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Transcript
```UNIT 7
Similarity and
Right Triangles
CONTENTS
COMMON
CORE
G-SRT.A.1a
G-SRT.A.2
G-SRT.A.2
G-SRT.A.3
COMMON
CORE
G-SRT.B.4
G-GPE.B.6
G-SRT.B.5
G-SRT.B.4
COMMON
CORE
G-SRT.C.6
G-SRT.C.6
G-SRT.C.8
G-SRT.C.8
F-TF.C.8
823A
Unit 7
MODULE 16
Similarity and Transformations
Lesson 16.1
Lesson 16.2
Lesson 16.3
Lesson 16.4
Dilations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Proving Figures are Similar Using Transformations .
Corresponding Parts of Similar Figures . . . . . . . .
AA Similarity of Triangles . . . . . . . . . . . . . . . . .
MODULE 17
Using Similar Triangles
Lesson 17.1
Lesson 17.2
Lesson 17.3
Lesson 17.4
Triangle Proportionality Theorem . . . . .
Subdividing a Segment in a Given Ratio .
Using Proportional Relationships . . . . .
Similarity in Right Triangles . . . . . . . . .
MODULE 18
Trigonometry with Right Triangles
Lesson 18.1
Lesson 18.2
Lesson 18.3
Lesson 18.4
Lesson 18.5
Tangent Ratio . . . . . . . . . . . . . . .
Sine and Cosine Ratios . . . . . . . . .
Special Right Triangles . . . . . . . . .
Problem Solving with Trigonometry
Using a Pythagorean Identity. . . . .
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UNIT 7
Unit Pacing Guide
45-Minute Classes
Module 16
DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 16.1
Lesson 16.1
Lesson 16.2
Lesson 16.3
Lesson 16.4
DAY 6
Module Review and
Module 17
DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 17.1
Lesson 17.2
Lesson 17.3
Lesson 17.3
Lesson 17.4
DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 18.1
Lesson 18.1
Lesson 18.2
Lesson 18.2
Lesson 18.3
DAY 6
DAY 7
DAY 8
DAY 9
DAY 10
Lesson 18.4
Lesson 18.4
Lesson 18.5
Module Review and
Unit Review and
DAY 6
Module Review and
Module 18
90-Minute Classes
Module 16
DAY 1
DAY 2
DAY 2
Lesson 16.1
Lesson 16.2
Lesson 16.3
Lesson 16.4
Module Review and
Module 17
DAY 1
DAY 2
DAY 3
Lesson 17.1
Lesson 17.2
Lesson 17.3
Lesson 17.4
Module Review and
DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 18.1
Lesson 18.2
Lesson 18.3
Lesson 18.4
Module Review and
Lesson 18.4
Lesson 18.5
Unit Review and
Module 18
Unit 7
823B
Program Resources
PLAN
ENGAGE AND EXPLORE
HMH Teacher App
Access a full suite of teacher resources online and
offline on a variety of devices. Plan present, and
manage classes, assignments, and activities.
Real-World Videos Engage
students with interesting and
relevant applications of the
mathematical content of each
module.
Explore Activities
Students interactively explore new concepts
using a variety of tools and approaches.
ePlanner Easily plan your classes, create
and view assignments, and access all
customizable planning tool.
Professional Development Videos
Authors Juli Dixon and Matt Larson
model successful teaching practices and strategies in actual
classroom settings.
QR Codes Scan with your smart
phone to jump directly from your
print book to online videos and
other resources.
DO NOT EDIT--Changes must be made through "File info"
CorrectionKey=NL-A;CA-A
Teacher’s Edition
Support students with point-of-use Questioning
Strategies, teaching tips, resources for differentiated instruction, additional activities, and more.
NOT
EDIT--Changes
must
through
"File
info"
DODO
NOT
EDIT--Changes
must
bebe
through
"File
info"
CorrectionKey=NL-A;CA-A
CorrectionKey=NL-A;CA-A
22.2
Name
Name
Isosceles and
Equilateral Triangles
NOT
EDIT--Changes
must
through
"File
info"
DODO
NOT
EDIT--Changes
must
bebe
through
"File
info"
CorrectionKey=NL-A;CA-A
CorrectionKey=NL-A;CA-A
Class
Class
straightedge
to draw
segment
to draw
lineline
segment
BCBC
. .
CCUseUsethethestraightedge
22.2 Isosceles
Isoscelesand
andEquilateral
Equilateral
22.2
Triangles
Triangles
Common Core Math Standards
Investigating Isosceles Triangles
INTEGRATE TECHNOLOGY
Resource
Resource
Locker
Locker
G-CO.C.10
Explore
Explore
Vertex
angle
Vertex
angle
MP.3 Logic
angles
have
base
a side
base
angles.
TheThe
angles
thatthat
have
thethe
base
as aasside
areare
thethe
base
angles.
ENGAGE
work
in the
space
provided.
a straightedge
to draw
angle.
in the
space
provided.
UseUse
a straightedge
to draw
an an
angle.
a different
each
time.
is aisdifferent
sizesize
each
time.
Label
your
angle
∠A,
as shown
in the
figure.
Label
your
angle
∠A,
as shown
in the
figure.
A A
Reflect
Reflect
© Houghton Mifflin Harcourt Publishing Company
Make
a Conjecture
Looking
at your
results,
what
conjecture
base
angles,
2. 2. Make
a Conjecture
Looking
at your
results,
what
conjecture
cancan
be be
thethe
base
angles,
∠C?
∠B∠B
andand
∠C?
The
base
angles
congruent.
The
base
angles
areare
congruent.
Using
a compass,
place
point
vertex
draw
intersects
a compass,
place
thethe
point
onon
thethe
vertex
andand
draw
an an
arcarc
thatthat
intersects
thethe
BBUsing
Explain
Explain
1 1 Proving
Provingthe
theIsosceles
Isosceles
Triangle
Theorem
Triangle
Theorem
sides
of the
angle.
Label
points
B and
sides
of the
angle.
Label
thethe
points
B and
C. C.
andItsItsConverse
Converse
and
A A
In the
Explore,
a conjecture
base
angles
of an
isosceles
triangle
congruent.
In the
Explore,
youyou
a conjecture
thatthat
thethe
base
angles
of an
isosceles
triangle
areare
congruent.
This
conjecture
proven
it can
stated
a theorem.
This
conjecture
cancan
be be
proven
so so
it can
be be
stated
as aastheorem.
C C
Isosceles
Triangle
Theorem
Isosceles
Triangle
Theorem
If two
sides
a triangle
congruent,
then
angles
opposite
sides
If two
sides
of aoftriangle
areare
congruent,
then
thethe
twotwo
angles
opposite
thethe
sides
areare
congruent.
congruent.
This
theorem
is sometimes
called
Base
Angles
Theorem
stated
as “Base
angles
This
theorem
is sometimes
called
thethe
Base
Angles
Theorem
andand
cancan
alsoalso
be be
stated
as “Base
angles
of an
isosceles
triangle
congruent.
of an
isosceles
triangle
areare
congruent.
” ”
Module
Module
22 22
must be
EDIT--Changes
A;CA-A
DO NOT
CorrectionKey=NL-
Lesson
Lesson
2 2
1097
1097
"File info"
The angles that have the base as a side are the base angles.
