𝟏𝟏 − √𝟕𝟕𝟕𝟕
√𝟑𝟑𝟑𝟑 + √𝟐𝟐
𝟔𝟔√𝟏𝟏𝟏𝟏 + √𝟐𝟐𝟐𝟐
The expressions that can be simplified using the distributive property are noted in red.
Example 1 (5 minutes)
The expressions in this example have been taken from the list that students completed in Exercise 6. Ask students who
circled this expression to explain why.
Explain how the expression 𝟖𝟖. 𝟑𝟑√𝟐𝟐 + 𝟕𝟕. 𝟗𝟗√𝟐𝟐 can be simplified using the distributive property.
Each term of the expression has a common factor, √𝟐𝟐. For that reason, the distributive property can be applied.
𝟖𝟖. 𝟑𝟑√𝟐𝟐 + 𝟕𝟕. 𝟗𝟗√𝟐𝟐 = (𝟖𝟖. 𝟑𝟑 + 𝟕𝟕. 𝟗𝟗)√𝟐𝟐
= 𝟏𝟏𝟏𝟏. 𝟐𝟐√𝟐𝟐
By the distributive property
Explain how the expression 𝟏𝟏𝟏𝟏√𝟕𝟕 − 𝟔𝟔√𝟕𝟕 + 𝟑𝟑√𝟐𝟐 can be simplified using the distributive property.
The expression can be simplified because the first two terms contain the expression √𝟕𝟕. Using the distributive property,
we get the following:
𝟏𝟏𝟏𝟏√𝟕𝟕 − 𝟔𝟔√𝟕𝟕 + 𝟑𝟑√𝟐𝟐 = (𝟏𝟏𝟏𝟏 − 𝟔𝟔)√𝟕𝟕 + 𝟑𝟑√𝟐𝟐
= 𝟓𝟓√𝟕𝟕 + 𝟑𝟑√𝟐𝟐
By the distributive property
Example 2 (4 minutes)
The expression in this example has been taken from the list that students completed in Exercise 6. Ask students, “Who
circled this expression?” Have those students explain to small groups why they believe the expression can be simplified.
MP.3
Then, allow students who had not selected the expression to circle it if they have been convinced that it can be
simplified. Finally, ask one of the students who changed his answer to explain how the expression can be simplified.
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Explain how the expression 𝟏𝟏𝟏𝟏√𝟐𝟐 + 𝟐𝟐√𝟖𝟖 can be simplified using the distributive property.
The expression can be simplified, but first the term 𝟐𝟐√𝟖𝟖 must be rewritten.
𝟏𝟏𝟏𝟏√𝟐𝟐 + 𝟐𝟐√𝟖𝟖 = 𝟏𝟏𝟏𝟏√𝟐𝟐 + 𝟐𝟐√𝟒𝟒 ⋅ √𝟐𝟐
= 𝟏𝟏𝟏𝟏√𝟐𝟐 + 𝟐𝟐 ⋅ 𝟐𝟐√𝟐𝟐
= 𝟏𝟏𝟏𝟏√𝟐𝟐 + 𝟒𝟒√𝟐𝟐
By Rule 1
By the distributive property
= (𝟏𝟏𝟏𝟏 + 𝟒𝟒)√𝟐𝟐
= 𝟐𝟐𝟐𝟐√𝟐𝟐
Example 3 (6 minutes)
The expressions in this example have been taken from the list that students completed in Exercise 6. Ask students who
circled this expression to explain why.
Can the expression √𝟕𝟕 + 𝟐𝟐√𝟏𝟏𝟏𝟏 be simplified using the distributive property?
No, the expression cannot be simplified because neither term can be rewritten in a way that the distributive property
could be applied.
To determine if an expression can be simplified, you must first simplify each of the terms within the expression. Then,
apply the distributive property, or other properties as needed, to simplify the expression.
MP.3
Have students return to Exercise 6 and discuss the remaining expressions in small groups. Any groups that cannot agree
on an expression should present their arguments to the class.
Closing (3 minutes)
Working in pairs, have students describe to a partner how to simplify the expressions below. Once students have
partner shared, ask the class how the work completed today is related to the structure of rational numbers they have
observed in the past. The expected response is that the distributive property can be applied to square roots because
they are numbers, too.

Describe how to simplify the expression 3√18 + 10√2.

To simplify the expression, we must first rewrite 3√18 so that is has a factor of √2.
3√18 = 3√9√2
= 3(3)√2
= 9√2
Now that both terms have a factor of √2, the distributive property can be applied to simplify.
9√2 + 10√2 = (9 + 10)√2
= 19√2
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
Describe how to simplify the expression 5√3 + √12.

To simplify the expression, we must first rewrite √12 so that is has a factor of √3.
√12 = √4√3
= 2√3
Now that both terms have a factor of √3, the distributive property can be applied to simplify.
5√3 + 2√3 = (5 + 2)√3
= 7√3
Exit Ticket (3 minutes)
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Name
Date
Exit Ticket
1.
Simplify 5√11 − 17√11.
2.
Simplify √8 + 5√2.
3.
Write a radical addition or subtraction problem that cannot be simplified, and explain why it cannot be simplified.
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Exit Ticket Sample Solutions
1.
Simplify 𝟓𝟓√𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏√𝟏𝟏𝟏𝟏.
𝟓𝟓√𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏√𝟏𝟏𝟏𝟏 = (𝟓𝟓 − 𝟏𝟏𝟏𝟏)√𝟏𝟏𝟏𝟏
= −𝟏𝟏𝟏𝟏√𝟏𝟏𝟏𝟏
2.
Simplify √𝟖𝟖 + 𝟓𝟓√𝟐𝟐.
√𝟖𝟖 + 𝟓𝟓√𝟐𝟐 = √𝟒𝟒√𝟐𝟐 + 𝟓𝟓√𝟐𝟐
= 𝟐𝟐√𝟐𝟐 + 𝟓𝟓√𝟐𝟐
= (𝟐𝟐 + 𝟓𝟓)√𝟐𝟐
= 𝟕𝟕√𝟐𝟐
3.
Write a radical addition or subtraction problem that cannot be simplified, and explain why it cannot be simplified.
Answers will vary. Students should state that their expression cannot be simplified because one or both terms
cannot be rewritten so that each has a common factor. Therefore, the distributive property cannot be applied.
Problem Set Sample Solutions
1.
2.
𝟏𝟏𝟏𝟏√𝟓𝟓 − 𝟏𝟏𝟏𝟏√𝟓𝟓 =
𝟏𝟏𝟏𝟏√𝟓𝟓 − 𝟏𝟏𝟏𝟏√𝟓𝟓 = (𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏)√𝟓𝟓
√𝟐𝟐𝟐𝟐 + 𝟒𝟒√𝟓𝟓𝟓𝟓 = √𝟒𝟒 ⋅ √𝟔𝟔 + 𝟒𝟒 ⋅ √𝟗𝟗 ⋅ √𝟔𝟔
= 𝟐𝟐√𝟔𝟔 + 𝟒𝟒 ⋅ 𝟑𝟑√𝟔𝟔
= 𝟔𝟔√𝟓𝟓
3.
𝟐𝟐√𝟕𝟕 + 𝟒𝟒√𝟔𝟔𝟔𝟔
√𝟐𝟐𝟐𝟐 + 𝟒𝟒√𝟓𝟓𝟓𝟓 =
= (𝟐𝟐 + 𝟏𝟏𝟏𝟏)√𝟔𝟔
= 𝟏𝟏𝟏𝟏√𝟔𝟔
=
𝟐𝟐√𝟕𝟕 + 𝟒𝟒√𝟔𝟔𝟔𝟔 = 𝟐𝟐√𝟕𝟕 + 𝟒𝟒√𝟗𝟗√𝟕𝟕
= 𝟐𝟐√𝟕𝟕 + 𝟒𝟒(𝟑𝟑)√𝟕𝟕
= (𝟐𝟐 + 𝟏𝟏𝟏𝟏)√𝟕𝟕
4.
= 𝟏𝟏𝟏𝟏√𝟕𝟕
What is the perimeter of the triangle shown below?
𝟐𝟐√𝟐𝟐 + √𝟏𝟏𝟏𝟏 + √𝟑𝟑𝟑𝟑 = 𝟐𝟐√𝟐𝟐 + √𝟒𝟒 ⋅ √𝟑𝟑 + √𝟏𝟏𝟏𝟏 ⋅ √𝟐𝟐
= 𝟐𝟐√𝟐𝟐 + 𝟐𝟐√𝟑𝟑 + 𝟒𝟒√𝟐𝟐
= (𝟐𝟐 + 𝟒𝟒)√𝟐𝟐 + 𝟐𝟐√𝟑𝟑
= 𝟔𝟔√𝟐𝟐 + 𝟐𝟐√𝟑𝟑
The perimeter of the triangle is 𝟔𝟔√𝟐𝟐 + 𝟐𝟐√𝟑𝟑 units.
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5.
Determine the area and perimeter of the triangle shown. Simplify as much as possible.
The perimeter of the triangle is
√𝟐𝟐𝟐𝟐 + 𝟓𝟓√𝟔𝟔 + √𝟏𝟏𝟏𝟏𝟏𝟏 = √𝟒𝟒√𝟔𝟔 + 𝟓𝟓√𝟔𝟔 + √𝟏𝟏𝟏𝟏𝟏𝟏
= 𝟐𝟐√𝟔𝟔 + 𝟓𝟓√𝟔𝟔 + √𝟏𝟏𝟏𝟏𝟏𝟏
= (𝟐𝟐 + 𝟓𝟓)√𝟔𝟔 + √𝟏𝟏𝟏𝟏𝟏𝟏
= 𝟕𝟕√𝟔𝟔 + √𝟏𝟏𝟏𝟏𝟏𝟏.
The area of the triangle is
√𝟐𝟐𝟐𝟐�𝟓𝟓√𝟔𝟔� 𝟐𝟐√𝟔𝟔�𝟓𝟓√𝟔𝟔� 𝟔𝟔𝟔𝟔
=
=
= 𝟑𝟑𝟑𝟑.
𝟐𝟐
𝟐𝟐
𝟐𝟐
The perimeter is 𝟕𝟕√𝟔𝟔 + √𝟏𝟏𝟏𝟏𝟏𝟏 units, and the area is 𝟑𝟑𝟑𝟑 square units.
6.
Determine the area and perimeter of the rectangle shown. Simplify as much as possible.
The perimeter of the rectangle is
𝟏𝟏𝟏𝟏√𝟑𝟑 + 𝟏𝟏𝟏𝟏√𝟑𝟑 + √𝟕𝟕𝟕𝟕 + √𝟕𝟕𝟕𝟕 = 𝟐𝟐�𝟏𝟏𝟏𝟏√𝟑𝟑� + 𝟐𝟐�√𝟐𝟐𝟐𝟐√𝟑𝟑�
= 𝟐𝟐𝟐𝟐√𝟑𝟑 + 𝟏𝟏𝟏𝟏√𝟑𝟑
= (𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏)√𝟑𝟑
= 𝟑𝟑𝟐𝟐√𝟑𝟑.
The area of the rectangle is
𝟏𝟏𝟏𝟏√𝟑𝟑�𝟓𝟓√𝟑𝟑� = 𝟓𝟓𝟓𝟓(𝟑𝟑)
= 𝟏𝟏𝟏𝟏𝟏𝟏.
The perimeter is 𝟑𝟑𝟑𝟑√𝟑𝟑 units, and the area is 𝟏𝟏𝟏𝟏𝟏𝟏 square units.
7.
Determine the area and perimeter of the triangle shown. Simplify as much as possible.
The perimeter of the triangle is
𝟖𝟖√𝟑𝟑 + 𝟖𝟖√𝟑𝟑 + √𝟑𝟑𝟑𝟑𝟑𝟑 = (𝟖𝟖 + 𝟖𝟖)√𝟑𝟑 + √𝟑𝟑𝟑𝟑𝟑𝟑
= 𝟏𝟏𝟏𝟏√𝟑𝟑 + √𝟑𝟑𝟑𝟑𝟑𝟑
= 𝟏𝟏𝟏𝟏√𝟑𝟑 + √𝟔𝟔𝟔𝟔√𝟔𝟔
The area of the triangle is
= 𝟏𝟏𝟏𝟏√𝟑𝟑 + 𝟖𝟖√𝟔𝟔.
𝟖𝟖𝟐𝟐 �√𝟑𝟑�
�𝟖𝟖√𝟑𝟑�
=
𝟐𝟐
𝟐𝟐
𝟔𝟔𝟔𝟔(𝟑𝟑)
=
𝟐𝟐
= 𝟑𝟑𝟑𝟑(𝟑𝟑)
= 𝟗𝟗𝟗𝟗.
The perimeter of the triangle is 𝟏𝟏𝟏𝟏√𝟑𝟑 + 𝟖𝟖√𝟔𝟔 units, and the area of the triangle is 𝟗𝟗𝟗𝟗 square units.
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8.
Determine the area and perimeter of the triangle shown. Simplify as much as possible.
The perimeter of the triangle is
𝟐𝟐𝟐𝟐 + 𝒙𝒙 + 𝒙𝒙√𝟑𝟑 = 𝟑𝟑𝟑𝟑 + 𝒙𝒙√𝟑𝟑.
The area of the triangle is
𝒙𝒙(𝒙𝒙√𝟑𝟑) 𝒙𝒙𝟐𝟐 √𝟑𝟑
=
.
𝟐𝟐
𝟐𝟐
The perimeter is 𝟑𝟑𝟑𝟑 + 𝒙𝒙√𝟑𝟑 units, and the area is
9.
𝒙𝒙𝟐𝟐 √𝟑𝟑
𝟐𝟐
square units.
The area of the rectangle shown in the diagram below is 𝟏𝟏𝟏𝟏𝟏𝟏 square units. Determine the area and perimeter of the
The length of the rectangle is 𝟖𝟖𝟖𝟖, and
the width is 𝟒𝟒𝟒𝟒. Using the given area
of the rectangle:
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥 × 𝐰𝐰𝐰𝐰𝐰𝐰𝐰𝐰𝐰𝐰
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟖𝟖𝟖𝟖 ⋅ 𝟒𝟒𝟒𝟒
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟑𝟑𝒙𝒙𝟐𝟐
𝟓𝟓 = 𝒙𝒙𝟐𝟐
√𝟓𝟓 = 𝒙𝒙
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫 = 𝑨𝑨𝟏𝟏 + 𝑨𝑨𝟐𝟐 + 𝑨𝑨𝟑𝟑 + 𝑨𝑨𝟒𝟒
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟏𝟏 = 𝒃𝒃𝒃𝒃
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟏𝟏 =
𝟏𝟏
⋅ 𝟑𝟑𝒙𝒙 ⋅ 𝟒𝟒𝒙𝒙
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟏𝟏 = 𝟔𝟔𝒙𝒙𝟐𝟐
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟐𝟐 = 𝒃𝒃𝒃𝒃
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟐𝟐 =
𝟏𝟏
⋅ 𝟒𝟒𝒙𝒙 ⋅ 𝟒𝟒𝒙𝒙
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟐𝟐 = 𝟖𝟖𝒙𝒙𝟐𝟐
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟏𝟏 = 𝟔𝟔�√𝟓𝟓�
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟑𝟑 =
𝟏𝟏
⋅ 𝟖𝟖𝒙𝒙 ⋅ 𝒙𝒙
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟑𝟑 = 𝟒𝟒𝒙𝒙𝟐𝟐
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟐𝟐 = 𝟖𝟖�√𝟓𝟓�
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟏𝟏 = 𝟑𝟑𝟑𝟑
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟑𝟑 = 𝒃𝒃𝒃𝒃
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟑𝟑 = 𝟒𝟒�√𝟓𝟓�
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟐𝟐 = 𝟒𝟒𝟒𝟒
𝐀𝐀𝐀𝐀𝐀𝐀𝐚𝐚𝟑𝟑 = 𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒 + 𝟐𝟐𝟐𝟐 + 𝑨𝑨𝟒𝟒
𝑨𝑨𝟒𝟒 = 𝟕𝟕𝟕𝟕
The area of the shaded triangle in the diagram is 𝟕𝟕𝟕𝟕 square units.
The perimeter of the shaded triangle requires use of the Pythagorean theorem to find the hypotenuses of right
triangles 1, 2, and 3. Let 𝒉𝒉𝟏𝟏 , 𝒉𝒉𝟐𝟐 , and 𝒉𝒉𝟑𝟑 represent the lengths of the hypotenuses of triangles 1, 2, and 3,
respectively.
𝟐𝟐
𝟐𝟐
𝟐𝟐
𝟐𝟐
𝟐𝟐
𝟐𝟐
�𝟑𝟑√𝟓𝟓� + �𝟒𝟒√𝟓𝟓� = (𝒄𝒄𝟏𝟏 )𝟐𝟐
�𝟒𝟒√𝟓𝟓� + �𝟒𝟒√𝟓𝟓� = (𝒄𝒄𝟐𝟐 )𝟐𝟐
�√𝟓𝟓� + �𝟖𝟖√𝟓𝟓� = (𝒄𝒄𝟑𝟑 )𝟐𝟐
𝟒𝟒𝟒𝟒 + 𝟖𝟖𝟖𝟖 = (𝒄𝒄𝟏𝟏 )𝟐𝟐
𝟖𝟖𝟖𝟖 + 𝟖𝟖𝟖𝟖 = (𝒄𝒄𝟐𝟐 )𝟐𝟐
𝟓𝟓 + 𝟑𝟑𝟑𝟑𝟑𝟑 = (𝒄𝒄𝟑𝟑 )𝟐𝟐
𝟓𝟓√𝟓𝟓 = 𝒄𝒄𝟏𝟏
𝟒𝟒√𝟏𝟏𝟏𝟏 = 𝒄𝒄𝟐𝟐
𝟓𝟓√𝟏𝟏𝟏𝟏 = 𝒄𝒄𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏 = (𝒄𝒄𝟏𝟏
)𝟐𝟐
𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏 = 𝒄𝒄𝟏𝟏 + 𝒄𝒄𝟐𝟐 + 𝒄𝒄𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏 = (𝒄𝒄𝟐𝟐
)𝟐𝟐
𝟑𝟑𝟑𝟑𝟑𝟑 = (𝒄𝒄𝟑𝟑 )𝟐𝟐
𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏 = 𝟓𝟓√𝟓𝟓 + 𝟒𝟒√𝟏𝟏𝟏𝟏 + 𝟓𝟓√𝟏𝟏𝟏𝟏
The perimeter of the shaded triangle is approximately 𝟒𝟒𝟒𝟒. 𝟗𝟗 units.
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10. Parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 has an area of 𝟗𝟗√𝟑𝟑 square units. 𝑫𝑫𝑫𝑫 = 𝟑𝟑√𝟑𝟑, and 𝑮𝑮 and 𝑯𝑯 are midpoints of ����
𝑫𝑫𝑫𝑫 and ����
𝑪𝑪𝑪𝑪,
Using the area of a parallelogram:
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = 𝒃𝒃𝒃𝒃
𝟗𝟗√𝟑𝟑 = 𝟑𝟑√𝟑𝟑 ⋅ 𝒉𝒉
𝟑𝟑 = 𝒉𝒉
The height of the parallelogram is 𝟑𝟑.
The area of the shaded region is the sum of
the areas of △ 𝑬𝑬𝑬𝑬𝑬𝑬 and △ 𝑭𝑭𝑭𝑭𝑭𝑭.
The given points 𝑮𝑮 and 𝑯𝑯 are midpoints of
����
𝑫𝑫𝑫𝑫 and ����
𝑪𝑪𝑪𝑪; therefore, by the triangle side
����, and thus, also parallel to ����
����� must be parallel to 𝑫𝑫𝑫𝑫
𝑨𝑨𝑨𝑨. Furthermore,
splitter theorem, 𝑮𝑮𝑮𝑮
𝟏𝟏
𝟏𝟏
𝟑𝟑
𝑮𝑮𝑮𝑮 = 𝑪𝑪𝑪𝑪 = 𝑨𝑨𝑨𝑨 = √𝟑𝟑.
𝟐𝟐
𝟐𝟐
𝟐𝟐
△ 𝑬𝑬𝑬𝑬𝑬𝑬 ~ △ 𝑬𝑬𝑬𝑬𝑬𝑬 by AA~ criterion with a scale factor of
𝟏𝟏
𝟐𝟐
𝟐𝟐
𝟏𝟏
𝟐𝟐
.
The areas of scale drawings are related by the square of
the scale factor; therefore, 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) = � � ⋅ 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬).
𝟏𝟏
⋅ 𝟑𝟑√𝟑𝟑 ⋅ 𝟑𝟑
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) =
𝟗𝟗
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) = √𝟑𝟑
𝟐𝟐
𝟏𝟏
𝟐𝟐
By a similar argument:
𝟗𝟗
𝟗𝟗
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) = � � ⋅ √𝟑𝟑
𝟐𝟐
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) =
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑭𝑭𝑭𝑭𝑭𝑭) = √𝟑𝟑
𝟖𝟖
𝟏𝟏 𝟗𝟗
⋅
𝟒𝟒 𝟐𝟐 √𝟑𝟑
𝟗𝟗
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) = √𝟑𝟑
𝟖𝟖
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬) = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) + 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑭𝑭𝑭𝑭𝑭𝑭)
𝟗𝟗
𝟗𝟗
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬) = √𝟑𝟑 + √𝟑𝟑
𝟖𝟖
𝟖𝟖
𝟗𝟗
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬) = √𝟑𝟑
𝟒𝟒
The area of the shaded region is
Lesson 23:
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𝟗𝟗
𝟒𝟒
√𝟑𝟑 square units.
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Lesson 24: Prove the Pythagorean Theorem Using Similarity
Student Outcomes

Students prove the Pythagorean theorem using similarity.

Students use similarity and the Pythagorean theorem to find the unknown side lengths of a right triangle.

Students are familiar with the ratios of the sides of special right triangles with angle measures 45–45–90 and
30–60–90.
Lesson Notes
In Grade 8, students proved the Pythagorean theorem using what they knew about similar triangles. The basis of this
proof is the same, but students are better prepared to understand the proof because of their work in Lesson 23. This
proof differs from what students did in Grade 8 because it uses knowledge of ratios within similar triangles and more
The proof of the Pythagorean theorem and the Exploratory Challenge addressing special right triangles are the essential
components of this lesson. Exercises 1–3 and the bullet points in the Discussion between the exercises can be moved to
the Problem Set if necessary.
Classwork
Opening (5 minutes)
Show the diagram below, and then ask the three questions that follow, which increase in difficulty. Have students
respond using a white board. Ask advanced students to attempt to show that 𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2 , without the scaffolded
questions. Discuss the methods used as a class.
Scaffolding:
Some groups of students may
respond better to a triangle with
numeric values.

Write an expression in terms of 𝑐𝑐 for 𝑥𝑥 and 𝑦𝑦.


𝑥𝑥 + 𝑦𝑦 = 𝑐𝑐
Write a similarity statement for the three right triangles in the image.

△ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝐶𝐶𝐶𝐶𝐶𝐶
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
Write a ratio based on the similarity statement from the previous question.

Several answers are acceptable. Ensure that students are writing ratios as they did in Lesson 21, that is,
ratios that compare shorter leg: hypotenuse, longer leg: hypotenuse, or shorter leg: longer leg.
Discussion (10 minutes)
In the Discussion that follows, students prove the Pythagorean theorem using similarity and the converse of the
Pythagorean theorem using SSS for congruent triangles.

Our goal is to prove the Pythagorean theorem using what we know about similar triangles. Consider the right
triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴 so that ∠𝐶𝐶 is a right angle. We label the side lengths 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 so that side length 𝑎𝑎 is
opposite ∠𝐴𝐴, side length 𝑏𝑏 is opposite ∠𝐵𝐵, and hypotenuse 𝑐𝑐 is opposite ∠𝐶𝐶, as shown.

Next, we draw the altitude, ℎ, from the right angle to the hypotenuse so that it splits the hypotenuse at point
𝐷𝐷 into lengths 𝑥𝑥 and 𝑦𝑦 so that 𝑥𝑥 + 𝑦𝑦 = 𝑐𝑐, as shown.
Scaffolding:

What do we know about the original triangle and the two sub-triangles?

Students may want to use the
cutouts that they created in
Lesson 21.
The three triangles are similar.
If necessary, show students the set of three triangles in the diagram below. The original triangle and the two triangles
formed by the altitude ℎ have been moved via rigid motions that make their corresponding sides and angles easier to
see.
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
We want to prove the Pythagorean theorem using what we know about similar triangles and the ratios we
wrote in Lesson 23, that is, shorter leg: hypotenuse, longer leg: hypotenuse, or shorter leg: longer leg.

We begin by identifying the two triangles that each have ∠𝐵𝐵 as one of their angles. Which are they?



Since they are similar, we can write equivalent ratios of the two similar triangles we just named in terms of
longer leg: hypotenuse.
𝑎𝑎 𝑥𝑥
=
𝑐𝑐 𝑎𝑎
𝑎𝑎2 = 𝑥𝑥𝑥𝑥
Now, identify two right triangles in the figure that each have ∠𝐴𝐴 as an acute angle.



The two triangles are △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝐶𝐶𝐶𝐶𝐶𝐶.
The two triangles are △ 𝐴𝐴𝐴𝐴𝐴𝐴 and △ 𝐴𝐴𝐴𝐴𝐴𝐴.
Write equivalent ratios in terms of shorter leg: hypotenuse, and simplify as we just did:
𝑏𝑏 𝑦𝑦
=
𝑐𝑐 𝑏𝑏
𝑏𝑏 2 = 𝑦𝑦𝑦𝑦
Our goal is to show that 𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2 . We know that 𝑎𝑎2 = 𝑥𝑥𝑥𝑥 and 𝑏𝑏 2 = 𝑦𝑦𝑦𝑦. By substitution,
𝑎𝑎2 + 𝑏𝑏 2 = 𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦
= (𝑥𝑥 + 𝑦𝑦)𝑐𝑐
= 𝑐𝑐𝑐𝑐
= 𝑐𝑐 2 .
(because 𝑥𝑥 + 𝑦𝑦 = 𝑐𝑐)
Therefore, 𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2 , and we have proven the Pythagorean theorem using similarity.
MP.6 Ask students to summarize the steps of the proof in writing or with a partner.

We next show that the converse of the Pythagorean theorem is true.
CONVERSE OF THE PYTHAGOREAN THEOREM: If a triangle has side lengths 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 so that 𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2 , then the
triangle is a right triangle with 𝑐𝑐 as the length of the hypotenuse (the side opposite the right angle).
In the diagram below, we have a triangle with side lengths 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐.
In this diagram, we have constructed a right triangle with side lengths 𝑎𝑎 and 𝑏𝑏.
If time permits, actually show the construction. Draw a segment of length 𝑎𝑎, construct a line through one of the
endpoints of segment 𝑎𝑎 that is perpendicular to 𝑎𝑎, mark a point along the perpendicular line that is equal to the length
of 𝑏𝑏, and draw the hypotenuse c.
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
Consider the right triangle with leg lengths of 𝑎𝑎 and 𝑏𝑏. By the Pythagorean theorem, we know that
𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2 . Taking the square root of both sides gives the length of the hypotenuse as √𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐. Then,
by SSS, the two triangles are congruent because both triangles have side lengths of 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐. Therefore, the
given triangle is a right triangle with a right angle that is opposite the side of length 𝑐𝑐.
Exercises 1–2 (5 minutes)
Students complete Exercise 1 independently. Exercise 2 is designed so that early finishers of Exercise 1 can begin
thinking about the topic of the next discussion. If time allows, consider asking students why they do not need to
consider the negative solution for 𝑐𝑐 2 = 12500, which is 𝑐𝑐 = ±50√5; they only consider 50√5 in our solution. The
desired response is that the context is length; therefore, there is no need to consider a negative length.
Exercises 1–2
1.
Find the length of the hypotenuse of a right triangle whose legs have lengths 𝟓𝟓𝟓𝟓 and 𝟏𝟏𝟏𝟏𝟏𝟏.
𝒄𝒄𝟐𝟐 = 𝟓𝟓𝟎𝟎𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟎𝟎𝟐𝟐
𝒄𝒄𝟐𝟐 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝒄𝒄𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
�𝒄𝒄𝟐𝟐 = √𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
Scaffolding:
Guide students’ thinking by
triangles make the work in
Exercise 1 easier?”
𝒄𝒄 = �𝟐𝟐𝟐𝟐 ⋅ 𝟓𝟓𝟓𝟓
𝒄𝒄 = 𝟐𝟐 ⋅ 𝟓𝟓𝟐𝟐 √𝟓𝟓
𝒄𝒄 = 𝟓𝟓𝟓𝟓√𝟓𝟓
2.
Can you think of a simpler method for finding the length of the hypotenuse in Exercise 1? Explain.
Accept any reasonable methods. Students may recall from Grade 8 that they can use what they know about similar
triangles and scale factors to make their computations easier.
Discussion (4 minutes)
In the Discussion that follows, students use what they know about similar triangles to simplify their work from
Exercise 1.

To simplify our work with large numbers, as in the leg lengths of 50 and 100 from Exercise 1, we can find the
greatest common factor (GCF) of 50 and 100 and then consider a similar triangle with those smaller side
lengths. Since GCF(50, 100) = 50, we can use that GCF to determine the side lengths of a dilated triangle.
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

Specifically, we can consider a triangle that has been dilated by a scale factor of
triangle with leg lengths 1 and 2.
1
50
, which produces a similar
A triangle with leg lengths 1 and 2 has a hypotenuse length of √5. The original triangle has side lengths that
are 50 times longer than this one; therefore, the length of the hypotenuse of the original triangle is 50√5.
Exercise 3 (2 minutes)
Students complete Exercise 3 independently.
Exercise 3
3.
Find the length of the hypotenuse of a right triangle whose legs have lengths 𝟕𝟕𝟕𝟕 and 𝟐𝟐𝟐𝟐𝟐𝟐.
A right triangle with leg lengths 𝟕𝟕𝟕𝟕 and 𝟐𝟐𝟐𝟐𝟐𝟐 has leg lengths that are 𝟕𝟕𝟕𝟕 times longer than a triangle with leg
lengths 𝟏𝟏 and 𝟑𝟑. A triangle with leg lengths 𝟏𝟏 and 𝟑𝟑 has a hypotenuse of length √𝟏𝟏𝟏𝟏. Therefore, the length of the
hypotenuse of a triangle with leg lengths 𝟕𝟕𝟕𝟕 and 𝟐𝟐𝟐𝟐𝟐𝟐 is 𝟕𝟕𝟕𝟕√𝟏𝟏𝟏𝟏.
Exploratory Challenge/Exercises 4–5 (10 minutes)
This challenge reveals the relationships of special right triangles with angle measures 30–60–90 and 45–45–90. Divide
the class so that one half investigates the 30–60–90 triangle and the other half investigates the 45–45–90 triangle.
Consider having pairs of students from each half become one small group to share their results. Another option is to
discuss the results as a whole class using the closing questions below.
Exploratory Challenge/Exercises 4–5
4.
An equilateral triangle has sides of length 𝟐𝟐 and angle measures of 𝟔𝟔𝟔𝟔°, as shown below. The altitude from one
vertex to the opposite side divides the triangle into two right triangles.
a.
Are those triangles congruent? Explain.
Yes, the two right triangles are congruent by ASA. Since the altitude is perpendicular to the base, then each
of the right triangles has angles of measure 𝟗𝟗𝟗𝟗° and 𝟔𝟔𝟔𝟔°. By the triangle sum theorem, the third angle has a
measure of 𝟑𝟑𝟑𝟑°. Then, each of the right triangles has corresponding angle measures of 𝟑𝟑𝟑𝟑° and 𝟔𝟔𝟔𝟔°, and the
included side length is 𝟐𝟐.
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b.
What is the length of the shorter leg of each of the right triangles? Explain.
Since the total length of the base of the equilateral triangle is 𝟐𝟐, and the two right triangles formed are
congruent, then the bases of each must be equal in length. Therefore, the length of the base of one right
triangle is 𝟏𝟏.
c.
Use the Pythagorean theorem to determine the length of the altitude.
Let 𝒉𝒉 represent the length of the altitude.
𝟏𝟏𝟐𝟐 + 𝒉𝒉𝟐𝟐 = 𝟐𝟐𝟐𝟐
𝒉𝒉𝟐𝟐 = 𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟐𝟐
𝒉𝒉𝟐𝟐 = 𝟑𝟑
𝒉𝒉 = √𝟑𝟑
d.
e.
f.
g.
Write the ratio that represents shorter leg: hypotenuse.
𝟏𝟏: 𝟐𝟐
Write the ratio that represents longer leg: hypotenuse.
√𝟑𝟑: 𝟐𝟐
Write the ratio that represents shorter leg: longer leg.
𝟏𝟏: √𝟑𝟑
By the AA criterion, any triangles with measures 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 will be similar to this triangle. If a 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗
triangle has a hypotenuse of length 𝟏𝟏𝟏𝟏, what are the lengths of the legs?
Consider providing the following picture for students:
Let 𝒂𝒂 represent the length of the shorter leg.
Let 𝒃𝒃 represent the length of the longer leg.
𝒂𝒂
𝟏𝟏
=
𝟏𝟏𝟏𝟏 𝟐𝟐
𝒂𝒂 = 𝟖𝟖
𝒃𝒃
√𝟑𝟑
=
𝟐𝟐
𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟏𝟏√𝟑𝟑
The length 𝒂𝒂 = 𝟖𝟖, and the length 𝒃𝒃 = 𝟖𝟖√𝟑𝟑.
𝒃𝒃 = 𝟖𝟖√𝟑𝟑
Note: After finding the length of one of the legs, some students may have used the ratio shorter leg: longer
leg to determine the length of the other leg.
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5.
An isosceles right triangle has leg lengths of 𝟏𝟏, as shown.
a.
What are the measures of the other two angles? Explain.
Base angles of an isosceles triangle are equal; therefore, the other two angles have a measure of 𝟒𝟒𝟒𝟒°.
b.
Use the Pythagorean theorem to determine the length of the hypotenuse of the right triangle.
Let 𝒄𝒄 represent the length of the hypotenuse.
𝟏𝟏𝟐𝟐 + 𝟏𝟏𝟐𝟐 = 𝒄𝒄𝟐𝟐
𝟐𝟐 = 𝒄𝒄𝟐𝟐
√𝟐𝟐 = 𝒄𝒄
c.
Is it necessary to write all three ratios: shorter leg: hypotenuse, longer leg: hypotenuse, and shorter leg:
longer leg? Explain.
No, it is not necessary to write all three ratios. The reason is that the shorter leg and the longer leg are the
same length. Therefore, the ratios shorter leg: hypotenuse and longer leg: hypotenuse will be the same.
Additionally, the shorter leg: longer leg ratio would be 𝟏𝟏: 𝟏𝟏, which is not useful since we are given that the
right triangle is an isosceles right triangle.
d.
Write the ratio that represents leg: hypotenuse.
𝟏𝟏: √𝟐𝟐
e.
By the AA criterion, any triangles with measures 𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 will be similar to this triangle. If a 𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗
triangle has a hypotenuse of length 𝟐𝟐𝟐𝟐, what are the lengths of the legs?
Let 𝒂𝒂 represent the length of the leg.
𝒂𝒂
𝟏𝟏
=
𝟐𝟐𝟐𝟐 √𝟐𝟐
𝒂𝒂√𝟐𝟐 = 𝟐𝟐𝟎𝟎
𝒂𝒂 =
𝒂𝒂 =
𝒂𝒂 =
𝟐𝟐𝟐𝟐
√𝟐𝟐
𝟐𝟐𝟐𝟐 √𝟐𝟐
� �
√𝟐𝟐 √𝟐𝟐
𝟐𝟐𝟐𝟐√𝟐𝟐
𝟐𝟐
𝒂𝒂 = 𝟏𝟏𝟏𝟏√𝟐𝟐
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Closing (4 minutes)
Ask students to summarize the key points of the lesson. Additionally, consider asking students the following questions.
Have students respond in writing, to a partner, or to the whole class.

Explain in your own words the proof of the Pythagorean theorem using similarity.

Triangles with angle measures 30–60–90 are a result of drawing an altitude from one angle of an equilateral
triangle to the opposite side. Explain how to use ratios of legs and the hypotenuse to find the lengths of any
30–60–90 triangle. Why does it work?

Triangles with angle measures 45–45–90 are isosceles right triangles. Explain how to use ratios of legs and the
hypotenuse to find the lengths of any 45–45–90 triangle. Why does it work?
Exit Ticket (5 minutes)
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Name
Date
Lesson 24: Prove the Pythagorean Theorem Using Similarity
Exit Ticket
A right triangle has a leg with a length of 18 and a hypotenuse with a length of 36. Bernie notices that the hypotenuse is
twice the length of the given leg, which means it is a 30–60–90 triangle. If Bernie is right, what should the length of the
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Exit Ticket Sample Solutions
A right triangle has a leg with a length of 𝟏𝟏𝟏𝟏 and a hypotenuse with a length of 𝟑𝟑𝟑𝟑. Bernie notices that the hypotenuse is
twice the length of the given leg, which means it is a 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 triangle. If Bernie is right, what should the length of the
A right angle and two given sides of a triangle determine a unique triangle. All 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 triangles are similar by AA
criterion, and the lengths of their sides are 𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐, and 𝒄𝒄√𝟑𝟑, for some positive number 𝒄𝒄. The given hypotenuse is twice
the length of the given leg of the right triangle, so Bernie’s conclusion is accurate. The ratio of the length of the short leg
to the length of the longer leg of any 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 triangle is 𝟏𝟏: √𝟑𝟑. The given leg has a length of 𝟏𝟏𝟏𝟏, which is
hypotenuse and, therefore, must be the shorter leg of the triangle.
𝟏𝟏
𝟏𝟏𝟏𝟏
=
√𝟑𝟑 𝐥𝐥𝐥𝐥𝐠𝐠𝟐𝟐
𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 = 𝟏𝟏𝟏𝟏√𝟑𝟑
𝟏𝟏
𝟐𝟐
of the
𝟏𝟏𝟖𝟖𝟐𝟐 + 𝐥𝐥𝐥𝐥𝐠𝐠𝟐𝟐 = 𝟑𝟑𝟔𝟔𝟐𝟐
𝟑𝟑𝟑𝟑𝟑𝟑 + 𝐥𝐥𝐥𝐥𝐠𝐠𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝐥𝐥𝐥𝐥𝐠𝐠𝟐𝟐 = 𝟗𝟗𝟗𝟗𝟗𝟗
𝐥𝐥𝐥𝐥𝐥𝐥 = √𝟗𝟗𝟗𝟗𝟗𝟗
𝐥𝐥𝐥𝐥𝐥𝐥 = √𝟑𝟑𝟑𝟑𝟑𝟑√𝟑𝟑
𝐥𝐥𝐥𝐥𝐥𝐥 = 𝟏𝟏𝟏𝟏√𝟑𝟑
Problem Set Sample Solutions
1.
In each row of the table below are the lengths of the legs and hypotenuses of different right triangles. Find the
missing side lengths in each row, in simplest radical form.
𝐋𝐋𝐋𝐋𝐠𝐠𝟏𝟏
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏
𝐋𝐋𝐋𝐋𝐠𝐠𝟐𝟐
𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐𝟐𝟐
𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇
𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑
𝟕𝟕
𝟏𝟏𝟏𝟏𝟏𝟏√𝟓𝟓
By the Pythagorean theorem:
𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐𝟏𝟏 + 𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐𝟏𝟏 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
𝟏𝟏𝟓𝟓𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟐𝟐𝟓𝟓𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟔𝟔𝟔𝟔𝟔𝟔
𝒚𝒚𝟐𝟐 = 𝟒𝟒𝟒𝟒𝟒𝟒
𝒚𝒚 = 𝟐𝟐𝟐𝟐 or 𝒚𝒚 = −𝟐𝟐𝟐𝟐
The case where 𝒚𝒚 = −𝟐𝟐𝟐𝟐 does not make sense since it represents a length of a side of a triangle and is, therefore,
disregarded.
Alternative strategy:
Divide each side length by the greatest common factor to get the side lengths of a similar right triangle. Find the
missing side length for the similar triangle, and multiply by the GCF.
𝐋𝐋𝐋𝐋𝐋𝐋: 𝟏𝟏𝟏𝟏, 𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇: 𝟐𝟐𝟐𝟐, 𝐆𝐆𝐆𝐆𝐆𝐆(𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐) = 𝟓𝟓
Consider the triangle with 𝐋𝐋𝐋𝐋𝐋𝐋: 𝟑𝟑 and 𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇: 𝟓𝟓. This is a 𝟑𝟑–𝟒𝟒–𝟓𝟓 right triangle. The missing leg length is
𝟓𝟓 ⋅ 𝟒𝟒 = 𝟐𝟐𝟐𝟐.
𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋: 𝟏𝟏𝟏𝟏𝟏𝟏 and 𝟐𝟐𝟐𝟐𝟐𝟐: 𝐆𝐆𝐆𝐆𝐆𝐆(𝟏𝟏𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐) = 𝟏𝟏𝟏𝟏𝟏𝟏
Consider the right triangle with 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋: 𝟏𝟏 and 𝟐𝟐. 𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇 = √𝟓𝟓. The missing hypotenuse length is 𝟏𝟏𝟏𝟏𝟏𝟏√𝟓𝟓.
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2.
Claude sailed his boat due south for 𝟑𝟑𝟑𝟑 miles and then due west for 𝟐𝟐𝟐𝟐 miles. Approximately how far is Claude
from where he began?
Claude’s path forms a right triangle since south and west are perpendicular to
each other. His distance from where he began is a straight line segment
represented by the hypotenuse of the triangle.
𝟑𝟑𝟖𝟖𝟐𝟐 + 𝟐𝟐𝟓𝟓𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟔𝟔𝟔𝟔𝟔𝟔 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
√𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐡𝐡
3.
Claude is approximately 𝟒𝟒𝟒𝟒. 𝟓𝟓 miles from where he began.
Find the lengths of the legs in the triangle given the hypotenuse with length 𝟏𝟏𝟏𝟏𝟏𝟏.
By the Pythagorean theorem:
𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 + 𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟎𝟎𝟐𝟐
𝟐𝟐 𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 = 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓
𝐥𝐥𝐥𝐥𝐥𝐥 = √𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓
𝐥𝐥𝐥𝐥𝐥𝐥 = √𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐√𝟐𝟐
Alternative strategy:
𝐥𝐥𝐥𝐥𝐥𝐥 = 𝟓𝟓𝟓𝟓√𝟐𝟐
The right triangle is an isosceles right triangle, so the leg lengths are equal. The hypotenuse of an isosceles right
triangle can be calculated as follows:
𝐡𝐡𝐡𝐡𝐡𝐡 = 𝐥𝐥𝐥𝐥𝐥𝐥 ⋅ √𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝐥𝐥𝐥𝐥𝐥𝐥√𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏
= 𝐥𝐥𝐥𝐥𝐥𝐥
√𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏√𝟐𝟐
= 𝟓𝟓𝟓𝟓√𝟐𝟐 = 𝐥𝐥𝐥𝐥𝐥𝐥
𝟐𝟐
4.
The legs of the 45–45–90 right triangle with a hypotenuse of 𝟏𝟏𝟏𝟏𝟏𝟏 are 𝟓𝟓𝟓𝟓√𝟐𝟐.
Find the length of the hypotenuse in the right triangle given that the legs have lengths of 𝟏𝟏𝟏𝟏𝟏𝟏.
By the Pythagorean theorem:
𝟏𝟏𝟏𝟏𝟎𝟎𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟎𝟎𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
√𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐡𝐡
√𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏√𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
Alternative strategy:
𝟏𝟏𝟏𝟏𝟏𝟏√𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐡𝐡
The given right triangle is a 𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 triangle. Therefore, the ratio of the length of its legs to the length of its
hypotenuse is 𝟏𝟏: √𝟐𝟐.
𝐡𝐡𝐡𝐡𝐡𝐡 = 𝐥𝐥𝐥𝐥𝐥𝐥 ⋅ √𝟐𝟐
𝐡𝐡𝐡𝐡𝐡𝐡 = 𝟏𝟏𝟏𝟏𝟏𝟏√𝟐𝟐
The hypotenuse of the right triangle with legs of length 𝟏𝟏𝟏𝟏𝟏𝟏 is 𝟏𝟏𝟎𝟎𝟎𝟎√𝟐𝟐.
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Lesson 24
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M2
GEOMETRY
Each row in the table below shows the side lengths of a different 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 right triangle. Complete the table with
the missing side lengths in simplest radical form. Use the relationships of the values in the first three rows to
complete the last row. How could the expressions in the last row be used?
Shorter Leg
𝟐𝟐
𝟏𝟏
Longer Leg
𝟐𝟐 √𝟑𝟑
𝟏𝟏 √𝟑𝟑
𝟑𝟑
𝟐𝟐
Hypotenuse
𝟓𝟓𝟓𝟓
𝟑𝟑
𝟐𝟐
𝟏𝟏
𝟑𝟑
𝟏𝟏
√𝟑𝟑
𝒙𝒙
𝟐𝟐√𝟑𝟑
𝟐𝟐
𝒙𝒙√𝟑𝟑
𝟐𝟐
The last row of the table shows that the sides of a 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 right triangle are multiples of 𝟏𝟏, 𝟐𝟐, and √𝟑𝟑 by some
constant 𝒙𝒙, with 𝟐𝟐𝟐𝟐 being the longest and, therefore, the hypotenuse. The expressions could be used to find two
unknown sides of a 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 triangle where only one of the sides is known.
.
5.
���� into
In right triangle 𝑨𝑨𝑨𝑨𝑨𝑨 with ∠𝑪𝑪 a right angle, an altitude of length 𝒉𝒉 is dropped to side ����
𝑨𝑨𝑨𝑨 that splits the side 𝑨𝑨𝑨𝑨
segments of length 𝒙𝒙 and 𝒚𝒚. Use the Pythagorean theorem to show 𝒉𝒉𝟐𝟐 = 𝒙𝒙𝒙𝒙.
By the Pythagorean theorem, 𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒄𝒄𝟐𝟐.
Since 𝒄𝒄 = 𝒙𝒙 + 𝒚𝒚, we have
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = (𝒙𝒙 + 𝒚𝒚)𝟐𝟐
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒙𝒙𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒚𝒚𝟐𝟐 .
Also by the Pythagorean theorem,
𝒂𝒂𝟐𝟐 = 𝒉𝒉𝟐𝟐 + 𝒚𝒚𝟐𝟐 , and 𝒃𝒃𝟐𝟐 = 𝒉𝒉𝟐𝟐 + 𝒙𝒙𝟐𝟐 , so
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒉𝒉𝟐𝟐 + 𝒚𝒚𝟐𝟐 + 𝒉𝒉𝟐𝟐 + 𝒙𝒙𝟐𝟐
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝟐𝟐𝒉𝒉𝟐𝟐 + 𝒙𝒙𝟐𝟐 + 𝒚𝒚𝟐𝟐 .
Thus, by substitution,
𝒙𝒙𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟐𝟐𝒉𝒉𝟐𝟐 + 𝒙𝒙𝟐𝟐 + 𝒚𝒚𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟐𝟐𝒉𝒉𝟐𝟐
𝒙𝒙𝒙𝒙 = 𝒉𝒉𝟐𝟐 .
6.
In triangle 𝑨𝑨𝑨𝑨𝑨𝑨, the altitude from ∠𝑪𝑪 splits side ����
𝑨𝑨𝑨𝑨 into two segments of lengths 𝒙𝒙 and 𝒚𝒚. If 𝒉𝒉 denotes the length
of the altitude and 𝒉𝒉𝟐𝟐 = 𝒙𝒙𝒙𝒙, use the Pythagorean theorem and its converse to show that triangle 𝑨𝑨𝑨𝑨𝑨𝑨 is a right
triangle with ∠𝑪𝑪 a right angle.
Let 𝒂𝒂, 𝒃𝒃, and 𝒄𝒄 be the lengths of the sides of the triangle opposite
∠𝑨𝑨, ∠𝑩𝑩, and ∠𝑪𝑪, respectively. By the Pythagorean theorem:
𝒂𝒂𝟐𝟐 = 𝒉𝒉𝟐𝟐 + 𝒚𝒚𝟐𝟐 and 𝒃𝒃𝟐𝟐 = 𝒉𝒉𝟐𝟐 + 𝒙𝒙𝟐𝟐 , so
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒉𝒉𝟐𝟐 + 𝒚𝒚𝟐𝟐 + 𝒉𝒉𝟐𝟐 + 𝒙𝒙𝟐𝟐
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒙𝒙𝟐𝟐 + 𝟐𝟐𝒉𝒉𝟐𝟐 + 𝒚𝒚𝟐𝟐
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒙𝒙𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒚𝒚𝟐𝟐
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = (𝒙𝒙 + 𝒚𝒚)𝟐𝟐
𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒄𝒄𝟐𝟐.
So, by the converse of the Pythagorean theorem, △ 𝑨𝑨𝑨𝑨𝑨𝑨 is a right triangle, 𝒂𝒂 and 𝒃𝒃 are the lengths of legs of the
triangle, and 𝒄𝒄 is the hypotenuse, which lies opposite the right angle. Therefore, ∠𝑪𝑪 is the right angle of the right
triangle.
Lesson 24:
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Prove the Pythagorean Theorem Using Similarity
386
New York State Common Core
Mathematics Curriculum
GEOMETRY • MODULE 2
Topic E
Trigonometry
G-SRT.C.6, G-SRT.C.7, G-SRT.C.8
Focus Standards:
Instructional Days:
G-SRT.C.6
Understand that by similarity, side ratios in right triangles are properties of the angles
in the triangle, leading to definitions of trigonometric ratios for acute angles.
G-SRT.C.7
Explain and use the relationship between the sine and cosine of complementary angles.
G-SRT.C.8
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in
applied problems.★
10
Lesson 25:
Incredibly Useful Ratios (E) 1
Lesson 26:
The Definition of Sine, Cosine, and Tangent (P)
Lesson 27:
Sine and Cosine of Complementary Angles and Special Angles (P)
Lesson 28:
Solving Problems Using Sine and Cosine (P)
Lesson 29:
Applying Tangents (P)
Lesson 30:
Trigonometry and the Pythagorean Theorem (P)
Lesson 31:
Using Trigonometry to Determine Area (P)
Lesson 32:
Using Trigonometry to Find Side Lengths of an Acute Triangle (S)
Lesson 33:
Applying the Laws of Sines and Cosines (P)
Lesson 34:
Unknown Angles (P)
Students begin the study of trigonometry in the final topic of the module. The emphasis in the module on
side length relationships within similar triangles (Topic C) and the specific emphasis on right triangles
(Topic D) help set the foundation for trigonometry. Lesson 25 is a last highlight of the side length ratios
within and between right triangles. Students are guided to the idea that the values of the ratios depend
solely on a given acute angle in the right triangle before the basic trigonometric ratios are explicitly defined in
Lesson 26 (G.SRT.C.6). After practice with ratios labeled as shorter leg: hypotenuse (Lesson 21) and
opp: hyp (Lesson 25), students are introduced to the trigonometric ratios sine, cosine, and tangent
(G-SRT.C.6) in Lesson 26. Students examine the relationship between sine and cosine in Lesson 27,
discovering that the sine and cosine of complementary angles are equal (G-SRT.C.7). They are also
1Lesson
Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
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Trigonometry
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NYS COMMON CORE MATHEMATICS CURRICULUM
Topic E
M2
GEOMETRY
introduced to the common sine and cosine values of angle measures frequently seen in trigonometry.
Students apply the trigonometric ratios to solve for unknown lengths in Lessons 28 and 29; students also
learn about the relationship between tangent and slope in Lesson 29 (G-SRT.C.8). In Lesson 30, students use
sin 𝜃𝜃
the Pythagorean theorem to prove the identity sin2 𝜃𝜃 + cos2 𝜃𝜃 = 1 and also show why tan 𝜃𝜃 = cos 𝜃𝜃. In
Lessons 31–33, students study the application of trigonometry to determine area and solve for unknown
lengths using the laws of sines and cosines (G-SRT.9, G-SRT.10, G-SRT.11). Finally, in Lesson 34, students
learn how to determine the unknown measure of an angle of a right triangle. Students are introduced to the
trigonometric functions arcsin, arccos, and arctan. These inverse functions are taught formally in Algebra II.
For now, students should understand the meaning of and how to use arcsin, arccos, and arctan to determine
unknown measures of angles.
Topic E:
GEO-M2-TE-1.3.0-07.2015
Trigonometry
388
Lesson 25
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Lesson 25: Incredibly Useful Ratios
Student Outcomes

For a given acute angle of a right triangle, students identify the opposite side, adjacent side, and hypotenuse.

Students understand that the values of the
and
opp
ratios for a given acute angle are constant.
and
opp
ratios in a right triangle depend solely on the
hyp
hyp
Lesson Notes
In Lesson 25, students discover that the values of the
hyp
hyp
measure of the acute angle by which the adjacent, opposite, and hypotenuse sides are identified. To do this, students
first learn how to identify these reference labels. Then, two groups take measurements and make calculations of the
values of the
hyp
and
opp
hyp
ratios for two sets of triangles, where each triangle in one set is similar to a triangle in the
other. This exploration leads to the conclusion regarding the “incredibly useful ratios.”
Classwork
Exercises 1–3 (4 minutes)
Exercises 1–3
Use the right triangle △ 𝑨𝑨𝑨𝑨𝑨𝑨 to answer Exercises 1–3.
1.
Name the side of the triangle opposite ∠𝑨𝑨.
����
𝑩𝑩𝑩𝑩
MP.7
2.
Name the side of the triangle opposite ∠𝑩𝑩.
����
𝑨𝑨𝑨𝑨
3.
Name the side of the triangle opposite ∠𝑪𝑪.
����
𝑨𝑨𝑨𝑨

Another common way of labeling the sides of triangles is to make use of the relationship between a vertex and
the side opposite the vertex and the lowercase letter of each vertex angle.
Lesson 25:
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Incredibly Useful Ratios
389
Lesson 25
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY


In the diagram above, label the side opposite ∠𝐴𝐴 as 𝑎𝑎. Similarly, label the side opposite ∠𝐵𝐵 as 𝑏𝑏 and the side
opposite ∠𝐶𝐶 as 𝑐𝑐.
Why is this way of labeling the triangle useful?

Using a lowercase letter to represent the side opposite a vertex immediately describes a relationship
within a triangle. It might be useful when we are sketching diagrams for problems that provide partial
Discussion (8 minutes)

Now we will discuss another set of labels. These labels are specific to right triangles.

The triangle below is right triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴. Denote ∠𝐵𝐵 as the right angle of the triangle.



Of the two acute angles, we are going to focus our attention on labeling ∠𝐴𝐴. Mark ∠𝐴𝐴 in the triangle with a
single arc.
Scaffolding:
With respect to acute angle ∠𝐴𝐴 of a right triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴, we identify the
opposite side, the adjacent side, and the hypotenuse as follows:
to create a poster of this
���� ,
- With respect to ∠𝐴𝐴, the opposite side, denoted 𝑜𝑜𝑜𝑜𝑜𝑜, is side 𝐵𝐵𝐵𝐵
diagram to display on the
- With respect to ∠𝐴𝐴, the adjacent side, denoted 𝑎𝑎𝑎𝑎𝑎𝑎, is side ����
𝐴𝐴𝐴𝐴 ,
classroom wall
(alternatively, a sample is
����
- The hypotenuse, denoted ℎ𝑦𝑦𝑦𝑦, is side 𝐴𝐴𝐴𝐴 and is opposite from the
provided in this lesson
90° angle.
before the Exit Ticket).
With respect to ∠𝐴𝐴, label the sides of the triangle as opp, adj, and hyp.
 Consider color coding the
sides and the definitions.
For example, write the
word hypotenuse in red,
and then trace that side of
the triangle in red.
 Rehearse the various
terms by pointing to each
responses of opposite,

Note that the sides are identified with respect to ∠𝐴𝐴, or using ∠𝐴𝐴 as a reference. The labels change if the
acute angle selected changes. How would the labels of the sides change if we were looking at acute angle ∠𝐶𝐶,
instead of ∠𝐴𝐴, as a reference?
 The hypotenuse would remain the same, but the opposite side of ∠𝐶𝐶 is ����
𝐴𝐴𝐴𝐴 , and the adjacent side of ∠𝐶𝐶
is ����
𝐵𝐵𝐵𝐵 .
Lesson 25:
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Incredibly Useful Ratios
390
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 25
M2
GEOMETRY

The label hypotenuse will always be the hypotenuse of the right triangle, that is, the side opposite from the
right angle, but the adjacent and opposite side labels are dependent upon which acute angle is used as a
reference.
Exercises 4–6 (4 minutes)
In Exercises 4–6, students practice identifying the adjacent side, the opposite side, and the hypotenuse of a right
triangle, with respect to a given acute angle.
Exercises 4–6
For each exercise, label the appropriate sides as adjacent, opposite, and hypotenuse, with respect to the marked acute
angle.
4.
5.
6.
Exploratory Challenge (10 minutes)
In the following Exploratory Challenge, assign one set of triangles to each half of the class. Every triangle in each set has
a corresponding, similar triangle in the other set. Students complete the table for missing angle and side length
measurements and values of the ratios
Lesson 25:
GEO-M2-TE-1.3.0-07.2015
hyp
and
opp
hyp
.
Incredibly Useful Ratios
391
Lesson 25
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Exploratory Challenge
Note: Angle measures are approximations.
For each triangle in your set, determine missing angle measurements and side lengths. Side lengths should be measured
to one decimal place. Make sure that each of the
rounded appropriately.
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
and
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
ratios are set up and missing values are calculated and
Group 1
Triangle
Angle Measures
Length Measures
1.
△ 𝑨𝑨𝑨𝑨𝑨𝑨
𝒎𝒎∠𝑨𝑨 ≈ 𝟔𝟔𝟔𝟔°
𝑨𝑨𝑨𝑨 = 𝟓𝟓 𝐜𝐜𝐜𝐜
𝑩𝑩𝑩𝑩 = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
2.
△ 𝑫𝑫𝑫𝑫𝑫𝑫
𝒎𝒎∠𝑫𝑫 ≈ 𝟓𝟓𝟓𝟓°
3.
△ 𝑮𝑮𝑮𝑮𝑮𝑮
𝒎𝒎∠𝑰𝑰 ≈ 𝟒𝟒𝟒𝟒°
4.
△ 𝑱𝑱𝑱𝑱𝑱𝑱
𝒎𝒎∠𝑳𝑳 ≈ 𝟑𝟑𝟑𝟑°
5.
△ 𝑴𝑴𝑴𝑴𝑴𝑴
𝒎𝒎∠𝑴𝑴 ≈ 𝟐𝟐𝟐𝟐°
Lesson 25:
GEO-M2-TE-1.3.0-07.2015
𝑫𝑫𝑫𝑫 = 𝟑𝟑 𝐜𝐜𝐜𝐜
𝑬𝑬𝑬𝑬 = 𝟒𝟒 𝐜𝐜𝐜𝐜
𝑫𝑫𝑫𝑫 = 𝟓𝟓 𝐜𝐜𝐜𝐜
𝑮𝑮𝑮𝑮 = 𝟓𝟓. 𝟑𝟑 𝐜𝐜𝐜𝐜
𝑯𝑯𝑯𝑯 = 𝟔𝟔 𝐜𝐜𝐜𝐜
𝑰𝑰𝑰𝑰 = 𝟖𝟖 𝐜𝐜𝐜𝐜
𝑱𝑱𝑱𝑱 = 𝟒𝟒 𝐜𝐜𝐜𝐜
𝑲𝑲𝑲𝑲 = 𝟔𝟔. 𝟗𝟗𝟗𝟗 𝐜𝐜𝐜𝐜
𝑱𝑱𝑱𝑱 = 𝟖𝟖 𝐜𝐜𝐜𝐜
𝑴𝑴𝑴𝑴 = 𝟕𝟕. 𝟓𝟓 𝐜𝐜𝐜𝐜
𝑵𝑵𝑵𝑵 = 𝟒𝟒 𝐜𝐜𝐜𝐜
𝑶𝑶𝑶𝑶 = 𝟖𝟖. 𝟓𝟓 𝐜𝐜𝐜𝐜
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
𝟏𝟏𝟏𝟏
≈ 𝟎𝟎. 𝟗𝟗𝟗𝟗
𝟏𝟏𝟏𝟏
𝟓𝟓
≈ 𝟎𝟎. 𝟑𝟑𝟑𝟑
𝟏𝟏𝟏𝟏
𝟒𝟒
= 𝟎𝟎. 𝟖𝟖
𝟓𝟓
𝟑𝟑
= 𝟎𝟎. 𝟔𝟔
𝟓𝟓
𝟓𝟓. 𝟑𝟑
≈ 𝟎𝟎. 𝟔𝟔𝟔𝟔
𝟖𝟖
𝟔𝟔/𝟖𝟖 = 𝟎𝟎. 𝟕𝟕𝟕𝟕
𝟒𝟒
= 𝟎𝟎. 𝟓𝟓
𝟖𝟖
𝟔𝟔. 𝟗𝟗𝟗𝟗
≈ 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝟖𝟖
𝟒𝟒
≈ 𝟎𝟎. 𝟒𝟒𝟒𝟒
𝟖𝟖. 𝟓𝟓
𝟕𝟕. 𝟓𝟓
≈ 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝟖𝟖. 𝟓𝟓
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M2
GEOMETRY
Group 2
Triangle
Angle Measures
Length Measures
1.
△ 𝑨𝑨′𝑩𝑩′𝑪𝑪′
𝒎𝒎∠𝑨𝑨′ ≈ 𝟔𝟔𝟔𝟔°
𝑨𝑨′ 𝑩𝑩′ = 𝟐𝟐. 𝟓𝟓 𝐜𝐜𝐜𝐜
𝑩𝑩′𝑪𝑪′ = 𝟔𝟔 𝐜𝐜𝐜𝐜
𝑨𝑨′ 𝑪𝑪′ = 𝟔𝟔. 𝟓𝟓 𝐜𝐜𝐜𝐜
2.
△ 𝑫𝑫′𝑬𝑬′𝑭𝑭′
𝒎𝒎∠𝑫𝑫′ ≈ 𝟓𝟓𝟓𝟓°
3.
△ 𝑮𝑮′𝑯𝑯′𝑰𝑰′
𝒎𝒎∠𝑰𝑰′ ≈ 𝟒𝟒𝟒𝟒°
4.
△ 𝑱𝑱′𝑲𝑲′𝑳𝑳′
𝒎𝒎∠𝑳𝑳′ ≈ 𝟑𝟑𝟑𝟑°
5.
△ 𝑴𝑴′𝑵𝑵′𝑶𝑶′
𝒎𝒎∠𝑴𝑴′ ≈ 𝟐𝟐𝟐𝟐°
𝑫𝑫′𝑬𝑬′ = 𝟔𝟔 𝐜𝐜𝐜𝐜
𝑬𝑬′𝑭𝑭′ = 𝟖𝟖 𝐜𝐜𝐜𝐜
𝑫𝑫′𝑭𝑭′ = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
𝑮𝑮′𝑯𝑯′ = 𝟕𝟕. 𝟗𝟗 𝐜𝐜𝐜𝐜
𝑯𝑯′𝑰𝑰′ = 𝟗𝟗 𝐜𝐜𝐜𝐜
𝑰𝑰′𝑮𝑮′ = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
𝑱𝑱′𝑲𝑲′ = 𝟔𝟔 𝐜𝐜𝐜𝐜
𝑲𝑲′ 𝑳𝑳′ = 𝟏𝟏𝟏𝟏. 𝟒𝟒 𝐜𝐜𝐜𝐜
𝑱𝑱′𝑳𝑳′ = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
𝑴𝑴′𝑵𝑵′ = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
𝑵𝑵′𝑶𝑶′ = 𝟖𝟖 𝐜𝐜𝐜𝐜
𝑶𝑶′𝑴𝑴′ = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
𝟔𝟔
≈ 𝟎𝟎. 𝟗𝟗𝟗𝟗
𝟔𝟔. 𝟓𝟓
𝟐𝟐. 𝟓𝟓
≈ 𝟎𝟎. 𝟑𝟑𝟑𝟑
𝟔𝟔. 𝟓𝟓
𝟖𝟖
= 𝟎𝟎. 𝟖𝟖
𝟏𝟏𝟏𝟏
𝟔𝟔
= 𝟎𝟎. 𝟔𝟔
𝟏𝟏𝟏𝟏
𝟕𝟕. 𝟗𝟗
≈ 𝟎𝟎. 𝟔𝟔𝟔𝟔
𝟏𝟏𝟏𝟏
𝟗𝟗
= 𝟎𝟎. 𝟕𝟕𝟕𝟕
𝟏𝟏𝟏𝟏
𝟔𝟔
= 𝟎𝟎. 𝟓𝟓
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏. 𝟒𝟒
≈ 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝟏𝟏𝟏𝟏
𝟖𝟖
≈ 𝟎𝟎. 𝟒𝟒𝟒𝟒
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
≈ 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝟏𝟏𝟏𝟏
Have each half of the class share all of the actual measurements and values of the ratios for both sets of triangles (on a
poster or on the board).
With a partner, discuss what you can conclude about each pair of triangles between the two sets.
MP.7
Each pair of triangles is similar by the AA criterion. The marked acute angle for each pair of corresponding triangles is the
same. The values of the
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
and
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
ratios are the same.

Since the triangles are similar, it is no surprise that corresponding angles are equal in measure and that the
values of the ratios of lengths are in constant proportion.

Use these observations to guide you through solving for unknown lengths in Exercises 7–10. Consider why it is
that you are able to actually find the missing lengths.
Exercises 7–10 (10 minutes)
Students approximate unknown lengths to one decimal place. The answers are approximations because the acute
angles are really only approximations, not exact measurements.
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M2
GEOMETRY
Exercises 7–10
For each question, round the unknown lengths appropriately. Refer back to your completed chart from the Exploratory
Challenge; each indicated acute angle is the same approximated acute angle measure as in the chart. Set up and label the
appropriate length ratios, using the terms opp, adj, and hyp in the setup of each ratio.
7.
𝐨𝐨𝐨𝐨𝐨𝐨
= 𝟎𝟎. 𝟖𝟖
𝐡𝐡𝐡𝐡𝐡𝐡
𝒚𝒚
= 𝟎𝟎. 𝟖𝟖
𝟕𝟕
𝒚𝒚 = 𝟓𝟓. 𝟔𝟔
8.
𝐚𝐚𝐚𝐚𝐚𝐚
= 𝟎𝟎. 𝟔𝟔
𝐡𝐡𝐡𝐡𝐡𝐡
𝒙𝒙
= 𝟎𝟎. 𝟔𝟔
𝟕𝟕
𝒙𝒙 = 𝟒𝟒. 𝟐𝟐
𝐨𝐨𝐨𝐨𝐨𝐨
= 𝟎𝟎. 𝟓𝟓
𝐡𝐡𝐡𝐡𝐡𝐡
𝟗𝟗. 𝟐𝟐
= 𝟎𝟎. 𝟓𝟓
𝒛𝒛
𝒛𝒛 = 𝟏𝟏𝟏𝟏. 𝟒𝟒
9.
𝐚𝐚𝐚𝐚𝐚𝐚
= 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝐡𝐡𝐡𝐡𝐡𝐡
𝒙𝒙
= 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝟏𝟏𝟏𝟏. 𝟒𝟒
𝒙𝒙 = 𝟏𝟏𝟏𝟏. 𝟎𝟎
𝐚𝐚𝐚𝐚𝐚𝐚
= 𝟎𝟎. 𝟑𝟑𝟑𝟑
𝐡𝐡𝐡𝐡𝐡𝐡
𝒙𝒙
= 𝟎𝟎. 𝟑𝟑𝟑𝟑
𝒛𝒛
𝒙𝒙
= 𝟎𝟎. 𝟑𝟑𝟑𝟑
𝟐𝟐𝟐𝟐. 𝟒𝟒𝟒𝟒
𝒙𝒙 = 𝟖𝟖. 𝟗𝟗
𝐨𝐨𝐨𝐨𝐨𝐨
= 𝟎𝟎. 𝟗𝟗𝟗𝟗
𝐡𝐡𝐡𝐡𝐡𝐡
𝟐𝟐𝟐𝟐. 𝟔𝟔
= 𝟎𝟎. 𝟗𝟗𝟗𝟗
𝒛𝒛
𝒛𝒛 = 𝟐𝟐𝟐𝟐. 𝟒𝟒𝟒𝟒
10. From a point 𝟏𝟏𝟏𝟏𝟏𝟏 𝐦𝐦 away from a building, Serena measures the angle between the ground and the top of a building
and finds it measures 𝟒𝟒𝟒𝟒°.
What is the height of the building? Round to the nearest meter.
𝐚𝐚𝐚𝐚𝐚𝐚
= 𝟎𝟎. 𝟕𝟕𝟕𝟕
𝐡𝐡𝐡𝐡𝐡𝐡
𝟏𝟏𝟏𝟏𝟏𝟏
= 𝟎𝟎. 𝟕𝟕𝟕𝟕
𝒛𝒛
𝒛𝒛 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝐨𝐨𝐨𝐨𝐨𝐨
= 𝟎𝟎. 𝟔𝟔𝟔𝟔
𝐡𝐡𝐡𝐡𝐡𝐡
𝒚𝒚
= 𝟎𝟎. 𝟔𝟔𝟔𝟔
𝟏𝟏𝟏𝟏𝟏𝟏
𝒚𝒚 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔
𝑧𝑧
The height of the building is approximately 𝟏𝟏𝟏𝟏𝟏𝟏 meters.
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M2
GEOMETRY
Closing (4 minutes)

What did you notice about the triangles in Exercises 7–10 and the exercises in the Exploratory Challenge?
Explain.


The triangles in the exercises were similar to those used in the Exploratory Challenge. We can be sure
of this because the triangles in the exercises and the Exploratory Challenge each had two angle
measures in common: each had a right angle and an acute angle equal in measure.
Why were you able to rely on the Exploratory Challenge chart to determine the unknown lengths in Exercises
7–10?

Because the triangles in Exercises 7–10 are similar to those triangles in the Exploratory Challenge, we
were able to use the values of the

opp
hyp
and
hyp
ratios to set up an equation for each question.
Since all the triangles used in the lesson are right triangles in general, what determines when two right
triangles have

opp
hyp
and
hyp
ratios equal in value?
Both right triangles must have corresponding acute angles of equal measure, and the same acute angle
must be referenced in both triangles for the values of the

opp
hyp
and
hyp
ratios to be equal.
In this lesson, you were provided with just a handful of ratio values, each of which corresponded to a given
acute angle measure in a right triangle. Knowing these ratio values for the given acute angle measure allowed
us to solve for unknown lengths in triangles similar to right triangles; this is what makes these ratios so
incredible. In the next lesson, we learn that we have easy access to these ratios for any angle measure.
Relevant Vocabulary
SIDES OF A RIGHT TRIANGLE: The hypotenuse of a right triangle is the side opposite the right angle; the other two sides of the
right triangle are called the legs. Let 𝜃𝜃 be the angle measure of an acute angle of the right triangle. The opposite side is
the leg opposite that angle. The adjacent side is the leg that is contained in one of the two rays of that angle (the
hypotenuse is contained in the other ray of the angle).
Exit Ticket (5 minutes)
Lesson 25:
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Lesson 25
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M2
GEOMETRY
Name
Date
Lesson 25: Incredibly Useful Ratios
Exit Ticket
1.
Use the chart from the Exploratory Challenge to approximate the unknown lengths 𝑦𝑦 and 𝑧𝑧 to one decimal place.
°
2.
Why can we use the chart from the Exploratory Challenge to approximate the unknown lengths?
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Lesson 25
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M2
GEOMETRY
Exit Ticket Sample Solutions
1.
Use the chart from the Exploratory Challenge to approximate the unknown lengths 𝒚𝒚 and 𝒛𝒛 to one decimal place.
𝐚𝐚𝐚𝐚𝐚𝐚
= 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝐡𝐡𝐡𝐡𝐡𝐡
𝟐𝟐𝟐𝟐. 𝟓𝟓
= 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝒛𝒛
𝒛𝒛 = 𝟐𝟐𝟐𝟐. 𝟔𝟔
2.
𝐨𝐨𝐨𝐨𝐨𝐨
= 𝟎𝟎. 𝟒𝟒𝟒𝟒
𝐡𝐡𝐡𝐡𝐡𝐡
𝒚𝒚
= 𝟎𝟎. 𝟒𝟒𝟒𝟒
𝟐𝟐𝟐𝟐. 𝟔𝟔
𝒚𝒚 ≈ 𝟏𝟏𝟏𝟏
°
Why can we use the chart from the Exploratory Challenge to approximate the unknown lengths?
The triangle in Problem 1 is similar to a triangle in the chart from the Exploratory Challenge. Since the triangles are
similar, the values of the
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
and
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
needed to solve for unknown lengths.
ratios in reference to the acute angle of 𝟐𝟐𝟐𝟐° can be used in the equations
Problem Set Sample Solutions
The table below contains the values of the ratios
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
and
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
for a variety of right triangles based on a given acute angle,
𝜽𝜽, from each triangle. Use the table and the diagram of the right triangle below to complete each problem.
𝜽𝜽
(degrees)
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
𝟎𝟎
𝟎𝟎
𝟏𝟏
GEO-M2-TE-1.3.0-07.2015
𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑
𝟒𝟒𝟒𝟒
𝟒𝟒𝟒𝟒
𝟓𝟓𝟓𝟓
𝟔𝟔𝟔𝟔
𝟕𝟕𝟕𝟕
𝟖𝟖𝟖𝟖
𝟏𝟏
= 𝟎𝟎. 𝟓𝟓 𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝟐𝟐
𝟏𝟏
𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
= 𝟎𝟎. 𝟓𝟓 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐
𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
Lesson 25:
𝟏𝟏𝟏𝟏
𝟗𝟗𝟗𝟗
𝟏𝟏
𝟎𝟎
Incredibly Useful Ratios
397
Lesson 25
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
For each problem, approximate the unknown lengths to one decimal place. Write the appropriate length ratios using the
terms opp, adj, and hyp in the setup of each ratio.
1.
Find the approximate length of the leg opposite the 𝟖𝟖𝟖𝟖° angle.
Using the value of the ratio
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
for an 𝟖𝟖𝟖𝟖° angle:
𝐨𝐨𝐨𝐨𝐨𝐨
= 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝟐𝟐𝟐𝟐
𝐨𝐨𝐨𝐨𝐨𝐨 = 𝟐𝟐𝟐𝟐. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
The length of the leg opposite the 𝟖𝟖𝟖𝟖° angle is
approximately 𝟐𝟐𝟐𝟐. 𝟔𝟔.
2.
Find the approximate length of the hypotenuse.
Using the value of the ratio
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
𝟕𝟕. 𝟏𝟏
= 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝐡𝐡𝐡𝐡𝐡𝐡
𝟕𝟕. 𝟏𝟏
= 𝐡𝐡𝐡𝐡𝐡𝐡
𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 ≈ 𝐡𝐡𝐡𝐡𝐡𝐡
for a 𝟒𝟒𝟒𝟒° angle:
The length of the hypotenuse is approximately 𝟏𝟏𝟏𝟏. 𝟎𝟎.
3.
Find the approximate length of the hypotenuse.
Using the value of the ratio
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
for a 𝟔𝟔𝟔𝟔° angle:
𝟎𝟎. 𝟕𝟕 𝟏𝟏
=
𝐡𝐡𝐡𝐡𝐡𝐡 𝟐𝟐
𝟎𝟎. 𝟕𝟕
= 𝐡𝐡𝐡𝐡𝐡𝐡
𝟏𝟏
𝟐𝟐
𝟏𝟏. 𝟒𝟒 = 𝐡𝐡𝐡𝐡𝐡𝐡
The length of the hypotenuse is 𝟏𝟏. 𝟒𝟒.
4.
Find the approximate length of the leg adjacent to the 𝟒𝟒𝟒𝟒° angle.
Using the value of the ratio
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
for a 𝟒𝟒𝟒𝟒° angle:
𝐚𝐚𝐚𝐚𝐚𝐚
= 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝟏𝟏𝟏𝟏
𝐚𝐚𝐚𝐚𝐚𝐚 = 𝟏𝟏𝟏𝟏. 𝟕𝟕𝟕𝟕𝟕𝟕
The length of the leg adjacent to the given 𝟒𝟒𝟒𝟒° angle is approximately
𝟏𝟏𝟏𝟏. 𝟖𝟖.
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5.
Find the length of both legs of the right triangle below. Indicate which leg is adjacent and which is opposite the
given angle of 𝟑𝟑𝟑𝟑°.
Using the value of the ratio
𝐨𝐨𝐨𝐨𝐨𝐨 𝟏𝟏
=
𝟐𝟐
𝟏𝟏𝟏𝟏
𝐨𝐨𝐨𝐨𝐨𝐨 = 𝟔𝟔
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
for a 𝟑𝟑𝟑𝟑° angle:
The length of the leg that is opposite the given 𝟑𝟑𝟑𝟑° angle is 𝟔𝟔.
Using the value of the ratio
𝐚𝐚𝐚𝐚𝐚𝐚
= 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝟏𝟏𝟏𝟏
𝐚𝐚𝐚𝐚𝐚𝐚 = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
for a 𝟑𝟑𝟑𝟑° angle:
The length of the leg that is adjacent to the given 𝟑𝟑𝟑𝟑° angle is approximately 𝟏𝟏𝟏𝟏. 𝟒𝟒.
6.
Three city streets form a right triangle. Main Street and State Street are perpendicular. Laura Street and State
Street intersect at a 𝟓𝟓𝟓𝟓° angle. The distance along Laura Street to Main Street is 𝟎𝟎. 𝟖𝟖 mile. If Laura Street is closed
between Main Street and State Street for a festival, approximately how far (to the nearest tenth) will someone have
to travel to get around the festival if they take only Main Street and State Street?
The leg opposite the 𝟓𝟓𝟓𝟓° angle represents the distance
along Main Street, the leg adjacent to the 𝟓𝟓𝟓𝟓° angle
represents the distance along State Street, and the
hypotenuse of the triangle represents the 𝟎𝟎. 𝟖𝟖 mile
distance along Laura Street.
Using the ratio
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
for 𝟓𝟓𝟎𝟎°:
𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 =
Using the ratio
𝐨𝐨𝐨𝐨𝐨𝐨
𝟎𝟎. 𝟖𝟖
𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 = 𝐨𝐨𝐨𝐨𝐨𝐨
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
for 𝟓𝟓𝟓𝟓°:
𝐚𝐚𝐚𝐚𝐚𝐚
𝟎𝟎. 𝟖𝟖
𝟎𝟎. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 = 𝐚𝐚𝐚𝐚𝐚𝐚
𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 =
The total distance of the detour:
𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 + 𝟎𝟎. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 = 𝟏𝟏. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
The total distance to travel around the festival along
State Street and Main Street is approximately 𝟏𝟏. 𝟏𝟏 miles.
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GEOMETRY
7.
A cable anchors a utility pole to the ground as shown in the picture. The cable forms an angle of 𝟕𝟕𝟕𝟕° with the
ground. The distance from the base of the utility pole to the anchor point on the ground is 𝟑𝟑. 𝟖𝟖 meters.
Approximately how long is the support cable?
The hypotenuse of the triangle represents the length of the support
cable in the diagram, and the leg adjacent to the given 𝟕𝟕𝟕𝟕° angle
represents the distance between the anchor point and the base of
the utility pole. Using the value of the ratio
𝟑𝟑. 𝟖𝟖
𝐡𝐡𝐡𝐡𝐡𝐡
𝟑𝟑. 𝟖𝟖
𝐡𝐡𝐡𝐡𝐡𝐡 =
𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝐡𝐡𝐡𝐡𝐡𝐡 = 𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 =
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
for 𝟕𝟕𝟕𝟕°:
The length of the support cable is approximately 𝟏𝟏𝟏𝟏. 𝟏𝟏 meters long.
8.
Indy says that the ratio of
𝐨𝐨𝐨𝐨𝐨𝐨
𝐚𝐚𝐚𝐚𝐚𝐚
for an angle of 𝟎𝟎° has a value of 𝟎𝟎 because the opposite side of the triangle has a
length of 𝟎𝟎. What does she mean?
Indy’s triangle is not actually a triangle since the opposite side does not have length, which means that it does not
exist. As the degree measure of the angle considered gets closer to 𝟎𝟎°, the value of the ratio gets closer to 𝟎𝟎.
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Lesson 25
M2
GEOMETRY
Group 1
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Lesson 25
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Group 2
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Identifying Sides of a Right Triangle with Respect to a Given Acute Angle
Poster
•
•
•
With respect to ∠𝐴𝐴, the opposite side, 𝑜𝑜𝑜𝑜𝑜𝑜, is side
����
𝐵𝐵𝐵𝐵 .
With respect to ∠𝐴𝐴, the adjacent side, 𝑎𝑎𝑎𝑎𝑎𝑎, is side
����
𝐴𝐴𝐴𝐴 .
The hypotenuse, ℎ𝑦𝑦𝑦𝑦, is side ����
𝐴𝐴𝐴𝐴 and is always
opposite from the 90° angle.
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•
•
•
With respect to ∠𝐶𝐶, the opposite side, 𝑜𝑜𝑜𝑜𝑜𝑜, is side
����
𝐴𝐴𝐴𝐴.
With respect to ∠𝐶𝐶, the adjacent side, 𝑎𝑎𝑎𝑎𝑎𝑎, is side
����
𝐵𝐵𝐵𝐵 .
The hypotenuse, ℎ𝑦𝑦𝑦𝑦, is side ����
𝐴𝐴𝐴𝐴 and is always
opposite from the 90° angle.
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Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Lesson 26: The Definition of Sine, Cosine, and Tangent
Student Outcomes