In this activity, you will construct isosceles triangles and investigate other potential
characteristics/properties of these special triangles.
How
know
triangles
constructed
isosceles
triangles?
1. 1. How
do do
youyou
know
thethe
triangles
youyou
constructed
areare
isosceles
triangles?
――.
――
The
compass
marks
equal
lengths
both
sides
∠A;
therefore,
ABAB
≅≅
ACAC
.
The
compass
marks
equal
lengths
onon
both
sides
of of
∠A;
therefore,
Check
students’
construtions.
Check
students’
construtions.
B B
The side opposite the vertex angle is the base.
How could you draw isosceles triangles
without using a compass? Possible answer:
Draw ∠A and plot point B on one side of ∠A. Then
_
use a ruler to measure AB and plot point C on the
other side of ∠A so that AC = AB.
Repeat
steps
A–D
at least
more
times
record
results
in the
table.
Make
sure
steps
A–D
at least
twotwo
more
times
andand
record
thethe
results
in the
table.
Make
sure
∠A∠A
EERepeat
Module
Module
22 22
1098
1098
Lesson
Lesson
2 2
Date
EXPLAIN 1

Proving the Isosceles Triangle
Theorem and Its Converse
Essential
COMMON
CORE
IN1_MNLESE389762_U8M22L2
IN1_MNLESE389762_U8M22L2
10971097
Question:
G-CO.C.10
relationships
the special
What are
triangles?
and equilateral
Prove theorems
triangle is
The congruent
The angle
The side
a triangle
sides are
formed by
opposite
with at least
called the
the legs is
the vertex
among
in isosceles
Resource
Locker
HARDCOVER
PAGES
10971110
HARDCOVER
PAGES
10971110
PROFESSIONALDEVELOPMENT
DEVELOPMENT
PROFESSIONAL
Investigating
Explore
An isosceles
sides
angles and
legs of the
the vertex
angle is the
Isosceles
two congruent
Triangles
sides.
Legs
Vertex angle
triangle.
Base
angle.
Base angles
base.
the
as a side are
base angles.
other potential
the base
and investigate
that have
triangles
isosceles
you will construct special triangles.
angle.
es of these
In this activity,
to draw an
characteristics/properti
Use a straightedge
space provided. figure.
work in the
in the
Do your
as shown
angle ∠A,
A
Label your
The angles

Check students’
construtions.
4/19/14
12:10
4/19/14
12:10
PM PM
Watch
the
hardcover
Watch
forfor
the
hardcover
student
edition
page
student
edition
page
numbers
this
lesson.
numbers
forfor
this
lesson.
IN1_MNLESE389762_U8M22L2
IN1_MNLESE389762_U8M22L2
10981098
LearningProgressions
Progressions
Learning
this
lesson,
students
their
prior
knowledge
isosceles
and
equilateral
InIn
this
lesson,
students
toto
their
prior
knowledge
ofof
isosceles
and
equilateral
4/19/14
12:10
4/19/14
12:10
PM PM
Do your work in the space provided. Use a straightedge to draw an angle.
Label your angle ∠A, as shown in the figure.
A
CONNECT VOCABULARY
Ask a volunteer to define isosceles triangle and have
students give real-world examples of them. If
possible, show the class a baseball pennant or other
flag in the shape of an isosceles triangle. Tell students
they will be proving theorems about isosceles
triangles and investigating their properties in this
lesson.
Class
al
and Equilater
22.2 Isosceles
Triangles
Name
Legs
The angle formed by the legs is the vertex angle.
What must be true about the triangles you
construct in order for them to be isosceles
triangles? They must have two congruent sides.
m∠B
m∠B
m∠C
m∠C
Vertex angle
The congruent sides are called the legs of the triangle.
QUESTIONING STRATEGIES
Possible
Triangle
m∠A
70°;
m∠B
∠55°;
m∠C
55°.
Possible
forfor
Triangle
1: 1:
m∠A
==
70°;
m∠B
==
∠55°;
m∠C
==
55°.
In this
activity,
construct
isosceles
triangles
investigate
other
potential
In this
activity,
youyou
willwill
construct
isosceles
triangles
andand
investigate
other
potential
characteristics/properties
of these
special
triangles.
characteristics/properties
of these
special
triangles.
© Houghton Mifflin Harcourt Publishing Company
Triangle
Triangle
4 4
© Houghton Mifflin Harcourt Publishing Company
View the Engage section online. Discuss the photo,
explaining that the instrument is a sextant and that
long ago it was used to measure the elevation of the
sun and stars, allowing one’s position on Earth’s
surface to be calculated. Then preview the Lesson
Triangle
Triangle
3 3
© Houghton Mifflin Harcourt Publishing Company
PREVIEW: LESSON
Triangle
Triangle
2 2
m∠mA∠A
Base
Base
Base
angles
Base
angles
opposite
vertex
angle
is the
base.
TheThe
sideside
opposite
thethe
vertex
angle
is the
base.
Explain to a partner what you can deduce about a triangle if it has two
sides with the same length.
In an isosceles triangle, the angles opposite the
congruent sides are congruent. In an equilateral
triangle, all the sides and angles are congruent, and
the measure of each angle is 60°.
Triangle
Triangle
1 1
angle
formed
is the
vertex
angle.
TheThe
angle
formed
by by
thethe
legslegs
is the
vertex
angle.
Language Objective
Essential Question: What are the
special relationships among angles and
sides in isosceles and equilateral
triangles?
Legs
Legs
congruent
sides
called
of the
triangle.
TheThe
congruent
sides
areare
called
thethe
legslegs
of the
triangle.
ing Company
COMMON
CORE
DD
InvestigatingIsosceles
Isosceles
Triangles
Investigating
Triangles
isosceles
triangle
a triangle
with
at least
congruent
sides.
AnAn
isosceles
triangle
is aistriangle
with
at least
twotwo
congruent
sides.
Mathematical Practices
Investigating Isosceles Triangles
An isosceles triangle is a triangle with at least two congruent sides.
Students have the option of completing the isosceles
triangle activity either in the book or online.
a protractor
to measure
each
angle.
Record
measures
in the
table
under
column
UseUse
a protractor
to measure
each
angle.
Record
thethe
measures
in the
table
under
thethe
column
Triangle
forfor
Triangle
1. 1.
Resource
Locker
Explore
C C
The student is expected to:
COMMON
CORE
COMMON
CORE
EXPLORE
A A
B B
Essential
Question:
What
special
relationships
among
angles
and
sides
in isosceles
Essential
Question:
What
areare
thethe
special
relationships
among
angles
and
sides
in isosceles
and
equilateral
triangles?
and
equilateral
triangles?
Date
Essential Question: What are the special relationships among angles and sides in isosceles
and equilateral triangles?
NOT
EDIT--Changes
must
through
"File
info"
DODO
NOT
EDIT--Changes
must
bebe
through
"File
info"
CorrectionKey=NL-A;CA-A
CorrectionKey=NL-A;CA-A
__
Date
Date
Class
22.2 Isosceles and Equilateral
Triangles
DONOT
NOTEDIT--Changes
EDIT--Changesmust
through"File
"Fileinfo"
info"
DO
CorrectionKey=NL-A;CA-A
CorrectionKey=NL-A;CA-A
DO NOT EDIT--Changes must be made through "File info"
CorrectionKey=NL-A;CA-A
LESSON
Name
Base
Base angles
PROFESSIONAL DEVELOPMENT
TEACH
ASSESSMENT AND INTERVENTION
Math On the Spot video tutorials, featuring
program authors Dr. Edward Burger and Martha
Sandoval-Martinez, accompany every example in
the textbook and give students step-by-step
instructions and explanations of key math
concepts.
Interactive Teacher Edition
Customize and present course materials with
collaborative activities and integrated formative
assessment.
C1
Lesson 19.2 Precision and Accuracy
Evaluate
1
Lesson XX.X ComparingLesson
Linear,
19.2 Precision and Accuracy
teacher Support
1
EXPLAIN Concept 1
Explain
The Personal Math Trainer provides
online practice, homework, assessments,
and intervention. Monitor student
Create and customize assignments aligned to specific
lessons or Common Core standards.
• Practice – With dynamic items and assignments, students
get unlimited practice on key concepts supported by
guided examples, step-by-step solutions, and video
tutorials.
• Assessments – Choose from course assignments or
customize your own based on course content, Common
Core standards, difficulty levels, and more.
• Homework – Students can complete online homework with
a wide variety of problem types, including the ability to
enter expressions, equations, and graphs. Let the system
automatically grade homework, so you can focus where
help the most!
• Intervention – Let
the Personal Math
Trainer automatically
prescribe a targeted,
personalized
intervention path for
2
3
4
Question 3 of 17
Concept 2
Determining Precision
ComPLEtINg thE SquArE wIth EXPrESSIoNS
Avoid Common Errors
Some students may not pay attention to
whether b is positive or negative, since c
is positive regardless of the sign of b. Have
student change the sign of b in some problems and compare the factored forms of both
expressions.
questioning Strategies
In a perfect square trinomial, is the last term
always positive?
Explain. es, a perfect square trinomial can be
either (a + b)2 or (a – b)2 which can be factored as (a + b)2 = a 2 + 2ab = b 2 and (a – b)2
= a 2 + 2ab = b 2. In both cases the last term is
positive.
reflect
3. The sign of b has no effect on the sign of
c because c = ( b __ 2 ) 2 and a nonzero
number squared is always positive. Thus, c is
always positive. c = ( b __ 2 ) 2 and a
nonzero number c = ( b __ 2 ) 2 and a nonzero number
5
6
7
View Step by Step
8
9
10
11 - 17
Video Tutor
Personal Math Trainer
Textbook
X2 Animated Math
Solve the quadratic equation by factoring.
7x + 44x = 7x − 10
As you have seen, measurements are given to a certain precision. Therefore,
x=
the value reported does not necessarily represent the actual value of the
measurement. For example, a measurement of 5 centimeters, which is
,
Check
given to the nearest whole unit, can actually range from 0.5 units below the
reported value, 4.5 centimeters, up to, but not including, 0.5 units above
it, 5.5 centimeters. The actual length, l, is within a range of possible values:
Save & Close
centimeters. Similarly, a length given to the nearest tenth can actually range
from 0.05 units below the reported value up to, but not including, 0.05 units
above it. So a length reported as 4.5 cm could actually be as low as 4.45 cm or
as high as nearly 4.55 cm.
?
!
Turn It In
Elaborate
Look Back
Focus on Higher Order Thinking
Raise the bar with homework and practice that incorporates
higher-order thinking and mathematical practices in every lesson.
Differentiated Instruction Resources
Support all learners with Differentiated
Instruction Resources, including
• Leveled Practice and Problem Solving
• Success for English Learners
• Challenge
the nearest square centimeters.
The width and length of a rectangle are 8 cm and 19.5 cm, respectively.
Prepare students for success on high
stakes tests for Integrated Mathematics 2
with practice at every module and unit
Find the range of values for the actual length and width of the rectangle.
Minimum width =
7.5
cm and maximum width <
8.5 cm
Find the range of values for the actual length and width of the rectangle.
Minimum length =
19.45
cm and maximum length < 19.55
Name ________________________________________ Date __________________ Class __________________
LESSON
1-1
Assessment Resources
cm
Name ____________
__________________
__________ Date
__________________
LESSON
Class ____________
______
Precision and Significant Digits
6-1
Success for English Learners
Linear Functions
Reteach
The graph of a linear
The precision of a measurement is determined
bythe
therange
smallest
unit or
Find
of values
for the actual length and width of the rectangle.
function is a straig
ht line.
fraction of a unit used.
Ax + By + C = 0
is the standard
form for the equat
ion of a linear functi
• A, B, and C are
on.
Problem 1
Minimum Area = Minimum width × Minimum length real numbers. A and B are not
both zero.
• The variables
x and y
Choose the more precise measurement.
=
7.5 cm × 19.45 cm
have exponents
of 1
are not multiplied
together
are not in denom
42.3 g is to the
42.27 g is to the
inators, exponents
nearest tenth.
nearest
Examples
These are NOT
hundredth.
linear functions:
2+4=6
no variable
x2 = 9
exponent on x ≥
1
xy = 8
x and y multiplied
42.3 g or 42.27 g
together
6
=3
Because a hundredth of a gram is smaller than a tenth of a gram, 42.27 g
x in denominator
x
is more precise.
2y = 8
y in exponent
Problem 2
In the above exercise, the location of the uncertainty in the linear y = 5
y in a square root
measurements results in different amounts of uncertainty in the calculated
Choose the more precise measurement: 36 inches or 3 feet. measurement. Explain how to fix this problem.
Tell whether each
function is linear
or not.
1. 14 = 2 x
2. 3xy = 27
3. 14 = 28
4. 6x 2 = 12
x
____________
Reflect
____
________________
_______________
The graph of y =
C is always a horiz
ontal line. The graph
always a vertical
line.
of x = C
_______________
is
Unit 7
Send to Notebook
_________________________________________________________________________________________
2. An object is weighed on three different scales. The results are shown
Explore
in the table. Which scale is the
Measurement
____________________________________________________________
• Tier 1, Tier 2, and Tier 3 Resources
Examples
1. When deciding which measurement is more precise, what should
you
Formula
consider?
Scale
Tailor assessments and response to intervention to meet the
needs of all your classes and students, including
• Leveled Module Quizzes
• Leveled Unit Tests
• Placement, Diagnostic, and Quarterly Benchmark Tests
y=1
T
x=2
y = −3
x=3
823D
Math Background
Transformations
COMMON
CORE
G-SRT.A.2
LESSONS 16.1 and 16.2
A transformation is a function that changes the position,
size, or shape of a figure. In this course, the emphasis is on
transformations that are most closely linked to congruence
and similarity: reflections, translations, rotations, and
dilations.
However, it is important to understand that there exist
many other transformations. Perhaps the simplest
transformation is the transformation that maps every point
to itself. This is known as the identity transformation. Another
simple transformation is the one that maps every point to
the origin.
Dilations
COMMON
CORE
G-SRT.A.1
LESSON 16.1
A dilation is a transformation that changes the size of a
figure, but not its shape. As such, a dilation is an example of
a transformation that is not an isometry (unless the scale
factor of the dilation is 1). Every dilation has exactly one
fixed point, which is the center of the dilation.