Students define sine, cosine, and tangent of 𝜃𝜃, where 𝜃𝜃 is the angle measure of an acute angle of a right
triangle. Students denote sine, cosine, and tangent as sin, cos, and tan, respectively.
If ∠𝐴𝐴 is an acute angle whose measure in degrees is 𝜃𝜃, then sin ∠𝐴𝐴 = sin𝜃𝜃, cos ∠𝐴𝐴 = cos𝜃𝜃, and
tan ∠𝐴𝐴 = tan𝜃𝜃.
Given the side lengths of a right triangle with acute angles, students find sine, cosine, and tangent of each
acute angle.
Lesson Notes
It is convenient, as adults, to use the notation “sin2 𝑥𝑥” to refer to the value of the square of the sine function. However,
rushing too fast to this abbreviated notation for trigonometric functions leads to incorrect understandings of how
functions are manipulated, which can lead students to think that sin 𝑥𝑥 is short for “sin ⋅ 𝑥𝑥” and to incorrectly divide out
the variable
sin 𝑥𝑥
𝑥𝑥
= sin.
To reduce these types of common notation-driven errors later, this curriculum is very deliberate about how and when to
use abbreviated function notation for sine, cosine, and tangent:
1.
2.
3.
In Geometry, sine, cosine, and tangent are thought of as the value of ratios of triangles, not as functions. No
attempt is made to describe the trigonometric ratios as functions of the real number line. Therefore, the notation is
just an abbreviation for the sine of an angle (sin ∠𝐴𝐴) or sine of an angle measure (sin 𝜃𝜃). Parentheses are used more
for grouping and clarity reasons than as symbols used to represent a function.
In Algebra II, to distinguish between the ratio version of sine in geometry, all sine functions are notated as functions:
sin(𝑥𝑥) is the value of the sine function for the real number 𝑥𝑥, just like 𝑓𝑓(𝑥𝑥) is the value of the function 𝑓𝑓 for the real
number 𝑥𝑥. In this grade, students maintain function notation integrity and strictly maintain parentheses as part of
𝜋𝜋
2
𝜋𝜋
2
function notation, writing, for example,sin � − 𝜃𝜃� = 𝑐𝑐𝑐𝑐𝑐𝑐(𝜃𝜃) instead of sin − 𝜃𝜃 = cos 𝜃𝜃.
By Precalculus and Advanced Topics, students have had two full years of working with sine, cosine, and tangent as
both ratios and functions. It is finally in this year that the distinction between ratio and function notations begins to
blur and students write, for example, sin2 (𝜃𝜃) as the value of the square of the sine function for the real number 𝜃𝜃,
which is how most calculus textbooks notate these functions.
Classwork
Exercises 1–3 (6 minutes)
The following exercises provide students with practice identifying specific ratios of sides based on the labels learned in
Lesson 25. The third exercise leads into an understanding of the relationship between sine and cosine, which is
observed in this lesson and formalized in Lesson 27.
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Exercises 1–3
1.
Identify the
For ∠𝑨𝑨:
For ∠𝑩𝑩:
2.
For ∠𝑩𝑩:
𝐡𝐡𝐡𝐡𝐡𝐡
ratios for ∠𝑨𝑨 and ∠𝑩𝑩.
𝟏𝟏𝟏𝟏
𝟓𝟓
𝟏𝟏𝟑𝟑
Identify the
For ∠𝑨𝑨:
3.
𝟏𝟏𝟏𝟏
𝐨𝐨𝐨𝐨𝐨𝐨
𝟓𝟓
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
Scaffolding:
 For students who are
struggling, model labeling
opp, hyp, and adj for ∠𝐴𝐴.
ratios for ∠𝑨𝑨 and ∠𝑩𝑩.
𝟏𝟏𝟏𝟏
 Allow students to use the
reference chart provided
with Lesson 25.
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
Describe the relationship between the ratios for ∠𝑨𝑨 and ∠𝑩𝑩.
The
𝐨𝐨𝐨𝐨𝐨𝐨
The
𝐨𝐨𝐨𝐨𝐨𝐨
𝐡𝐡𝐡𝐡𝐡𝐡
𝐡𝐡𝐡𝐡𝐡𝐡
ratio for ∠𝑨𝑨 is equal to the
ratio for ∠𝑩𝑩 is equal to the
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
ratio for ∠𝑩𝑩.
ratio for ∠𝑨𝑨.
Discussion (6 minutes)
The Discussion defines sine, cosine, and tangent. As the names opp, adj, and hyp are
relatively new to students, it is important that students have a visual of the triangle to
reference throughout the Discussion. Following the Discussion is an exercise
that can be used to informally assess students’ understanding of these
definitions.

include an extension:
Draw another triangle
𝐴𝐴𝐴𝐴𝐴𝐴 with different side
lengths that have the
same ratios for ∠𝐴𝐴 and
∠𝐵𝐵. Explain how you
know the ratios is the
same, and justify your
In everyday life, we reference objects and people by name, especially
when we use the object or see the person frequently. Imagine
always calling your friend hey or him or her or friend! We want to be
able to easily distinguish and identify a person, so we use a name.
The same reasoning can be applied to the fractional expressions that
we have been investigating:
,
hyp hyp
, and
opp
need names.
Normally, these fractional expressions are said to be values of the ratios.
However, to avoid saying “value of the ratio opp: hyp as
opp
hyp
” all of the time,
these fractional expressions are called, collectively, the trigonometric ratios,
based upon historical precedence, even though they really are not defined as ratios in the standards. Consider sharing
this fact with students.

These incredibly useful ratios were discovered long ago and have had several names. The names we currently
use are translations of Latin words. You will learn more about the history behind these ratios in Algebra II.

Presently, mathematicians have agreed upon the names sine, cosine, and tangent.
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
If 𝜃𝜃 is the angle measure of ∠𝐴𝐴, as shown, then we define:
The sine of 𝜃𝜃 is the value of the ratio of the length of the opposite side to the length of the hypotenuse. As a
formula,
opp
sin 𝜃𝜃 =
.
hyp
We also say sin ∠𝐴𝐴 = sin 𝜃𝜃. Then, sin ∠𝐴𝐴 =
𝐵𝐵𝐵𝐵
.
𝐴𝐴𝐴𝐴
The cosine of 𝜃𝜃 is the value of the ratio of the length of the adjacent side to the length of the hypotenuse. As a
formula,
cos 𝜃𝜃 =
.
hyp
We also say cos ∠𝐴𝐴 = cos 𝜃𝜃. Then, cos ∠𝐴𝐴 =
𝐴𝐴𝐴𝐴
.
𝐴𝐴𝐴𝐴
We also say tan ∠𝐴𝐴 = tan 𝜃𝜃 . Then, tan ∠𝐴𝐴 =
𝐵𝐵𝐵𝐵
.
𝐴𝐴𝐴𝐴
The tangent of 𝜃𝜃 is the value of the ratio of the length of the opposite side to the length of the adjacent side.
As a formula,
opp
tan 𝜃𝜃 =
.

There are still three possible combinations of quotients of the side lengths; we briefly introduce them here.
The secant of 𝜃𝜃 is the value of the ratio of the length of the hypotenuse to the length of the adjacent side. As a
formula,
hyp
sec 𝜃𝜃 =
.
The cosecant of 𝜃𝜃 is the value of the ratio of the length of the hypotenuse to the length of the opposite side.
As a formula,
hyp
csc 𝜃𝜃 =
.
opp

The cotangent of 𝜃𝜃 is the value of the ratio of the length of the adjacent side to the length of the opposite side.
As a formula,
cot 𝜃𝜃 =
.
opp
We have little need in this course for secant, cosecant, and cotangent because, given any two sides, it is
possible to write the quotient so as to get sine, cosine, and tangent. In more advanced courses, secant,
cosecant, and cotangent are more useful.
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Exercises 4–9 (15 minutes)
Have students complete Exercises 4–6 independently, or divide the work among students, and have them share their
results. Once Exercises 4–6 have been completed, encourage students to discuss in small groups the relationships they
notice between the sine of the angle and the cosine of its complement. Also, encourage them to discuss the relationship
they notice about the tangent of both angles. Finally, have students share their observations with the whole class and
then complete Exercises 7–8. Note that because students are not being asked to rationalize denominators, the
relationships are clearer.
Scaffolding:
Exercises 4–9
4.
 Consider having students
draw two right triangles
and then color-code
and/or label each with
to the angle they are
looking at.
In △ 𝑷𝑷𝑷𝑷𝑷𝑷, 𝒎𝒎∠𝑷𝑷 = 𝟓𝟓𝟓𝟓. 𝟐𝟐° and 𝒎𝒎∠𝑸𝑸 = 𝟑𝟑𝟑𝟑. 𝟖𝟖°. Complete the following table.
students draw two right
triangles, △ 𝐷𝐷𝐷𝐷𝐷𝐷 and
△ 𝐺𝐺𝐺𝐺𝐺𝐺, such that they are
not congruent but so that
sin 𝐸𝐸 = sin 𝐻𝐻 . Then, have
students explain how they
know.
Measure of Angle
𝟓𝟓𝟓𝟓. 𝟐𝟐
𝟑𝟑𝟑𝟑. 𝟖𝟖
Lesson 26:
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𝐨𝐨𝐨𝐨𝐨𝐨
Sine �
𝐡𝐡𝐡𝐡𝐡𝐡
�
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓. 𝟐𝟐 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑. 𝟖𝟖 =
𝟒𝟒
𝟓𝟓
𝟑𝟑
𝟓𝟓
Cosine �
𝐚𝐚𝐚𝐚𝐚𝐚
𝐡𝐡𝐡𝐡𝐡𝐡
𝐜𝐜𝐜𝐜𝐜𝐜 𝟓𝟓𝟓𝟓. 𝟐𝟐 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑. 𝟖𝟖 =
�
𝟑𝟑
𝟓𝟓
𝟒𝟒
𝟓𝟓
𝐨𝐨𝐨𝐨𝐨𝐨
Tangent �
𝐚𝐚𝐚𝐚𝐚𝐚
𝐭𝐭𝐭𝐭𝐭𝐭 𝟓𝟓𝟓𝟓. 𝟐𝟐 =
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑. 𝟖𝟖 =
�
𝟒𝟒
𝟑𝟑
𝟑𝟑
𝟒𝟒
The Definition of Sine, Cosine, and Tangent
407
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
In the triangle below, 𝒎𝒎∠𝑨𝑨 = 𝟑𝟑𝟑𝟑. 𝟕𝟕° and 𝒎𝒎∠𝑩𝑩 = 𝟓𝟓𝟓𝟓. 𝟑𝟑°. Complete the following table.
=
𝟐𝟐
𝟑𝟑
𝟑𝟑
𝟓𝟓 .
𝟐𝟐
𝟑𝟑𝟑𝟑
𝐭𝐭
𝟑𝟑
𝐭𝐭𝐭𝐭
𝟑𝟑
√𝟏𝟏𝟏𝟏
. 𝟕𝟕 =
𝟓𝟓
=
𝐭𝐭
𝐭𝐭𝐭𝐭
𝟓𝟓 .
√𝟏𝟏𝟏𝟏
𝟐𝟐
𝟑𝟑𝟑𝟑
𝐜𝐜
𝟑𝟑
. 𝟕𝟕 =
𝟓𝟓
√𝟏𝟏𝟏𝟏
𝐜𝐜
𝐜𝐜𝐜𝐜
𝟐𝟐
=
𝟑𝟑
𝟑𝟑𝟑𝟑
𝟓𝟓
𝟑𝟑𝟑𝟑
√𝟏𝟏𝟏𝟏
Tangent
In the triangle below, let 𝒆𝒆 be the measure of ∠𝑬𝑬 and 𝒅𝒅 be the measure of ∠𝑫𝑫. Complete the following table.
Lesson 26:
√
𝟕𝟕
𝐜𝐜
𝒆𝒆 =
𝟒𝟒
𝟕𝟕
𝒅𝒅 =
𝐭𝐭
𝒆𝒆 =
𝟒𝟒
√
𝟑𝟑𝟑𝟑
𝐭𝐭
𝐭𝐭𝐭𝐭
√
𝟕𝟕
𝟑𝟑𝟑𝟑
𝒅𝒅 =
𝐜𝐜𝐜𝐜
𝐜𝐜
√
𝟒𝟒
𝟑𝟑𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝒆𝒆 =
𝟒𝟒
𝟕𝟕
Tangent
𝐭𝐭𝐭𝐭
𝒆𝒆
𝐬𝐬𝐬𝐬𝐬𝐬 𝒅𝒅 =
Cosine
𝐜𝐜𝐜𝐜
𝒅𝒅
Sine
𝟑𝟑𝟑𝟑
Measure of Angle
GEO-M2-TE-1.3.0-07.2015
Cosine
. 𝟕𝟕 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓 .
𝟑𝟑
𝟓𝟓
6.
𝐬𝐬𝐬𝐬𝐬𝐬
. 𝟕𝟕
𝟓𝟓 .
Sine
𝐜𝐜𝐜𝐜
Measure of Angle
𝟑𝟑
5.
The Definition of Sine, Cosine, and Tangent
408
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
In the triangle below, let 𝒙𝒙 be the measure of ∠𝑿𝑿 and 𝒚𝒚 be the measure of ∠𝒀𝒀. Complete the following table.
√𝟏𝟏𝟏𝟏
𝟏𝟏
√𝟏𝟏𝟏𝟏
𝐭𝐭
𝒚𝒚 =
𝟏𝟏
𝟑𝟑
𝒙𝒙 =
𝟑𝟑
𝒚𝒚 =
𝐭𝐭
𝐭𝐭𝐭𝐭
𝐜𝐜
√𝟏𝟏𝟏𝟏
𝟏𝟏
Tamer did not finish completing the table below for a diagram similar to the previous problems that the teacher had
on the board where 𝒑𝒑 was the measure of ∠𝑷𝑷 and 𝒒𝒒 was the measure of ∠𝑸𝑸. Use any patterns you notice from
Exercises 1–4 to complete the table for Tamer.
𝟏𝟏𝟏𝟏
𝐭𝐭
√𝟏𝟏
𝟏𝟏𝟏𝟏
𝒒𝒒 =
𝟔𝟔
𝒒𝒒 =
𝒑𝒑 =
𝟔𝟔
𝐜𝐜
𝐭𝐭
√𝟏𝟏
𝐭𝐭𝐭𝐭
𝒑𝒑 =
𝐭𝐭𝐭𝐭
√𝟏𝟏
𝐜𝐜
𝟔𝟔
𝐬𝐬𝐬𝐬𝐬𝐬 𝒒𝒒 =
√𝟏𝟏
𝟏𝟏𝟏𝟏
𝒒𝒒
𝟏𝟏𝟏𝟏
Tangent
𝟏𝟏𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬 𝒑𝒑 =
𝐜𝐜𝐜𝐜
𝒑𝒑
Cosine
𝐜𝐜𝐜𝐜
MP.8
Sine
𝟏𝟏𝟏𝟏
Measure of Angle
9.
𝒙𝒙 =
𝐭𝐭𝐭𝐭
𝐬𝐬𝐬𝐬𝐬𝐬 𝒚𝒚 =
𝐜𝐜
Tangent
𝟑𝟑
𝒚𝒚
𝟏𝟏
√𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜
𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙 =
𝐜𝐜𝐜𝐜
𝒙𝒙
Cosine
𝟔𝟔
8.
Sine
𝟑𝟑
Measure of Angle
𝟏𝟏𝟏𝟏
7.
𝟏𝟏𝟏𝟏
Explain how you were able to determine the sine, cosine, and tangent of ∠𝑸𝑸 in Exercise 8.
I was able to complete the table for Tamer by observing the patterns of previous problems. For example, I noticed
that the sine of one angle was always equal to the ratio that represented the cosine of the other angle. Since I was
given 𝐬𝐬𝐢𝐢𝐢𝐢 𝒑𝒑, I knew the ratio
𝟏𝟏𝟏𝟏
√𝟏𝟏𝟏𝟏𝟏𝟏
would be the 𝐜𝐜𝐜𝐜𝐜𝐜 𝒒𝒒. Similarly, 𝐜𝐜𝐜𝐜𝐜𝐜 𝒑𝒑 = 𝐬𝐬𝐬𝐬𝐬𝐬 𝒒𝒒 =
𝟔𝟔
�𝟏𝟏𝟏𝟏𝟏𝟏
Finally, I noticed that the
tangents of the angles were always reciprocals of each other. Since I was given the 𝐭𝐭𝐭𝐭𝐭𝐭 𝒑𝒑 =
𝐭𝐭𝐭𝐭𝐭𝐭 𝒒𝒒 must be equal to
𝟔𝟔
.
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
, I knew that the
𝟔𝟔
Discussion (8 minutes)
The sine, cosine, and tangent of an angle can be used to find unknown lengths of other triangles using within-figure
ratios of similar triangles. The discussion that follows begins by posing a question to students. Provide time for students
to discuss the answer in pairs or small groups, and then have them share their thoughts with the class.


If 0 < 𝜃𝜃 < 90, we can define sine, cosine, and tangent of 𝜃𝜃: Take a right triangle that has an acute angle with
angle degree measure 𝜃𝜃, and use the appropriate side lengths.
If we use different right triangles, why will we get the same value for sin 𝜃𝜃, cos 𝜃𝜃, and tan 𝜃𝜃?
Lesson 26:
GEO-M2-TE-1.3.0-07.2015
The Definition of Sine, Cosine, and Tangent
409
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Provide time for students to talk to a partner or small group. If necessary, use the questions and diagrams below to
MP.3 guide students’ thinking. The decimal values for the side lengths are used to make less obvious the fact that the ratios of
the side lengths are equal.

For example, consider the following two triangles.
The triangles below contain approximations for all lengths. If the Pythagorean theorem is used to verify that the
triangles are right triangles, then you will notice that the values are slightly off. For example, the length of the
hypotenuse of △ 𝐷𝐷𝐷𝐷𝐷𝐷 contains 11 decimal digits, not the 5 shown.

Find the sin 𝐴𝐴, cos 𝐴𝐴, and tan 𝐴𝐴. Compare those ratios to the sin 𝐷𝐷, cos 𝐷𝐷, and tan 𝐷𝐷. What do you notice?
The task of finding the ratios can be divided among students and shared with the group.


Under what circumstances have we observed ratios within one figure being equal to ratios within another
figure?



Students should notice that sin 𝐴𝐴 = sin 𝐷𝐷, cos 𝐴𝐴 = cos 𝐷𝐷, and tan 𝐴𝐴 = tan 𝐷𝐷 .
Within-figure ratios are equal when the figures are similar.
Two right triangles, each having an acute angle of angle measure 𝜃𝜃, are similar by the AA criterion. So, we
know that the values of corresponding ratios of side lengths are equal. That means sin 𝜃𝜃, cos 𝜃𝜃, and tan 𝜃𝜃 do
not depend on which right triangle we use.
The ratios we write for the sine, cosine, and tangent of an angle are useful because they allow us to solve for
two sides of a triangle when we know only the length of one side.
Closing (5 minutes)
Ask students the following questions. Have students respond in writing, to a partner, or to the whole class.

Describe the ratios that we used to calculate sine, cosine, and tangent.


Given an angle, 𝜃𝜃, sin 𝜃𝜃 =
𝑜𝑜𝑜𝑜𝑜𝑜
𝑎𝑎𝑎𝑎𝑎𝑎
𝑜𝑜𝑜𝑜𝑜𝑜
, cos 𝜃𝜃 =
, and tan 𝜃𝜃 =
.
ℎ𝑦𝑦𝑦𝑦
ℎ𝑦𝑦𝑦𝑦
𝑎𝑎𝑎𝑎𝑎𝑎
Given any two right triangles that each have an acute angle with measure 𝜃𝜃, why would we get the same value
for sin 𝜃𝜃, cos 𝜃𝜃, and tan 𝜃𝜃 using either triangle?

Since the two right triangles each have an acute angle with measure 𝜃𝜃, they are similar by the AA
criterion. Similar triangles have corresponding side lengths that are equal in ratio. Additionally, based
on our investigations in Lesson 25, we know that the value of the ratios of corresponding sides for a
particular angle size are equal to the same constant.
Lesson 26:
GEO-M2-TE-1.3.0-07.2015
The Definition of Sine, Cosine, and Tangent
410
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 26
M2
GEOMETRY

Given a right triangle, describe the relationship between the sine of one acute angle and the cosine of the
other acute angle.

The sine of one acute angle of a right triangle is equal to the cosine of the other acute angle in the
triangle.
Exit Ticket (5 minutes)
Lesson 26:
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The Definition of Sine, Cosine, and Tangent
411
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Name
Date
Lesson 26: The Definition of Sine, Cosine, and Tangent
Exit Ticket
1.
Given the diagram of the triangle, complete the following table.
Angle Measure
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝒔𝒔
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽
𝒕𝒕
2.
a.
Which values are equal?
b.
How are tan 𝑠𝑠 and tan 𝑡𝑡 related?
If 𝑢𝑢 and 𝑣𝑣 are the measures of complementary angles such that sin 𝑢𝑢 =
�21
2
and tan 𝑣𝑣 =
, label the sides and
5
2
angles of the right triangle in the diagram below with possible side lengths.
Lesson 26:
GEO-M2-TE-1.3.0-07.2015
The Definition of Sine, Cosine, and Tangent
412
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Exit Ticket Sample Solutions
1.
Given the diagram of the triangle, complete the following table.
Angle Measure
𝒔𝒔
𝒕𝒕
a.
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝟔𝟔√𝟔𝟔𝟔𝟔
=
𝟔𝟔𝟔𝟔
√𝟔𝟔𝟔𝟔
√𝟔𝟔𝟔𝟔
√𝟔𝟔𝟔𝟔
𝟔𝟔
=
𝟔𝟔√𝟔𝟔𝟔𝟔
𝟔𝟔𝟔𝟔
𝟔𝟔
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
𝟓𝟓√𝟔𝟔𝟔𝟔
=
𝟔𝟔𝟔𝟔
√𝟔𝟔𝟔𝟔
𝟓𝟓
𝟓𝟓
=
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽
𝟓𝟓
𝟔𝟔
𝟔𝟔
𝟓𝟓
𝟓𝟓√𝟔𝟔𝟔𝟔
𝟔𝟔𝟔𝟔
Which values are equal?
𝐬𝐬𝐬𝐬𝐬𝐬 𝒔𝒔 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝒕𝒕 and 𝐜𝐜𝐜𝐜𝐜𝐜 𝒔𝒔 = 𝐬𝐬𝐬𝐬𝐬𝐬 𝒕𝒕
b.
How are 𝐭𝐭𝐭𝐭𝐭𝐭 𝒔𝒔 and 𝐭𝐭𝐭𝐭𝐭𝐭 𝒕𝒕 related?
They are reciprocals:
2.
𝟓𝟓 𝟔𝟔
⋅
𝟔𝟔 𝟓𝟓
= 𝟏𝟏.
If 𝒖𝒖 and 𝒗𝒗 are the measures of complementary angles such that 𝐬𝐬𝐬𝐬𝐬𝐬 𝒖𝒖 =
�𝟐𝟐𝟐𝟐
𝟐𝟐
and 𝐭𝐭𝐭𝐭𝐭𝐭 𝒗𝒗 =
, label the sides and
𝟐𝟐
𝟓𝟓
angles of the right triangle in the diagram below with possible side lengths:
A possible solution is shown below; however, any similar triangle having a shorter leg with length of 𝟐𝟐𝟐𝟐, longer leg
with length of 𝒙𝒙√𝟐𝟐𝟐𝟐, and hypotenuse with length of 𝟓𝟓𝟓𝟓, for some positive number 𝒙𝒙, is also correct.
Problem Set Sample Solutions
1.
Given the triangle in the diagram, complete the following table.
Angle Measure
𝜶𝜶
𝜷𝜷
Lesson 26:
GEO-M2-TE-1.3.0-07.2015
𝐬𝐬𝐬𝐬𝐬𝐬
𝟐𝟐√𝟓𝟓
𝟔𝟔
𝟒𝟒 𝟐𝟐
=
𝟔𝟔 𝟑𝟑
𝐜𝐜𝐜𝐜𝐜𝐜
𝟒𝟒 𝟐𝟐
=
𝟔𝟔 𝟑𝟑
𝟐𝟐√𝟓𝟓
𝟔𝟔
𝐭𝐭𝐭𝐭𝐭𝐭
𝟐𝟐√𝟓𝟓 √𝟓𝟓
=
𝟒𝟒
𝟐𝟐
𝟒𝟒
𝟐𝟐√𝟓𝟓
=
𝟐𝟐
√𝟓𝟓
The Definition of Sine, Cosine, and Tangent
413
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
2.
Given the table of values below (not in simplest radical form), label the sides and angles in the right triangle.
Angle Measure
3.
𝐬𝐬𝐬𝐬𝐬𝐬
𝜶𝜶
𝟐𝟐√𝟏𝟏𝟏𝟏
𝜷𝜷
𝟐𝟐√𝟏𝟏𝟏𝟏
𝐭𝐭𝐭𝐭𝐭𝐭
𝟒𝟒
𝟐𝟐√𝟔𝟔
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟐𝟐√𝟔𝟔
𝟒𝟒
𝟐𝟐√𝟔𝟔
𝟒𝟒
𝟐𝟐√𝟔𝟔
𝟐𝟐√𝟏𝟏𝟏𝟏
Given 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 and 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷, complete the missing values in the table. You may draw a diagram to help you.
Angle Measure
4.
𝐜𝐜𝐜𝐜𝐜𝐜
𝟒𝟒
𝐬𝐬𝐬𝐬𝐬𝐬
𝐜𝐜𝐜𝐜𝐜𝐜
𝐭𝐭𝐭𝐭𝐭𝐭
𝟓𝟓
𝜶𝜶
√𝟐𝟐
𝟑𝟑√𝟑𝟑
𝟑𝟑√𝟑𝟑
𝜷𝜷
𝟑𝟑√𝟑𝟑
𝟓𝟓
𝟑𝟑√𝟑𝟑
√𝟐𝟐
𝟓𝟓
𝟓𝟓
√𝟐𝟐
√𝟐𝟐
Given the triangle shown to the right, fill in the missing values in the table.
Using the Pythagorean theorem:
𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐 = 𝟐𝟐 + 𝟔𝟔𝟐𝟐
𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐 = 𝟒𝟒 + 𝟑𝟑𝟑𝟑
𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐 = 𝟒𝟒𝟒𝟒
𝐡𝐡𝐡𝐡𝐡𝐡 = √𝟒𝟒𝟒𝟒
𝐡𝐡𝐡𝐡𝐡𝐡 = 𝟐𝟐√𝟏𝟏𝟏𝟏
Angle Measure
𝜶𝜶
𝜷𝜷
5.
𝟔𝟔
𝐬𝐬𝐬𝐬𝐬𝐬
𝐜𝐜
𝐜𝐜𝐜𝐜
𝟑𝟑√𝟏𝟏𝟏𝟏
=
𝟏𝟏𝟏𝟏
𝟐𝟐√𝟏𝟏𝟏𝟏
√𝟏𝟏𝟏𝟏
=
𝟏𝟏𝟏𝟏
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟐𝟐
=
√𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝟐𝟐
𝟔𝟔
=
𝟑𝟑√𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝐭𝐭
𝐭𝐭𝐭𝐭
𝟔𝟔
= 𝟑𝟑
𝟐𝟐
𝟐𝟐 𝟏𝟏
=
𝟔𝟔 𝟑𝟑
Jules thinks that if 𝜶𝜶 and 𝜷𝜷 are two different acute angle measures, then 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 ≠ 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷. Do you agree or disagree?
Explain.
I agree. If 𝜶𝜶 and 𝜷𝜷 are different acute angle measures, then either 𝜶𝜶 > 𝜷𝜷 or 𝜷𝜷 > 𝜶𝜶. A right triangle with acute
angle 𝜶𝜶 cannot be similar to a right triangle with acute angle 𝜷𝜷 (unless 𝜶𝜶 + 𝜷𝜷 = 𝟗𝟗𝟗𝟗) because the triangles fail the
AA criterion. If the triangles are not similar, then their corresponding sides are not in proportion, meaning their
within-figure ratios are not in proportion; therefore, 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 ≠ 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷. In the case where 𝜶𝜶 + 𝜷𝜷 = 𝟗𝟗𝟗𝟗, the given right
triangles are similar; however, 𝜶𝜶 and 𝜷𝜷 must be alternate acute angles, meaning 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷, and 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶,
but 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 ≠ 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷.
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6.
Given the triangle in the diagram, complete the following table.
Angle Measure
𝜶𝜶
𝜷𝜷
𝐬𝐬𝐬𝐬𝐬𝐬
𝟑𝟑√𝟔𝟔
𝟗𝟗
𝟑𝟑√𝟑𝟑
𝟗𝟗
𝐜𝐜𝐜𝐜𝐜𝐜
𝟑𝟑√𝟑𝟑
𝟗𝟗
𝟑𝟑√𝟔𝟔
𝟗𝟗
𝐭𝐭𝐭𝐭𝐭𝐭
𝟑𝟑√𝟔𝟔
𝟑𝟑√𝟑𝟑
𝟑𝟑√𝟑𝟑
𝟑𝟑√𝟔𝟔
Rewrite the values from the table in simplest terms.
Angle Measure
𝜶𝜶
𝜷𝜷
𝐬𝐬𝐬𝐬𝐬𝐬
√𝟔𝟔
𝟑𝟑
√𝟑𝟑
𝟑𝟑
𝐜𝐜𝐜𝐜𝐜𝐜
√𝟑𝟑
𝟑𝟑
√𝟔𝟔
𝟑𝟑
𝐭𝐭𝐭𝐭𝐭𝐭
√𝟔𝟔
= √𝟐𝟐
√𝟑𝟑
√𝟑𝟑 √𝟐𝟐
=
𝟐𝟐
√𝟔𝟔
Draw and label the sides and angles of a right triangle using the simplified values of the ratios 𝐬𝐬𝐬𝐬𝐬𝐬 and 𝐜𝐜𝐜𝐜𝐜𝐜. How is
the new triangle related to the original triangle?
The triangles are similar by SSS criterion because the new triangle has sides
that are
7.
𝟏𝟏
𝟑𝟑
of the length of their corresponding sides in the original triangle.
Given 𝐭𝐭𝐭𝐭𝐭𝐭 𝜶𝜶 and 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷, in simplest terms, find the missing side lengths of the right triangle if one leg of the triangle
has a length of 𝟒𝟒. Draw and label the sides and angles of the right triangle.
Angle Measure
𝜶𝜶
𝜷𝜷
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽
𝟏𝟏
𝟐𝟐
√𝟑𝟑
𝟐𝟐
√𝟑𝟑
𝟑𝟑
√𝟑𝟑
𝟐𝟐
𝟏𝟏
𝟐𝟐
√𝟑𝟑
The problem does not specify which leg is 𝟒𝟒, so there are two possible solutions to this problem. The values given in
the table do not represent the actual lengths of the sides of the triangles; however, they do represent the lengths of
the sides of a similar triangle, which is a 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 right triangle with side lengths 𝟏𝟏, 𝟐𝟐, and √𝟑𝟑.
Case 1: The short leg of the right triangle is 𝟒𝟒:
Lesson 26:
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Case 2: The long leg of the right triangle is 𝟒𝟒:
The Definition of Sine, Cosine, and Tangent
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GEOMETRY
8.
Eric wants to hang a rope bridge over a small ravine so that it is easier to cross. To hang the bridge, he needs to
know how much rope is needed to span the distance between two trees that are directly across from each other on
either side of the ravine. Help Eric devise a plan using sine, cosine, and tangent to determine the approximate
distance from tree A to tree B without having to cross the ravine.
Student solutions will vary. Possible solution:
If Eric walks a path parallel to the ravine to a
point 𝑷𝑷 at a convenient distance from 𝑨𝑨, he
could measure the angle formed by his line of
sight to both trees. Using the measured angle
and distance, he could use the value of the
tangent ratio of the angle to determine the
length of the opposite leg of the triangle. The
length of the opposite leg of the triangle
represents the distance between the two trees.
9.
A fisherman is at point 𝑭𝑭 on the open sea and has three favorite fishing locations. The locations are indicated by
points 𝑨𝑨, 𝑩𝑩, and 𝑪𝑪. The fisherman plans to sail from 𝑭𝑭 to 𝑨𝑨, then to 𝑩𝑩, then to 𝑪𝑪, and then back to 𝑭𝑭. If the
fisherman is 𝟏𝟏𝟏𝟏 miles from ����
𝑨𝑨𝑨𝑨, find the total distance that he will sail.
𝑭𝑭𝑭𝑭 = 𝟏𝟏𝟏𝟏 and can be considered the
adjacent side to the 𝟑𝟑𝟑𝟑° angle shown
in triangle 𝑨𝑨𝑨𝑨𝑨𝑨.
Using cosine:
𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 =
𝑨𝑨𝑨𝑨 =
𝟏𝟏𝟏𝟏
𝑨𝑨𝑨𝑨
𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑
𝑨𝑨𝑨𝑨 ≈ 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎
Using tangent:
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 =
𝑨𝑨𝑨𝑨
𝟏𝟏𝟏𝟏
𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑
𝑨𝑨𝑨𝑨 ≈ 𝟗𝟗. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
����
𝑷𝑷𝑷𝑷 is the leg opposite angle 𝑷𝑷𝑷𝑷𝑷𝑷 in triangle 𝑷𝑷𝑷𝑷𝑷𝑷 and has a degree measure of 𝟒𝟒𝟒𝟒. 𝟓𝟓.
Using tangent:
𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒𝟒𝟒. 𝟓𝟓 =
𝑷𝑷𝑷𝑷
𝟏𝟏𝟏𝟏
𝑷𝑷𝑷𝑷 = 𝟏𝟏𝟏𝟏 𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒𝟒𝟒. 𝟓𝟓
𝑷𝑷𝑷𝑷 ≈ 𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
The total distance that the fisherman will sail:
Using cosine:
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒. 𝟓𝟓 =
𝟏𝟏𝟏𝟏
𝑭𝑭𝑭𝑭
𝟏𝟏𝟏𝟏
𝐅𝐅𝐅𝐅 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒.𝟓𝟓
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 = 𝑨𝑨𝑨𝑨 + 𝑨𝑨𝑨𝑨 + 𝑷𝑷𝑷𝑷 + 𝑭𝑭𝑭𝑭
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
+ 𝟏𝟏𝟏𝟏 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 + 𝟏𝟏𝟏𝟏 𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒𝟒𝟒. 𝟓𝟓 +
𝐜𝐜𝐜𝐜 𝐬𝐬 𝟑𝟑𝟑𝟑
𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒.𝟓𝟓
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 ≈ 𝟓𝟓𝟓𝟓. 𝟕𝟕
𝑭𝑭𝑭𝑭 ≈ 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
The total distance that the fisherman will sail is approximately 𝟓𝟓𝟓𝟓. 𝟕𝟕 miles.
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Lesson 27: Sine and Cosine of Complementary Angles and
Special Angles
Student Outcomes


Students understand that if 𝛼𝛼 and 𝛽𝛽 are the measurements of complementary angles, then sin 𝛼𝛼 = cos 𝛽𝛽.
Students solve triangle problems using special angles.
Scaffolding:
Lesson Notes
Students examine the sine and cosine relationship more closely and find that the sine and
cosine of complementary angles are equal. Students become familiar with the values
associated with sine and cosine of special angles. Once familiar with these common
values, students use them to find unknown values in problems.
 Use the cutouts from
Lesson 21.
Classwork
Example 1 (8 minutes)
Students discover why cosine has the prefix co-. It may be necessary to remind students
why we know alpha and beta are complementary.
Example 1
If 𝜶𝜶 and 𝜷𝜷 are the measurements of complementary angles, then we are going to show that
𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷.
In right triangle 𝑨𝑨𝑨𝑨𝑨𝑨, the measurement of acute angle ∠𝑨𝑨 is denoted by 𝜶𝜶, and the measurement
of acute angle ∠𝑩𝑩 is denoted by 𝜷𝜷.
Determine the following values in the table:
MP.2
 If students are struggling
to see the connection, use
a right triangle with side
lengths 3, 4, and 5 to help
make the values of the
ratios more apparent.
𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶
𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷
𝐨𝐨𝐨𝐨𝐨𝐨 𝒂𝒂
=
𝐡𝐡𝐡𝐡𝐡𝐡 𝒄𝒄
𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷 =
𝐨𝐨𝐨𝐨𝐨𝐨 𝒃𝒃
=
𝐡𝐡𝐡𝐡𝐡𝐡 𝒄𝒄
𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶
𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶 =
𝐚𝐚𝐚𝐚𝐚𝐚 𝒃𝒃
=
𝐡𝐡𝐡𝐡𝐡𝐡 𝒄𝒄
values of sine and cosine
for the acute angles (by
them, “What do you
notice?”
students to write a letter
to a middle school student
explaining why the sine of
an angle is equal to the
cosine of its
complementary angle.
𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷
𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷 =
𝐚𝐚𝐚𝐚𝐚𝐚 𝒂𝒂
=
𝐡𝐡𝐡𝐡𝐡𝐡 𝒄𝒄
What can you conclude from the results?
Since the ratios for 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 and 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷 are the same, 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷, and the ratios for 𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶
and 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷 are the same; additionally, 𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶 = 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷. The sine of an angle is equal to the
cosine of its complementary angle, and the cosine of an angle is equal to the sine of its
complementary angle.
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
Therefore, we conclude for complementary angles 𝛼𝛼 and 𝛽𝛽 that sin 𝛼𝛼 = cos 𝛽𝛽, or in other words, when
0 < 𝜃𝜃 < 90 that sin(90 − 𝜃𝜃) = cos 𝜃𝜃, and sin 𝜃𝜃 = cos (90 − 𝜃𝜃). Any two angles that are complementary can
be realized as the acute angles in a right triangle. Hence, the co- prefix in cosine is a reference to the fact that
the cosine of an angle is the sine of its complement.
Exercises 1–3 (7 minutes)
Students apply what they know about the sine and cosine of complementary angles to solve for unknown angle values.
Exercises 1–3
1.
Consider the right triangle 𝑨𝑨𝑨𝑨𝑨𝑨 so that ∠𝑪𝑪 is a right angle, and the degree measures of ∠𝑨𝑨 and ∠𝑩𝑩 are 𝜶𝜶 and 𝜷𝜷,
respectively.
a.
Find 𝜶𝜶 + 𝜷𝜷.
𝟗𝟗𝟗𝟗°
b.
Use trigonometric ratios to describe
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨 =
c.
d.
e.
2.
Use trigonometric ratios to describe
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑩𝑩 =
𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨
, 𝐜𝐜𝐜𝐜𝐜𝐜 ∠𝑨𝑨 =
𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨
two different ways.
𝑨𝑨𝑨𝑨
two different ways.
𝑨𝑨𝑨𝑨
What can you conclude about 𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 and 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷?
𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷
What can you conclude about 𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶 and 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷?
𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶 = 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷
Find values for 𝜽𝜽 that make each statement true.
a.
b.
c.
d.
3.
𝑩𝑩𝑩𝑩
𝑩𝑩𝑩𝑩
, 𝐜𝐜𝐜𝐜𝐜𝐜 ∠𝑩𝑩 =
𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨
𝑩𝑩𝑩𝑩
𝑨𝑨𝑨𝑨
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = 𝐜𝐜𝐜𝐜𝐜𝐜 (𝟐𝟐𝟐𝟐)
𝜽𝜽 = 𝟔𝟔𝟔𝟔
𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
𝜽𝜽 = 𝟏𝟏𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = 𝐜𝐜𝐜𝐜𝐜𝐜 (𝜽𝜽 + 𝟏𝟏𝟏𝟏)
𝜽𝜽 = 𝟒𝟒𝟒𝟒
𝐬𝐬𝐬𝐬𝐬𝐬 (𝜽𝜽 − 𝟒𝟒𝟒𝟒) = 𝐜𝐜𝐜𝐜𝐜𝐜 (𝜽𝜽)
𝜽𝜽 = 𝟔𝟔𝟔𝟔. 𝟓𝟓
For what angle measurement must sine and cosine have the same value? Explain how you know.
Sine and cosine have the same value for 𝜽𝜽 = 𝟒𝟒𝟒𝟒. The sine of an angle is equal to the cosine of its complement.
Since the complement of 𝟒𝟒𝟒𝟒 is 𝟒𝟒𝟒𝟒, 𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒.
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Example 2 (8 minutes)
Students begin to examine special angles associated with sine and cosine, starting with the angle measurements of 0°
and 90°. Consider modeling this on the board by drawing a sketch of the following figure and using a meter stick to
represent 𝑐𝑐.
Example 2
What is happening to 𝒂𝒂 and 𝒃𝒃 as 𝜽𝜽 changes? What happens to 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 and
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽?



There are values for sine and cosine commonly known for certain angle measurements. Two such angle
measurements are when 𝜃𝜃 = 0° and 𝜃𝜃 = 90°.
To better understand sine and cosine values, imagine a right triangle whose hypotenuse has a fixed length 𝑐𝑐 of
1 unit. We illustrate this by imagining the hypotenuse as the radius of a circle, as in the image.
What happens to the value of the sine ratio as 𝜃𝜃 approaches 0°? Consider what is happening to the opposite
side, 𝑎𝑎.
With one end of the meter stick fixed at 𝐴𝐴, rotate it like the hands of a clock, and show how 𝑎𝑎 decreases as 𝜃𝜃 decreases.
Demonstrate the change in the triangle for each case.


Similarly, what happens to the value of the cosine ratio as 𝜃𝜃 approaches 0°? Consider what is happening to


Now, consider what happens to the value of the sine ratio as 𝜃𝜃 approaches 90°. Consider what is happening to
the opposite side, 𝑎𝑎.
𝑎𝑎
1
As 𝜃𝜃 gets closer to 90°, 𝑎𝑎 increases and becomes closer to 1. Since sin 𝜃𝜃 = , the value of sin 𝜃𝜃 is also
approaching 1.
What happens to the value of the cosine ratio as 𝜃𝜃 approaches 90°? Consider what is happening to the


𝑏𝑏
1
As 𝜃𝜃 gets closer to 0°, 𝑏𝑏 increases and becomes closer to 1. Since cos 𝜃𝜃 = , the value of cos 𝜃𝜃 is
approaching 1.


𝑎𝑎
1
As 𝜃𝜃 gets closer to 0°, 𝑎𝑎 decreases. Since sin 𝜃𝜃 = , the value of sin 𝜃𝜃 is also approaching 0.
𝑏𝑏
1
As 𝜃𝜃 gets closer to 90°, 𝑏𝑏 decreases and becomes closer to 0. Since cos 𝜃𝜃 = , the value of cos 𝜃𝜃 is
approaching 0.
Remember, because there are no right triangles with an acute angle of 0° or of 90°, in the above thought
experiment, we are really defining sin 0 = 0 and cos 0 = 1.
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
Similarly, we define sin 90 = 1 and cos 90 = 0; notice that this falls in line with our conclusion that the sine of
an angle is equal to the cosine of its complementary angle.
Example 3 (10 minutes)
Students examine the remaining special angles associated with sine and cosine in Example 3. Consider assigning parts
(b) and (c) to two halves of the class and having students present a share-out of their findings.
Example 3
There are certain special angles where it is possible to give the exact value of sine and cosine. These are the angles that
measure 𝟎𝟎°, 𝟑𝟑𝟑𝟑°, 𝟒𝟒𝟒𝟒°, 𝟔𝟔𝟔𝟔°, and 𝟗𝟗𝟗𝟗°; these angle measures are frequently seen.
You should memorize the sine and cosine of these angles with quick recall just as you did your arithmetic facts.
a.
MP.7
Learn the following sine and cosine values of the key angle measurements.
𝜽𝜽
𝟎𝟎°
Sine
𝟎𝟎
Cosine
𝟏𝟏
𝟑𝟑𝟑𝟑°
𝟏𝟏
𝟐𝟐
√𝟑𝟑
𝟐𝟐
𝟒𝟒𝟒𝟒°
𝟔𝟔𝟔𝟔°
√𝟐𝟐
𝟐𝟐
√𝟑𝟑
𝟐𝟐
𝟏𝟏
𝟐𝟐
√𝟐𝟐
𝟐𝟐
𝟗𝟗𝟗𝟗°
𝟏𝟏
𝟎𝟎
We focus on an easy way to remember the entries in the table. What do you notice about the table values?
The entries for cosine are the same as the entries for sine but in the reverse order.
This is easily explained because the pairs (𝟎𝟎, 𝟗𝟗𝟗𝟗), (𝟑𝟑𝟑𝟑, 𝟔𝟔𝟔𝟔), and (𝟒𝟒𝟒𝟒, 𝟒𝟒𝟒𝟒) are the measures of complementary angles.
So, for instance, 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔.
√𝟎𝟎 √𝟏𝟏 √𝟐𝟐 √𝟑𝟑 √𝟒𝟒
𝟏𝟏 �𝟐𝟐 �𝟑𝟑
, , 𝟏𝟏 may be easier to remember as the sequence
,
,
,
, .
𝟐𝟐 𝟐𝟐 𝟐𝟐
𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐
The sequence 𝟎𝟎, ,
b.
△ 𝑨𝑨𝑨𝑨𝑨𝑨 is equilateral, with side length 𝟐𝟐; 𝑫𝑫 is the midpoint of side ����
𝑨𝑨𝑨𝑨. Label all side lengths and angle
measurements for △ 𝑨𝑨𝑨𝑨𝑨𝑨. Use your figure to determine the sine and cosine of 𝟑𝟑𝟑𝟑 and 𝟔𝟔𝟔𝟔.
Provide students with a hint, if necessary, by suggesting they construct the angle bisector of ∠𝐵𝐵, which is also the
���� .
altitude to 𝐴𝐴𝐴𝐴
𝐬𝐬𝐬𝐬𝐬𝐬 (𝟑𝟑𝟑𝟑) =
𝑨𝑨𝑨𝑨 𝟏𝟏
𝑩𝑩𝑩𝑩 √𝟑𝟑
𝑩𝑩𝑩𝑩 √𝟑𝟑
𝑨𝑨𝑨𝑨 𝟏𝟏
= , 𝐜𝐜𝐜𝐜𝐜𝐜 (𝟑𝟑𝟑𝟑) =
=
, 𝐬𝐬𝐬𝐬𝐬𝐬 (𝟔𝟔𝟔𝟔) =
=
, 𝐜𝐜𝐜𝐜𝐜𝐜 (𝟔𝟔𝟔𝟔) =
=
𝟐𝟐
𝟐𝟐
𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨 𝟐𝟐
𝑨𝑨𝑨𝑨 𝟐𝟐
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c.
Draw an isosceles right triangle with legs of length 𝟏𝟏. What are the measures of the acute angles of the
triangle? What is the length of the hypotenuse? Use your triangle to determine sine and cosine of the acute
angles.
𝐬𝐬𝐬𝐬𝐬𝐬 (𝟒𝟒𝟒𝟒) =
𝑨𝑨𝑨𝑨
𝟏𝟏
𝟏𝟏
𝑩𝑩𝑩𝑩
=
, 𝐜𝐜𝐜𝐜𝐜𝐜 (𝟒𝟒𝟒𝟒) =
=
𝑨𝑨𝑨𝑨 √𝟐𝟐
𝑨𝑨𝑨𝑨 √𝟐𝟐
Parts (b) and (c) demonstrate how the sine and cosine values of the mentioned special angles can be found. These
triangles are common to trigonometry; we refer to the triangle in part (b) as a 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 triangle and the triangle in part
(c) as a 𝟒𝟒𝟓𝟓–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 triangle.

Remind students that the values of the sine and cosine ratios of triangles similar to each of these are the same.
Highlight the length ratios for 30–60–90 and 45–45–90 triangles. Consider using a setup like the table below to begin
the conversation. Ask students to determine side lengths of three different triangles similar to each of the triangles
provided above. Remind them that the scale factor determines side length. Then, have them generalize the length
relationships.
Scaffolding:
𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 Triangle,
side length ratio 𝟏𝟏: 𝟐𝟐: √𝟑𝟑
𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 Triangle,
side length ratio 𝟏𝟏: 𝟏𝟏: √𝟐𝟐
𝟑𝟑: 𝟔𝟔: 𝟑𝟑√𝟑𝟑
𝟑𝟑: 𝟑𝟑: 𝟑𝟑√𝟐𝟐
𝟐𝟐: 𝟒𝟒: 𝟐𝟐√𝟑𝟑
𝟒𝟒: 𝟖𝟖: 𝟒𝟒√𝟑𝟑
𝒙𝒙: 𝟐𝟐𝒙𝒙: 𝒙𝒙√𝟑𝟑
𝟐𝟐: 𝟐𝟐: 𝟐𝟐√𝟐𝟐
𝟒𝟒: 𝟒𝟒: 𝟒𝟒√𝟐𝟐
𝒙𝒙: 𝒙𝒙: 𝒙𝒙√𝟐𝟐
 For the 1: 2: √3 triangle,
students may develop the
misconception that the
last value is the length of
the hypotenuse, the
longest side of the right
triangle. Help students
correct this misconception
by comparing √3 and √4
to show that √4 > √3,
and √4 = 2, so 2 > √3.
 The ratio 1: 2: √3 is easier
to remember because of
the numbers 1, 2, 3.
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Exercises 4–5 (5 minutes)
Exercises 4–5
4.
Find the missing side lengths in the triangle.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 =
5.
𝒂𝒂 𝟏𝟏
𝟑𝟑
= , 𝒂𝒂 =
𝟑𝟑 𝟐𝟐
𝟐𝟐
𝒃𝒃 √𝟑𝟑
𝟑𝟑√𝟑𝟑
=
, 𝒃𝒃 =
𝟑𝟑
𝟐𝟐
𝟐𝟐
Find the missing side lengths in the triangle.
𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 =
𝟑𝟑 √𝟑𝟑
𝟔𝟔
=
, 𝒄𝒄 =
= 𝟐𝟐√𝟑𝟑
𝟐𝟐
𝒄𝒄
√𝟑𝟑
𝒂𝒂
𝟐𝟐√𝟑𝟑
=
𝟏𝟏
, 𝒂𝒂 = √𝟑𝟑
𝟐𝟐
Closing (2 minutes)
Ask students to respond to these questions about the key ideas of the lesson independently in writing, with a partner, or
as a class.

What is remarkable about the sine and cosine of a pair of angles that are complementary?
The sine of an angle is equal to the cosine of its complementary angle, and the cosine of an angle is
equal to the sine of its complementary angle.


Why is sin 90 = 1? Similarly, why is sin 0 = 0, cos 90 = 0, and cos 0 = 1?
We can see that sin 𝜃𝜃 approaches 1 as 𝜃𝜃 approaches 90. The same is true for the other sine and cosine
values for 0 and 90.


What do you notice about the sine and cosine of the following special angle values?
The entries for cosine are the same as the entries for sine, but the values are in reverse order. This is
explained by the fact the special angles can be paired up as complements, and we already know that
the sine and cosine values of complementary angles are equal.

Sine
𝟎𝟎°
𝜽𝜽
0
Cosine
1
𝟑𝟑𝟑𝟑°
𝟒𝟒𝟒𝟒°
𝟔𝟔𝟔𝟔°
𝟗𝟗𝟗𝟗°
√3
2
√2
2
1
2
0
1
2
√2
2
√3
2
1
Exit Ticket (5 minutes)
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Name
Date
Lesson 27: Sine and Cosine of Complementary Angles and Special
Angles
Exit Ticket
1.
2.
Find the values for 𝜃𝜃 that make each statement true.
a.
sin 𝜃𝜃 = cos 32
b.
cos 𝜃𝜃 = sin(𝜃𝜃 + 20)
△ 𝐿𝐿𝐿𝐿𝐿𝐿 is a 30–60–90 right triangle. Find the unknown lengths 𝑥𝑥 and 𝑦𝑦.
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Exit Ticket Sample Solutions
1.
Find the values for 𝜽𝜽 that make each statement true.
a.
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑
𝜽𝜽 = 𝟗𝟗𝟗𝟗 − 𝟑𝟑𝟑𝟑
𝜽𝜽 = 𝟓𝟓𝟓𝟓
b.
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 = 𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽 + 𝟐𝟐𝟐𝟐)
𝐬𝐬𝐬𝐬𝐬𝐬(𝟗𝟗𝟗𝟗 − 𝜽𝜽) = 𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽 + 𝟐𝟐𝟐𝟐)
𝟗𝟗𝟗𝟗 − 𝜽𝜽 = 𝜽𝜽 + 𝟐𝟐𝟐𝟐
𝟕𝟕𝟕𝟕 = 𝟐𝟐𝜽𝜽
𝟑𝟑𝟑𝟑 = 𝜽𝜽
2.
△ 𝑳𝑳𝑳𝑳𝑳𝑳 is a 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 right triangle. Find the unknown lengths 𝒙𝒙 and 𝒚𝒚.
𝟏𝟏
√𝟑𝟑
𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 =
𝟐𝟐
𝟐𝟐
𝟏𝟏 𝒙𝒙
√𝟑𝟑 𝒚𝒚
=
=
𝟐𝟐 𝟕𝟕
𝟐𝟐
𝟕𝟕
𝟕𝟕 = 𝟐𝟐𝒙𝒙
𝟕𝟕√𝟑𝟑 = 𝟐𝟐𝒚𝒚
𝟕𝟕
= 𝒙𝒙
𝟕𝟕√𝟑𝟑
𝟐𝟐
𝒚𝒚 =
𝟐𝟐
Problem Set Sample Solutions
1.
Find the value of 𝜽𝜽 that makes each statement true.
a.
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = 𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽 + 𝟑𝟑𝟑𝟑)
𝐜𝐜𝐜𝐜𝐜𝐜(𝟗𝟗𝟗𝟗 − 𝜽𝜽) = 𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽 + 𝟑𝟑𝟑𝟑)
𝟗𝟗𝟗𝟗 − 𝜽𝜽 = 𝜽𝜽 + 𝟑𝟑𝟑𝟑
𝟓𝟓𝟓𝟓 = 𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐 = 𝜽𝜽
b.
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 = 𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽 − 𝟑𝟑𝟑𝟑)
𝐬𝐬𝐬𝐬𝐬𝐬(𝟗𝟗𝟗𝟗 − 𝜽𝜽) = 𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽 − 𝟑𝟑𝟑𝟑)
𝟗𝟗𝟗𝟗 − 𝜽𝜽 = 𝜽𝜽 − 𝟑𝟑𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟐𝟐𝜽𝜽
𝟔𝟔𝟔𝟔 = 𝜽𝜽
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c.
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = 𝐜𝐜𝐜𝐜𝐜𝐜(𝟑𝟑𝟑𝟑 + 𝟐𝟐𝟐𝟐)
𝐜𝐜𝐜𝐜𝐜𝐜(𝟗𝟗𝟗𝟗 − 𝜽𝜽) = 𝐜𝐜𝐜𝐜𝐜𝐜(𝟑𝟑𝜽𝜽 + 𝟐𝟐𝟐𝟐)
𝟗𝟗𝟗𝟗 − 𝜽𝜽 = 𝟑𝟑𝜽𝜽 + 𝟐𝟐𝟐𝟐
𝟕𝟕𝟕𝟕 = 𝟒𝟒𝜽𝜽
𝟏𝟏𝟏𝟏. 𝟓𝟓 = 𝜽𝜽
d.
𝜽𝜽
𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 � + 𝟏𝟏𝟏𝟏� = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
𝜽𝜽
𝐬𝐬𝐬𝐬𝐬𝐬 � + 𝟏𝟏𝟏𝟏� = 𝐬𝐬𝐬𝐬𝐬𝐬(𝟗𝟗𝟗𝟗 − 𝜽𝜽)
𝟑𝟑
𝜽𝜽
+ 𝟏𝟏𝟏𝟏 = 𝟗𝟗𝟗𝟗 − 𝜽𝜽
𝟑𝟑
𝟒𝟒𝜽𝜽
= 𝟖𝟖𝟖𝟖
𝟑𝟑
𝜽𝜽 = 𝟔𝟔𝟔𝟔
2.
a.
Make a prediction about how the sum 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔 will relate to the sum 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑.
Answers will vary; however, some students may believe that the sums will be equal. This is explored in
Problems 3 through 5.
b.
Use the sine and cosine values of special angles to find the sum: 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 =
𝟏𝟏
𝟏𝟏
𝟏𝟏 𝟏𝟏
and 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔 = Therefore, 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔 = + = 𝟏𝟏.
𝟐𝟐
𝟐𝟐
𝟐𝟐 𝟐𝟐
Alternative strategy:
𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔° = 𝐬𝐬𝐬𝐬𝐬𝐬 (𝟗𝟗𝟗𝟗 − 𝟔𝟔𝟔𝟔)° = 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑°
𝟏𝟏
𝟐𝟐
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑° + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔° = 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑° + 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑° = 𝟐𝟐(𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑°) = 𝟐𝟐 � � = 𝟏𝟏
c.
Find the sum: 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 =
d.
�𝟑𝟑
𝟐𝟐
and 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 =
�𝟑𝟑
𝟐𝟐
Therefore, 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 =
�𝟑𝟑
𝟐𝟐
+
�𝟑𝟑
𝟐𝟐
= √𝟑𝟑.
Was your prediction a valid prediction? Explain why or why not.
3.
Langdon thinks that the sum 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 + 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 is equal to 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔. Do you agree with Langdon? Explain what this
means about the sum of the sines of angles.
I disagree. Explanations may vary. It was shown in the solution to Problem 3 that 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 + 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 = 𝟏𝟏, and it is
known that 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 =
the angles.
Lesson 27:
GEO-M2-TE-1.3.0-07.2015
�𝟑𝟑
𝟐𝟐
≠ 𝟏𝟏. This shows that the sum of the sines of angles is not equal to the sine of the sum of
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GEOMETRY
4.
A square has side lengths of 𝟕𝟕√𝟐𝟐. Use sine or cosine to find the length of the diagonal of the square. Confirm your
The diagonal of a square cuts the square into two congruent
𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 right triangles. Let 𝒅𝒅 represent the length of the
diagonal of the square:
𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒 =
√𝟐𝟐
𝟐𝟐
√𝟐𝟐 𝟕𝟕√𝟐𝟐
=
𝟐𝟐
𝒅𝒅
𝒅𝒅√𝟐𝟐 = 𝟏𝟏𝟏𝟏√𝟐𝟐
𝒅𝒅 = 𝟏𝟏𝟏𝟏
Confirmation using Pythagorean theorem:
𝟐𝟐
𝟐𝟐
�𝟕𝟕√𝟐𝟐� + �𝟕𝟕√𝟐𝟐� = 𝐡𝐡𝐲𝐲𝐩𝐩𝟐𝟐
𝟗𝟗𝟗𝟗 + 𝟗𝟗𝟗𝟗 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
√𝟏𝟏𝟏𝟏𝟏𝟏 = 𝐡𝐡𝐡𝐡𝐡𝐡
𝟏𝟏𝟏𝟏 = 𝐡𝐡𝐡𝐡𝐡𝐡
5.
Given an equilateral triangle with sides of length 𝟗𝟗, find the length of the altitude. Confirm your answer using the
Pythagorean theorem.
An altitude drawn within an equilateral triangle cuts the
equilateral triangle into two congruent 𝟑𝟑𝟑𝟑–𝟔𝟔𝟔𝟔–𝟗𝟗𝟗𝟗 right
triangles. Let 𝒉𝒉 represent the length of the altitude:
√𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 =
𝟐𝟐
√𝟑𝟑 𝒉𝒉
=
𝟐𝟐
𝟗𝟗
𝟗𝟗√𝟑𝟑 = 𝟐𝟐𝒉𝒉
𝟗𝟗√𝟑𝟑
= 𝒉𝒉
𝟐𝟐
The altitude of the triangle has a length of
Confirmation using Pythagorean theorem:
𝟗𝟗√𝟑𝟑
𝟐𝟐
.
𝟗𝟗 𝟐𝟐
� � + 𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 = 𝟗𝟗𝟐𝟐
𝟐𝟐
𝟖𝟖𝟖𝟖
+ 𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 = 𝟖𝟖𝟖𝟖
𝟒𝟒
𝟐𝟐𝟐𝟐𝟐𝟐
𝐥𝐥𝐥𝐥𝐠𝐠 𝟐𝟐 =
𝟒𝟒
𝟐𝟐𝟐𝟐𝟐𝟐
𝐥𝐥𝐥𝐥𝐥𝐥 = �
𝟒𝟒
√𝟐𝟐𝟐𝟐𝟐𝟐
𝟐𝟐
𝟗𝟗√𝟑𝟑
𝐥𝐥𝐥𝐥𝐥𝐥 =
𝟐𝟐
𝐥𝐥𝐥𝐥𝐥𝐥 =
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Lesson 28
M2
GEOMETRY
Lesson 28: Solving Problems Using Sine and Cosine
Student Outcomes



Students use graphing calculators to find the values of sin 𝜃𝜃 and cos 𝜃𝜃 for 𝜃𝜃 between 0° and 90°.
Students solve for missing sides of a right triangle given the length of one side and the measure of one of the
acute angles.
Students find the length of the base of a triangle with acute base angles given the lengths of the other two
sides and the measure of each of the base angles.
Lesson Notes
Students need access to a graphing calculator to calculate the sine and cosine of given angle measures. It is likely
necessary to show students how to set the calculator in degree mode and to perform these operations. Encourage
students to make one computation on the calculator and then approximate their answer as opposed to making
intermediate approximations throughout the solution process. Intermediate approximations lead to a less accurate
answer than doing the approximation once.
Classwork
Exercises 1–4 (12 minutes)
Allow students to work in pairs to complete Exercise 1. It may be necessary to demonstrate how to use a graphing
calculator to perform the following calculations. Ensure that all calculators are in degree mode, not radian. Consider
telling students that radian is a measure they encounter in Module 5 and use in Algebra II. For now, the unit of angle
measure is degree. After completing the exercises, debrief by having students share their explanations in Exercise 4.
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GEOMETRY
Exercises 1–4
1.
a.
The bus drops you off at the corner of H Street and 1st Street, approximately 𝟑𝟑𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. from school. You plan to
walk to your friend Janneth’s house after school to work on a project. Approximately how many feet will you
have to walk from school to Janneth’s house? Round your answer to the nearest foot. (Hint: Use the ratios
you developed in Lesson 25.)
Let 𝒙𝒙 represent the distance from school to Janneth’s house.
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
𝟑𝟑𝟑𝟑𝟑𝟑
𝟓𝟓.𝟑𝟑
𝐨𝐨𝐨𝐨𝐨𝐨
, then 𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒 = 𝟓𝟓.𝟑𝟑
. Then,
=
and
𝐡𝐡𝐡𝐡𝐡𝐡
𝟖𝟖
𝒙𝒙
𝟖𝟖
𝒙𝒙 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 …
The distance I will have to walk from school to Janneth’s house is
approximately 𝟒𝟒𝟒𝟒𝟒𝟒 𝒇𝒇𝒇𝒇.
b.
In real life, it is unlikely that you would calculate the distance between school and Janneth’s house in this
manner. Describe a similar situation in which you might actually want to determine the distance between
two points using a trigonometric ratio.
Accept any reasonable responses. Some may include needing to calculate the distance to determine if a
vehicle has enough fuel to make the trip or the need to determine the length prior to attempting the walk
because a friend is on crutches and cannot easily get from one location to the next when the distance is too
long.
2.
Use a calculator to find the sine and cosine of 𝜽𝜽. Give your answer rounded to the ten-thousandth place.
𝜽𝜽
𝟎𝟎
𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐
𝟎𝟎
𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
𝟏𝟏
𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
MP.8
3.
𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝟑𝟑𝟑𝟑
𝟏𝟏
= 𝟎𝟎. 𝟓𝟓
𝟐𝟐
𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝟒𝟒𝟒𝟒
𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝟓𝟓𝟓𝟓
𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
𝟔𝟔𝟔𝟔
𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝟏𝟏
= 𝟎𝟎. 𝟓𝟓
𝟐𝟐
𝟕𝟕𝟕𝟕
𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝟖𝟖𝟖𝟖
𝟗𝟗𝟗𝟗
𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟎𝟎
𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝟏𝟏
What do you notice about the numbers in the row 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 compared with the numbers in the row 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽?
The numbers are the same but reversed in order.
4.
Provide an explanation for what you noticed in Exercise 2.
The pattern exists because the sine and cosine of complementary angles are equal.
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Example 1 (8 minutes)
Students find the missing side length of a right triangle using sine and cosine.
Example 1
Find the values of 𝒂𝒂 and 𝒃𝒃.

Now that we can calculate the sine and cosine of a given angle using a calculator, we can use the decimal value
of the ratio to determine the unknown side length of a triangle.
Consider the following triangle.

What can we do to find the length of side 𝑎𝑎?


Let’s begin by using the sin 40. We expect sin 40 =



We can find the sin 40 or cos 50.
By definition of sine; sin 𝜃𝜃 =
opp
.
hyp
𝑎𝑎
. Why?
26
To calculate the length of 𝑎𝑎, we must determine the value of 26 sin 40 because 𝑎𝑎 = 26 sin 40. We will round
our answer to two decimal places.
Using the decimal approximation of sin 40 ≈ 0.6428, we can write
26(0.6428) ≈ 𝑎𝑎

16.71 ≈ 𝑎𝑎
Now let’s use cos 50, which is approximately 0.6428. What do you expect the result to be? Explain.

I expect the result to be the same. Since the approximation of sin 40 is equal to the approximation of
cos 50, the computation should be the same.
Note that students may say that sin 40 = cos 50. Ensure that students know that once decimal approximations are used
in place of the functions, they are no longer looking at two quantities that are equal because the decimals are
approximations. To this end, ask students to recall that in Exercise 1 they were only taking the first four decimal digits of
the number; that is, they are using approximations of those values. Therefore, they cannot explicitly claim that
sin 40 = cos 50, rather that their approximations are extremely close in value to one another.
If necessary, show the computation below that verifies the claim made above.
cos 50 =
𝑎𝑎
26
26 cos 50 = 𝑎𝑎
26(0.6428) ≈ 𝑎𝑎
16.71 ≈ 𝑎𝑎
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GEOMETRY

Now, calculate the length of side 𝑏𝑏.


Side 𝑏𝑏 can be determined using sin 50 or cos 40.
26(0.7660) ≈ 𝑏𝑏
19.92 ≈ 𝑏𝑏
Could we have used another method to determine the length of side 𝑏𝑏?

Yes. Because this is a right triangle and two sides are known, we could use the Pythagorean theorem to
determine the length of the third side.
The points below are to make clear that the calculator gives approximations of the ratios we desire when using
trigonometric functions.

When we use a calculator to compute, what we get is a decimal approximation of the ratio
𝑎𝑎
26
. Our calculators
are programmed to know which number 𝑎𝑎 is needed, relative to 26, so that the value of the ratio

𝑎𝑎
26
is equal
𝑎𝑎
and sin 40 ≈ 0.6428. Our calculators give us the number 𝑎𝑎
to the value of sin 40. For example, sin 40 =
26
that, when divided by 26, is closest to the actual value of sin 40.
Here is a simpler example illustrating this fact. Consider a right triangle with an acute angle of 30° and
𝑎𝑎
9
hypotenuse length of 9 units. Then, sin 30 = . We know that sin 30 =
find the number 𝑎𝑎 so that
𝑎𝑎
9
=
1
2
= 0.5, which is 𝑎𝑎 = 4.5.
1
= 0.5. What our calculators do is
2
Exercise 5 (5 minutes)
Students complete Exercise 5 independently. All students should be able to complete part
(a) in the allotted time. Consider assigning part (b) to only those students who finish part
(a) quickly. Once completed, have students share their solutions with the class.
Exercise 5
5.
A shipmate set a boat to sail exactly 𝟐𝟐𝟐𝟐° NE from the dock. After traveling 𝟏𝟏𝟏𝟏𝟏𝟏 miles, the
shipmate realized he had misunderstood the instructions from the captain; he was
supposed to set sail going directly east!
Scaffolding:
summarize the situation
with a partner.
 English language learners
may benefit from labeling
the horizontal distance E
for east and the vertical
distance S for south.
 Consider simplifying the
problem by drawing only
the triangle and labeling
the measures of the angle
and the hypotenuse and
find the unknown lengths.
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a.
How many miles will the shipmate have to travel directly south before he is directly east of the dock?
Let 𝑺𝑺 represent the distance they traveled directly south.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟐𝟐 =
𝑺𝑺
𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟐𝟐 = 𝑺𝑺
𝟓𝟓𝟓𝟓. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 … = 𝑺𝑺
He traveled approximately 𝟓𝟓𝟓𝟓 𝐦𝐦𝐦𝐦. south.
b.
How many extra miles does the shipmate travel by going the wrong direction compared to going directly
Solutions may vary. Some students may use the Pythagorean theorem while others may use the cosine
function. Both are acceptable strategies. If students use different strategies, make sure to share them with
the class, and discuss the benefits of each.
Let 𝑬𝑬 represent the distance the boat is when it is directly east of the dock.
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 =
𝑬𝑬
𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 = 𝑬𝑬
𝟏𝟏𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 … = 𝑬𝑬
𝟏𝟏𝟏𝟏𝟏𝟏 ≈ 𝑬𝑬
The total distance traveled by the boat is 𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟓𝟓𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏. They ended up exactly 𝟏𝟏𝟏𝟏𝟏𝟏 miles east of the
dock. 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟔𝟔𝟔𝟔, so they traveled an extra 𝟔𝟔𝟔𝟔 miles.
Example 2 (8 minutes)
Students find the missing side length of a triangle using sine and cosine.
Example 2
Johanna borrowed some tools from a friend so that she could precisely, but not exactly, measure the corner space in her
backyard to plant some vegetables. She wants to build a fence to prevent her dog from digging up the seeds that she
plants. Johanna returned the tools to her friend before making the most important measurement: the one that would
give the length of the fence!
Johanna decided that she could just use the Pythagorean theorem to find the length of the fence she would need. Is the
Pythagorean theorem applicable in this situation? Explain.
MP.1
No The corner of her backyard is not a 𝟗𝟗𝟗𝟗° angle; therefore, the Pythagorean theorem cannot be applied in this situation.
The Pythagorean theorem will, however, provide an approximation since the given angle has a measure that is close to
𝟗𝟗𝟗𝟗°.
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
What can we do to help Johanna figure out the length of fence she needs?
MP.1 Provide time for students to discuss this in pairs or small groups. Allow them to make sense of the problem and
persevere in solving it. It may be necessary to guide their thinking using the prompts below.



If we dropped an altitude from the angle with measure 95°, could that help? How?
Would we be able to use the Pythagorean theorem now? Explain.
If we denote the side opposite the 95° angle as 𝑥𝑥 and 𝑦𝑦, as shown, can we use what we know about sine and
cosine? Explain.

The missing side length is equal to 𝑥𝑥 + 𝑦𝑦. The length 𝑥𝑥 is equal to 100 cos 35, and the length 𝑦𝑦 is equal
to 74.875 cos 50. Therefore, the length of
𝑥𝑥 + 𝑦𝑦 = 100 cos 35 + 74.875 cos 50 ≈ 81.92 + 48.12872 ≈ 130.05.
Note: The Pythagorean theorem provides a reasonable approximation of 124.93.
Exercise 6 (4 minutes)
Students complete Exercise 6 independently.
Exercise 6
6.
The measurements of the triangle shown below are rounded to the nearest hundredth. Calculate the missing side
length to the nearest hundredth.
Drop an altitude from the angle that measures 𝟗𝟗𝟗𝟗°.
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Then, the length of the missing side is 𝒙𝒙 + 𝒚𝒚, which can be found by
𝟒𝟒. 𝟎𝟎𝟎𝟎 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 + 𝟑𝟑. 𝟖𝟖𝟖𝟖 𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒 ≈ 𝟑𝟑. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟐𝟐. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 = 𝟔𝟔. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 ≈ 𝟔𝟔. 𝟎𝟎𝟎𝟎.
Closing (3 minutes)
Ask students to discuss the answers to the following questions with a partner, and then select students to share with the
class. For the first question, elicit as many acceptable responses as possible.


Explain how to find the unknown length of a side of a right triangle.

If two sides are known, then the Pythagorean theorem can be used to determine the length of the third
side.

If one side is known and the measure of one of the acute angles is known, then sine, cosine, or tangent
can be used.

If the triangle is known to be similar to another triangle where the side lengths are given, then
corresponding ratios or knowledge of the scale factor can be used to determine the unknown length.

Direct measurement can be used.
Explain when and how you can find the unknown length of a side of a triangle that does not have a right angle.