Although dilations do not preserve distance, they do
preserve other properties of a figure. For example, dilations
preserve angle measure. In other words, under a dilation,
an angle in the preimage is congruent to the corresponding
angle in the image.
Dilations also preserve parallel lines. That is, two lines that
are parallel in the preimage are mapped to two parallel lines
in the image.
If two figures are congruent, then there is an isometry that
maps one figure onto the other. If two figures are similar,
one may be mapped onto the other through a combination
of a dilation and an isometry.
A dilation with a scale factor greater than 1 is an enlargement.
A dilation with a scale factor greater than 0 but less than 1 is a
reduction.
It is also possible to extend the definition of a dilation to
allow a scale factor of 0 (in this case, the entire preimage
823E
Unit 7
is collapsed to a single point, the center of dilation) and
negative scale factors.
A dilation with a scale factor of –k , where k > 0, is
equivalent to the dilation with scale factor k followed by a
rotation of 180° about the center of dilation.
Similarity
COMMON
CORE
G-SRT.A.2
LESSONS 16.2 to 16.4
Recall that two figures can be defined as congruent if there
is a sequence of isometries—reflections, translations,
and/or rotations—that maps one figure onto the
other. Likewise, similarity may be defined in terms of
transformations. In particular, two figures are similar if one
may be obtained from the other through a combination of
a dilation and one or more isometries. Dilations transform
figures by enlarging or reducing them.
Finally, it is worth noting that similarity is an equivalence
relation; that is, similarity is reflexive, symmetric, and
transitive.
For figures F 1, F 2, and F 3, F 1 ∼ F 1 (reflexivity);
if F 1 ∼ F 2, then F 2 ∼ F 1 (symmetry); and
if F 1 ∼ F 2 and F 2 ∼ F 3, then F 1 ∼ F 3 (transitivity).
Triangle Proportionality
COMMON
CORE
G-SRT.B.4
LESSON 17.1
The Triangle Proportionality Theorem states that if a line
is parallel to a side of a triangle and intersects the other
two sides, then it divides those sides proportionally. In the
‹ › _
−
AE
AF
figure, EF ∥ BC. Therefore, __
= __
.
EB
FC
A
E
B
F
C
PROFESSIONAL DEVELOPMENT
The proof, which uses similarity and facts about proportions,
runs as follows. Because they are corresponding angles,
∠AEF ≅ ∠ABC. Since ∠A ≅ ∠A, ▵AEF ∼ ▵ABC by AA
Similarity.
Similarity in Right
Triangles
AE
AF
Thus, __
= __
, so the reciprocals are also equal,
AB
AC
In 1816, the French mathematician August Leopold Crelle
stated, “It is indeed wonderful that so simple a figure as the
triangle is so inexhaustible in properties.” It is possible to
develop and prove theorems that are increasingly elegant
and unexpected, for example, the rather surprising fact that
the three altitudes of any triangle are concurrent.
AC
AB
__
= __
.
AE
AF
By the Segment Addition Postulate, AB = AE + EB and
AE + EB
AF + AC
AC = AF + FC. By substitution, ______
= ______
or
AE
AF
FC
FC
EB
EB
1 + __
= 1 + __
. Therefore, __
= __
, which is equivalent to
AF
AE
AF
AE
the required proportion.
Proportional
Relationships
COMMON
CORE
G-SRT.B.5
LESSONS 17.2 and 17.3
A proportion is an equation that states that two ratios are
equal. Students should realize that the steps for solving a
proportion such as __3x = __58 are the same as those for solving
any other type of equation; that is, the equation must be
transformed into simpler equivalent equations with the
goal of isolating the variable on one side of the equation.
A possible first step in solving __3x = __58 is to clear the fractions
by multiplying both sides of the equation by 3 ⋅ 8 or 24. This
results in the equivalent equation 8x = 15, which may be
solved by dividing both sides by 8.
Students may already be familiar with the shortcut that
results from performing the initial multiplication on a
general proportion—the Cross Product Property. This
property states that if __ba = __dc , where b ≠ 0 and d ≠ 0,
To see why the Cross Product Property is true, begin by
multiplying both sides of the proportion __ba = __dc by bd.
a _
c
_
= ,
b d
a ( )_
c
(bd)_
= bd ,
b
d
bda _
bdc
_
=
b
d
Multiplication Property of Equality
Multiply.
da = bc
Simplify.
Commutative Property of Mult.
COMMON
CORE
G-SRT.B.4
LESSON 17.4
This lesson contains the following important theorems:
• The altitude to the hypotenuse of a right triangle forms
two triangles that are similar to each other and to the
original triangle.
• The length of the altitude to the hypotenuse of a right
triangle is the geometric mean of the lengths of the
two segments of the hypotenuse.
• The length of a leg of a right triangle is the geometric
mean of the lengths of the hypotenuse and segment of
the hypotenuse adjacent to that leg.
• The Pythagorean Theorem
The Pythagorean Theorem, one of the best-known
relationships in mathematics, can be traced back almost as
far as recorded history itself. Most of the early appearances
of the theorem occur in the form of Pythagorean triples
(sets of three nonzero whole numbers a, b, and c such
that a 2 + b 2 = c 2). One example is the clay tablet known
as Plimpton 322, which was written in Babylonia around
1800 B.C.E.; it contains 15 rows of numbers based on
Pythagorean triples.
Such archaeological findings show that the relationship was
known long before the time of the Greek mathematician
Pythagoras (c. 582 B.C.E.–507 B.C.E.); it is because of the work
of later Greek and Roman historians that the theorem has
come to bear Pythagoras’s name.
Regardless of its beginnings, the theorem has continued
to inspire both professional and amateur mathematicians
because of its elegance and adaptability to diverse methods
of proof.
In fact, The Pythagorean Proposition by Elisha Scott Loomis
presents more than 350 different proofs of the Pythagorean
Theorem!
Unit 7
823F
UNIT
7
UNIT 7
Similarity and Right
Triangles
MODULE
Similarity and
Right Triangles
MATH IN CAREERS
Unit Activity Preview
16
Similarity and
Transformations
MODULE
17
Using Similar Triangles
MODULE
18
Trigonometry with Right
Triangles
After completing this unit, students will complete a
Math in Careers task by using similarity to make a
calculation for a special effects engineer. Critical
skills include modeling real-world situations and
using similar figures to find missing measurements.
Pictures & Touchstone Pictures/Everett Collection, Inc.
as well as various mathematics appreciation topics,
visit The American Mathematical Society at
http://www.ams.org.
MATH IN CAREERS
Special Effects Engineer Special
effects engineers make movies come
to life. With the use of math and some
creative camera angles, special effects
engineers can make big things appear
small and vice versa.
If you’re interested in a career as a special
effects engineer, you should study these
mathematical subjects:
• Algebra
• Geometry
• Trigonometry
Research other careers that require
the use of engineering to understand
real-world scenarios. See the related
Career Activity at the end of this unit.