You can find the length of an unknown side length of a triangle when you know two of the side lengths
and the missing side is between two acute angles. Split the triangle into two right triangles, and find
the lengths of two pieces of the missing side.
Exit Ticket (5 minutes)
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Name
Date
Lesson 28: Solving Problems Using Sine and Cosine
Exit Ticket
1.
Given right triangle 𝐴𝐴𝐴𝐴𝐴𝐴 with hypotenuse 𝐴𝐴𝐴𝐴 = 8.5 and 𝑚𝑚∠𝐴𝐴 = 55°, find 𝐴𝐴𝐴𝐴 and 𝐵𝐵𝐵𝐵 to the nearest hundredth.
2.
Given triangle 𝐷𝐷𝐷𝐷𝐷𝐷, 𝑚𝑚∠𝐷𝐷 = 22°, 𝑚𝑚∠𝐹𝐹 = 91°, 𝐷𝐷𝐷𝐷 = 16.55, and 𝐸𝐸𝐸𝐸 = 6.74, find 𝐷𝐷𝐷𝐷 to the nearest hundredth.
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Exit Ticket Sample Solutions
1.
Given right triangle 𝑨𝑨𝑨𝑨𝑨𝑨 with hypotenuse 𝑨𝑨𝑨𝑨 = 𝟖𝟖. 𝟓𝟓 and 𝒎𝒎∠𝑨𝑨 = 𝟓𝟓𝟓𝟓°, find 𝑨𝑨𝑨𝑨 and 𝑩𝑩𝑩𝑩 to the nearest hundredth.
𝑩𝑩𝑩𝑩 = 𝟖𝟖. 𝟓𝟓(𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓)
𝑩𝑩𝑩𝑩 ≈ 𝟔𝟔. 𝟗𝟗𝟗𝟗
𝑨𝑨𝑨𝑨 = 𝟖𝟖. 𝟓𝟓 (𝐜𝐜𝐜𝐜𝐜𝐜 𝟓𝟓𝟓𝟓)
𝑨𝑨𝑨𝑨 ≈ 𝟒𝟒. 𝟖𝟖𝟖𝟖
2.
Given triangle 𝑫𝑫𝑫𝑫𝑫𝑫, 𝒎𝒎∠𝑫𝑫 = 𝟐𝟐𝟐𝟐°, 𝒎𝒎∠𝑭𝑭 = 𝟗𝟗𝟗𝟗°, 𝑫𝑫𝑫𝑫 = 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓, and 𝑬𝑬𝑬𝑬 = 𝟔𝟔. 𝟕𝟕𝟕𝟕, find 𝑫𝑫𝑫𝑫 to the nearest hundredth.
���� at point 𝑷𝑷. Cosines can be used on angles 𝑫𝑫 and 𝑬𝑬 to determine the lengths of �����
Draw altitude from 𝑭𝑭 to 𝑫𝑫𝑫𝑫
𝑫𝑫𝑫𝑫 and
����, which together compose 𝑫𝑫𝑫𝑫
����.
𝑷𝑷𝑷𝑷
𝑷𝑷𝑷𝑷 = 𝟔𝟔. 𝟕𝟕𝟕𝟕(𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)
𝑷𝑷𝑷𝑷 ≈ 𝟐𝟐. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
𝑫𝑫𝑫𝑫 = 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓(𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐)
𝑫𝑫𝑫𝑫 ≈ 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝑫𝑫𝑫𝑫 = 𝑫𝑫𝑫𝑫 + 𝑷𝑷𝑷𝑷
𝑫𝑫𝑫𝑫 ≈ 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 + 𝟐𝟐. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
𝑫𝑫𝑫𝑫 ≈ 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗
Note to teacher: Answers of 𝑫𝑫𝑫𝑫 ≈ 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗 result from rounding to the nearest hundredth too early in the problem.
Problem Set Sample Solutions
1.
Given right triangle 𝑮𝑮𝑮𝑮𝑮𝑮, with right angle at 𝑯𝑯, 𝑮𝑮𝑮𝑮 = 𝟏𝟏𝟏𝟏. 𝟐𝟐, and 𝒎𝒎∠𝑮𝑮 = 𝟐𝟐𝟐𝟐°, find the measures of the remaining
sides and angle to the nearest tenth.
𝟏𝟏𝟏𝟏. 𝟐𝟐
𝑮𝑮𝑮𝑮
𝟏𝟏𝟏𝟏. 𝟐𝟐
𝑮𝑮𝑮𝑮 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 =
𝑮𝑮𝑮𝑮 ≈ 𝟏𝟏𝟏𝟏. 𝟖𝟖
𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐 =
𝑰𝑰𝑰𝑰
𝟏𝟏𝟏𝟏. 𝟐𝟐
𝑰𝑰𝑰𝑰 = 𝟏𝟏𝟏𝟏. 𝟐𝟐 𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐
𝑰𝑰𝑰𝑰 ≈ 𝟔𝟔. 𝟓𝟓
𝟐𝟐𝟐𝟐° + 𝒎𝒎∠𝑰𝑰 = 𝟗𝟗𝟗𝟗°
𝒎𝒎∠𝑰𝑰 = 𝟔𝟔𝟔𝟔°
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2.
The Occupational Safety and Health Administration (OSHA) provides standards for safety at the
workplace. A ladder is leaned against a vertical wall according to OSHA standards and forms an
angle of approximately 𝟕𝟕𝟕𝟕° with the floor.
a.
If the ladder is 𝟐𝟐𝟐𝟐 𝐟𝐟𝐭𝐭. long, what is the distance from the base of the ladder to the base
of the wall?
Let 𝒃𝒃 represent the distance of the base of the ladder from the wall in feet.
𝒃𝒃 = 𝟐𝟐𝟐𝟐(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒃𝒃 ≈ 𝟔𝟔. 𝟓𝟓
The base of the ladder is approximately 𝟔𝟔 𝐟𝐟𝐟𝐟. 𝟔𝟔 𝐢𝐢𝐢𝐢. from the wall.
b.
How high on the wall does the ladder make contact?
Let 𝒉𝒉 represent the height on the wall where the ladder makes contact in feet.
𝒉𝒉 = 𝟐𝟐𝟐𝟐(𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕)
𝒉𝒉 ≈ 𝟐𝟐𝟐𝟐. 𝟏𝟏
The ladder contacts the wall just over 𝟐𝟐𝟐𝟐 𝐟𝐟t. above the ground.
c.
Describe how to safely set a ladder according to OSHA standards without using a protractor.
The horizontal distance of the base of the ladder to the point of contact of the ladder should be approximately
𝟏𝟏
𝟒𝟒
3.
of the length of the ladder.
A regular pentagon with side lengths of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 is inscribed in a circle. What is the radius of the circle?
Draw radii from center 𝑪𝑪 of the circle to two
consecutive vertices of the pentagon, 𝑨𝑨 and 𝑩𝑩, and
draw an altitude from the center 𝑪𝑪 to 𝑫𝑫 on ����
𝑨𝑨𝑨𝑨.
The interior angles of a regular pentagon have
𝑩𝑩𝑩𝑩 bisect the
measure of 𝟏𝟏𝟏𝟏𝟏𝟏°, and ����
𝑨𝑨𝑨𝑨 and ����
interior angles at 𝑨𝑨 and 𝑩𝑩.
𝑨𝑨𝑨𝑨 = 𝑩𝑩𝑩𝑩 = 𝟕𝟕 𝐜𝐜𝐜𝐜
���� in centimeters.
Let 𝒙𝒙 represent the lengths of 𝑨𝑨𝑨𝑨
𝟕𝟕
𝒙𝒙
Using cosine, 𝐜𝐜𝐜𝐜𝐜𝐜 𝟓𝟓𝟓𝟓 = , and thus:
𝟕𝟕
𝐜𝐜𝐜𝐜𝐜𝐜 𝟓𝟓𝟓𝟓
𝒙𝒙 ≈ 𝟏𝟏𝟏𝟏. 𝟗𝟗.
𝒙𝒙 =
����
𝑨𝑨𝑨𝑨 is a radius of the circle and has a length of
approximately 𝟏𝟏𝟏𝟏. 𝟗𝟗 𝐜𝐜𝐜𝐜.
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4.
The circular frame of a Ferris wheel is suspended so that it sits 𝟒𝟒 𝐟𝐟𝐟𝐟. above the ground and has a radius of 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. A
segment joins center 𝑪𝑪 to point 𝑺𝑺 on the circle. If ����
𝑪𝑪𝑪𝑪 makes an angle of 𝟒𝟒𝟒𝟒° with the horizon, what is the distance of
point 𝑺𝑺 to the ground?
Note to teacher: There are two correct answers
to this problem since the segment can make an
angle of 𝟒𝟒𝟒𝟒° above or below the horizon in four
distinct locations, providing two different heights
above the ground.
C
There are four locations at which the segment
makes an angle of 𝟒𝟒𝟒𝟒° with the horizon. In each
𝑪𝑪𝑪𝑪 is the hypotenuse of a right triangle with
case, ����
acute angles with measures of 𝟒𝟒𝟒𝟒° and 𝟒𝟒𝟒𝟒°.
Horizontal Center
Line
Let 𝒅𝒅 represent the distance in feet from point 𝑺𝑺 to the horizon (applies to each case):
𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒 =
𝒅𝒅
𝟑𝟑𝟑𝟑
𝟑𝟑𝟑𝟑 (𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒) = 𝒅𝒅
𝟐𝟐𝟐𝟐. 𝟑𝟑 ≈ 𝒅𝒅
The center of the Ferris wheel is 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. above the ground; therefore, the distance from points 𝑺𝑺𝟏𝟏 and 𝑺𝑺𝟒𝟒 to the
ground in feet is
𝟑𝟑𝟑𝟑 − 𝟐𝟐𝟐𝟐. 𝟑𝟑 = 𝟏𝟏𝟏𝟏. 𝟕𝟕.
Points 𝑺𝑺𝟐𝟐 and 𝑺𝑺𝟑𝟑 are approximately 𝟐𝟐𝟐𝟐. 𝟑𝟑 𝐟𝐟𝐟𝐟. above the center of the Ferris wheel, so the distance from 𝑺𝑺𝟐𝟐 and 𝑺𝑺𝟑𝟑 to
the ground in feet is
𝟑𝟑𝟑𝟑 + 𝟐𝟐𝟐𝟐. 𝟑𝟑 = 𝟓𝟓𝟓𝟓. 𝟑𝟑.
When ����
𝑪𝑪𝑪𝑪 forms a 𝟒𝟒𝟒𝟒° angle with the horizon, point 𝑺𝑺 is either approximately 𝟏𝟏𝟏𝟏. 𝟕𝟕 𝐟𝐟𝐟𝐟. above the ground or
approximately 𝟓𝟓𝟓𝟓. 𝟑𝟑 𝐟𝐟𝐟𝐟. above the ground.
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5.
Tim is a contractor who is designing a wheelchair ramp for handicapped access to a business. According to the
Americans with Disabilities Act (ADA), the maximum slope allowed for a public wheelchair ramp forms an angle of
approximately 𝟒𝟒. 𝟕𝟕𝟕𝟕° to level ground. The length of a ramp’s surface cannot exceed 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. without including a flat
𝟓𝟓 𝐟𝐟𝐟𝐟.× 𝟓𝟓 𝐟𝐟𝐟𝐟. platform (minimum dimensions) on which a person can rest, and such a platform must be included at the
bottom and top of any ramp.
Tim designs a ramp that forms an angle of 𝟒𝟒° to the level ground to reach the entrance of the building. The entrance
of the building is 𝟐𝟐 𝐟𝐟𝐟𝐟. 𝟗𝟗 𝐢𝐢𝐢𝐢. above the ground. Let 𝒙𝒙 and 𝒚𝒚 as shown in Tim’s initial design below be the indicated
distances in feet.
a.
Assuming that the ground in front of the building’s entrance is flat, use Tim’s measurements and the ADA
requirements to complete and/or revise his wheelchair ramp design.
Note to teacher: Student designs will vary; however, the length of the ramp’s surface is greater than 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟.,
which requires at least one resting platform along the ramp. Furthermore, Tim’s design does not include a
platform at the top of the ramp as required by the guidelines, rendering his design incorrect.
Possible student solution:
𝟐𝟐 𝐟𝐟𝐟𝐟. 𝟗𝟗 𝐢𝐢𝐢𝐢. = 𝟐𝟐. 𝟕𝟕𝟕𝟕 𝐟𝐟𝐟𝐟.
Using tangent, 𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒 =
𝟐𝟐.𝟕𝟕𝟕𝟕
, and thus
𝒙𝒙
𝒙𝒙 =
𝟐𝟐. 𝟕𝟕𝟕𝟕
𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒
𝒙𝒙 ≈ 𝟑𝟑𝟑𝟑. 𝟑𝟑𝟑𝟑.
The ramp begins approximately 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. 𝟒𝟒 𝐢𝐢𝐢𝐢. from the building; thus, the ramp’s surface is greater than 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟.
in length. The hypotenuse of the triangle represents the sloped surface of the ramp and must be longer than
the legs. Tim’s design will not meet the ADA guidelines because it does not include a flat resting section along
the ramp‘s slope, nor does it include a platform at the top of the ramp. (The bottom of the ramp is flat
ground. The student’s design may or may not include a platform at the bottom.)
The vertical distance from the ground to the entrance is 𝟐𝟐. 𝟕𝟕𝟕𝟕 𝐟𝐟𝐟𝐟. Using sine, 𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒 =
𝒚𝒚 =
𝟐𝟐. 𝟕𝟕𝟕𝟕
𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒
𝟐𝟐.𝟕𝟕𝟕𝟕
, and thus,
𝒚𝒚
𝒚𝒚 ≈ 𝟑𝟑𝟑𝟑. 𝟒𝟒𝟒𝟒.
The total length of the ramp surface is approximately 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. 𝟓𝟓 𝐢𝐢𝐢𝐢.; however, because of its length, it requires a
resting platform somewhere in the first 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. and another platform at the top.
Resting
Platform 1

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b.
What is the total distance from the start of the ramp to the entrance of the building in your design?
If each platform is 𝟓𝟓 𝐟𝐟𝐟𝐟. in length, then the total distance along the ramp from the ground to the building is
approximately 𝟒𝟒𝟒𝟒 𝐟𝐟𝐟𝐟. 𝟓𝟓 𝐢𝐢𝐢𝐢.
6.
Tim is designing a roof truss in the shape of an isosceles triangle. The design shows the base angles of the truss to
have measures of 𝟏𝟏𝟏𝟏. 𝟓𝟓°. If the horizontal base of the roof truss is 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. across, what is the height of the truss?
Let 𝒉𝒉 represent the height of the truss in feet. Using tangent, 𝐭𝐭𝐭𝐭𝐭𝐭 𝟏𝟏𝟏𝟏. 𝟓𝟓 =
𝒉𝒉 = 𝟏𝟏𝟏𝟏(𝐭𝐭𝐭𝐭𝐭𝐭 𝟏𝟏𝟏𝟏. 𝟓𝟓)
The height of the truss is approximately 𝟔𝟔 𝐟𝐟𝐟𝐟.
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𝒉𝒉
, and thus,
𝟏𝟏𝟏𝟏
𝒉𝒉 ≈ 𝟔𝟔.
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Lesson 29: Applying Tangents
Student Outcomes

Students understand that the value of the tangent ratio of the angle of elevation or depression of a line is
equal to the slope of the line.

Students use the value of the tangent ratio of the angle of elevation or depression to solve real-world
problems.
Lesson Notes
Lesson 26 introduced students to the tangent of 𝜃𝜃 as the ratio of the length of the opposite side to the length of the
adjacent side. In Lesson 29, students use tan 𝜃𝜃 in two different contexts: (1) the value of the ratio as it has to do with
slope and (2) its use in solving real-world problems involving angles of elevation and depression. The focus of this lesson
is modeling with mathematics, MP.4, as students work on a series of real-world applications with tangent.
Classwork
Opening Exercise (7 minutes)
For these exercises, consider asking half the class to work on part (a) and the other half to go directly to part (b), and
then share results. Alternately, have students work in small groups. Students can work in pairs (or fours to split the
work a little more); a student divides the values while the partner finds the tangent values, and they compare their
results. Then, the class can debrief as a whole.
Opening Exercise
𝟎𝟎
𝟏𝟏𝟏𝟏𝟏𝟏
𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕
𝟓𝟓. 𝟔𝟔
𝟔𝟔𝟔𝟔𝟔𝟔
𝟑𝟑𝟑𝟑𝟑𝟑
𝟎𝟎. 𝟗𝟗
𝟓𝟓. 𝟔𝟔
𝟗𝟗
𝟎𝟎. 𝟏𝟏
𝟖𝟖
𝟗𝟗
𝟔𝟔𝟔𝟔𝟔𝟔
𝟏𝟏. 𝟕𝟕
𝟎𝟎. 𝟑𝟑
𝟕𝟕
𝟏𝟏. 𝟏𝟏𝟏𝟏𝟏𝟏
𝟖𝟖
𝟗𝟗𝟗𝟗𝟗𝟗
𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝟎𝟎. 𝟓𝟓
𝟗𝟗
𝟏𝟏. 𝟕𝟕
𝟖𝟖𝟖𝟖𝟖𝟖
𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗
𝟕𝟕𝟕𝟕𝟕𝟕
𝟏𝟏. 𝟏𝟏𝟏𝟏𝟏𝟏
𝟕𝟕
𝟔𝟔
𝟓𝟓
𝟕𝟕𝟕𝟕𝟕𝟕
𝟔𝟔𝟔𝟔𝟔𝟔
𝟎𝟎. 𝟔𝟔
𝟕𝟕
𝟎𝟎. 𝟖𝟖
𝟕𝟕𝟕𝟕𝟕𝟕
𝟎𝟎. 𝟖𝟖
𝟎𝟎. 𝟕𝟕
𝟏𝟏
𝟒𝟒
𝟔𝟔𝟔𝟔𝟔𝟔
𝟕𝟕𝟕𝟕𝟕𝟕
𝟎𝟎. 𝟖𝟖
𝟔𝟔
𝟏𝟏
𝟎𝟎. 𝟓𝟓
𝟎𝟎. 𝟕𝟕
𝟖𝟖𝟖𝟖𝟖𝟖
𝟑𝟑
𝟖𝟖𝟖𝟖𝟖𝟖
𝟐𝟐
𝟑𝟑𝟑𝟑𝟑𝟑
𝟎𝟎. 𝟑𝟑
𝟎𝟎. 𝟓𝟓
𝟎𝟎. 𝟔𝟔
𝟓𝟓
𝟏𝟏
undefined
undefined
Note to the teacher: Dividing the values in the first two rows provides a different answer than using the full
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𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑
𝟎𝟎. 𝟖𝟖
𝟓𝟓𝟓𝟓𝟓𝟓
𝟗𝟗𝟗𝟗𝟗𝟗
𝟎𝟎. 𝟏𝟏
𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗
𝟎𝟎. 𝟓𝟓
𝟒𝟒
𝟖𝟖𝟖𝟖𝟖𝟖
𝟎𝟎
𝟏𝟏𝟏𝟏𝟏𝟏
𝟎𝟎. 𝟏𝟏
𝟎𝟎. 𝟑𝟑
𝟓𝟓𝟓𝟓𝟓𝟓
𝐭𝐭𝐭𝐭 𝜽𝜽
𝟎𝟎
𝟎𝟎. 𝟗𝟗
𝟏𝟏𝟏𝟏𝟏𝟏
𝐜𝐜
𝐜𝐜
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝐜𝐜𝐜𝐜 𝜽𝜽
𝟎𝟎. 𝟏𝟏
𝟑𝟑
𝟗𝟗
𝟏𝟏
𝟎𝟎
𝟐𝟐
𝟑𝟑
𝐜𝐜𝐜𝐜 𝜽𝜽
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝟏𝟏
𝟑𝟑𝟑𝟑𝟑𝟑
𝟎𝟎
𝟏𝟏
𝜽𝜽
𝐭𝐭
MP.8
Use a calculator to find the tangent of 𝜽𝜽. Enter the values, correct to four decimal places, in the last row of
the table.
𝟏𝟏𝟏𝟏𝟏𝟏
a.
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b.
The table from Lesson 29 is provided here for you. In the row labeled
cosine values. What do you notice?
MP.8
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
, divide the sine values by the
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
For each of the listed degree values, dividing the value 𝒐𝒐𝒐𝒐 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 by the corresponding value of 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 yields a
value very close to the value of 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽. The values divided to obtain
values might be exactly the same.
Note to the teacher: Consider showing the calculation of
sin 𝜃𝜃
cos 𝜃𝜃
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
were approximations, so the actual
using the calculator to reveal that the quotients are not
just close to the value of tan 𝜃𝜃, but are, in fact, equal to the value of tan 𝜃𝜃.
Depending on student readiness, consider asking students to speculate what the relationship between sine, cosine, and
tangent is. The relationship is summarized as
opp
hyp
=
opp
hyp
×
hyp
=
opp
but is expanded upon in Lesson 30. This level
of explanation makes it clear why tan 0 = 0 and why tan 90 is undefined.
Discussion (5 minutes)
Lead students through a discussion that ties together the concepts of angle of elevation/depression in a real-world
sense, and then, using the coordinate plane, tying it to tangent and slope.

Consider the image below. How would you describe the angle of elevation? How would you describe the
angle of depression?
Allow students a moment to come up with their own definitions and then share out
responses before presenting the following definition:
 The angle of elevation or depression is the angle between the horizontal
(parallel with the earth’s surface) and the line of sight.
 In a case where two viewers can observe each other, such as in the diagram
below, what do you notice about the measure of the angle of depression and the
angle of elevation? Why?

Scaffolding:
 Consider placing a poster
of the following in a
prominent place in the
classroom:
They have the same measure since the horizontal from each viewer
forms a pair of parallel lines, and the line of sight acts as a transversal;
the angle of depression and angle of elevation are alternate interior
angles of equal measure.
 Students may benefit from
choral recitation of the
description of each term.
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Example 1 (7 minutes)
Example 1 is a real-world application of the angle of elevation and tangent. Encourage students to label the diagram as
they read through the problem; consider using the same labels for the key vertices as a whole class. If students struggle
to begin the problem, consider drawing the triangle for the class and then asking students to write or speak about how
the triangle models the situation.

Consider standing near a tree and observing a bird at the top of the tree. We can use measurements we know
and the angle of elevation to help determine how high off the ground the bird is.
Please encourage students to attempt to solve first on their own or with a partner.
Example 1
Scott, whose eye level is 𝟏𝟏. 𝟓𝟓 𝐦𝐦 above the ground, stands 𝟑𝟑𝟑𝟑 𝐦𝐦 from a tree. The angle of elevation of a bird at the top of
the tree is 𝟑𝟑𝟑𝟑°. How far above ground is the bird?
MP.1

With respect to your diagram, think of the measurement you are looking for.

How will you find 𝐵𝐵𝐵𝐵?


In our diagram, we are looking for 𝐵𝐵𝐵𝐵.
I can use the tangent to determine 𝐵𝐵𝐵𝐵 in meters.
𝐵𝐵𝐵𝐵
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
tan 36 =
30
tan 36 =
30 tan 36 = 𝐵𝐵𝐵𝐵

𝐵𝐵𝐵𝐵 ≈ 21.8
Have we found the height at which the bird is off the ground?


No. The full height must be measured from the ground, so the distance from the ground to Scott’s eye
level must be added to 𝐵𝐵𝐵𝐵.
The height of the bird off of the ground is 1.5 m + 21.8 m = 23.3 m.
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MP.1

So, between the provided measurements, including the angle of elevation, and the use of the tangent ratio, we
were able to find the height of the bird.
Example 2 (5 minutes)
Example 2 is a real-world application of the angle of depression and tangent. Encourage students to label the diagram as
they read through the problem; consider having the whole class use common labels for the key vertices. Consider
having students work with a partner for this example.
Example 2
From an angle of depression of 𝟒𝟒𝟒𝟒°, John watches his friend approach his building while standing on the rooftop.
The rooftop is 𝟏𝟏𝟏𝟏 𝐦𝐦 from the ground, and John’s eye level is at about 𝟏𝟏. 𝟖𝟖 𝐦𝐦 from the rooftop. What is the distance
between John’s friend and the building?
Make sure to point out the angle of depression in the diagram below. Emphasize that the 40° angle of depression is the
angle between the line of sight and the line horizontal (to the ground) from the eye.

Use the diagram to describe the distance we must determine.

We are going to find 𝐵𝐵𝐵𝐵 in meters.
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
How will we find 𝐵𝐵𝐵𝐵?


We can use the tangent: tan 40 =
𝐴𝐴𝐴𝐴
.
𝐵𝐵𝐵𝐵
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
17.8
tan 40 =
𝐵𝐵𝐵𝐵
17.8
𝐵𝐵𝐵𝐵 =
tan 40
tan 40 =
𝐵𝐵𝐵𝐵 ≈ 21.2
Again, with the assistance of a few measurements, including the angle of depression, we were able to
determine the distance between John’s friend and the building, which is 21.2 meters.
Exercise 1 (4 minutes)
In the following problem, the height of the person is not provided as a piece of information. This is because the relative
height of the person does not matter with respect to the other heights mentioned.
Exercise 1
Standing on the gallery of a lighthouse (the deck at the top of a lighthouse), a person spots a ship at an angle of
depression of 𝟐𝟐𝟐𝟐°. The lighthouse is 𝟐𝟐𝟐𝟐 𝐦𝐦 tall and sits on a cliff 𝟒𝟒𝟒𝟒 𝐦𝐦 tall as measured from sea level. What is the
horizontal distance between the lighthouse and the ship? Sketch a diagram to support your answer.
Approximately 𝟐𝟐𝟐𝟐𝟐𝟐 𝐦𝐦
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Discussion (6 minutes)

Now, let’s examine the angle of elevation and depression on the coordinate plane.

Observe the following line, which has a positive slope and crosses the 𝑥𝑥-axis.

The indicated angle is the angle of elevation.

Notice that the slope of the line is the tangent of the angle of elevation. Why is this true?

By building a right triangle (so that one leg is parallel to the 𝑥𝑥-axis and the other leg is parallel to the
𝑦𝑦-axis) around any two points selected on the line, we can determine the slope by finding the value of
rise
run
. This is equal to the tangent of the angle of elevation because, with respect to the angle of
elevation, the tangent ratio is
opp
, which is equivalent to the value of
rise
run
.

Similar to the angle of elevation, there is also an angle of depression. The indicated angle between the line
declining to the right and the 𝑥𝑥-axis is the angle of depression.

Notice that the tangent of the angle of depression will be positive; to accurately capture the slope of the line,
we must take the negative of the tangent.
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Exercise 2 (4 minutes)
Exercise 2
A line on the coordinate plane makes an angle of depression of 𝟑𝟑𝟑𝟑°. Find the slope of the line correct to four decimal
places.
Choose a segment on the line, and construct legs
of a right triangle parallel to the 𝒙𝒙- and 𝒚𝒚-axes,
as shown. If 𝒎𝒎 is the length of the vertical leg
and 𝒏𝒏 is the length of the horizontal leg, then
𝒎𝒎
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 = . The line decreases to the right, so
𝒏𝒏
the value of the slope must be negative.
Therefore,
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 = −
𝒎𝒎
= − 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 ≈ −𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕.
𝒏𝒏
Closing (2 minutes)
Ask students to summarize the key points of the lesson. Additionally, consider asking students the following questions
independently in writing, to a partner, or to the whole class.

Review the relationship between tangent and slope.

The slope of a line increasing to the right is the same as the tangent of the angle of elevation of the line.
The slope of a line decreasing to the right is the same as the negative tangent of the angle of
depression of the line.

Review the relationship between tangent and slope.

Review what the angle of elevation and depression means in a real-world context.

The angle of elevation or depression is the angle between the horizontal and the line of sight.
Exit Ticket (5 minutes)
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Name
Date
Lesson 29: Applying Tangents
Exit Ticket
1.
The line on the coordinate plane makes an angle of depression of 24°. Find the slope of the line correct to four
decimal places.
2.
Samuel is at the top of a tower and will ride a trolley down a zip line to a lower tower. The total vertical drop of the
zip line is 40 ft. The zip line’s angle of elevation from the lower tower is 11.5°. What is the horizontal distance
between the towers?
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Exit Ticket Sample Solutions
1.
A line on the coordinate plane makes an angle of depression of 𝟐𝟐𝟐𝟐°. Find the slope of the line correct to four
decimal places.
Choose a segment on the line, and
construct legs of a right triangle parallel to
the 𝒙𝒙- and 𝒚𝒚-axes, as shown. If 𝒎𝒎 is the
length of the vertical leg and 𝒏𝒏 is the
length of the horizontal leg, then 𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐 =
𝒎𝒎
. The line decreases to the right, so the
𝒏𝒏
value of the slope must be negative.
Therefore,
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 = −
2.
𝒎𝒎
= − 𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐 ≈ −𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒.
𝒏𝒏
Samuel is at the top of a tower and will ride a trolley down a zip line to a lower tower. The total vertical drop of the
zip line is 𝟒𝟒𝟒𝟒 𝐟𝐟𝐟𝐟. The zip line’s angle of elevation from the lower tower is 𝟏𝟏𝟏𝟏. 𝟓𝟓°. What is the horizontal distance
between the towers?
A right triangle is formed by the zip line’s path, the vertical drop along the upper tower, and the horizontal distance
between the towers. Let 𝒙𝒙 represent the horizontal distance between the towers in feet. Using tangent:
𝐭𝐭𝐭𝐭𝐭𝐭 𝟏𝟏𝟏𝟏. 𝟓𝟓 =
𝒙𝒙 =
𝟒𝟒𝟒𝟒
𝒙𝒙
𝟒𝟒𝟒𝟒
𝐭𝐭𝐭𝐭𝐭𝐭 𝟏𝟏𝟏𝟏. 𝟓𝟓
𝒙𝒙 ≈ 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔
The horizontal distance between the towers is approximately 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔 𝐟𝐟𝐟𝐟.
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Problem Set Sample Solutions
1.
A line in the coordinate plane has an angle of elevation of 𝟓𝟓𝟓𝟓°. Find the slope of the line correct to 𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟 decimal
places.
Since parallel lines have the same slope, we can
consider the line that passes through the origin with
an angle of inclination of 𝟓𝟓𝟓𝟓°. Draw a vertex at
(𝟓𝟓, 𝟎𝟎) on the 𝒙𝒙-axis, and draw a segment from
(𝟓𝟓, 𝟎𝟎) parallel to the 𝒚𝒚-axis to the intersection with
the line to form a right triangle.
If the base of the right triangle is 𝟓𝟓-units long, let
the height of the triangle be represented by 𝒚𝒚.
𝐭𝐭𝐭𝐭𝐭𝐭 𝟓𝟓𝟓𝟓 =
𝒙𝒙
𝟓𝟓
𝟓𝟓(𝐭𝐭𝐭𝐭𝐭𝐭 𝟓𝟓𝟓𝟓) = 𝒙𝒙
The slope of the line is
𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫
𝐫𝐫𝐫𝐫𝐫𝐫
𝟓𝟓(𝐭𝐭𝐭𝐭𝐭𝐭 𝟓𝟓𝟓𝟓)
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 =
𝟓𝟓
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 =
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 = 𝐭𝐭𝐭𝐭𝐭𝐭 𝟓𝟓𝟓𝟓 ≈ 𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑.
2.
A line in the coordinate plane has an angle of depression of 𝟐𝟐𝟐𝟐°. Find the slope of the line correct to four decimal
places.
A line crosses the 𝒙𝒙-axis, decreasing to the right with
an angle of depression of 𝟐𝟐𝟐𝟐°. Using a similar
method as in Problem 1, a right triangle is formed.
If the leg along the 𝒙𝒙-axis represents the base of the
triangle, let 𝒉𝒉 represent the height of the triangle.
Using tangent:
𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐 =
𝒉𝒉
𝟓𝟓
𝟓𝟓(𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐) = 𝒉𝒉
The slope of the line is negative since the line
decreases to the right:
𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫
𝐫𝐫𝐫𝐫𝐫𝐫
𝒉𝒉
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 = −
𝟓𝟓
𝟓𝟓(𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐)
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 = −
𝟓𝟓
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 =
3.
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 = − 𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐 ≈ −𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
In Problems 1 and 2, why do the lengths of the legs of the right triangles formed not affect the slope of the line?
When using the tangent ratio, the length of one leg of a right triangle can be determined in terms of the other leg.
Let 𝒙𝒙 represent the length of the horizontal leg of a slope triangle. The vertical leg of the triangle is then 𝒙𝒙 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽,
where 𝜽𝜽 is the measure of the angle of inclination or depression. The slope of the line is
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𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫
𝐫𝐫𝐫𝐫𝐫𝐫
=
𝒙𝒙 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽
𝒙𝒙
= 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽.
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GEOMETRY
Given the angles of depression below, determine the slope of the line with the indicated angle correct to four
decimal places.
𝟑𝟑 ° angle of depression
d.
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 ≈ 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝐭𝐭𝐭𝐭𝐭𝐭 𝟖𝟖𝟖𝟖 ≈ 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 ≈ −𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
b.
e.
𝟒𝟒𝟒𝟒° angle of depression
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 ≈ −𝟏𝟏. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
f.
𝟖𝟖𝟖𝟖° angle of depression
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 ≈ −𝟓𝟓𝟓𝟓. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝟖𝟖𝟖𝟖. 𝟗𝟗° angle of depression
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 ≈ −𝟓𝟓𝟓𝟓𝟓𝟓. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 ≈ −𝟓𝟓. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
h.
𝟖𝟖𝟖𝟖° angle of depression
𝐭𝐭𝐭𝐭𝐭𝐭 𝟖𝟖𝟖𝟖. 𝟗𝟗 ≈ 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝐭𝐭𝐭𝐭𝐭𝐭 𝟖𝟖𝟖𝟖 ≈ 𝟓𝟓. 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
g.
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 ≈ −𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝐭𝐭𝐭𝐭𝐭𝐭 𝟖𝟖𝟖𝟖 ≈ 𝟓𝟓𝟓𝟓. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒𝟒𝟒 ≈ 𝟏𝟏. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
c.
𝟖𝟖 ° angle of depression
𝟖𝟖
a.
𝟑𝟑
4.
What appears to be happening to the slopes (and tangent values) as the angles of depression get closer to
𝟗𝟗𝟗𝟗°?
As the angles get closer to 𝟗𝟗𝟗𝟗°, their slopes (and tangent values) get much farther from zero.
Find the slopes of angles of depression that are even closer to 𝟗𝟗𝟗𝟗° than 𝟖𝟖𝟖𝟖. 𝟗𝟗°. Can the value of the tangent
of 𝟗𝟗𝟗𝟗° be defined? Why or why not?
Choices of angles will vary. The closer an angle is in measure to 𝟗𝟗𝟗𝟗°, the greater the tangent value of that
angle and the farther the slope of the line determined by that angle is from zero. An angle of depression of
𝟗𝟗𝟗𝟗° would be a vertical line, and vertical lines have 𝟎𝟎 run; therefore, the value of the ratio 𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓: 𝒓𝒓𝒓𝒓𝒓𝒓 is
undefined. The value of the tangent of 𝟗𝟗𝟗𝟗° would have a similar outcome because the adjacent leg of the
“triangle” would have a length of 𝟎𝟎, so the ratio
5.
𝐚𝐚𝐚𝐚𝐚𝐚
=
𝐨𝐨𝐨𝐨𝐨𝐨
𝟎𝟎
, which is undefined.
For the indicated angle, express the quotient in terms of sine, cosine, or tangent. Then, write the quotient in
simplest terms.
a.
𝟒𝟒
𝟐𝟐√𝟏𝟏𝟏𝟏
; 𝜶𝜶
𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶 =
b.
𝟔𝟔
; 𝜶𝜶
𝟒𝟒
𝐭𝐭𝐭𝐭𝐭𝐭 𝜶𝜶 =
c.
𝟒𝟒
𝟐𝟐√𝟏𝟏𝟏𝟏
d.
𝟒𝟒
𝟔𝟔
; 𝜷𝜷
𝐭𝐭𝐭𝐭𝐭𝐭 𝜷𝜷 =
𝟒𝟒
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟒𝟒
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟐𝟐
√𝟏𝟏𝟏𝟏
=
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
=
𝟐𝟐√𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝟒𝟒 𝟐𝟐
=
𝟔𝟔 𝟑𝟑
Lesson 29:
=
𝟔𝟔 𝟑𝟑
=
𝟒𝟒 𝟐𝟐
; 𝜷𝜷
𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷 =
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GEOMETRY
6.
The pitch of a roof on a home is expressed as a ratio of 𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯 𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫: 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡 𝐫𝐫𝐫𝐫𝐫𝐫 where the run has a length of
𝟏𝟏𝟏𝟏 units. If a given roof design includes an angle of elevation of 𝟐𝟐𝟐𝟐. 𝟓𝟓° and the roof spans 𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. as shown in the
diagram, determine the pitch of the roof. Then, determine the distance along one of the two sloped surfaces of the
roof.
The diagram as shown is an isosceles
triangle since the base angles have
equal measure. The altitude, 𝒂𝒂, of the
triangle is the vertical rise of the roof.
The right triangles formed by drawing
the altitude of the given isosceles
triangle have a leg of length 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟.
𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐. 𝟓𝟓 =
𝒂𝒂
𝟏𝟏𝟏𝟏
𝒂𝒂 = 𝟏𝟏𝟏𝟏 𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐. 𝟓𝟓
𝒂𝒂 ≈ 𝟕𝟕. 𝟓𝟓
𝐜𝐜
𝒔𝒔 =
𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏 ⋅ 𝟕𝟕. 𝟓𝟓
𝒔𝒔
𝒉𝒉 = 𝟓𝟓
𝟏𝟏
𝟏𝟏
𝒔𝒔
𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐. 𝟓𝟓 =
𝟕𝟕. 𝟓𝟓
𝒉𝒉
=
𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐. 𝟓𝟓
≈ 𝟏𝟏 . 𝟓𝟓
𝟏𝟏
Roof pitch:
The pitch of the roof is 𝟓𝟓: 𝟏𝟏𝟏𝟏.
The sloped surface of the roof has a distance of approximately 𝟏𝟏𝟏𝟏. 𝟓𝟓 𝐟𝐟𝐟𝐟.
7.
An anchor cable supports a vertical utility pole forming a 𝟓𝟓𝟓𝟓° angle with the ground. The cable is attached to the
top of the pole. If the distance from the base of the pole to the base of the cable is 𝟓𝟓 meters, how tall is the pole?
Let 𝒉𝒉 represent the height of the pole in meters.
Using tangent:
𝐭𝐭𝐭𝐭𝐭𝐭 𝟓𝟓𝟓𝟓 =
𝒉𝒉
𝟓𝟓
𝟓𝟓(𝐭𝐭𝐭𝐭𝐭𝐭 𝟓𝟓𝟓𝟓) = 𝒉𝒉
𝟔𝟔. 𝟏𝟏𝟏𝟏 ≈ 𝒉𝒉
The height of the utility pole is approximately 𝟔𝟔. 𝟏𝟏𝟏𝟏 meters.
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GEOMETRY
8.
A winch is a tool that rotates a cylinder, around which a cable is wound. When the winch rotates in one direction, it
draws the cable in. Joey is using a winch and a pulley (as shown in the diagram) to raise a heavy box off the floor
and onto a cart. The box is 𝟐𝟐 𝐟𝐟𝐟𝐟. tall, and the winch is 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟. horizontally from where cable drops
down vertically from the pulley. The angle of elevation to the pulley is 𝟒𝟒𝟒𝟒°. What is the
approximate length of cable required to connect the winch and the box?
Let 𝒉𝒉 represent the length of cable in the distance from the winch to the pulley along the
hypotenuse of the right triangle shown in feet, and let 𝒚𝒚 represent the distance from
the pulley to the floor in feet.
Using tangent:
𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒𝟒𝟒 =
𝒚𝒚
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏(𝐭𝐭𝐭𝐭𝐭𝐭 𝟒𝟒𝟒𝟒) = 𝒚𝒚
𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔 ≈ 𝒚𝒚
Using cosine:
𝟏𝟏𝟏𝟏
𝒉𝒉
𝟏𝟏𝟏𝟏
𝒉𝒉 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒
2 ft.
𝐜𝐜𝐜𝐜𝐜𝐜 𝟒𝟒𝟒𝟒 =
14 ft.
𝒉𝒉 ≈ 𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖
Let 𝒕𝒕 represent the total amount of cable from the winch to the box in feet:
𝒕𝒕 ≈ 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔 + 𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖 − 𝟐𝟐
𝒕𝒕 ≈ 𝟐𝟐𝟐𝟐. 𝟒𝟒𝟒𝟒
The total length of cable from the winch to the box is approximately 𝟐𝟐𝟐𝟐. 𝟒𝟒𝟒𝟒 𝐟𝐟𝐟𝐟.
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GEOMETRY
Lesson 30: Trigonometry and the Pythagorean Theorem
Student Outcomes

Students rewrite the Pythagorean theorem in terms of sine and cosine ratios and use it in this form to solve
problems.