Unit 7
811
IN2_MNLESE389847_U7UO.indd 811
4/12/14 1:44 PM
Before
In this Unit
After
Students understand:
• parallel lines, transversals, and angle relationships
• perpendicular lines and bisectors
• slopes and equations of parallel and
perpendicular lines
• properties of parallelograms, rectangles, rhombuses,
and squares
• special segments of triangles
• similarity and dilations
• similarity of circles
• corresponding parts of similar figures
• proving triangles similar
• the triangle proportionality theorem
• dividing segments in a given ratio
• geometric means theorems
• Using tangents, sine, and cosine
Students will study:
• central and inscribed angles
• chords, secants, tangent lines, and arcs
• segment lengths in circles
• angles formed by intersecting lines of a circle
• formulas for circumference and area of a circle
• area of a sector
823
Unit 7
Visualize Vocabulary
Use the ✔ words to complete the main idea web. Write the review words
in the squares and include definitions.
Translation
A translation
slides all points
of a figure the
same distance
in the same
direction.
Images formed by rigid
transformations
are congruent to their preimages.
Reflection
Rotation
A reflection across a
line maps each point
to its image such that
the line forms the
perpendicular bisector
of the point and its
image. If the point is
on the line, then the
image is as well.
A rotation moves
center such that the
distance from the
center doesn’t
change and all angles
formed by a point
and its image have
the same measures.
Complete the sentences using the preview words.
2.
3.
The image formed by a(n) similarity transformation/dilation has the
same shape as its pre-image.
The scale factor indicates the ratio of the lengths of corresponding sides
of two similar figures.
geometric mean .
x
a = __
In the proportion __
x b , x is called the
Review Words
✔ betweenness
(intermediación)
✔ congruent (congruente)
✔ hypotenuse (hipotenusa)
✔ legs (catetos)
✔ orientation (orientación)
✔ parallel (paralelo)
✔ reflection (reflejo)
✔ rotation (rotación)
✔ transformation
(transformación)
✔ translation (traslación)
working alone or with others.
VISUALIZE VOCABULARY
The word web graphic helps students review
vocabulary associated with transformations. If time
allows, brainstorm other connections among the
review words.
Preview Words
center of dilation (centro de la
dilatación)
cosine (coseno)
dilation (dilatación)
geometric mean (media
geométrica)
indirect measurement
(medición indirecta)
scale factor (factor de escala)
similar (similar)
similarity transformation
(transformación de
semejanza)
sine (seno)
tangent (tangente)
trigonometric ratio (razón
trigonométrica)
UNDERSTAND VOCABULARY
Use the following explanations to help students learn
the preview words.
A dilation or similarity transformation is a
transformation that changes the size of a figure but
not its shape. The image of a dilation is a figure
similar to the preimage. The center of dilation is
the intersection of the lines that connect each
point of the image with the corresponding point of
the preimage. Indirect measurement is a method
of measurement that uses similar figures. The
multiplier used on each dimension to change one
figure into a similar figure is the scale factor.
© Houghton Mifflin Harcourt Publishing Company
Understand Vocabulary
1.
Vocabulary
Students can use these reading and note-taking
strategies to help them organize and understand the
new concepts and vocabulary. Encourage students to
and associated concepts that seem unclear.
Emphasize the importance of understanding precise
mathematical language to accurately work through
problems.
Double-Door Fold Create a Double-Door Fold prior to
starting the unit. Write characteristics of congruency
under one flap. Fill out the other flap with corresponding
characteristics of similarity so that the two topics can be compared
more easily.
Unit 7
IN2_MNLESE389847_U7UO.indd 812
812
4/12/14 1:43 PM
Differentiated Instruction
Unit 7
824
MODULE
16
Similarity and
Transformations
Similarity and
Transformations
Essential Question: How can you use similarity
ESSENTIAL QUESTION:
and transformations to solve real-world problems?
Answer: Similarity and transformations are useful
tools in any real-world things that have the same
shape but different sizes.
16
MODULE
LESSON 16.1
Dilations
LESSON 16.2
Proving Figures
Are Similar Using
Transformations
LESSON 16.3
Corresponding Parts of
Similar Figures
This version is for
Algebra 1 and
Geometry only
PROFESSIONAL DEVELOPMENT
VIDEO
LESSON 16.4
AA Similarity
of Triangles
Professional Development Video
Professional
Development
my.hrw.com
Schwartz/Digital Vision/Getty Images
Author Juli Dixon models successful
teaching practices in an actual
high-school classroom.
REAL WORLD VIDEO
Check out how properties of similarity
and transformations can be used to
create scale models of large, real-world
structures like monuments.
Modeling the Washington Monument
In this module, you will be challenged to create a plan for a scale model of the Washington
Monument. How can you use similarity and dilations to help you produce an accurate model?
Let’s find out.
Module 16
DIGITAL TEACHER EDITION
IN2_MNLESE389847_U7M16MO 825
Access a full suite of teaching resources when and
where you need them:
• Access content online or offline
• Customize lessons to share with your class
• Communicate with your students in real-time
• View student grades and data instantly to target
your instruction where it is needed most
825
Module 16
825
PERSONAL MATH TRAINER
Assessment and Intervention
tests, and intervention activities. Prepare your
students with updated, Common Core-aligned
practice tests.
18/04/14 7:47 PM
Complete these exercises to review skills you will need
for this module.
Properties of Transformations
Example 1
Stretch △ABC with points A(1, 2), B(3, 2), and C(3, –1)
horizontally and vertically by a factor of 4.
(x, y) → (4x, 4y)
students need strategic or intensive intervention for
the module’s prerequisite skills.
• Online Homework
• Hints and Help
• Extra Practice
Write the transformation rule.
A′(4, 8), B(12, 8), C′(12, –4)
Use the transformation to
write each transformed point.
Describe the transformation.
1.
Stretch △DEF with points D(–2, 1), E(–1, –1), and F(–2, –2) horizontally and vertically by a
factor of –3.
D′(6, –3), E′(3, 3), F′(6, 6)
2.
Is the stretch a rigid motion?
ASSESSMENT AND INTERVENTION
No
3.
Is it true that △DEF ≅ △D′E′F′?
No
3
2
1
Personal Math Trainer will automatically create a
standards-based, personalized intervention
assignment for your students, targeting each student’s
individual needs!
Similar Figures
Example 2
Transform △ABC with points A(3, 4), B(–1, 6), and C(0, 1) by shifting it 2 units
to the right and 1 unit up.
(x, y) → (x + 2, y + 1)
Write the transformation rule.
A′(5, 5), B′(1, 7), C′(2, 2)
Write each transformed point.
5.
4
)
Write the rule used to transform △ABC.
y
x + 1, _
(x, y) → _
–4
(2
2
Describe in words the transformation shown in the figure.
Answers may vary. Sample: Compress △ABC vertically and
IN2_MNLESE389847_U7M16MO 826
Tier 1
Lesson Intervention
Worksheets
Reteach 16.1
Reteach 16.2
Reteach 16.3
Reteach 16.4
B
2
-4
-2
horizontally by a factor of 2, and then shift it to the right 1
unit and down 4 units.
Module 16
y A
0
-2
-4
C
2
A′
x
4
B′
C′
© Houghton Mifflin Harcourt Publishing Company
Describe the transformation shown in the graph.
4.
TIER 1, TIER 2, TIER 3 SKILLS
See the table below for a full list of intervention
resources available for this module.