Students write tangent as an identity in terms of sine and cosine and use it in this form to solve problems.
Lesson Notes
Students discover the Pythagorean theorem in terms of sine and cosine ratios and demonstrate why tan 𝜃𝜃 =
begin solving problems where any one of the values of sin 𝜃𝜃, cos 𝜃𝜃, tan 𝜃𝜃 are provided.
sin 𝜃𝜃
. They
cos 𝜃𝜃
In this Geometry course, trigonometry is anchored in right triangle trigonometry as evidenced by standards G-SRT.C.6–8
in the cluster that states: Define trigonometric ratios and solve problems involving right triangles. The focus is on the
values of ratios for a given right triangle. This is an important distinction from trigonometry studied in Algebra II, which
is studied from the perspective of functions; sine, cosine, and tangent are functions of any real number, not just those of
right triangles. The language in G-SRT.C.8 juxtaposes trigonometric ratios and the Pythagorean theorem in the same
standard, which leads directly to the Pythagorean identity for right triangles. In Algebra II, students prove that the
identity holds for all real numbers. Presently, this lesson offers an opportunity for students to develop deep
understanding of the relationship between the trigonometric ratios and the Pythagorean theorem.
Classwork
Exercises 1–2 (4 minutes)

In this lesson, we will use a fact well known to us, the Pythagorean theorem, and tie it to trigonometry.
Exercises 1–2
1.
Draw a diagram as part of your response.
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 =
2.
√𝟑𝟑
𝟐𝟐
𝟕𝟕
𝟗𝟗
In a right triangle with acute angle of measure 𝜽𝜽, 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = . What is the value of 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽?
Draw a diagram as part of your response.
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝟕𝟕
𝟒𝟒√𝟐𝟐
=
𝟕𝟕√𝟐𝟐
𝟖𝟖
Lesson 30:
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Scaffolding:
𝟏𝟏
𝟐𝟐
In a right triangle with acute angle of measure 𝜽𝜽, 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = . What is the value of 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽?
For students who are
struggling, begin with an image
of a right triangle with one
acute angle marked and the
value of opp: hypotenuse ratio
is 3: 5, what is the value of
Trigonometry and the Pythagorean Theorem
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Lesson 30
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M2
GEOMETRY

How did you apply the Pythagorean theorem to answer Exercises 1–2?
2
Since the triangles are right triangles, we used the relationship between side lengths, 𝑎𝑎2 + 𝑏𝑏 = 𝑐𝑐2 , to
solve for the missing side length and then used the missing side length to determine the value of the
appropriate ratio.

MP.3
Example 1 (13 minutes)


The Great Pyramid of Giza in Egypt was constructed around 2600 B.C.E. out of limestone blocks weighing
several tons each. The angle measure between the base and each of the four triangular faces of the pyramid is
roughly 53°.
Observe Figure 1, a model of the Great Pyramid, and Figure 2, which isolates the right triangle formed by the
height, the slant height, and the segment that joins the center of the base to the bottom of the slant height.
Example 1
a.
What common right triangle was probably modeled in the construction of the triangle in Figure 2? Use
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓° ≈ 𝟎𝟎. 𝟖𝟖.
Figure 2
Figure 1
𝟒𝟒
𝟓𝟓
Right triangle with side lengths 𝟑𝟑, 𝟒𝟒, and 𝟓𝟓, since 𝟎𝟎. 𝟖𝟖 = .

What common right triangle was probably modeled in the construction of the triangle in Figure 2?
Though it may not be immediately obvious to students, part (a) is the same type of question as they completed in
Exercises 1–2. The difference is the visual appearance of the value of sin 53 in decimal form versus in fraction form.
Allow students time to sort through what they must do to answer part (a). Offer guiding questions and comments as
needed such as the following:

Revisit Exercises 1–2. What similarities and differences do you notice between Example 1, part (a), and
Exercises 1–2?

What other representation of 0.8 may be useful in this question?
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GEOMETRY
Alternatively, students should also see that the value of sin 53° can be thought of as
opp
hyp
=
0.8
1
part (a) using this fraction. If students have responses to share, share them as a whole class, or proceed to share the
following solution:



To determine the common right triangle that was probably modeled in the construction of the triangle in
Figure 2, using the approximation sin 53° ≈ 0.8 means we are looking for a right triangle with side-length
relationships that are well known.
Label the triangle with given acute angle measure of approximately 53° as
is labeled in the following figure. The hypotenuse has length 1, the
opposite side has length 0.8, and the side adjacent to the marked angle is
labeled as 𝑥𝑥.
How can we determine the value of 𝑥𝑥?
We can apply the Pythagorean theorem.


Solve for 𝑥𝑥.
(0.8)2 + 𝑥𝑥 2 = (1)2

0.64 + 𝑥𝑥 2 = 1
𝑥𝑥 2 = 0.36

𝑥𝑥 = 0.6
The side lengths of this triangle are 0.6, 0.8, and 1. What well-known right triangle matches these lengths?
Even though the calculations to determine the lengths of the triangle have been made, determining that this triangle is a
3– 4–5 triangle is still a jump. Allow time for students to persevere after the answer. Offer guiding questions and
comments as needed such as the following:


Sketch a right triangle with side lengths 6, 8, and 10, and ask how that triangle is related to the one in the
problem.
List other triangle side lengths that are in the same ratio as a 6–8–10 triangle.
Students should conclude part (a) with the understanding that a triangle with an acute angle measure of approximately
53° is a 3–4–5 triangle.
b.
The actual angle between the base and lateral faces of the pyramid is actually closer to 𝟓𝟓𝟓𝟓°. Considering the
age of the pyramid, what could account for the difference between the angle measure in part (a) and the
actual measure?
The Great Pyramid is approximately 𝟒𝟒, 𝟓𝟓𝟓𝟓𝟓𝟓 years old, and the weight of each block is several tons. It is
conceivable that over time, the great weight of the blocks caused the pyramids to settle and shift the lateral
faces enough so that the angle is closer to 𝟓𝟓𝟐𝟐° than to 𝟓𝟓𝟓𝟓°.
c.
Why do you think the architects chose to use a 𝟑𝟑–𝟒𝟒–𝟓𝟓 as a model for the triangle?
Answers may vary. Perhaps they used it (1) because it is the right triangle with the shortest whole-number
side lengths to satisfy the converse of the Pythagorean theorem and (2) because of the aesthetic it offers.
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Discussion (10 minutes)


Let 𝜃𝜃 be the angle such that sin 𝜃𝜃 = 0.8. Before, we were using an
approximation. We do not know exactly what the angle measure is, but we will
assign the angle measure that results in the sine of the angle as 0.8 the label 𝜃𝜃.
Then, we know (draw the following image):
In the diagram, rewrite the leg lengths in terms of sin 𝜃𝜃 and cos 𝜃𝜃.
Allow students a few moments to struggle with this connection. If needed, prompt them,
and ask what the values of sin 𝜃𝜃 and cos 𝜃𝜃 are in this right triangle.


Since the value of sin 𝜃𝜃 is 0.8 and the value of cos 𝜃𝜃 = 0.6, the leg
lengths can be rewritten as:
The Pythagorean theorem states that for a right triangle with lengths 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐, where 𝑐𝑐 is the hypotenuse,
the relationship between the side lengths is 𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2 . Apply the Pythagorean theorem to this triangle.

(sin 𝜃𝜃) 2 + (cos 𝜃𝜃)2 = 1

This statement is called the Pythagorean identity. This relationship is easy to show in general.

Referencing the diagram above, we can say
opp2 + adj2 = hyp2 .
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
Divide both sides by hyp2 .
hyp2
+
=
2
2
hyp
hyp2
hyp
(sin 𝜃𝜃) 2 + (cos 𝜃𝜃) 2 = 1
Explain to students that they might see the Pythagorean identity written in the following way on the Web and in other
texts:
sin2 𝜃𝜃 + cos 2 𝜃𝜃 = 1
Let them know that in Precalculus and Advanced Topics, they use this notation but not to worry about such notation
now.
Example 2 (7 minutes)
Students discover a second trigonometric identity; this identity describes a relationship between sine, cosine, and
tangent.

Recall Opening Exercise part (b) from Lesson 29. We found that the tangent values of the listed angle
measurements were equal to dividing sine by cosine for the same angle measurements. We discover why this
is true in this example.

Use the provided diagram to reason why the trigonometric identity tan 𝜃𝜃 =
Example 2
Show why 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 =
sin 𝜃𝜃
.
cos 𝜃𝜃
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
.
𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜
MP.7
Allow students time to work through the reasoning independently before guiding them through an explanation. To
provide more support, consider having the diagram on the board and then writing the following to start students off:
𝑎𝑎
𝑐𝑐
𝑏𝑏
cos 𝜃𝜃 =
𝑐𝑐
𝑎𝑎
tan 𝜃𝜃 =
𝑏𝑏
sin 𝜃𝜃 =
sin 𝜃𝜃 =
cos 𝜃𝜃 =
tan 𝜃𝜃 =
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
MP.7
tan 𝜃𝜃 =
sin 𝜃𝜃
because
cos 𝜃𝜃
Then,

𝑎𝑎
𝑏𝑏
sin 𝜃𝜃 =
𝑎𝑎
𝑏𝑏
and cos 𝜃𝜃 = .
𝑐𝑐
𝑐𝑐
tan 𝜃𝜃 =
tan 𝜃𝜃 = , which is what we found earlier.
sin 𝜃𝜃
cos 𝜃𝜃
If you are given one of the values sin 𝜃𝜃, cos 𝜃𝜃, or tan 𝜃𝜃, we can find the other two values using the identities
sin2 𝜃𝜃 + cos 2 𝜃𝜃 = 1 and tan 𝜃𝜃 =
Exercises 3–4 (5 minutes)
sin 𝜃𝜃
or by using the Pythagorean theorem.
cos 𝜃𝜃
Exercises 3–4 are the same as Exercise 1–2; however, students answer them now by applying the Pythagorean identity.
Exercises 3–4
3.
𝟏𝟏
𝟐𝟐
In a right triangle with acute angle of measure 𝜽𝜽, 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = , use the Pythagorean identity to determine the value of
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽.
𝐬𝐬𝐬𝐬𝐬𝐬𝟐𝟐 𝜽𝜽 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟏𝟏 𝟐𝟐
� � + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟐𝟐
𝟏𝟏
+ 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟒𝟒
𝟑𝟑
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 =
𝟒𝟒
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 =
4.
√𝟑𝟑
𝟐𝟐
𝟕𝟕
𝟗𝟗
Given a right triangle with acute angle of measure 𝜽𝜽, 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = , use the Pythagorean identity to determine the value
of 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽.
𝐬𝐬𝐬𝐬𝐬𝐬𝟐𝟐 𝜽𝜽 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟕𝟕 𝟐𝟐
� � + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟗𝟗
𝟒𝟒𝟒𝟒
+ 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟖𝟖𝟖𝟖
𝟑𝟑𝟑𝟑
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 =
𝟖𝟖𝟖𝟖
𝟒𝟒√𝟐𝟐
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 =
𝟗𝟗
𝟕𝟕
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝟕𝟕
= 𝟗𝟗 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 𝟒𝟒√𝟐𝟐 𝟒𝟒√𝟐𝟐
𝟗𝟗
𝟕𝟕√𝟐𝟐
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝟖𝟖
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 =
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Closing (1 minute)
Ask students to summarize the key points of the lesson. Additionally, consider asking students the following questions
independently in writing, to a partner, or to the whole class.

What is the Pythagorean identity?


What are the ways the tangent can be represented?



sin2 𝜃𝜃 + cos 2 𝜃𝜃 = 1
opp
sinθ
tan 𝜃𝜃 =
cosθ
tan 𝜃𝜃 =
If one of the values sin 𝜃𝜃, cos 𝜃𝜃, or tan 𝜃𝜃 is provided to us, we can find the other two values by using the
identities sin2 𝜃𝜃 + cos 2 𝜃𝜃 = 1, tan 𝜃𝜃 =
sin 𝜃𝜃
, or the Pythagorean theorem.
cos 𝜃𝜃
Exit Ticket (5 minutes)
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Name
Date
Lesson 30: Trigonometry and the Pythagorean Theorem
Exit Ticket
4�29
, use trigonometric identities to find cos 𝛽𝛽 and tan 𝛽𝛽.
29
1.
If sin 𝛽𝛽 =
2.
Find the missing side lengths of the following triangle using sine, cosine, and/or tangent. Round your answer to four
decimal places.
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GEOMETRY
Exit Ticket Sample Solutions
𝜷𝜷 + 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜷𝜷 = 𝟏𝟏
𝐭𝐭𝐭𝐭 𝜷𝜷 =
𝟐𝟐
𝟒𝟒√𝟐𝟐𝟐𝟐
�
� + 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜷𝜷 = 𝟏𝟏
𝟐𝟐𝟐𝟐
𝟒𝟒√𝟐𝟐𝟐𝟐
𝐭𝐭𝐭𝐭𝐭𝐭 𝜷𝜷 = 𝟐𝟐𝟐𝟐
√𝟑𝟑𝟕𝟕𝟕𝟕
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏
+ 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜷𝜷 = 𝟏𝟏
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜷𝜷 =
𝟐𝟐𝟐𝟐
𝐭𝐭𝐭𝐭𝐭𝐭 𝜷𝜷 =
𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷 = �
𝟐𝟐𝟐𝟐
𝟐𝟐
𝐭𝐭
𝐭𝐭𝐭𝐭 𝜷𝜷 =
𝟐𝟐
𝐜𝐜
√𝟏𝟏𝟏𝟏
𝐭𝐭𝐭𝐭𝐭𝐭 𝜷𝜷 =
√𝟑𝟑𝟑𝟑𝟑𝟑
𝐜𝐜𝐜𝐜 𝜷𝜷 =
=
𝟐𝟐
√𝟐𝟐
2.
𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷
𝐜𝐜𝐜𝐜 𝜷𝜷
𝟒𝟒√𝟐𝟐𝟐𝟐
√𝟑𝟑𝟑𝟑𝟑𝟑
𝟒𝟒
√𝟏𝟏𝟏𝟏
𝟒𝟒√𝟏𝟏𝟏𝟏
𝟏𝟏
𝟏𝟏
𝟐𝟐
𝐭𝐭
𝐬𝐬𝐬𝐬
𝐜𝐜
𝟒𝟒�𝟐𝟐𝟐𝟐
, use trigonometric identities to find 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷 and 𝐭𝐭𝐭𝐭𝐭𝐭 𝜷𝜷.
𝟐𝟐𝟐𝟐
If 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷 =
𝐧𝐧
1.
Find the missing side lengths of the following triangle using sine, cosine, and/or tangent. Round your answer to four
decimal places.
𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕 =
𝒚𝒚 =
𝟑𝟑
𝒚𝒚
𝟑𝟑
≈ 𝟖𝟖. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕
𝐭𝐭𝐭𝐭𝐭𝐭 𝟕𝟕𝟕𝟕 =
𝒙𝒙
𝟑𝟑
𝒙𝒙 = 𝟑𝟑(𝐭𝐭𝐭𝐭𝐭𝐭 𝟕𝟕𝟕𝟕) ≈ 𝟖𝟖. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
Problem Set Sample Solutions
𝐧𝐧
Using the identity 𝐬𝐬𝐬𝐬
𝟐𝟐
𝜽𝜽 + 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 = 𝟏𝟏:
𝟒𝟒 𝟐𝟐
𝐬𝐬𝐬𝐬𝐧𝐧𝟐𝟐 𝜽𝜽 + � � = 𝟏𝟏
𝟓𝟓
Using the identity 𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
:
𝐜𝐜𝐜𝐜 𝜽𝜽
𝐜𝐜
𝟒𝟒
𝟓𝟓
If 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 = , find 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 and 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽.
𝐭𝐭
1.
𝟑𝟑
𝟑𝟑
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 = 𝟓𝟓 =
𝟒𝟒 𝟒𝟒
𝟓𝟓
𝟒𝟒 𝟐𝟐
𝐬𝐬𝐬𝐬𝐧𝐧𝟐𝟐 𝜽𝜽 = 𝟏𝟏 − � �
𝟓𝟓
𝟏𝟏𝟏𝟏
𝟐𝟐
𝐬𝐬𝐬𝐬𝐧𝐧 𝜽𝜽 = 𝟏𝟏 − � �
𝟐𝟐𝟐𝟐
𝟗𝟗
𝟐𝟐
𝐬𝐬𝐬𝐬𝐧𝐧 𝜽𝜽 =
𝟐𝟐𝟐𝟐
𝟗𝟗
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 = �
𝟐𝟐𝟐𝟐
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
Lesson 30:
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𝟑𝟑
𝟓𝟓
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GEOMETRY
𝜽𝜽 = 𝟏𝟏
𝟒𝟒𝟒𝟒
�
� + 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝜽𝜽 = 𝟏𝟏 −
𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏
𝟒𝟒𝟒𝟒
÷
=
𝟏𝟏
𝟏𝟏
𝟏𝟏𝟏𝟏
𝟏𝟏
𝟐𝟐
𝟏𝟏
𝟐𝟐
𝟏𝟏𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬𝟐𝟐 𝜽𝜽 + 𝐜𝐜𝐜𝐜
𝟏𝟏𝟏𝟏
𝟒𝟒𝟒𝟒
, find 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 and 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽.
𝟏𝟏𝟏𝟏𝟏𝟏
𝐭𝐭
If 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
𝐜𝐜
2.
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
=
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
3.
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝐜𝐜
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜 𝜽𝜽 = �
=
𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏
If 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 = 𝟓𝟓, find 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 and 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽.
𝟓𝟓
𝟏𝟏
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 = 𝟓𝟓 = , so the legs of a right triangle can be considered to have lengths of 𝟓𝟓 and 𝟏𝟏. Using the Pythagorean
theorem:
𝟓𝟓𝟐𝟐 + 𝟏𝟏𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
𝟐𝟐𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐩𝐩𝟐𝟐
√𝟐𝟐𝟐𝟐 = 𝐡𝐡𝐡𝐡𝐡𝐡
�𝟓𝟓
𝟓𝟓
𝟓𝟓√𝟐𝟐𝟐𝟐
𝟏𝟏
√𝟐𝟐𝟐𝟐
; 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 =
=
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
√𝟐𝟐𝟐𝟐
, find 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 and 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽.
𝐧𝐧
Using the identity 𝐬𝐬𝐬𝐬
𝟐𝟐
𝟐𝟐
𝜽𝜽 + 𝐜𝐜𝐜𝐜
√𝟓𝟓
� � + 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟓𝟓
𝟐𝟐
𝜽𝜽 = 𝟏𝟏:
Using the identity 𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
:
𝐜𝐜𝐜𝐜 𝜽𝜽
𝐜𝐜
If 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
=
𝐭𝐭
4.
𝟓𝟓
√𝟐𝟐𝟐𝟐
𝐜𝐜
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
√𝟓𝟓
𝟏𝟏
√𝟓𝟓
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 = 𝟓𝟓 =
=
𝟐𝟐√𝟓𝟓 𝟐𝟐√𝟓𝟓 𝟐𝟐
𝟓𝟓
𝟐𝟐
√𝟓𝟓
𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 = 𝟏𝟏 − � �
𝟓𝟓
𝟓𝟓
𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 = 𝟏𝟏 − � �
𝟐𝟐𝟐𝟐
𝟏𝟏 𝟒𝟒
𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 = 𝟏𝟏 − =
𝟓𝟓 𝟓𝟓
𝟒𝟒 √𝟒𝟒
𝟐𝟐
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 = � =
=
𝟓𝟓 √𝟓𝟓 √𝟓𝟓
𝐜𝐜
𝐜𝐜𝐜𝐜 𝜽𝜽 =
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𝟐𝟐√𝟓𝟓
𝟓𝟓
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5.
Find the missing side lengths of the following triangle using sine, cosine, and/or tangent. Round your answer to four
decimal places.
𝒙𝒙
= 𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏
𝒙𝒙 = 𝟏𝟏𝟏𝟏 𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐 ≈ 𝟔𝟔. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
= 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔
𝒚𝒚
𝒚𝒚 =
6.
𝟏𝟏𝟏𝟏
≈ 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔
A surveying crew has two points 𝑨𝑨 and 𝑩𝑩 marked along a roadside at a distance of 𝟒𝟒𝟒𝟒𝟒𝟒 𝐲𝐲𝐲𝐲. A third point 𝑪𝑪 is
marked at the back corner of a property along a perpendicular to the road at 𝑩𝑩. A straight path joining 𝑪𝑪 to 𝑨𝑨 forms
a 𝟐𝟐𝟐𝟐° angle with the road. Find the distance from the road to point 𝑪𝑪 at the back of the property and the distance
from 𝑨𝑨 to 𝑪𝑪 using sine, cosine, and/or tangent. Round your answer to three decimal places.
𝑩𝑩𝑩𝑩
𝟒𝟒𝟒𝟒𝟒𝟒
𝑩𝑩𝑩𝑩 = 𝟒𝟒𝟒𝟒𝟒𝟒(𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐)
𝐭𝐭𝐭𝐭𝐭𝐭 𝟐𝟐𝟐𝟐 =
𝑩𝑩𝑩𝑩 ≈ 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟔𝟔𝟔𝟔𝟔𝟔
The distance from the road to the back of the property
is approximately 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟔𝟔𝟔𝟔𝟔𝟔 𝐲𝐲𝐲𝐲.
𝟒𝟒𝟒𝟒𝟒𝟒
𝑨𝑨𝑨𝑨
𝟒𝟒𝟒𝟒𝟒𝟒
𝑨𝑨𝑨𝑨 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐
𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 =
𝑨𝑨𝑨𝑨 ≈ 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟎𝟎𝟎𝟎𝟎𝟎
The distance from point 𝑪𝑪 to point 𝑨𝑨 is approximately 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟎𝟎𝟎𝟎𝟎𝟎 𝐲𝐲𝐲𝐲.
7.
The right triangle shown is taken from a slice of a right rectangular pyramid with a square base.
a.
Find the height of the pyramid (to the nearest tenth).
𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 =
𝒉𝒉
𝟗𝟗
𝒉𝒉 = 𝟗𝟗(𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔)
𝒉𝒉 ≈ 𝟖𝟖. 𝟐𝟐
The height of the pyramid is approximately 𝟖𝟖. 𝟐𝟐
units.
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M2
GEOMETRY
b.
Find the lengths of the sides of the base of the pyramid (to the nearest tenth).
The lengths of the sides of the base of the pyramid are twice the length of the short leg of the right triangle
shown.
𝒏𝒏
𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔 =
𝟗𝟗
𝒏𝒏 = 𝟗𝟗(𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)
𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥 = 𝟐𝟐(𝟗𝟗 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)
𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥 = 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔
𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥 ≈ 𝟕𝟕. 𝟑𝟑
The lengths of the sides of the base are approximately 𝟕𝟕. 𝟑𝟑 units.
c.
Find the lateral surface area of the right rectangular pyramid.
The faces of the prism are congruent isosceles triangles having bases of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔 and height of 𝟗𝟗.
𝟏𝟏
𝒃𝒃𝒃𝒃
𝟐𝟐
𝟏𝟏
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = (𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)(𝟗𝟗)
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟖𝟖𝟖𝟖 𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 ≈ 𝟑𝟑𝟑𝟑. 𝟗𝟗
The lateral surface area of the right rectangular pyramid is approximately 𝟑𝟑𝟑𝟑. 𝟗𝟗 square units.
8.
A machinist is fabricating a wedge in the shape of a right triangular prism. One acute angle of the right triangular
base is 𝟑𝟑𝟑𝟑°, and the opposite side is 𝟔𝟔. 𝟓𝟓 𝐜𝐜𝐜𝐜. Find the length of the edges labeled 𝒍𝒍 and 𝒎𝒎 using sine, cosine, and/or
𝟔𝟔. 𝟓𝟓
𝒍𝒍
𝟔𝟔. 𝟓𝟓
𝒍𝒍 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 =
𝒍𝒍 ≈ 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗
Distance 𝒍𝒍 is approximately 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗 𝐜𝐜𝐜𝐜.
𝟔𝟔. 𝟓𝟓
𝒎𝒎
𝟔𝟔. 𝟓𝟓
𝒎𝒎 =
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 =
𝒎𝒎 ≈ 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎
Distance 𝒎𝒎 is approximately 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎 𝐜𝐜𝐜𝐜.
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𝐬𝐬𝐬𝐬
𝟐𝟐
𝜽𝜽 + 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝟐𝟐
𝒍𝒍
� � + 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 = 𝟏𝟏
𝒎𝒎
𝒍𝒍 𝟐𝟐
𝐜𝐜𝐜𝐜𝐬𝐬 𝜽𝜽 = 𝟏𝟏 − � �
𝒎𝒎
𝟐𝟐
𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 =
𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐 𝜽𝜽 =
𝐜𝐜
Lesson 30:
GEO-M2-TE-1.3.0-07.2015
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝐜𝐜𝐜𝐜 𝜽𝜽
By substituting the previous result for 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽,
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝒎𝒎𝟐𝟐
𝒍𝒍𝟐𝟐
− 𝟐𝟐
𝟐𝟐
𝒎𝒎
𝒎𝒎
𝒎𝒎𝟐𝟐 − 𝒍𝒍𝟐𝟐
𝒎𝒎𝟐𝟐
𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝒎𝒎𝟐𝟐 − 𝒍𝒍𝟐𝟐
𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 = �
𝒎𝒎𝟐𝟐
𝐜𝐜𝐜𝐜 𝜽𝜽 =
𝐭𝐭𝐭𝐭 𝜽𝜽 =
𝐜𝐜
𝒍𝒍
, where 𝒍𝒍, 𝒎𝒎 > 𝟎𝟎. Express 𝐭𝐭𝐭𝐭𝐭𝐭 𝜽𝜽 and 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 in terms of 𝒍𝒍 and 𝒎𝒎.
𝒎𝒎
𝐭𝐭
Let 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
𝐧𝐧
9.
𝒍𝒍
𝒎𝒎
√𝒎𝒎𝟐𝟐 − 𝒍𝒍𝟐𝟐
𝒎𝒎
𝒍𝒍
√𝒎𝒎𝟐𝟐 − 𝒍𝒍𝟐𝟐
√𝒎𝒎𝟐𝟐 − 𝒍𝒍𝟐𝟐
𝒎𝒎
Trigonometry and the Pythagorean Theorem
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M2
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
GEOMETRY
Lesson 31: Using Trigonometry to Determine Area
Student Outcomes

Students prove that the area of a triangle is one-half times the product of two side lengths times the sine of
the included angle and solve problems using this formula.

Students find the area of an isosceles triangle given the base length and the measure of one angle.
Lesson Notes
Students discover how trigonometric ratios can help with area calculations in cases where the measurement of the
height is not provided. In order to determine the height in these cases, students must draw an altitude to create right
triangles within the larger triangle. With the creation of the right triangles, students can then set up the necessary
trigonometric ratios to express the height of the triangle (G-SRT.C.8). Students carefully connect the meanings of
1
2
formulas to the diagrams they represent (MP.2 and 7). In addition, this lesson introduces the formula Area = 𝑎𝑎𝑎𝑎 sin 𝐶𝐶
as described by G-SRT.D.9.
Classwork
Opening Exercise (5 minutes)
Opening Exercise
Three triangles are presented below. Determine the areas for each triangle, if possible. If it is not possible to find the
area with the provided information, describe what is needed in order to determine the area.
The area of △ 𝑨𝑨𝑨𝑨𝑨𝑨 is
𝟏𝟏
𝟐𝟐
𝟏𝟏
(𝟓𝟓)(𝟏𝟏𝟏𝟏), or 𝟑𝟑𝟑𝟑 square units, and the area of △ 𝑫𝑫𝑫𝑫𝑫𝑫 is (𝟖𝟖)(𝟐𝟐𝟐𝟐), or 𝟖𝟖𝟖𝟖 square units.
𝟐𝟐
There is
not enough information to find the height of △ 𝑮𝑮𝑮𝑮𝑮𝑮 and, therefore, the area of the triangle.
Is there a way to find the missing information?
Without further information, there is no way to calculate the area.
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Example 1 (13 minutes)

What if the third side length of the triangle were provided? Is it possible to determine the area of the triangle
now?
Example 1
Find the area of △ 𝑮𝑮𝑮𝑮𝑮𝑮.
Allow students the opportunity and the time to determine what they must find (the height) and how to locate it (one
MP.1 option is to drop an altitude from vertex 𝐻𝐻 to side ���
𝐺𝐺𝐺𝐺 ). For students who are struggling, consider showing just the
altitude and allowing them to label the newly divided segment lengths and the height.

How can the height be calculated?

By applying the Pythagorean theorem to both of the created right triangles to find 𝑥𝑥,
ℎ2 = 49 − 𝑥𝑥 2
49 − 𝑥𝑥 2 = 144 − (15 − 𝑥𝑥)2
ℎ2 = 144 − (15 − 𝑥𝑥)2
49 − 𝑥𝑥 2 = 144 − 225 + 30𝑥𝑥 − 𝑥𝑥 2
𝐼𝐼𝐼𝐼
130 = 30𝑥𝑥
13
𝑥𝑥 =
3
13
32
𝐻𝐻𝐻𝐻 =
, =
3
3

The value of 𝑥𝑥 can then be substituted into either of the expressions equivalent to ℎ2 to find ℎ.
13 2
�
3
169
ℎ2 = 49 −
9
ℎ2 = 49 − �
ℎ=
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4√17
3
Using Trigonometry to Determine Area
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Lesson 31
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M2
GEOMETRY

What is the area of the triangle?
1
4√17
Area = � � (15) �
�
3
2
Area = 10√17
Discussion (10 minutes)
Scaffolding:
Now, consider △ 𝐴𝐴𝐴𝐴𝐴𝐴, which is set up similarly to the triangle in Example 1:

 For students who are
struggling, use a numerical
example, such as the
6–8–10 triangle in Lesson
24, to demonstrate how
ℎ can be found using sin 𝜃𝜃.
𝑎𝑎
Write an equation that describes the area of this triangle.

1
2
 For students who are
the unstructured question,
“How could you write a
formula for area that uses
trigonometry, if we didn’t
know ℎ?”
Area = 𝑎𝑎ℎ
Write the left-hand column on the board, and elicit the anticipated student response on the right-hand side after writing
the second line; then, elicit the anticipated student response after writing the third line.

We will rewrite this equation. Describe what you see.
1
Area = 𝑎𝑎ℎ
2
𝑏𝑏
1
Area = 𝑎𝑎ℎ � �
2
1
ℎ
Area = 𝑎𝑎𝑎𝑎 � �
2
The statement is multiplied by 1.
The last statement is rearranged, but the value remains the
same.
𝑏𝑏
MP.2
&
MP.7
Create a discussion around the third line of the newly written formula. Modify △ 𝐴𝐴𝐴𝐴𝐴𝐴: Add in an arc mark at vertex 𝐶𝐶,
and label it 𝜃𝜃.

ℎ
What do you notice about the structure of ? Can we think of this newly written area formula in a different
𝑏𝑏
way using trigonometry?

The value of
ℎ
𝑏𝑏
is equivalent to sin 𝜃𝜃; the
newly written formula can be written as

1
2
Area = 𝑎𝑎𝑎𝑎 sin 𝜃𝜃.
1
If the area can be expressed as 𝑎𝑎𝑎𝑎 sin 𝜃𝜃, which
2
part of the expression represents the height?

ℎ = 𝑏𝑏 sin 𝜃𝜃
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
MP.2
&
MP.7
Compare the information provided to find the area of △ 𝐺𝐺𝐺𝐺𝐺𝐺 in the Opening Exercise, how the information
changed in Example 1, and then changed again in the triangle above, △ ABC.


Only two side lengths were provided in the Opening Exercise, and the area could not be found because
there was not enough information to determine the height. In Example 1, all three side lengths were
provided, and then the area could be determined because the height could be determined by applying
the Pythagorean theorem in two layers. In the most recent figure, we only needed two sides and the
included angle to find the area.
If you had to determine the area of a triangle and were given the option to have three side lengths of a triangle
or two side lengths and the included measure, which set of measurements would you rather have? Why?
The response to this question is a matter of opinion. Considering the amount of work needed to find the area when
provided three side lengths, students should opt for the briefer, trigonometric solution!
Example 2 (5 minutes)
Example 2
A farmer is planning how to divide his land for planting next year’s crops. A triangular plot of land is left with two known
side lengths measuring 𝟓𝟓𝟓𝟓𝟓𝟓 𝐦𝐦 and 𝟏𝟏, 𝟕𝟕𝟕𝟕𝟕𝟕 𝐦𝐦.
What could the farmer do next in order to find the area of the plot?

With just two side lengths known of the plot of land, what are the farmer’s options to determine the area of
his plot of land?


He can either measure the third side length, apply the Pythagorean theorem to find the height of the
triangle, and then calculate the area, or he can find the measure of the included angle between the
known side lengths and use trigonometry to express the height of the triangle and then determine the
area of the triangle.
Suppose the included angle measure between the known side lengths is 30°. What is the area of the plot of
land? Sketch a diagram of the plot of land.

1
2
Area = (1700)(500) sin 30
Area = 212 500
The area of the plot of land is
212,500 square meters.
Exercise 1 (5 minutes)
Exercise 1
A real estate developer and her surveyor are searching for their next piece of land to build on. They each examine a plot
���� and ����
𝑨𝑨𝑨𝑨 and finds them to both be
of land in the shape of △ 𝑨𝑨𝑨𝑨𝑪𝑪. The real estate developer measures the length of 𝑨𝑨𝑨𝑨
approximately 𝟒𝟒, 𝟎𝟎𝟎𝟎𝟎𝟎 feet, and the included angle has a measure of approximately 𝟓𝟓𝟓𝟓°. The surveyor measures the
length of ����
𝑨𝑨𝑨𝑨 and ����
𝑩𝑩𝑩𝑩 and finds the lengths to be approximately 𝟒𝟒, 𝟎𝟎𝟎𝟎𝟎𝟎 feet and 𝟑𝟑, 𝟒𝟒𝟒𝟒𝟒𝟒 feet, respectively, and measures
the angle between the two sides to be approximately 𝟔𝟔𝟔𝟔°.
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a.
Draw a diagram that models the situation, labeling all lengths and angle measures.
b.
The real estate developer and surveyor each calculate the area of the plot of land and both find roughly the
same area. Show how each person calculated the area; round to the nearest hundred. Redraw the diagram
with only the relevant labels for both the real estate agent and surveyor.
c.
square feet.
𝟔𝟔
The area is approximately 𝟔𝟔, 𝟏𝟏
, 𝟖𝟖𝟖𝟖 square feet.
𝟖𝟖
, 𝟑𝟑
) 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔
𝟒𝟒𝟒𝟒𝟒𝟒
𝟓𝟓
𝑨𝑨 =
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
The area is approximately 𝟔𝟔, 𝟏𝟏
𝟏𝟏
(𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑)(𝟒𝟒
𝟐𝟐
𝑨𝑨 ≈ 𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖𝟖𝟖
) 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓
𝟑𝟑𝟑𝟑
𝟒𝟒𝟒𝟒𝟒𝟒
𝟒𝟒𝟒𝟒𝟒𝟒
𝟏𝟏
(𝟒𝟒
)(𝟒𝟒
𝟐𝟐
𝑨𝑨 ≈ 𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑𝟑𝟑
𝑨𝑨 =
What could possibly explain the difference between the real estate agent’s and surveyor’s calculated areas?
The difference in the area of measurements can be accounted for by the approximations of the measurements
Closing (2 minutes)
Ask students to summarize the key points of the lesson. Additionally, consider asking students to respond to the
following questions independently in writing, to a partner, or to the whole class.

For a triangle with side lengths 𝑎𝑎 and 𝑏𝑏 and included angle of measure 𝜃𝜃, when will we need to use the area
1
2
formula Area = 𝑎𝑎𝑎𝑎 sin 𝜃𝜃 ?