Response to Intervention Resources also includes:
• Tier 2 Skill Pre-Tests for each Module
• Tier 2 Skill Post-Tests for each skill
826
Response to Intervention
Tier 2
Strategic Intervention
Skills Intervention
Worksheets
35 Properties of
Dilations
41 Similar Figures
29 Geometric Drawings
Differentiated
Instruction
18/04/14 7:47 PM
Tier 3
Intensive Intervention
Worksheets available
online
Building Block Skills
36, 50, 80, 86, 95, 99,
103, 104, 112
Challenge
worksheets
Extend the Math
Lesson Activities
in TE
Module 16
826
LESSON
16.1
Name
Dilations
Class
Date
16.1 Dilations
Essential Question: How does a dilation transform a figure?
Common Core Math Standards
The student is expected to:
COMMON
CORE
G-SRT.A.1a, G-SRT.A.1.b
Verify experimentally the properties of dilations given by a center and a
scale factor … . Also G-CO.A.2
Use △ABC and its image △A'B'C' after a dilation to answer the following questions.
MP.5 Using Tools
B
Language Objective
B'
Work with a partner to identify the center of dilation and scale factor in
a dilation.
A
A
ENGAGE
View the Engage section online. Discuss the photo.
Ask students to describe the relationship between
the forms created by the hands and the shapes that
appear on the wall. Then preview the Lesson
© Houghton Mifflin Harcourt Publishing Company
PREVIEW: LESSON
A'
C
B
Use a ruler to measure the following
lengths. Measure to the nearest
tenth of a centimeter.
Essential Question: How does a
dilation transform a figure?
A dilation changes the size of a figure without
changing its shape, so the image is similar to the
pre-image but not congruent.
Resource
Locker
A dilation is a transformation that can change the size of a polygon but leaves the shape
unchanged. A dilation has a center of dilation and a scale factor which together determine
the position and size of the image of a figure after the dilation.
Mathematical Practices
COMMON
CORE
Investigating Properties of Dilations
Explore 1
C
AB = 6.0 cm
A'B' = 3.0 cm
AC = 4.0 cm
A'C' = 2.0 cm
BC = 3.0 cm
B'C' = 1.5 cm
C'
Use a protractor to measure the
corresponding angles.
m∠A = 22°
m∠A' = 22°
m∠B = 33°
m∠B' = 33°
m∠C = 125°
m∠C' = 125°
Complete the following ratios
_
3.0
A'B' = _ = 1
_
AB
2
6.0
_
_
2.0
A'C' = _ = 1
_
AC
2
4.0
1.5
B'C' = _ = 1
_
BC
2
3.0
Reflect
1.
What do you notice about the corresponding sides of the figures? What do you
The ratios of the lengths of corresponding sides are equal. Corresponding angles are
congruent because the measures of the corresponding angles are the same.
2.
Discussion What similarities are there between reflections, translations, rotations,
and dilations? What is the difference?
Reflections, translations, rotations, and dilations all preserve angle measure. Reflections,
translations, and rotations all preserve distance but dilations do not preserve distance.
Module 16
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 1
827
gh "File info"
Date
Class
Name
ons
16.1 Dilati
IN2_MNLESE389847_U7M16L1 827
a figure?
a
transform
center and
a dilation
given by a
of dilations
How does
properties
Question:
entally the
Essential
Verify experim
G-SRT.A.1.b
Dilations
.2
COMMON
G-SRT.A.1a,
CORE
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the positio
after a dilatio
△A'B'C'
B
and its image
Use △ABC
Resource
Locker
HARDCOVER PAGES 827836
Watch for the hardcover
student edition page
numbers for this lesson.
B'
C
A

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Harcour t
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Lesson 16.1
m∠A =
22°
m∠B =
33°
m∠C =
125°
the
m∠A' =
22°
m∠B' =
33°
m∠C' =
125°
_
1
1.5
B'C' = _ = 2
_
BC
3.0
you
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the figure
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correspondi
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sides are
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Lesson 1
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827
Module 16
827
_
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Use a protra angles.
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Reflect
1.
6L1 827
47_U7M1
ESE3898
IN2_MNL

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to measu
nearest
Use a ruler
re to the
lengths. Measu eter.
centim
tenth of a
3.0 cm
A'B' =
cm
AB = 6.0
= 2.0 cm
A'C'
4.0 cm
AC =
1.5 cm
B'C' =
cm
BC = 3.0
ing ratios
the follow
1
2.0
Complete
A'C' = _ = 2
_
1
3.0
AC
4.0
A'B' = _ = 2
_
AB
6.0

C'
A'
18/04/14
7:51 PM
18/04/14 7:53 PM
Explore 2
Dilating a Line Segment
EXPLORE 1
The dilation of a line segment (the pre-image) is a line segment whose length is the product
of the scale factor and the length of the pre-image.
‹ ›
−
Use the following steps to apply a dilation by a factor of 3, with center at the point O, to AC .
O
A
Investigating Properties of Dilations
C
INTEGRATE TECHNOLOGY
B
Students have the option of doing the dilation activity
either in the book or online.
QUESTIONING STRATEGIES
A
To locate the point A', draw a ray from O through A. Place A' on this ray so that the
distance from O to A' is three times the distance from O to A.
B
To locate point B′, draw a ray from O through B. Place B′ on this ray so that the distance
from O to B′ is three times the distance from O to B.
C
To locate point C′, draw a ray from O through C. Place C′ on this ray so that the distance
from O to C′ is three times the distance from O to C.
D
Draw a line through A', B', and C'.
E
_ _
_
_ _
_
Measure AB, AC, and BC. Measure A'B', A'C', and B'C'. Make a conjecture about the lengths
of segments that have been dilated.
Do dilations appear to preserve angle
measure? yes
Do dilations appear to preserve distance? no
What happens when you dilate with a scale
factor of 1? The image and pre-image
coincide.
EXPLORE 2
The length of a segment that has been dilated is the original length times the scale factor.
3.
Make a conjecture about the length of the image of a 4 cm segment after a dilation
with scale factor k. Can the image ever be shorter than the preimage?
The image will be 4k cm long. Yes, if k is between 0 and 1, the image will be shorter than
the preimage.
4.
depend upon the location of the segment? Explain
The image of segment m is parallel to m. The only exception is when line m passes through
© Houghton Mifflin Harcourt Publishing Company
Reflect
Dilating a Line Segment
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Use an overhead projector to project a shape
onto the board and then trace the shape. Move the
projector closer and trace the shape again to
demonstrate a reduction. Move it farther away and
trace it again to demonstrate an enlargement.
the center of dilation. In that case, the dilation lies along line m.
Module 16
828
Lesson 1
QUESTIONING STRATEGIES
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M16L1 828
Math Background
Dilations are one of the major transformations that students study. Unlike earlier
transformations (reflections, translations, and rotations), dilations are not rigid
motions. That is, they do not preserve both the shape and the size of a figure.
However, dilations do preserve the shape of a figure. Thus, every dilation is either
an enlargement or a reduction. The image that results from a dilation is generally
not congruent to the pre-image, so a dilation is not an isometry. An exception is a
dilation with a scale factor of –1, which is equivalent to a 180° rotation about the
center of dilation.
18/04/14 7:53 PM
Is it ever possible for the image of a point to
coincide exactly with the pre-image?
Explain. yes; when the scale factor is 1
Dilations
828
Explain 1
EXPLAIN 1
Applying Properties of Dilations
The center of dilation is the fixed point about which all other points are transformed by a dilation.