We will need it when we are determining the area of a triangle and are not provided a height.
1
2
1
2
Recall how to transform the equation Area = 𝑏𝑏ℎ to Area = 𝑎𝑎𝑎𝑎 sin 𝜃𝜃.
Exit Ticket (5 minutes)
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Name
Date
Lesson 31: Using Trigonometry to Determine Area
Exit Ticket
1.
Given two sides of the triangle shown, having lengths of 3 and 7 and their included angle of 49°, find the area of the
triangle to the nearest tenth.
2.
In isosceles triangle 𝑃𝑃𝑃𝑃𝑃𝑃, the base 𝑄𝑄𝑄𝑄 = 11, and the base angles have measures of 71.45°. Find the area of △ 𝑃𝑃𝑃𝑃𝑃𝑃
to the nearest tenth.
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Exit Ticket Sample Solutions
1.
Given two sides of the triangle shown, having lengths of 𝟑𝟑 and 𝟕𝟕 and their included angle of 𝟒𝟒𝟒𝟒°, find the area of the
triangle to the nearest tenth.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝟏𝟏
(𝟑𝟑)(𝟕𝟕)(𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒)
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟏𝟏𝟏𝟏. 𝟓𝟓(𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒) ≈ 𝟕𝟕. 𝟗𝟗
The area of the triangle is approximately 𝟕𝟕. 𝟗𝟗 square units.
1.
In isosceles triangle 𝑷𝑷𝑷𝑷𝑷𝑷, the base 𝑸𝑸𝑸𝑸 = 𝟏𝟏𝟏𝟏, and the base angles have measures of 𝟕𝟕𝟕𝟕. 𝟒𝟒𝟒𝟒°. Find the area of
△ 𝑷𝑷𝑷𝑷𝑷𝑷 to the nearest tenth.
Drawing an altitude from 𝑷𝑷 to midpoint 𝑴𝑴 on ����
𝑸𝑸𝑸𝑸 cuts
the isosceles triangle into two right triangles with
𝑸𝑸𝑸𝑸 = 𝑴𝑴𝑴𝑴 = 𝟓𝟓. 𝟓𝟓. Using tangent, solve the following:
𝐭𝐭𝐭𝐭𝐭𝐭 𝟕𝟕𝟕𝟕. 𝟒𝟒𝟒𝟒 =
𝑷𝑷𝑷𝑷
𝟓𝟓. 𝟓𝟓
𝑷𝑷𝑷𝑷 = 𝟓𝟓. 𝟓𝟓(𝐭𝐭𝐭𝐭𝐭𝐭 𝟕𝟕𝟕𝟕. 𝟒𝟒𝟒𝟒)
𝟏𝟏
𝒃𝒃𝒃𝒃
𝟐𝟐
𝟏𝟏
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = (𝟏𝟏𝟏𝟏)�𝟓𝟓. 𝟓𝟓(𝐭𝐭𝐭𝐭𝐭𝐭 𝟕𝟕𝟕𝟕. 𝟒𝟒𝟒𝟒)�
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟑𝟑𝟎𝟎. 𝟐𝟐𝟐𝟐(𝐭𝐭𝐭𝐭𝐭𝐭 𝟕𝟕𝟕𝟕. 𝟒𝟒𝟒𝟒) ≈ 𝟗𝟗𝟗𝟗. 𝟏𝟏
The area of the isosceles triangle is approximately 𝟗𝟗𝟗𝟗. 𝟏𝟏 square units.
Problem Set Sample Solutions
Find the area of each triangle. Round each answer to the nearest tenth.
1.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝟏𝟏
(𝟏𝟏𝟏𝟏)(𝟗𝟗) (𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟏𝟏)
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟓𝟓𝟓𝟓(𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟐𝟐) ≈ 𝟏𝟏𝟏𝟏. 𝟒𝟒
The area of the triangle is approximately 𝟏𝟏𝟏𝟏. 𝟒𝟒 square
units.
2.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝟏𝟏
(𝟐𝟐)(𝟏𝟏𝟏𝟏)(𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑)
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟏𝟏𝟏𝟏(𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑) ≈ 𝟔𝟔. 𝟐𝟐
The area of the triangle is approximately 𝟔𝟔. 𝟐𝟐 square units.
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3.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝟏𝟏
𝟏𝟏
(𝟖𝟖) �𝟔𝟔 � (𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓)
𝟐𝟐
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟐𝟐𝟐𝟐(𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓) ≈ 𝟐𝟐𝟐𝟐. 𝟑𝟑
The area of the triangle is approximately 𝟐𝟐𝟐𝟐. 𝟑𝟑 square units.
4.
The included angle is 𝟔𝟔𝟔𝟔° by the angle sum of a triangle.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝟏𝟏
(𝟏𝟏𝟏𝟏)�𝟔𝟔 + 𝟔𝟔√𝟑𝟑� 𝐬𝐬𝐬𝐬𝐧𝐧 𝟔𝟔𝟔𝟔
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟔𝟔�𝟔𝟔 + 𝟔𝟔√𝟑𝟑� �
√𝟑𝟑
�
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = �𝟑𝟑𝟑𝟑 + 𝟑𝟑𝟑𝟑√𝟑𝟑� �
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟏𝟏𝟏𝟏√𝟑𝟑 + 𝟏𝟏𝟏𝟏(𝟑𝟑)
√𝟑𝟑
�
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟏𝟏𝟏𝟏√𝟑𝟑 + 𝟓𝟓𝟓𝟓 ≈ 𝟖𝟖𝟖𝟖. 𝟐𝟐
The area of the triangle is approximately 𝟖𝟖𝟖𝟖. 𝟐𝟐 square units.
5.
In △ 𝑫𝑫𝑫𝑫𝑫𝑫, 𝑬𝑬𝑬𝑬 = 𝟏𝟏𝟏𝟏, 𝑫𝑫𝑫𝑫 = 𝟐𝟐𝟐𝟐, and 𝒎𝒎∠𝑭𝑭 = 𝟔𝟔𝟔𝟔°. Determine the area of the triangle. Round to the nearest tenth.
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = (𝟐𝟐𝟐𝟐)(𝟏𝟏𝟏𝟏)𝐬𝐬𝐬𝐬𝐬𝐬(𝟔𝟔𝟔𝟔) ≈ 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟕𝟕
The area of △ 𝑫𝑫𝑫𝑫𝑫𝑫 is 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟕𝟕 units2.
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6.
A landscape designer is designing a flower garden for a triangular area that is bounded on two sides by the client’s
house and driveway. The length of the edges of the garden along the house and driveway are 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟. and 𝟖𝟖 𝐟𝐟𝐟𝐟.,
respectively, and the edges come together at an angle of 𝟖𝟖𝟖𝟖°. Draw a diagram, and then find the area of the garden
to the nearest square foot.
The garden is in the shape of a triangle in which the lengths of two sides and the included angle have been provided.
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨) =
𝟏𝟏
(𝟖𝟖 𝐟𝐟𝐟𝐟. )(𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟. ) 𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑪𝑪) = (𝟕𝟕𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖) 𝐟𝐟𝐭𝐭𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨) ≈ 𝟕𝟕𝟕𝟕 𝐟𝐟𝐭𝐭 𝟐𝟐
7.
A right rectangular pyramid has a square base with sides of length 𝟓𝟓. Each lateral face of the pyramid is an isosceles
triangle. The angle on each lateral face between the base of the triangle and the adjacent edge is 𝟕𝟕𝟕𝟕°. Find the
surface area of the pyramid to the nearest tenth.
Using tangent, the altitude of the triangle to the base of length 𝟓𝟓 is equal to 𝟐𝟐. 𝟓𝟓 𝐭𝐭𝐭𝐭𝐭𝐭 𝟕𝟕𝟕𝟕.
𝟏𝟏
𝒃𝒃𝒃𝒃
𝟐𝟐
𝟏𝟏
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = (𝟓𝟓)(𝟐𝟐. 𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕)
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟔𝟔. 𝟐𝟐𝟐𝟐(𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕)
The total surface area of the pyramid is the sum of the four lateral faces and the area of the
square base:
𝐒𝐒𝐒𝐒 = 𝟒𝟒�𝟔𝟔. 𝟐𝟐𝟐𝟐(𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕)� + 𝟓𝟓𝟐𝟐
𝐒𝐒𝐒𝐒 = 𝟐𝟐𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕 + 𝟐𝟐𝟐𝟐
𝐒𝐒𝐒𝐒 ≈ 𝟒𝟒𝟒𝟒. 𝟏𝟏
The surface area of the right rectangular pyramid is approximately 𝟒𝟒𝟒𝟒. 𝟏𝟏 square units.
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8.
The Pentagon building in Washington, DC, is built in the shape of a regular pentagon. Each side of the pentagon
measures 𝟗𝟗𝟗𝟗𝟗𝟗 𝐟𝐟𝐟𝐟. in length. The building has a pentagonal courtyard with the same center. Each wall of the center
courtyard has a length of 𝟑𝟑𝟑𝟑𝟑𝟑 𝐟𝐟𝐟𝐟. What is the approximate area of the roof of the Pentagon building?
Let 𝑨𝑨𝟏𝟏 represent the area within the outer perimeter of the
Pentagon building in square feet.
𝑨𝑨𝟏𝟏 =
𝑨𝑨𝟏𝟏 =
𝑨𝑨𝟏𝟏 =
𝒏𝒏𝒃𝒃𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭 �
�
𝒏𝒏
𝟓𝟓 ⋅ (𝟗𝟗𝟗𝟗𝟗𝟗)𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭 �
�
𝟓𝟓
𝟒𝟒 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐
≈ 𝟏𝟏 𝟒𝟒𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑𝟑𝟑
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭(𝟑𝟑𝟑𝟑)
The area within the outer perimeter of the Pentagon building is
approximately 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒, 𝟑𝟑𝟑𝟑𝟑𝟑 𝐟𝐟𝐭𝐭 𝟐𝟐.
Let 𝑨𝑨𝟐𝟐 represent the area within the perimeter of the
courtyard of the Pentagon building in square feet.
𝑨𝑨𝟐𝟐 =
𝟔𝟔𝟔𝟔𝟔𝟔 𝟔𝟔𝟔𝟔𝟔𝟔
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑
𝑨𝑨𝑻𝑻 = 𝑨𝑨𝟏𝟏 − 𝑨𝑨𝟐𝟐
𝟓𝟓(𝟑𝟑𝟑𝟑𝟑𝟑)𝟐𝟐
𝑨𝑨𝟐𝟐 =
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑
𝟎𝟎𝟎𝟎𝟎𝟎
𝟐𝟐𝟐𝟐
𝟐𝟐
𝟒𝟒𝟐𝟐
≈ 𝟐𝟐
𝟑𝟑
𝟒𝟒 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒𝟒𝟒
−
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭(𝟑𝟑𝟑𝟑)
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑
𝑨𝑨𝑻𝑻 =
𝟑𝟑 𝟔𝟔𝟔𝟔𝟔𝟔 𝟓𝟓𝟓𝟓𝟓𝟓
≈ 𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑
𝟒𝟒 𝐭𝐭 𝟑𝟑
𝑨𝑨𝑻𝑻 =
𝟑𝟑
𝟏𝟏𝟏𝟏
𝟏𝟏
𝐭𝐭
𝐭𝐭𝐭𝐭
𝑨𝑨𝟐𝟐 =
𝑨𝑨𝑻𝑻 =
𝟒𝟒 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟔𝟔𝟔𝟔𝟔𝟔 𝟔𝟔𝟔𝟔𝟔𝟔
−
𝟒𝟒 𝐭𝐭𝐚𝐚𝐚𝐚 𝟑𝟑𝟑𝟑
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑
𝟑𝟑
𝒏𝒏𝒃𝒃𝟐𝟐
𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑
𝐭𝐭𝐭𝐭
𝑨𝑨𝟐𝟐 =
Let 𝑨𝑨𝑻𝑻 represent the total area of the roof of the
Pentagon building in square feet.
The area of the roof of the Pentagon building is approximately 𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐, 𝟑𝟑𝟑𝟑𝟑𝟑 𝐟𝐟𝐭𝐭 𝟐𝟐.
9.
A regular hexagon is inscribed in a circle with a radius of 𝟕𝟕. Find the perimeter and area of the hexagon.
The regular hexagon can be divided into six equilateral triangular regions with
each side of the triangles having a length of 𝟕𝟕. To find the perimeter of the
hexagon, solve the following:
𝟔𝟔 ⋅ 𝟕𝟕 = 𝟒𝟒𝟒𝟒 , so the perimeter of the hexagon is 𝟒𝟒𝟒𝟒 units.
To find the area of one equilateral triangle:
𝟏𝟏
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = (𝟕𝟕)(𝟕𝟕) 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝟒𝟒𝟒𝟒 √𝟑𝟑
� �
𝟐𝟐 𝟐𝟐
𝟒𝟒𝟒𝟒√𝟑𝟑
𝟒𝟒
The area of the hexagon is six times the area of the equilateral triangle.
𝟒𝟒𝟒𝟒√𝟑𝟑
𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟔𝟔 �
�
𝟒𝟒
𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝟏𝟏𝟏𝟏𝟏𝟏√𝟑𝟑
≈ 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑
𝟐𝟐
The total area of the regular hexagon is approximately 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑 square units.
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Lesson 31
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GEOMETRY
𝟏𝟏
𝟐𝟐
10. In the figure below, ∠𝑨𝑨𝑨𝑨𝑨𝑨 is acute. Show that 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 (△ 𝑨𝑨𝑨𝑨𝑨𝑨) = 𝑨𝑨𝑨𝑨 ⋅ 𝑩𝑩𝑬𝑬 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨𝑨𝑨𝑨𝑨.
Let 𝜽𝜽 represent the degree measure of angle 𝑨𝑨𝑨𝑨𝑨𝑨, and let 𝒉𝒉 represent the
altitude of △ 𝑨𝑨𝑨𝑨𝑨𝑨 (and △ 𝑨𝑨𝑨𝑨𝑨𝑨).
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨) =
𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
𝟏𝟏
⋅ 𝑨𝑨𝑨𝑨 ⋅ 𝒉𝒉
𝟐𝟐
𝒉𝒉
, which implies that 𝒉𝒉 = 𝑩𝑩𝑩𝑩 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽.
𝑩𝑩𝑩𝑩
Therefore, by substitution:
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨) = 𝑨𝑨𝑨𝑨 ⋅ 𝑩𝑩𝑩𝑩 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨𝑨𝑨𝑨𝑨.
11. Let 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 be a quadrilateral. Let 𝒘𝒘 be the measure of the acute angle formed by diagonals ����
𝑨𝑨𝑨𝑨 and �����
𝑩𝑩𝑩𝑩. Show that
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = 𝑨𝑨𝑨𝑨 ⋅ 𝑩𝑩𝑩𝑩 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘.
(Hint: Apply the result from Problem 10 to △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑨𝑨𝑨𝑨.)
Let the intersection of ����
𝑨𝑨𝑨𝑨 and �����
𝑩𝑩𝑩𝑩 be called point 𝑷𝑷.
Using the results from Problem 10, solve the following:
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨) = 𝑨𝑨𝑨𝑨 ⋅ 𝑩𝑩𝑩𝑩 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘
and
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑨𝑨𝑨𝑨𝑨𝑨) = 𝑨𝑨𝑨𝑨 ⋅ 𝑷𝑷𝑷𝑷 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘
𝟏𝟏
𝟐𝟐
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = � 𝑨𝑨𝑨𝑨 ⋅ 𝑩𝑩𝑩𝑩 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘� + � 𝑨𝑨𝑨𝑨 ⋅ 𝑷𝑷𝑷𝑷 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘�
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = � 𝑨𝑨𝑨𝑨 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘� ⋅ (𝑩𝑩𝑩𝑩 + 𝑷𝑷𝑷𝑷)
𝟏𝟏
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = � 𝑨𝑨𝑨𝑨 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘� ⋅ (𝑩𝑩𝑩𝑩)
And commutative addition gives us 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) =
Lesson 31:
GEO-M2-TE-1.3.0-07.2015
Distributive property
𝟏𝟏
⋅ 𝑨𝑨𝑨𝑨 ⋅ 𝑩𝑩𝑩𝑩 ⋅ 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘.
𝟐𝟐
Using Trigonometry to Determine Area
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Lesson 32
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M2
GEOMETRY
Lesson 32: Using Trigonometry to Find Side Lengths of an
Acute Triangle
Student Outcomes

Students find missing side lengths of an acute triangle given one side length and the measures of two angles.

Students find the missing side length of an acute triangle given two side lengths and the measure of the
included angle.
Lesson Notes
In Lesson 32, students learn how to determine unknown lengths in acute triangles. Once again, they drop an altitude in
the given triangle to create right triangles and use trigonometric ratios and the Pythagorean theorem to solve triangle
problems (G-SRT.C.8). This lesson and the next introduce the law of sines and cosines (G-SRT.D.10 and
G-SRT.D.11).
Based on availability of time in the module, teachers may want to divide the lesson into two parts by addressing
everything until Exercises 1–2 on one day and the remaining content on the following day.
Classwork
Opening Exercise (3 minutes)
The objective for part (b) is that students realize that 𝑥𝑥 and 𝑦𝑦 cannot be found using the method they know with
trigonometric ratios.
Opening Exercise
a.
Find the lengths of 𝒅𝒅 and 𝒆𝒆.
𝟓𝟓
𝟏𝟏𝟏𝟏
; 𝒆𝒆 =
𝒆𝒆
√𝟑𝟑
𝒅𝒅
𝟓𝟓
𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔 =
; 𝒅𝒅 =
𝟏𝟏𝟏𝟏
√𝟑𝟑
√𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔 =
b.
Find the lengths of 𝒙𝒙 and 𝒚𝒚. How is this different from part (a)?
Accept any reasonable answer explaining that the triangle is
not a right triangle; therefore, the trigonometric ratios used in
part (a) are not applicable here.
Lesson 32:
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Lesson 32
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Discussion (10 minutes)
Lead students through an explanation of the law of sines for acute triangles.
Scaffolding:

Today we will show how two facts in trigonometry aid us to find unknown
measurements in triangles that are not right triangles; we can use these facts for
acute and obtuse triangles, but today we will specifically study acute triangles.
The facts are called the law of sines and the law of cosines. We begin with the
law of sines.

LAW OF SINES: For an acute triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴 with ∠𝐴𝐴, ∠𝐵𝐵, and ∠𝐶𝐶 and the sides
opposite them 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐, the law of sines states:
sin ∠𝐴𝐴 sin ∠𝐵𝐵 sin ∠𝐶𝐶
=
=
𝑎𝑎
𝑏𝑏
𝑐𝑐
Restate the law of sines in your own words with a partner.

Show students that the law of
sines holds true by calculating
the ratios using the following
triangles:
This may be difficult for students to articulate generally, without reference to a specific
angle. State it for students if they are unable to state it generally.
The ratio of the sine of an angle in a triangle to the side opposite the
angle is the same for each angle in the triangle.
Consider △ 𝐴𝐴𝐴𝐴𝐴𝐴 with an altitude drawn from 𝐵𝐵 to ����
𝐴𝐴𝐴𝐴 . What is sin ∠𝐶𝐶?






What is sin 𝐴𝐴?
MP.3

sin ∠𝐴𝐴 =
ℎ
𝑐𝑐
Therefore, ℎ = 𝑐𝑐 sin ∠𝐴𝐴.
What can we conclude so far?


ℎ
𝑎𝑎
Therefore, ℎ = 𝑎𝑎 sin ∠𝐶𝐶.


sin ∠𝐶𝐶 =
Since ℎ = 𝑎𝑎 sin ∠𝐶𝐶 and ℎ = 𝑐𝑐 sin ∠𝐴𝐴, then 𝑎𝑎 sin ∠𝐶𝐶 = 𝑐𝑐 sin ∠𝐴𝐴.
With a little algebraic manipulation, we can rewrite 𝑎𝑎 sin ∠𝐶𝐶 = 𝑐𝑐 sin ∠𝐴𝐴 as
sin ∠𝐴𝐴
𝑎𝑎
=
sin ∠𝐶𝐶
𝑐𝑐
.
We have partially shown why the law of sines is true. What do we need to show in order to complete the
proof, and how can we go about determining this?
Allow students a few moments to try and develop this argument independently.
Lesson 32:
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Lesson 32
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY

We need to show that
achieve this drawing:
sin ∠𝐵𝐵
𝑏𝑏
is equal to

An altitude from 𝐴𝐴 gives us sin ∠𝐵𝐵 =

Since

Also, sin ∠𝐶𝐶 =
sin ∠𝐴𝐴
𝑎𝑎
=
sin ∠𝐴𝐴
𝑎𝑎
and to
sin ∠𝐶𝐶
𝑐𝑐
. If we draw a different altitude, we can
ℎ
or ℎ = 𝑐𝑐 sin ∠𝐵𝐵.
𝑐𝑐
sin ∠𝐵𝐵
sin ∠𝐶𝐶
ℎ
or ℎ = 𝑏𝑏 sin ∠𝐶𝐶. Therefore, 𝑐𝑐 sin ∠𝐵𝐵 = 𝑏𝑏 sin ∠𝐶𝐶 or
=
.
𝑏𝑏
𝑏𝑏
𝑐𝑐
sin ∠𝐶𝐶
sin ∠𝐵𝐵
sin ∠𝐶𝐶
sin ∠𝐴𝐴
sin ∠𝐵𝐵
sin ∠𝐶𝐶
𝑐𝑐
As soon as it has been established that
and
sin ∠𝐵𝐵
𝑏𝑏
𝑏𝑏
=
=
sin ∠𝐶𝐶
𝑐𝑐
𝑐𝑐
, then
𝑎𝑎
=
𝑏𝑏
=
𝑐𝑐
.
, the proof is really done, as the angles selected are arbitrary
and, therefore, apply to any angle within the triangle. This can be explained for students if they are ready for the
explanation.
Example 1 (4 minutes)
Students apply the law of sines to determine unknown measurements within a triangle.
Example 1
A surveyor needs to determine the distance between two points 𝑨𝑨 and 𝑩𝑩 that lie on opposite banks of a river. A point 𝑪𝑪
is chosen 𝟏𝟏𝟏𝟏𝟏𝟏 meters from point 𝑨𝑨, on the same side of the river as 𝑨𝑨. The measures of ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑨𝑨𝑨𝑨𝑨𝑨 are 𝟒𝟒𝟒𝟒° and
𝟓𝟓𝟓𝟓°, respectively. Approximate the distance from 𝑨𝑨 to 𝑩𝑩 to the nearest meter.
Allow students a few moments to begin the problem before assisting them.
Lesson 32:
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Using Trigonometry to Find Side Lengths of an Acute Triangle
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Lesson 32
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY

What measurement can we add to the diagram based on the provided information?


Use the law of sines to set up all possible ratios applicable to the diagram.


sin 41
𝑎𝑎
=
160
=
sin 84
160
=
sin 55
𝑐𝑐
Which ratios will be relevant to determining the distance from 𝐴𝐴 to 𝐵𝐵?


The measurement of ∠𝐵𝐵 must be 84° by the triangle sum theorem.
sin 84
Solve for 𝑐𝑐.

𝑐𝑐 =
sin 55
𝑐𝑐
160 sin 55
sin 84
𝑐𝑐 = 132
The distance from 𝐴𝐴 to 𝐵𝐵 is 132 m.
Exercises 1–2 (6 minutes)
Depending on the time available, consider having students move directly to Exercise 2.
Exercises 1–2
1.
In △ 𝑨𝑨𝑨𝑨𝑨𝑨, 𝒎𝒎∠𝑨𝑨 = 𝟑𝟑𝟑𝟑°, 𝒂𝒂 = 𝟏𝟏𝟏𝟏, and 𝒃𝒃 = 𝟏𝟏𝟏𝟏. Find 𝐬𝐬𝐬𝐬𝐬𝐬∠𝑩𝑩. Include a diagram in your answer.
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 𝐬𝐬𝐬𝐬𝐬𝐬∠𝑩𝑩
=
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝟓𝟓
𝐬𝐬𝐬𝐬𝐬𝐬∠𝑩𝑩 =
𝟏𝟏𝟏𝟏
2.
A car is moving toward a tunnel carved out of the base of a hill. As the accompanying diagram shows, the top of the
hill, 𝑯𝑯, is sighted from two locations, 𝑨𝑨 and 𝑩𝑩. The distance between 𝑨𝑨 and 𝑩𝑩 is 𝟐𝟐𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟. What is the height, 𝒉𝒉, of
the hill to the nearest foot?
Let 𝒙𝒙 represent 𝑩𝑩𝑩𝑩, in feet. Applying the law of sines,
𝐬𝐬𝐬𝐬𝐬𝐬 𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑
=
𝒙𝒙
𝟐𝟐𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑
𝒙𝒙 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏
𝒙𝒙 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟏𝟏𝟏𝟏
𝒙𝒙 ≈ 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟗𝟗𝟗𝟗
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�����
𝑩𝑩𝑩𝑩 is the hypotenuse of a 𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 triangle whose sides are in the ratio 𝟏𝟏: 𝟏𝟏: √𝟐𝟐, or 𝒉𝒉: 𝒉𝒉: 𝒉𝒉√𝟐𝟐.
𝒉𝒉√𝟐𝟐 = 𝒙𝒙
𝟏𝟏𝟏𝟏𝟏𝟏
𝒉𝒉√𝟐𝟐 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏
𝒉𝒉 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟏𝟏𝟏𝟏 ⋅ √𝟐𝟐
𝒉𝒉 ≈ 𝟑𝟑𝟑𝟑𝟑𝟑
The height of the hill is approximately 𝟑𝟑𝟑𝟑𝟑𝟑 feet.
Discussion (10 minutes)
Lead students through an explanation of the law of cosines for acute triangles. If more time is needed to cover the
following Discussion, skip Exercise 1 as suggested earlier to allow for more time here.

The next fact we will examine is the law of cosines.

LAW OF COSINES: For an acute triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴 with ∠𝐴𝐴, ∠𝐵𝐵, and ∠𝐶𝐶 and the sides opposite them 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐, the
law of cosines states

𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎 cos ∠𝐶𝐶.
State the law of cosines in your own words.

The square of one side of the triangle is equal to the sum of the squares of the other two sides minus
twice the product of the other two sides and the cosine of the angle between them.
The objective of being able to state the law in words is to move the focus away from specific letters and generalize the
formula for any situation.

The law of cosines is a generalization of the Pythagorean theorem, which can only be used with right triangles.

Substitute 90° for 𝑚𝑚∠𝐶𝐶 into the law of cosines formula, and observe the result.


What is the value of cos 90? What happens to the equation?


𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎 cos 90
2
2
2
cos 90 = 0; the equation simplifies to 𝑐𝑐 = 𝑎𝑎 + 𝑏𝑏 .
This explains why the law of cosines is a generalization of the Pythagorean
theorem. We use the law of cosines for acute triangles in order to determine
the side length of the third side of a triangle, provided two side lengths and the
included angle measure.
Lesson 32:
GEO-M2-TE-1.3.0-07.2015
Scaffolding:
them to substitute 90° for ∠𝐶𝐶,
with the law of cosines being
reduced to the Pythagorean
theorem.
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
���� , but this time the point where the
We return to △ 𝐴𝐴𝐴𝐴𝐴𝐴 from Example 1, with an altitude drawn from 𝐵𝐵 to 𝐴𝐴𝐴𝐴
altitude meets ����
𝐴𝐴𝐴𝐴 , point 𝐷𝐷, divides ����
𝐴𝐴𝐴𝐴 into lengths 𝑑𝑑 and 𝑒𝑒.

Express 𝑒𝑒 and ℎ using trigonometry with respect to ∠𝐶𝐶.



By the Pythagorean theorem, the length relationship in △ 𝐴𝐴𝐴𝐴𝐴𝐴 is 𝑐𝑐 𝟐𝟐 = 𝑑𝑑 𝟐𝟐 + ℎ𝟐𝟐 .
Substitute the trigonometric expressions for 𝑑𝑑 and ℎ into this statement. Notice you will need length 𝑏𝑏.


𝑒𝑒 = 𝑎𝑎 cos∠𝐶𝐶
Now we turn to right triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴. What length relationship can be concluded between the sides of the
triangle?


ℎ = 𝑎𝑎 sin∠𝐶𝐶
Then, the statement becomes 𝑐𝑐 𝟐𝟐 = (𝑏𝑏 − 𝑎𝑎 cos∠𝐶𝐶)𝟐𝟐 + (𝑎𝑎 sin∠ 𝐶𝐶)2.
Simplify this statement as much as possible.

The statement becomes:
𝑐𝑐 2 = 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎cos∠𝐶𝐶 + 𝑎𝑎2 cos 2 ∠𝐶𝐶 + 𝑎𝑎2 sin2 ∠𝐶𝐶
𝑐𝑐 2 = 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎cos∠𝐶𝐶 + 𝑎𝑎2 ((cos ∠𝐶𝐶)2 + (sin ∠𝐶𝐶)2 )
𝑐𝑐 2 = 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎cos∠𝐶𝐶 + 𝑎𝑎2 (1)
𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎cos∠𝐶𝐶

Notice that the right-hand side is composed of two side lengths and the cosine of the included angle. This will
remain true if the labeling of the triangle is rearranged.

What are all possible arrangements of the law of cosines for △ 𝐴𝐴𝐴𝐴𝐴𝐴?


𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2(𝑏𝑏𝑏𝑏)cos∠𝐴𝐴
b2 = a2 + c 2 − 2(𝑎𝑎𝑎𝑎)cos∠𝐵𝐵
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Example 2 (4 minutes)
Example 2
Our friend the surveyor from Example 1 is doing some further work. He has already found the distance between points 𝑨𝑨
and 𝑩𝑩 (from Example 1). Now he wants to locate a point 𝑫𝑫 that is equidistant from both 𝑨𝑨 and 𝑩𝑩 and on the same side of
the river as 𝑨𝑨. He has his assistant mark the point 𝑫𝑫 so that ∠𝑨𝑨𝑨𝑨𝑨𝑨 and ∠𝑩𝑩𝑩𝑩𝑩𝑩 both measure 𝟕𝟕𝟕𝟕°. What is the distance
between 𝑫𝑫 and 𝑨𝑨 to the nearest meter?

What do you notice about △ 𝐴𝐴𝐴𝐴𝐴𝐴 right away?


△ 𝐴𝐴𝐴𝐴𝐴𝐴 must be an isosceles triangle since it has two angles of equal measure.
We must keep this in mind going forward. Add all relevant labels to the diagram.
Students should add the distance of 132 m between 𝐴𝐴 and 𝐵𝐵 and add the label of 𝑎𝑎 and 𝑏𝑏 to the appropriate sides.

Set up an equation using the law of cosines. Remember, we are trying to find the distance between 𝐷𝐷 and 𝐴𝐴
or, as we have labeled it, 𝑏𝑏.


𝑏𝑏 2 = 1322 + 𝑎𝑎2 − 2(132)(𝑎𝑎) cos 75
Recall that this is an isosceles triangle; we know that 𝑎𝑎 = 𝑏𝑏. To reduce variables, we will substitute 𝑏𝑏 for 𝑎𝑎.
Rewrite the equation, and solve for 𝑏𝑏.

Sample solution:
𝑏𝑏 2 = 1322 + (𝑏𝑏)2 − 2(132)(𝑏𝑏) cos 75
𝑏𝑏 2 = 1322 + (𝑏𝑏)2 − 264(𝑏𝑏) cos 75
0 = 1322 − 264(𝑏𝑏) cos 75
264(𝑏𝑏) cos 75 = 1322
𝑏𝑏 =
1322
264 cos 75
𝑏𝑏 ≈ 255 m
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Exercise 3 (2 minutes)
Exercise 3
Parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 has sides of lengths 𝟒𝟒𝟒𝟒 𝐦𝐦𝐦𝐦 and 𝟐𝟐𝟐𝟐 𝐦𝐦𝐦𝐦, and one of the angles has a measure of 𝟏𝟏𝟏𝟏𝟏𝟏°.
Approximate the length of diagonal ����
𝑨𝑨𝑨𝑨 to the nearest millimeter.
3.
In parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨, 𝒎𝒎∠𝑪𝑪 = 𝟏𝟏𝟏𝟏𝟏𝟏°; therefore, 𝒎𝒎∠𝑫𝑫 = 𝟖𝟖𝟖𝟖°.
Let 𝒅𝒅 represent the length of ����
𝑨𝑨𝑨𝑨.
𝒅𝒅𝟐𝟐 = 𝟒𝟒𝟒𝟒𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟐𝟐(𝟒𝟒𝟒𝟒)(𝟐𝟐𝟐𝟐) 𝐜𝐜𝐜𝐜𝐜𝐜 𝟖𝟖𝟖𝟖
𝒅𝒅 = 𝟒𝟒𝟒𝟒
The length of ����
𝑨𝑨𝑨𝑨 is 𝟒𝟒𝟒𝟒 millimeters.
Closing (1 minute)
questions independently in writing, to a partner, or to the whole class.

In what kinds of cases are we applying the laws of sines and cosines?


We apply the laws of sines and cosines when we do not have right triangles to work with. We used the
laws of sines and cosines for acute triangles.
State the law of sines. State the law of cosines.

For an acute triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴 with ∠𝐴𝐴, ∠𝐵𝐵, and ∠𝐶𝐶 and the sides opposite them 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐,
-
-
The law of cosines states: 𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎 cos ∠𝐶𝐶.
The law of sines states:
Exit Ticket (5 minutes)
Lesson 32:
GEO-M2-TE-1.3.0-07.2015
sin ∠𝐴𝐴
𝑎𝑎
=
sin ∠𝐵𝐵
𝑏𝑏
=
sin ∠𝐶𝐶
𝑐𝑐
.
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Name
Date
Lesson 32: Using Trigonometry to Find Side Lengths of an Acute
Triangle
1.
Use the law of sines to find lengths 𝑏𝑏 and 𝑐𝑐 in the triangle below. Round answers to the nearest tenth as necessary.
2.
Given △ 𝐷𝐷𝐷𝐷𝐷𝐷, use the law of cosines to find the length of the side marked 𝑑𝑑 to the nearest tenth.
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GEOMETRY
Exit Ticket Sample Solutions
Use the law of sines to find lengths 𝒃𝒃 and 𝒄𝒄 in the triangle below. Round answers to the nearest tenth as necessary.
𝒎𝒎∠𝑪𝑪 = 𝟖𝟖𝟖𝟖°
𝐬𝐬𝐬𝐬𝐬𝐬 𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬 𝑩𝑩 𝐬𝐬𝐬𝐬𝐬𝐬 𝑪𝑪
=
=
𝒃𝒃
𝒄𝒄
𝒂𝒂
𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖
=
=
𝟏𝟏𝟏𝟏
𝒃𝒃
𝒄𝒄
𝒃𝒃 =
𝟏𝟏𝟏𝟏(𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓)
≈ 𝟐𝟐𝟐𝟐. 𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒
𝒄𝒄 =
𝟏𝟏𝟏𝟏(𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖)
≈ 𝟐𝟐𝟐𝟐. 𝟔𝟔
𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒
Given △ 𝑫𝑫𝑫𝑫𝑫𝑫, use the law of cosines to find the length of the side marked 𝒅𝒅 to the nearest tenth.
𝒅𝒅𝟐𝟐 = 𝟔𝟔𝟐𝟐 + 𝟗𝟗𝟐𝟐 − 𝟐𝟐(𝟔𝟔)(𝟗𝟗) (𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)
𝒅𝒅𝟐𝟐 = 𝟑𝟑𝟑𝟑 + 𝟖𝟖𝟖𝟖 − 𝟏𝟏𝟏𝟏𝟏𝟏(𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)
𝒅𝒅𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏 (𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)
𝒅𝒅 = �𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏(𝐜𝐜𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔)
𝒅𝒅 ≈ 𝟖𝟖. 𝟒𝟒
Problem Set Sample Solutions
1.
Given △ 𝑨𝑨𝑨𝑨𝑨𝑨, 𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏, 𝒎𝒎∠𝑨𝑨 = 𝟓𝟓𝟓𝟓. 𝟐𝟐°, and 𝒎𝒎∠𝑪𝑪 = 𝟕𝟕𝟕𝟕. 𝟒𝟒°, calculate the measure of angle 𝑩𝑩 to the nearest tenth
���� and ����
of a degree, and use the law of sines to find the lengths of 𝑨𝑨𝑨𝑨
𝑩𝑩𝑩𝑩 to the nearest tenth.
By the angle sum of a triangle, 𝒎𝒎∠𝑩𝑩 = 𝟒𝟒𝟒𝟒. 𝟒𝟒°.
𝐬𝐬𝐬𝐬𝐬𝐬 𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬 𝑩𝑩 𝐬𝐬𝐬𝐬𝐬𝐬 𝑪𝑪
=
=
𝒃𝒃
𝒄𝒄
𝒂𝒂
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓. 𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒. 𝟒𝟒 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕. 𝟒𝟒
=
=
𝒂𝒂
𝒃𝒃
𝟏𝟏𝟏𝟏
𝒂𝒂 =
𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓. 𝟐𝟐
≈ 𝟏𝟏𝟏𝟏. 𝟎𝟎
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕. 𝟒𝟒
𝒃𝒃 =
𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒. 𝟒𝟒
≈ 𝟏𝟏𝟏𝟏. 𝟎𝟎
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕. 𝟒𝟒
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GEOMETRY
Calculate the area of △ 𝑨𝑨𝑨𝑨𝑨𝑨 to the nearest square unit.
𝟏𝟏
𝒃𝒃𝒃𝒃 𝐬𝐬𝐬𝐬𝐬𝐬 𝑨𝑨
𝟐𝟐
𝟏𝟏
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = (𝟏𝟏𝟏𝟏)(𝟏𝟏𝟏𝟏) 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓. 𝟐𝟐
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟕𝟕𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓. 𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 ≈ 𝟓𝟓𝟓𝟓
2.
Given △ 𝑫𝑫𝑫𝑫𝑫𝑫, 𝒎𝒎∠𝑭𝑭 = 𝟑𝟑𝟑𝟑°, and 𝑬𝑬𝑬𝑬 = 𝟏𝟏𝟏𝟏, calculate the measure of ∠𝑬𝑬, and use the law of sines to find the lengths
𝑫𝑫𝑫𝑫 to the nearest hundredth.
of ����
𝑫𝑫𝑫𝑫 and ����
By the angle sum of a triangle, 𝒎𝒎∠𝑬𝑬 = 𝟓𝟓𝟓𝟓°.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
=
𝟏𝟏𝟏𝟏
𝒆𝒆
𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
𝒆𝒆 =
𝐬𝐬𝐬𝐬𝐬𝐬
𝒇𝒇 =
𝟖𝟖𝟖𝟖
𝟔𝟔
𝟏𝟏
𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖
𝒇𝒇 ≈ 𝟖𝟖. 𝟐𝟐𝟐𝟐
𝒆𝒆 ≈ 𝟏𝟏 . 𝟔𝟔
3.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑
𝒇𝒇
𝟑𝟑
=
𝟖𝟖𝟖𝟖
𝐬𝐬𝐬𝐬𝐬𝐬
𝟏𝟏
𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬
𝒇𝒇
𝑭𝑭
=
𝑬𝑬
𝐬𝐬𝐬𝐬𝐬𝐬 𝑫𝑫 𝐬𝐬𝐬𝐬𝐬𝐬
=
𝒅𝒅
𝒆𝒆
Does the law of sines apply to a right triangle? Based on △ 𝑨𝑨𝑨𝑨𝑨𝑨, the following ratios were set up according to the
law of sines.
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑩𝑩 𝐬𝐬𝐬𝐬𝐬𝐬 𝟗𝟗𝟗𝟗
=
=
𝒃𝒃
𝒄𝒄
𝒂𝒂
Fill in the partially completed work below:
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨 =
𝟏𝟏
𝒄𝒄
𝐚𝐚
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨
=
𝒂𝒂
𝒄𝒄
What conclusions can we draw?
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑩𝑩 𝐬𝐬𝐬𝐬𝐬𝐬 𝟗𝟗
=
𝒃𝒃
𝒄𝒄
𝟗𝟗
𝟗𝟗
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬 𝟗𝟗
=
𝒂𝒂
𝒄𝒄
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑩𝑩
=
𝒃𝒃
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑩𝑩 =
𝟏𝟏
𝒄𝒄
𝐛𝐛
𝒄𝒄
The law of sines does apply to a right triangle. We get the formulas that are equivalent to 𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑨𝑨 =
𝐨𝐨𝐨𝐨𝐨𝐨
𝐬𝐬𝐬𝐬𝐬𝐬 ∠𝑩𝑩 =
, where 𝑨𝑨 and 𝑩𝑩 are the measures of the acute angles of the right triangle.
𝐡𝐡𝐡𝐡𝐡𝐡
Lesson 32:
GEO-M2-TE-1.3.0-07.2015
𝐨𝐨𝐨𝐨𝐨𝐨
and
𝐡𝐡𝐡𝐡𝐡𝐡
Using Trigonometry to Find Side Lengths of an Acute Triangle
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Lesson 32
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M2
GEOMETRY
4.
Given quadrilateral 𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮, 𝒎𝒎∠𝑯𝑯 = 𝟓𝟓𝟓𝟓°, m∠𝑯𝑯𝑯𝑯𝑯𝑯 = 𝟖𝟖𝟖𝟖°, 𝒎𝒎∠𝑲𝑲𝑲𝑲𝑲𝑲 = 𝟓𝟓𝟓𝟓°, ∠𝑱𝑱 is a right angle, and 𝑮𝑮𝑮𝑮 = 𝟗𝟗 𝐢𝐢𝐢𝐢., use
𝑮𝑮𝑮𝑮 and ����
𝑱𝑱𝑱𝑱 to the nearest tenth of an inch.
the law of sines to find the length of �����
𝑮𝑮𝑮𝑮, and then find the lengths of ���
By the angle sum of a triangle, 𝒎𝒎∠𝑯𝑯𝑯𝑯𝑯𝑯 = 𝟓𝟓𝟓𝟓°; therefore,
△ 𝑮𝑮𝑮𝑮𝑮𝑮 is an isosceles triangle since its base ∠’s have equal
measure.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖
=
𝟗𝟗
𝒉𝒉
𝟗𝟗 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
𝒉𝒉 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟖𝟖𝟖𝟖
𝒌𝒌 = 𝟕𝟕 𝐜𝐜𝐜𝐜𝐜𝐜 𝟓𝟓𝟓𝟓 ≈ 𝟒𝟒. 𝟓𝟓
𝒉𝒉 ≈ 𝟕𝟕. 𝟎𝟎
𝒈𝒈 = 𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 ≈ 𝟓𝟓. 𝟒𝟒
5.
Given triangle 𝑳𝑳𝑳𝑳𝑳𝑳, 𝑳𝑳𝑳𝑳 = 𝟏𝟏𝟏𝟏, 𝑳𝑳𝑳𝑳 = 𝟏𝟏𝟏𝟏, and 𝒎𝒎∠𝑳𝑳 = 𝟑𝟑𝟑𝟑°, use the law of cosines to find the length of �����
𝑴𝑴𝑴𝑴 to the
nearest tenth.
𝒍𝒍𝟐𝟐 = 𝟏𝟏𝟎𝟎𝟐𝟐 + 𝟏𝟏𝟓𝟓𝟐𝟐 − 𝟐𝟐(𝟏𝟏𝟏𝟏)(𝟏𝟏𝟏𝟏) 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑
𝒍𝒍𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑
𝒍𝒍𝟐𝟐 = 𝟑𝟑𝟑𝟑𝟑𝟑 − 𝟑𝟑𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑
𝒍𝒍 = √𝟑𝟑𝟑𝟑𝟑𝟑 − 𝟑𝟑𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑
𝒍𝒍 ≈ 𝟗𝟗. 𝟒𝟒
𝑴𝑴𝑴𝑴 = 𝟗𝟗. 𝟒𝟒
The length of �����
𝑴𝑴𝑴𝑴 is approximately 𝟗𝟗. 𝟒𝟒 units.
6.
Given triangle 𝑨𝑨𝑨𝑨𝑨𝑨, 𝑨𝑨𝑨𝑨 = 𝟔𝟔, 𝑨𝑨𝑨𝑨 = 𝟖𝟖, and 𝒎𝒎∠𝑨𝑨 = 𝟕𝟕𝟕𝟕°, draw a diagram of triangle 𝑨𝑨𝑨𝑨𝑨𝑨, and use the law of cosines
to find the length of ����
𝑩𝑩𝑩𝑩.
𝒂𝒂𝟐𝟐 = 𝟔𝟔𝟐𝟐 + 𝟖𝟖𝟐𝟐 − 𝟐𝟐(𝟔𝟔)(𝟖𝟖)(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒂𝒂𝟐𝟐 = 𝟑𝟑𝟑𝟑 + 𝟔𝟔𝟔𝟔 − 𝟗𝟗𝟗𝟗(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒂𝒂𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟗𝟗𝟗𝟗 𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕
𝒂𝒂 = √𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟗𝟗𝟗𝟗 𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕
𝒂𝒂 ≈ 𝟖𝟖. 𝟗𝟗
The length of ����
𝑩𝑩𝑩𝑩 is approximately 𝟖𝟖. 𝟗𝟗 units.
Calculate the area of triangle 𝑨𝑨𝑨𝑨𝑨𝑨.
𝟏𝟏
𝒃𝒃𝒃𝒃(𝐬𝐬𝐬𝐬𝐬𝐬 𝑨𝑨)
𝟐𝟐
𝟏𝟏
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = (𝟔𝟔)(𝟖𝟖)(𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕)
𝟐𝟐
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟐𝟐𝟐𝟐. 𝟓𝟓(𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕)
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 ≈ 𝟐𝟐𝟐𝟐. 𝟓𝟓
The area of triangle 𝑨𝑨𝑨𝑨𝑨𝑨 is approximately 𝟐𝟐𝟐𝟐. 𝟓𝟓 square units.
Lesson 32:
GEO-M2-TE-1.3.0-07.2015
Using Trigonometry to Find Side Lengths of an Acute Triangle
488
Lesson 33
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Lesson 33: Applying the Laws of Sines and Cosines
Student Outcomes

Students understand that the law of sines can be used to find missing side lengths in a triangle when the
measures of the angles and one side length are known.