The ratio of the lengths of corresponding sides in the image and the preimage is called the scale factor.
Applying Properties of Dilations
Properties of Dilations
• Dilations preserve angle measure.
• Dilations preserve betweenness.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Have students review the characteristics of
• Dilations preserve collinearity.
• Dilations preserve orientation.
• Dilations map a line segment (the pre-image) to another line segment whose
length is the product of the scale factor and the length of the pre-image.
• Dilations map a line not passing through the center of dilation to a parallel
line and leave a line passing through the center unchanged.
similar figures. Help them see that a dilation is a
transformation that changes the size of a figure but
not the shape. The image and the pre-image of a
figure under a dilation are similar.
Example 1

Determine if the transformation on the coordinate plane is a dilation.
If it is, give the scale factor.
Preserves angle measure: yes
6
Preserves betweenness: yes
D'
A' 4
Preserves collinearity: yes
y
A
D
2
Preserves orientation: no
Ratio of corresponding sides: 1 : 1
B
B'
C'
-6
-4
-2
0
2
Cx
4
6
8
Is this transformation a dilation? No, it does
not preserve orientation.

Preserves angle measure (Y/N)
© Houghton Mifflin Harcourt Publishing Company
Preserves betweenness (Y/N)
Preserves collinearity (Y/N)
Preserves orientation (Y/N)
Scale Factor
Is this transformation a dilation?
Module 16
yes
y
4 C'
yes
2
yes
yes
x
-4
829
-2
A
2
Yes
C
A'
0
-2
2
B
B'
-4
Lesson 1
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M16L1 829
Peer-to-Peer Activity
Divide students into pairs. Have one student draw a pre-image and an image, and
then have the other determine whether the image is a dilation and explain why or
why not. If it is a dilation, they should find the scale factor. Then have students
switch roles.
829
Lesson 16.1
18/04/14 7:53 PM
QUESTIONING STRATEGIES
Determine if the transformations are dilations.
5.
If a transformation preserves angle measure,
betweenness, and collinearity, is the
transformation a dilation? Explain. Not necessarily;
rigid transformations such as translations also
preserve these things.
y
4
2
-4
0
-2
D
The transformation preserves angle measures,
betweenness, collinearity, and orientation but not
A'
C' C
D'
B'
B
2
-2
A 4
x
3
same. Therefore, it is a dilation. The scale factor is __
.
2
E
E'
-4
6.
AVOID COMMON ERRORS
C
4
A
The transformation preserves betweenness and
x
B
-6
A'
-4
B'
C'
Explain 2
0
-2
-2
Some students may think that any larger or smaller
figure is a dilation. Use examples to show that the
shape and angles of the pre-image and image must be
the same, and the lengths of the corresponding sides
proportional.
y
2
-8
distance. The ratios of corresponding lengths are all the
collinearity, angle measure, or ratio of corresponding
sides. It does not preserve orientation. This is not a
dilation.
EXPLAIN 2
-4
Determining the Center and Scale of
a Dilation
Determining the Center and Scale
of a Dilation
When you have a figure and its image after dilation, you can find the center of dilation by drawing
lines that connect corresponding vertices. These lines will intersect at the center of dilation.

Determine the center of dilation and the scale factor of the dilation of the
triangles.
‹ › −
−
‹ ›
‹ ›
−
Draw AA', BB' , and CC'. The point where the lines cross is the center
of dilation. Label the intersection O. Measure to find the scale factor.
OA = 25 mm
OB = 13 mm
OC = 19 mm
OA′ = 50 mm
OB′ = 26 mm
OC′ = 38 mm
A'
B'
C'
A
The scale factor is 2 to 1.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Discuss the reason that dilation does not
© Houghton Mifflin Harcourt Publishing Company
Example 2
affect a line through the center of dilation, but does
affect a line not through the center. Remind students
that the image of a point on a line through the center
of dilation will remain on the line, and move only
along it.
B
C
O
Module 16
830
Lesson 1
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M16L1 830
Critical Thinking
18/04/14 7:53 PM
Students may wonder how to construct the dilation image of a figure when
the scale factor is not a whole number. While it is not possible to use a compass
and straightedge to construct the image for every scale factor, it is possible for
certain scale factors, using a combination of the construction for copying a
segment, the construction of a segment bisector, and the construction for dividing
a segment into a given ratio. You may want to challenge students to use some
combination of these constructions to draw the image of a triangle after a dilation
with center of dilation O and scale factor 1.75.
Dilations 830
‹ › −
−
‹ ›
‹ ›
−
Draw AA', BB' , and CC'. Measure
from each point to the intersection
O to the nearest millimeter.
B
QUESTIONING STRATEGIES
How can you use the given scale factor to
predict whether the image is an
enlargement? A scale factor >1 produces an
enlargement.
OA′ = 30 mm
OB′ =
19 mm
OC =
52 mm
B'
The scale factor is
O
A'
A
26 mm
OC′ =
QUESTIONING STRATEGIES
C'
B
OB = 38 mm
ELABORATE
1 to 2 .
Reflect
How can a dilation produce an image in which
the pre-image and the image are two different
congruent figures? The image under a dilation with
a scale factor of –1 is congruent to the pre-image
and is rotated 180°.
7.
For the dilation in Your Turn 5, what is the center of dilation? Explain how
you can tell without drawing lines.
The origin; several of the points and their images lie on axes, which intersect at the origin.
For the points that are not on axes, you can use slopes to check that the lines connecting
the point and image go through the origin.
A
8.
SUMMARIZE THE LESSON
Determine the center of dilation and the scale factor of the
dilation.
Measurements may vary but scale factor
should not.
© Houghton Mifflin Harcourt Publishing Company
How can you determine whether the image of
a figure is the result of a dilation? How might
you determine the scale factor? If the lengths of the
sides of the image are proportional to the lengths
of the sides of the pre-image, and the
corresponding angles are congruent, the image is
the result of a dilation. If the figures are on a
coordinate gird, you could divide the coordinates
of the points of the image by the coordinates of the
corresponding points of the pre-image to find the
scale factor.
C
OA = 60 mm
OA' =
18 mm
cm, OA =
54 mm
The scale factor of the dilation is
1 to 3
A'
Elaborate
9.
B
C
.
C'
B'
How is the length of the image of a line segment
O
under a dilation related to the length of its preimage?
The ratio of the length of the image to the length of the preimage is the scale factor.
10. Discussion What is the result of dilating a figure using a scale factor of 1? For
this dilation, does the center of dilation affect the position of the image relative
to the preimage? Explain.
A dilation by a scale factor of 1 will leave the figure unchanged. It will remain in the same
position no matter what point is used as the center of dilation.
Module 16
831
Lesson 1
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M16L1 831
Connect Vocabulary
Relate the word dilation to its meaning of becoming larger or smaller by using the
example of the human eye. Pupils dilate in response to changes in light; in
darkness, they get larger to let in more light, and in bright light, pupils get smaller
to reduce the light entering the eye. Explain to students that when people talk
about pupils dilating, they are using the mathematical term correctly.
831
Lesson 16.1
18/04/14 7:53 PM
11. Essential Question Check-In In general how does a dilation transform a figure?
A dilation changes the size of a figure without changing the shape. Dilations preserve
EVALUATE
angle measures and orientation, but not length.