Students understand that the law of cosines can be used to find a missing side length in a triangle when the
angle opposite the side and the other two side lengths are known.

Students solve triangle problems using the laws of sines and cosines.
Lesson Notes
In this lesson, students apply the laws of sines and cosines learned in the previous lesson to find missing side lengths of
triangles. The goal of this lesson is to clarify when students can apply the law of sines and when they can apply the law
of cosines. Students are not prompted to use one law or the other; they must determine that on their own.
Classwork
Opening Exercise (10 minutes)
Scaffolding:
Once students have made their decisions, have them turn to a partner and compare their
choices. Pairs that disagree should discuss why and, if necessary, bring their arguments to
MP.3
the whole class so they may be critiqued. Ask students to provide justification for their
choices.
Opening Exercise
For each triangle shown below, identify the method (Pythagorean theorem, law of sines, law of
cosines) you would use to find each length 𝒙𝒙.
 Consider having students
make a graphic organizer
to clearly distinguish
between the law of sines
and the law of cosines.
write a letter to a younger
student that explains the
law of sines and the law of
cosines and how to apply
them to solve problems.
law of sines
Lesson 33:
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Applying the Laws of Sines and Cosines
489
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 33
M2
GEOMETRY
law of cosines
Pythagorean
theorem
law of sines
or
law of cosines
law of sines
Lesson 33:
GEO-M2-TE-1.3.0-07.2015
Applying the Laws of Sines and Cosines
490
Lesson 33
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Example 1 (5 minutes)
Students use the law of sines to find missing side lengths in a triangle.
Example 1
Find the missing side length in △ 𝑨𝑨𝑨𝑨𝑨𝑨.

Which method should we use to find length 𝐴𝐴𝐴𝐴 in the triangle shown below? Explain.
Provide a minute for students to discuss in pairs.

Yes. The law of sines can be used because we are given information about two of the angles and one
side. We can use the triangle sum theorem to find the measure of ∠𝐶𝐶, and then we can use that
information about the value of the pair of ratios
solve to find the missing length.

sin 𝐴𝐴
𝑎𝑎
=
Why can’t we use the Pythagorean theorem with this problem?

sin 𝐶𝐶
𝑐𝑐
. Since the values are equivalent, we can
We can only use the Pythagorean theorem with right triangles. The triangle in this problem is not a
right triangle.

Why can’t we use the law of cosines with this problem?

The law of cosines requires that we know the lengths of two sides of the triangle. We are only given
information about the length of one side.
Write the equation that allows us to find the length of ����
𝐴𝐴𝐴𝐴.



Let 𝑥𝑥 represent the length of ����
𝐴𝐴𝐴𝐴 .
sin 75 sin 23
=
2.93
𝑥𝑥
2.93 sin 23
𝑐𝑐𝑐𝑐 =
sin 75
We want to perform one calculation to determine the answer so that it is most accurate and rounding errors
are avoided. In other words, we do not want to make approximations at each step. Perform the calculation,
and round the length to the tenths place.
 The length of ����
𝐴𝐴𝐴𝐴 is approximately 1.2.
Lesson 33:
GEO-M2-TE-1.3.0-07.2015
Applying the Laws of Sines and Cosines
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Lesson 33
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Example 2 (5 minutes)
Students use the law of cosines to find missing side lengths in a triangle.
Example 2
Find the missing side length in △ 𝑨𝑨𝑨𝑨𝑨𝑨.

���� in the triangle shown below? Explain.
Which method should we use to find side 𝐴𝐴𝐴𝐴
Provide a minute for students to discuss in pairs.
We do not have enough information to use the law of sines because we do not have enough
information to write any of the ratios related to the law of sines. However, we can use
𝑏𝑏 2 = 𝑎𝑎2 + 𝑐𝑐 2 − 2𝑎𝑎𝑎𝑎 cos 𝐵𝐵 because we are given the lengths of sides 𝑎𝑎 and 𝑐𝑐 and we know the angle
measure for ∠𝐵𝐵.
Write the equation that can be used to find the length of ����
𝐴𝐴𝐴𝐴 , and determine the length using one calculation.



𝑥𝑥 2 = 5.812 + 5.952 − 2(5.81)(5.95) cos 16
𝑥𝑥 = �5.812 + 5.952 − 2(5.81)(5.95) cos 16
𝑥𝑥 ≈ 1.6
Scaffolding:
It may be necessary to
demonstrate to students how
to use a calculator to
step.
Exercises 1–6 (16 minutes)
All students should be able to complete Exercises 1 and 2 independently. These exercises can be used to informally
assess students’ understanding in how to apply the laws of sines and cosines. Information gathered from these
problems can inform how to present the next two exercises. Exercises 3–4 are challenging, and students should be
allowed to work in pairs or small groups to discuss strategies to solve them. These problems can be simplified by having
students remove the triangle from the context of the problem. For some groups of students, they may need to see a
model of how to simplify the problem before being able to do it on their own. It may also be necessary to guide
students to finding the necessary pieces of information, for example, angle measures, from the context or the diagram.
Students who need a challenge should have the opportunity to make sense of the problems and persevere in solving
them. The last two exercises are debriefed as part of the Closing.
Lesson 33:
GEO-M2-TE-1.3.0-07.2015
Applying the Laws of Sines and Cosines
492
Lesson 33
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Exercises 1–6
Use the laws of sines and cosines to find all missing side lengths for each of the triangles in the exercises below. Round
1.
Use the triangle to the right to complete this exercise.
a.
Identify the method (Pythagorean theorem, law of sines, law of
cosines) you would use to find each of the missing lengths of the
triangle. Explain why the other methods cannot be used.
Law of sines
The Pythagorean theorem requires a right angle, which is not
applicable to this problem. The law of cosines requires information
about two side lengths, which is not given. Applying the law of sines
requires knowing the measure of two angles and the length of one
side.
b.
���� and ����
Find the lengths of 𝑨𝑨𝑨𝑨
𝑨𝑨𝑨𝑨.
By the triangle sum theorem, 𝒎𝒎∠𝑨𝑨 = 𝟓𝟓𝟓𝟓°.
����.
Let 𝒃𝒃 represent the length of side 𝑨𝑨𝑨𝑨
2.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
=
𝒃𝒃
𝟑𝟑. 𝟑𝟑𝟑𝟑
𝟑𝟑. 𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
𝒃𝒃 =
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
𝒃𝒃 ≈ 𝟑𝟑. 𝟗𝟗
Let 𝒄𝒄 represent the length of side ����
𝑨𝑨𝑨𝑨.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
=
𝒄𝒄
𝟑𝟑. 𝟑𝟑𝟑𝟑
𝟑𝟑. 𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
𝒄𝒄 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
𝒄𝒄 ≈ 𝟑𝟑. 𝟐𝟐
Your school is challenging classes to compete in a triathlon. The race begins with a swim along the shore and then
continues with a bike ride for 𝟒𝟒 miles. School officials want the race to end at the place it began, so after the 𝟒𝟒-mile
bike ride, racers must turn 𝟑𝟑𝟑𝟑° and run 𝟑𝟑. 𝟓𝟓 miles directly back to the starting point. What is the total length of the
a.
Identify the method (Pythagorean theorem, law of sines, law
of cosines) you would use to find the total length of the race.
Explain why the other methods cannot be used.
Law of cosines
The Pythagorean theorem requires a right angle, which is not
applicable to this problem because we do not know if we have
a right triangle. The law of sines requires information about
two angle measures, which is not given. Applying the law of
cosines requires knowing the measure of two sides and the
included angle measure.
b.
Determine the total length of the race. Round your answer to the tenths place.
Let 𝒂𝒂 represent the length of the swim portion of the triathalon.
𝒂𝒂𝟐𝟐 = 𝟒𝟒𝟐𝟐 + 𝟑𝟑. 𝟓𝟓𝟐𝟐 − 𝟐𝟐(𝟒𝟒)(𝟑𝟑. 𝟓𝟓) 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟎𝟎
𝒂𝒂 = �𝟒𝟒𝟐𝟐 + 𝟑𝟑. 𝟓𝟓𝟐𝟐 − 𝟐𝟐(𝟒𝟒)(𝟑𝟑. 𝟓𝟓) 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑
𝒂𝒂 = 𝟐𝟐. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 …
𝒂𝒂 ≈ 𝟐𝟐
Total length of the race: 𝟒𝟒 + 𝟑𝟑. 𝟓𝟓 + (𝟐𝟐) ≈ 𝟗𝟗. 𝟓𝟓
The total length of the race is approximately 𝟗𝟗. 𝟓𝟓 miles.
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GEOMETRY
3.
Two lighthouses are 𝟑𝟑𝟑𝟑 miles apart on each side of shorelines running north and south, as shown. Each lighthouse
keeper spots a boat in the distance. One lighthouse keeper notes the location of the boat as 𝟒𝟒𝟒𝟒° east of south, and
the other lighthouse keeper marks the boat as 𝟑𝟑𝟑𝟑° west of south. What is the distance from the boat to each of the
lighthouses at the time it was spotted? Round your answers to the nearest mile.
Students must begin by identifying the angle formed by one lighthouse, the boat, and the other lighthouse. This may be
accomplished by drawing auxiliary lines and using facts about parallel lines cut by a transversal and the triangle sum
theorem (or knowledge of exterior angles of a triangle). Once students know the angle is 72°, then the other angles in
the triangle formed by the lighthouses and the boat can be found. The following calculations lead to the solution.
Let 𝒙𝒙 be the distance from the southern lighthouse to the boat.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
=
𝒙𝒙
𝟑𝟑𝟑𝟑
𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
𝒙𝒙 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
𝒙𝒙 = 𝟐𝟐𝟐𝟐. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑𝟑𝟑…
The southern lighthouse is approximately 𝟐𝟐𝟐𝟐 𝐦𝐦𝐦𝐦. from the boat.
With this information, students may choose to use the law of sines or the law of cosines to find the other distance.
Shown below are both options.
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Let 𝒚𝒚 be the distance from the northern lighthouse to the boat.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
=
𝒚𝒚
𝟑𝟑𝟑𝟑
𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓
𝒚𝒚 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
𝒚𝒚 = 𝟐𝟐𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕𝟕𝟕 𝟏𝟏𝟏𝟏…
The northern lighthouse is approximately 𝟐𝟐𝟐𝟐 𝐦𝐦𝐦𝐦. from the boat.
Let 𝒂𝒂 be the distance from the northern lighthouse to the boat.
𝒂𝒂𝟐𝟐 = 𝟑𝟑𝟑𝟑𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟐𝟐(𝟑𝟑𝟑𝟑)(𝟐𝟐𝟐𝟐) ⋅ 𝐜𝐜𝐜𝐜𝐜𝐜 𝟓𝟓𝟓𝟓
𝒂𝒂 = �𝟑𝟑𝟑𝟑𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟐𝟐(𝟑𝟑𝟑𝟑)(𝟐𝟐𝟐𝟐) ⋅ 𝐜𝐜𝐜𝐜𝐜𝐜 𝟓𝟓𝟓𝟓
𝒂𝒂 = 𝟐𝟐𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕 𝟒𝟒𝟒𝟒𝟒𝟒 𝟕𝟕𝟕𝟕…
The northern lighthouse is approximately 𝟐𝟐𝟐𝟐 𝐦𝐦𝐦𝐦. from the boat.
If groups of students work the problem both ways using the law of sines and the law of cosines, it may be a good place to
have a discussion about why the answers were close but not exactly the same. When rounded to the nearest mile, the
answers are the same, but if asked to round to the nearest tenths place, the results would be slightly different. The
reason for the difference is that in the solution using the law of cosines, one of the values had already been
approximated (24), leading to an even more approximated and less precise answer.
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GEOMETRY
4.
A pendulum 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. in length swings 𝟕𝟕𝟕𝟕° from right to left. What is the difference between the
highest and lowest point of the pendulum? Round your answer to the hundredths place, and
explain how you found it.
Scaffolding:
Allow students time to struggle
with the problem and, if
necessary, guide them to
drawing the horizontal and
vertical dashed lines shown
below.
At the bottom of a swing, the pendulum is perpendicular to the
ground, and it bisects the 𝟕𝟕𝟕𝟕° angle; therefore, the pendulum
currently forms an angle of 𝟑𝟑𝟑𝟑° with the vertical. By the
triangle sum theorem, the angle formed by the pendulum and
the horizontal is 𝟓𝟓𝟓𝟓°. The 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 will give us the length from the
top of the pendulum to where the horizontal and vertical lines
intersect, 𝒙𝒙, which happens to be the highest point of the
pendulum.
𝒙𝒙
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 = 𝒙𝒙
𝐬𝐬𝐬𝐬𝐬𝐬 𝟓𝟓𝟓𝟓 =
𝒙𝒙
5.
𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓 𝟑𝟑𝟑𝟑𝟑𝟑 𝟗𝟗… = 𝒙𝒙
𝟓𝟓𝟓𝟓°
The lowest point would be when the pendulum is perpendicular
to the ground, which would be exactly 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. Then, the
difference between those two points is approximately 𝟑𝟑. 𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢.
What appears to be the minimum amount of information about a triangle that must be given in order to use the law
of sines to find an unknown length?
To use the law of sines, you must know the measures of two angles and at least the length of one side.
6.
What appears to be the minimum amount of information about a triangle that must be given in order to use the law
of cosines to find an unknown length?
To use the law of cosines, you must know at least the lengths of two sides and the angle measure of the included
angle.
Closing (4 minutes)
Ask students to summarize the key points of the lesson. Additionally, consider asking students the following questions.
Have students respond in writing, to a partner, or to the whole class.

What is the minimum amount of information that must be given about a triangle in order to use the law of
sines to find missing lengths? Explain.

What is the minimum amount of information that must be given about a triangle in order to use the law of
cosines to find missing lengths? Explain.
Exit Ticket (5 minutes)
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Lesson 33
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M2
GEOMETRY
Name
Date
Lesson 33: Applying the Laws of Sines and Cosines
Exit Ticket
1.
Given triangle 𝑀𝑀𝑀𝑀𝑀𝑀, 𝐾𝐾𝐾𝐾 = 8, 𝐾𝐾𝐾𝐾 = 7, and 𝑚𝑚∠𝐾𝐾 = 75°, find the length of the unknown side to the nearest tenth.
2.
Given triangle 𝐴𝐴𝐴𝐴𝐴𝐴, 𝑚𝑚∠𝐴𝐴 = 36°, 𝑚𝑚∠𝐵𝐵 = 79°, and 𝐴𝐴𝐴𝐴 = 9, find the lengths of the unknown sides to the nearest
tenth.
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Lesson 33
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Exit Ticket Sample Solutions
1.
Given triangle 𝑴𝑴𝑴𝑴𝑴𝑴, 𝑲𝑲𝑲𝑲 = 𝟖𝟖, 𝑲𝑲𝑲𝑲 = 𝟕𝟕, and 𝒎𝒎∠𝑲𝑲 = 𝟕𝟕𝟕𝟕°, find the length of the unknown side to the nearest tenth.
The triangle provides the lengths of two sides and their included angles,
so using the law of cosines:
𝒌𝒌𝟐𝟐 = 𝟖𝟖𝟐𝟐 + 𝟕𝟕𝟐𝟐 − 𝟐𝟐(𝟖𝟖)(𝟕𝟕)(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒌𝒌𝟐𝟐 = 𝟔𝟔𝟔𝟔 + 𝟒𝟒𝟒𝟒 − 𝟏𝟏𝟏𝟏𝟏𝟏(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒌𝒌𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒌𝒌 = �𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒌𝒌 ≈ 𝟗𝟗. 𝟐𝟐
2.
Given triangle 𝑨𝑨𝑨𝑨𝑨𝑨, 𝒎𝒎∠𝑨𝑨 = 𝟑𝟑𝟑𝟑°, ∠𝑩𝑩 = 𝟕𝟕𝟕𝟕°, and 𝑨𝑨𝑨𝑨 = 𝟗𝟗, find the lengths of the unknown sides to the nearest
tenth.
By the angle sum of a triangle, 𝒎𝒎∠𝑪𝑪 = 𝟔𝟔𝟔𝟔°.
Using the law of sines:
𝐬𝐬𝐬𝐬𝐬𝐬 𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬 𝑩𝑩 𝐬𝐬𝐬𝐬𝐬𝐬 𝑪𝑪
=
=
𝒃𝒃
𝒄𝒄
𝒂𝒂
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔
=
=
𝟗𝟗
𝒄𝒄
𝒂𝒂
𝒂𝒂 =
𝟗𝟗 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
𝒂𝒂 ≈ 𝟓𝟓. 𝟒𝟒
𝑨𝑨𝑨𝑨 ≈ 𝟖𝟖. 𝟑𝟑 and 𝑩𝑩𝑩𝑩 ≈ 𝟓𝟓. 𝟒𝟒.
𝒄𝒄 =
𝟗𝟗 𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
𝒄𝒄 ≈ 𝟖𝟖. 𝟑𝟑
Problem Set Sample Solutions
1.
Given triangle 𝑬𝑬𝑬𝑬𝑬𝑬, 𝑭𝑭𝑭𝑭 = 𝟏𝟏𝟏𝟏, angle 𝑬𝑬 has a measure of 𝟑𝟑𝟑𝟑°, and angle 𝑭𝑭 has a measure of 𝟕𝟕𝟕𝟕°, find the measures
of the remaining sides and angle to the nearest tenth. Justify your method.
Using the angle sum of a triangle, the remaining angle 𝑮𝑮 has a measure of 𝟕𝟕𝟕𝟕°.
The given triangle provides two angles and one side opposite a given angle, so it is
appropriate to apply the law of sines.
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
=
𝟏𝟏𝟏𝟏
𝑬𝑬𝑬𝑬
𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
𝑬𝑬𝑬𝑬 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑
𝑬𝑬𝑬𝑬 ≈ 𝟐𝟐𝟐𝟐. 𝟐𝟐
Lesson 33:
GEO-M2-TE-1.3.0-07.2015
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
=
=
𝑬𝑬𝑬𝑬
𝑬𝑬𝑬𝑬
𝟏𝟏𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
=
𝟏𝟏𝟓𝟓
𝑬𝑬𝑬𝑬
𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕
𝑬𝑬𝑬𝑬 =
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑
𝑬𝑬𝑬𝑬 ≈ 𝟐𝟐𝟐𝟐. 𝟗𝟗
Applying the Laws of Sines and Cosines
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Lesson 33
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M2
GEOMETRY
2.
Given triangle 𝑨𝑨𝑨𝑨𝑨𝑨, angle 𝑨𝑨 has a measure of 𝟕𝟕𝟕𝟕°, 𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏. 𝟐𝟐, and 𝑨𝑨𝑨𝑨 = 𝟐𝟐𝟐𝟐, find 𝑩𝑩𝑩𝑩 to the nearest tenth. Justify
The given information provides the lengths of the sides of the triangle and
an included angle, so it is appropriate to use the law of cosines.
𝑩𝑩𝑪𝑪𝟐𝟐 = 𝟐𝟐𝟒𝟒𝟐𝟐 + 𝟏𝟏𝟏𝟏. 𝟐𝟐 − 𝟐𝟐(𝟐𝟐𝟐𝟐)(𝟏𝟏𝟏𝟏. 𝟐𝟐)(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝑩𝑩𝑪𝑪𝟐𝟐 = 𝟓𝟓𝟓𝟓𝟓𝟓 + 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎 − 𝟕𝟕𝟕𝟕𝟕𝟕. 𝟔𝟔 𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕
𝑩𝑩𝑪𝑪𝟐𝟐 = 𝟖𝟖𝟖𝟖𝟖𝟖. 𝟎𝟎𝟎𝟎 − 𝟕𝟕𝟕𝟕𝟕𝟕. 𝟔𝟔 𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕
𝑩𝑩𝑩𝑩 = √𝟖𝟖𝟖𝟖𝟖𝟖. 𝟎𝟎𝟎𝟎 − 𝟕𝟕𝟕𝟕𝟕𝟕. 𝟔𝟔 𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕
𝑩𝑩𝑩𝑩 ≈ 𝟐𝟐𝟐𝟐. 𝟗𝟗
3.
James flies his plane from point 𝑨𝑨 at a bearing of 𝟑𝟑𝟑𝟑° east of north, averaging a speed of 𝟏𝟏𝟏𝟏𝟏𝟏 miles per hour for
𝟑𝟑 hours, to get to an airfield at point 𝑩𝑩. He next flies 𝟔𝟔𝟔𝟔° west of north at an average speed of 𝟏𝟏𝟏𝟏𝟏𝟏 miles per hour
for 𝟒𝟒. 𝟓𝟓 hours to a different airfield at point 𝑪𝑪.
a.
Find the distance from 𝑨𝑨 to 𝑩𝑩.
distance = rate ⋅ time
𝒅𝒅 = 𝟏𝟏𝟏𝟏𝟏𝟏 ⋅ 𝟑𝟑
𝒅𝒅 = 𝟒𝟒𝟒𝟒𝟒𝟒
b.
The distance from 𝑨𝑨 to 𝑩𝑩 is 𝟒𝟒𝟒𝟒𝟒𝟒 𝐦𝐦𝐦𝐦.
Find the distance from 𝑩𝑩 to 𝑪𝑪.
distance = rate ⋅ time
𝒅𝒅 = 𝟏𝟏𝟏𝟏𝟏𝟏 ⋅ 𝟒𝟒. 𝟓𝟓
𝒅𝒅 = 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓
c.
d.
The distance from 𝑩𝑩 to 𝑪𝑪 is 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓 𝐦𝐦𝐦𝐦.
Find the measure of angle 𝑨𝑨𝑨𝑨𝑨𝑨.
All lines pointing to the north/south are parallel;
therefore, by alternate interior ∠’s, the return path
from 𝑩𝑩 to 𝑨𝑨 is 𝟑𝟑𝟑𝟑° west of south. Using angles on a
line, this mentioned angle, the measure of the angle
formed by the path from 𝑩𝑩 to 𝑪𝑪 with north (𝟔𝟔𝟔𝟔°)
and 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 sum to 𝟏𝟏𝟏𝟏𝟏𝟏; thus, 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 = 𝟕𝟕𝟕𝟕°.
Find the distance from 𝑪𝑪 to 𝑨𝑨.
The triangle shown provides two sides and the included
angle, so using the law of cosines,
𝒃𝒃𝟐𝟐 = 𝒂𝒂𝟐𝟐 + 𝒄𝒄𝟐𝟐 − 𝟐𝟐𝒂𝒂𝒂𝒂(𝐜𝐜𝐜𝐜𝐜𝐜 𝐁𝐁)
𝒃𝒃𝟐𝟐 = (𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓)𝟐𝟐 + (𝟒𝟒𝟒𝟒𝟗𝟗)𝟐𝟐 − 𝟐𝟐(𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓)(𝟒𝟒𝟒𝟒𝟒𝟒)(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒃𝒃𝟐𝟐 = 𝟑𝟑𝟑𝟑𝟑𝟑 𝟗𝟗𝟗𝟗𝟗𝟗. 𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎 − 𝟒𝟒𝟒𝟒𝟒𝟒 𝟎𝟎𝟎𝟎𝟎𝟎(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒃𝒃𝟐𝟐 = 𝟓𝟓𝟓𝟓𝟓𝟓 𝟎𝟎𝟎𝟎𝟎𝟎. 𝟐𝟐𝟐𝟐 − 𝟒𝟒𝟒𝟒𝟒𝟒 𝟎𝟎𝟎𝟎𝟗𝟗(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒃𝒃 = �𝟓𝟓𝟓𝟓𝟓𝟓 𝟎𝟎𝟎𝟎𝟎𝟎. 𝟐𝟐𝟐𝟐 − 𝟒𝟒𝟒𝟒𝟒𝟒 𝟎𝟎𝟎𝟎𝟎𝟎(𝐜𝐜𝐜𝐜𝐜𝐜 𝟕𝟕𝟕𝟕)
𝒃𝒃 ≈ 𝟔𝟔𝟔𝟔𝟔𝟔. 𝟕𝟕
The distance from 𝑪𝑪 to 𝑨𝑨 is approximately 𝟔𝟔𝟔𝟔𝟔𝟔. 𝟕𝟕 𝐦𝐦𝐦𝐦.
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GEOMETRY
e.
What length of time can James expect the return trip from 𝑪𝑪 to 𝑨𝑨 to take?
distance = rate ⋅ time
𝟔𝟔𝟔𝟔𝟔𝟔. 𝟕𝟕 = 𝟏𝟏𝟏𝟏𝟏𝟏 ⋅ 𝒕𝒕𝟏𝟏
𝟓𝟓. 𝟏𝟏 ≈ 𝒕𝒕𝟏𝟏
distance = rate ⋅ time
𝟔𝟔𝟔𝟔𝟔𝟔. 𝟕𝟕 = 𝟏𝟏𝟏𝟏𝟏𝟏𝒕𝒕𝟐𝟐
𝟒𝟒. 𝟔𝟔 ≈ 𝒕𝒕𝟐𝟐
James can expect the trip from 𝑪𝑪 to 𝑨𝑨 to take between 𝟒𝟒. 𝟔𝟔 and 𝟓𝟓. 𝟏𝟏 hours.
4.
Mark is deciding on the best way to get from point 𝑨𝑨 to point 𝑩𝑩 as shown on the map of Crooked Creek to go fishing.
He sees that if he stays on the north side of the creek, he would have to walk around a triangular piece of private
���� and ����
property (bounded by 𝑨𝑨𝑨𝑨
𝑩𝑩𝑩𝑩). His other option is to cross the creek at 𝑨𝑨 and take a straight path to 𝑩𝑩, which
he knows to be a distance of 𝟏𝟏. 𝟐𝟐 𝐦𝐦𝐦𝐦. The second option requires crossing the water, which is too deep for his boots
and very cold. Find the difference in distances to help Mark decide which path is his better choice.
����
𝑨𝑨𝑨𝑨 is 𝟒𝟒. 𝟖𝟖𝟖𝟖° north of east, and ����
𝑨𝑨𝑨𝑨 is 𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑° east of
north. The directions north and east are
perpendicular, so the angles at point 𝑨𝑨 form a right
angle. Therefore, 𝒎𝒎∠𝑪𝑪𝑪𝑪𝑪𝑪 = 𝟔𝟔𝟔𝟔. 𝟕𝟕𝟕𝟕°.
By the angle sum of a triangle, 𝒎𝒎∠𝑷𝑷𝑷𝑷𝑷𝑷 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔°.
∠𝑷𝑷𝑷𝑷𝑷𝑷 and ∠𝑩𝑩𝑩𝑩𝑩𝑩 are angles on a line with a sum of
𝟏𝟏𝟏𝟏𝟏𝟏°, so 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑°.
Also, by the angle sum of a triangle,
𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 = 𝟒𝟒𝟒𝟒. 𝟖𝟖𝟖𝟖°.
Using the law of sines:
𝐬𝐬𝐬𝐬𝐬𝐬 𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬 𝑩𝑩 𝐬𝐬𝐬𝐬𝐬𝐬 𝑪𝑪
=
=
𝒃𝒃
𝒄𝒄
𝒂𝒂
𝐬𝐬𝐬𝐬𝐬𝐬 𝟔𝟔𝟔𝟔. 𝟕𝟕𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒. 𝟖𝟖𝟖𝟖 𝐬𝐬𝐬𝐬𝐧𝐧 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑
=
=
𝟏𝟏. 𝟐𝟐
𝒃𝒃
𝒂𝒂
𝒃𝒃 =
𝟏𝟏. 𝟐𝟐(𝐬𝐬𝐬𝐬𝐬𝐬 𝟒𝟒𝟒𝟒. 𝟖𝟖𝟖𝟖)
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑
𝒃𝒃 ≈ 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝒂𝒂 =
Let 𝒅𝒅 represent the distance from point 𝑨𝑨 to 𝑩𝑩 through point 𝑪𝑪 in miles.
. (𝐬𝐬𝐬𝐬𝐬𝐬 . 𝟕𝟕 )
𝐬𝐬𝐬𝐬𝐬𝐬 𝟕𝟕 . 𝟑𝟑
𝟏𝟏
𝟐𝟐
𝟔𝟔𝟔𝟔
𝟕𝟕
𝟕𝟕
𝟑𝟑
𝒂𝒂 ≈ . 𝟎𝟎𝟎𝟎𝟎𝟎
𝟏𝟏
𝟎𝟎
𝒅𝒅 = 𝑨𝑨𝑨𝑨 + 𝑩𝑩𝑩𝑩
𝒅𝒅 ≈ 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 + 𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝒅𝒅 ≈ 𝟏𝟏. 𝟗𝟗
The distance from point 𝑨𝑨 to 𝑩𝑩 through point 𝑪𝑪 is approximately 𝟏𝟏. 𝟗𝟗 𝐦𝐦𝐦𝐦. This distance is approximately 𝟎𝟎. 𝟕𝟕 𝐦𝐦𝐦𝐦.
longer than walking a straight line from 𝑨𝑨 to 𝑩𝑩.
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GEOMETRY
5.
If you are given triangle 𝑨𝑨𝑨𝑨𝑨𝑨 and the measures of two of its angles and two of its sides, would it be appropriate to
apply the law of sines or the law of cosines to find the remaining side? Explain.
Case 1:
Given two angles and two sides, one of the angles being
an included angle, it is appropriate to use either the law
of sines or the law of cosines.
Lesson 33:
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Case 2:
Given two angles and the sides opposite those angles,
the law of sines can be applied as the remaining angle
can be calculated using the angle sum of a triangle.
The law of cosines then can also be applied as the
previous calculation provides an included angle.
Applying the Laws of Sines and Cosines
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Lesson 34
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M2
GEOMETRY
Lesson 34: Unknown Angles
Student Outcomes

Students develop an understanding of how to determine a missing angle in a right triangle diagram and apply
this to real-world situations.
Lesson Notes
This lesson introduces students to the use of reasoning based on trigonometric ratios to determine an unknown angle in
a right triangle (G-SRT.C.8). At this stage, students are limited to understanding trigonometry in terms of ratios rather
than functions. However, the concept of an inverse is not dealt with in this course; this is left until Algebra II
(F-BF.B.4). Therefore, these ideas are introduced carefully and without the formalism of inverses, based on students’
existing understanding of trigonometric ratios. It is important not to introduce the idea of inverses at this juncture,
without care, which is why the lesson refers more generally to arcsin, arccos, and arctan as “words” that mathematicians
have used to name, identify, or refer to the degree measures that give a certain trigonometric ratio.
Opening Exercise (12 minutes)
Ask students to complete this exercise independently or with a partner. Circulate and then discuss strategies.
Opening Exercise
a.
Dan was walking through a forest when he came upon a sizable tree. Dan estimated he was about 𝟒𝟒𝟒𝟒 meters
away from the tree when he measured the angle of elevation between the horizontal and the top of the tree
to be 𝟑𝟑𝟑𝟑 degrees. If Dan is about 𝟐𝟐 meters tall, about how tall is the tree?
Let 𝒙𝒙 represent the vertical distance from Dan’s eye level to the top of the tree.
𝒙𝒙
𝟒𝟒𝟒𝟒
𝟒𝟒𝟒𝟒 𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 = 𝒙𝒙
𝟐𝟐𝟐𝟐 ≈ 𝒙𝒙
𝐭𝐭𝐭𝐭𝐭𝐭 𝟑𝟑𝟑𝟑 =
The height of the tree is approximately 𝟑𝟑𝟑𝟑 𝐦𝐦.
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GEOMETRY
b.
Dan was pretty impressed with this tree until he turned around and saw a bigger one, also 𝟒𝟒𝟒𝟒 meters away in
the other direction. “Wow,” he said. “I bet that tree is at least 𝟓𝟓𝟓𝟓 meters tall!” Then, he thought a moment.
“Hmm … if it is 𝟓𝟓𝟓𝟓 meters tall, I wonder what angle of elevation I would measure from my eye level to the top
of the tree?” What angle will Dan find if the tree is 𝟓𝟓𝟓𝟓 meters tall? Explain your reasoning.
Let 𝒙𝒙 represent the angle measure from the horizontal to the top of the tree.
𝟓𝟓𝟓𝟓
𝟒𝟒𝟒𝟒
𝟓𝟓
𝐭𝐭𝐭𝐭𝐭𝐭 𝒙𝒙 =
𝟒𝟒
𝐭𝐭𝐭𝐭𝐭𝐭 𝒙𝒙 = 𝟏𝟏. 𝟐𝟐𝟐𝟐
𝐭𝐭𝐭𝐭𝐭𝐭 𝒙𝒙 =
Encourage students to develop conjectures about what the number of degrees are. Consider selecting some or all of the
measures, placing them in a central visible location, and discussing which are the most reasonable and which are the
least reasonable. Ideally, some students reason that measures less than 35° would be unreasonable; some may draw
hypothetical angles and use these to compute the tangent ratios; still others may use the table of values for tangent
given in Lesson 29 to see that the angle would be slightly greater than 50°.
Discussion (15 minutes)
Just like in the second exercise, sometimes we are confronted with diagrams or problems where we are curious about
what an angle measure might be.
In the same way that mathematicians have named certain ratios within right triangles, they have also developed
terminology for identifying angles in a right triangle, given the ratio of the sides. Mathematicians often use the prefix arc
to define these. The prefix arc is used because of how angles were measured, not just as an angle but also as the length
of an arc on the unit circle. We will learn more about arc lengths in Module 5.
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Lesson 34
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GEOMETRY

Write ratios for sin, cos, and tan of angle 𝐶𝐶:


𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
𝐴𝐴𝐴𝐴
, cos 𝐶𝐶 =
, tan 𝐶𝐶 =
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
Write ratios for sin, cos, and tan of angle 𝐴𝐴:


sin 𝐶𝐶 =
sin 𝐴𝐴 =
𝐵𝐵𝐵𝐵
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
, cos 𝐴𝐴 =
, tan 𝐴𝐴 =
𝐴𝐴𝐶𝐶
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
Mathematicians have developed some additional terms to describe the angles and side ratios in right triangles.
Examine the statements below, and see if you can determine the meaning of each one.
One by one, show each statement. Ask students to make and explain a guess about what these statements mean.



𝐴𝐴𝐴𝐴
� = 𝑚𝑚∠𝐶𝐶
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
arccos � � = 𝑚𝑚∠𝐶𝐶
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
arctan � � = 𝑚𝑚∠𝐶𝐶
𝐵𝐵𝐵𝐵
arcsin �
Once students have shared their guesses, formalize the ideas with a discussion:

Mathematicians use arcsin, arccos, and arctan to refer to the angle measure that results in the given sin, cos,
or tan ratio. For example, for this triangle, mathematicians would say, “arcsin �
meaning of this in your own words.


Explain the meaning of arccos �


𝐵𝐵𝐵𝐵
� = 𝑚𝑚∠𝐶𝐶.
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
This means that the angle that has a cosine ratio equal to
Explain the meaning of arctan �


This means that the angle that has a sine ratio equal to
𝐴𝐴𝐴𝐴
� = 𝑚𝑚∠𝐶𝐶.
𝐵𝐵𝐵𝐵
is 𝑚𝑚∠𝐶𝐶.
𝐵𝐵𝐵𝐵
𝐴𝐴𝐴𝐴
This means that the angle that has a tangent ratio equal to
𝐴𝐴𝐴𝐴
� = 𝑚𝑚∠𝐶𝐶.” Explain the
𝐴𝐴𝐴𝐴
is 𝑚𝑚∠𝐶𝐶.
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
is 𝑚𝑚∠𝐶𝐶.
We can use a calculator to help us determine the values of arcsin, arccos, and arctan. On most calculators,
these are represented by buttons that look like “sin-1,” “cos-1,” and “tan-1.”
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Lesson 34
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M2
GEOMETRY

Let’s revisit the example from the opening. How could we determine the angle of elevation that Dan would
measure if he is 40 meters away and the tree is 50 meters tall?
Let 𝑥𝑥 represent the angle measure from the horizontal to the top of the tree.

48
40
6
tan 𝑥𝑥 =
5
tan 𝑥𝑥 = 1.2
𝑥𝑥 ≈ 50
tan 𝑥𝑥 =
Exercises 1–5 (10 minutes)
Students complete the exercises independently or in pairs.
Exercises 1–5
1.
Find the measure of angles 𝒂𝒂 through 𝒅𝒅 to the nearest degree.
a.
𝟏𝟏𝟏𝟏
𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚 � � ≈ 𝟒𝟒𝟒𝟒
𝟐```