Evaluate: Homework and Practice
1.
• Online Homework
• Hints and Help
• Extra Practice
Consider the definition of a dilation. A dilation is a transformation that can change
the size of a polygon but leaves the shape unchanged. In a dilation, how are the ratios
of the measures of the corresponding sides related?
ASSIGNMENT GUIDE
The ratios of the lengths of the corresponding sides are equal.
Tell whether one figure appears to be a dilation of the other figure
Explain.
2.
3.
No, this is not a dilation, since the ratios
of the lengths of corresponding sides are
not equal. Some sides are about the same
length and others are not.
It appears to be a dilation that preserves
angle measure, betweenness, collinearity
and orientation. Also, the ratios of the
lengths of corresponding sides appears
to be equal.
4.
1 ? Explain.
Is the scale factor of the dilation of △ABC equal to _
2
6
4
A'
A
5.
B'
B
Exercise
IN2_MNLESE389847_U7M16L1 832
6
8
B
28
Exercises 1–5
Explore 2
Dilating a Line Segment
Exercises 6–11
Example 1
Applying Properties of Dilations
Exercises 12–13
Example 2
Finding the Center and Scale of a
Dilation
Exercises 14–15
similar? Why or why not? Yes; if figures are
congruent, then there is a sequence of rigid motions
that maps one to the other. Rigid motions are also
similarity transformations, so the figures must also
be similar.
Lesson 1
832
Depth of Knowledge (D.O.K.)
x
4
A
_ _
Module 16
2
35
Square A is a dilation of square B.
What is the scale factor?
1
5
35
a. _
=
The answer is (c). The ratio is
7
4
28
4
b. _
5
5
c. _
4
d. 7
25
e. _
16
Explore 1
Investigating Properties of Dilations
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 If two figures are congruent, are they also
C
2
0
Practice
C'
© Houghton Mifflin Harcourt Publishing Company
No, the scale factor is 2. The center of dilation is at (0, 0). The
measure from the center of dilation to A' is twice the measure
from the center of dilation to A. The image is larger than the
pre-image so the scale factor must be greater than 1.
y
Concepts and Skills
COMMON
CORE
Mathematical Practices
1–5
2 Skills/Concepts
MP.2 Reasoning
6–15
2 Skills/Concepts
MP.5 Using Tools
16–17
2 Skills/Concepts
MP.3 Logic
18
2 Skills/Concepts
MP.4 Modeling
19
3 Strategic Thinking
MP.1 Problem Solving
20
3 Strategic Thinking
MP.1 Problem Solving
18/04/14 7:53 PM
Dilations 832
6.
AVOID COMMON ERRORS
_
Apply a dilation to AC with a scale factor of 2 and
center at the point O.
7.
_
1
Apply a dilation to AC with a scale factor of _
3
and center at the point O.
O
Some students may have trouble determining scale
factors. In particular, they may have difficulty
distinguishing a dilation with a scale factor of k from
a dilation with a scale factor of __1k . Remind students
that they should always compare the image to the
pre-image. Also, remind students that if the image is
an enlargement of the pre-image, then the scale factor
must be greater than 1.
O
C
C'
B
A
A'
B'
C
B'
B
A'
8.
C'
A
What happens when a triangle is dilated using
one of the vertices as the center of dilation?
9.
Draw an image of WXYZ. The center of the
dilation is O, and the scale factor is 2.
The sides of the triangles adjacent to the
center of dilation will be collinear. The
third sides of the preimage and image will
be parallel. The vertex used as the center
of dilation will be in the same location
in both triangles.
X'
Y'
X
Y
O
Z
W
Z'
W'
10. Draw an image of △ABC . The center of
dilation is C, and the scale factor is 1.5.
11. Compare dilations to rigid motions. How are
they similar? How are they different?
Rigid motions preserve angle
measure, betweenness, and
collinearity. Dilations preserve all of
these except distance. The dilation of
a line segment (preimage) is another
line segment whose length is the
product of the scale factor and the
length of the preimage.
B'
B
© Houghton Mifflin Harcourt Publishing Company
C'
C
A
A'
Determine if the transformation of figure A to figure B on the coordinate plane is a
dilation. Verify ratios of corresponding side lengths for a dilation.
12.
13.
y
6
y
8
4
6
B
4
B
2
A
0
2
x
A
2
4
6
x
0
2
4
6
8
10
It is a dilation. The ratio for corresponding
It is a dilation. The ratio for corresponding
2
.
side lengths is _
1
side lengths is _
.
2
1
Module 16
IN2_MNLESE389847_U7M16L1 833
833
Lesson 16.1
833
Lesson 1
18/04/14 7:53 PM
Determine the center of dilation and the scale factor of the dilation.
14.
15.
E
A'
F
D
C'
E'
B'
A
F'
D'
B
C
O
O
The scale factor is 1 to 2 .
The scale factor is 3 to 1 .
16. You work at a photography store. A customer has a picture
that is 4.5 inches tall. The customer wants a reduced copy
of the picture to fit a space of 1.8 inches tall on a postcard.
What scale factor should you use to reduce the picture to
the correct size?
1.8 in _
2
_
=
5
4.5 in
2
A scale factor of should be used.
5
_
y
The length of the pre-image is ⎜5 - 2⎟ = 3 units.
14
C '(15, 12)
B'(6, 12)
12
The height of the pre-image is ⎜4 - 2⎟ = 2 units.
10
The height of the image is ⎜12 - 6⎟ = 6 units.
8
6
4
2
The length of the image is ⎜15 - 6⎟ = 9 units.
B(2, 4)
C(5, 4)
Module 16
IN2_MNLESE389847_U7M16L1 834
6
3
lengths is _
=_
. The scale factor is 3 to 1.
1
2
D(5, 2)
A(2, 2)
2
3
9
The ratio of the heights is _
=_
. The ratio of the
1
3
D '(15, 6)
A'(6, 6)
x
4
6
8
10
12
14
16
18
834
Vision/Getty Images
17. Computer Graphics An artist uses a computer program to enlarge a design, as shown. What is the scale
factor of the dilation?
Lesson 1
18/04/14 7:53 PM
Dilations 834
18. Explain the Error What mistakes did the student make when trying to determine
the center of dilation? Determine the center of dilation.
JOURNAL
Have students write an explanation of how, given a
triangle and its image under a dilation, you could use
a ruler to find the scale factor of the dilation.
P
P'
O
O
R'
Q'
R
The lines were constructed
‹ ›
−
incorrectly. PO should go through the
‹ ›
−
points P and P' to form PP' . The lines
must go through the vertex and the
corresponding vertex to meet at the
center of dilation O.
Q
F' y
H.O.T. Focus on Higher Order Thinking
19. Draw △DEF with vertices D (3, 1) E (3, 5) F (0, 5).
12
a. Determine the perimeter and the area of △DEF.
10
Perimeter is 12 units, Area is 6 square units
8
b. Draw an image of △DEF after a dilation having a scale factor of
3, with the center of dilation at the origin (0, 0). Determine the
perimeter and area of the image.
6
E
F
4
Perimeter is 36 units, Area is 54 square units
c.
E'
14
D'
2
perimeter △D'E'F'
How is the scale factor related to the ratios _____________
perimeter △DEF
area △D'E'F'
and __________
?
x
D
area △DEF
0
2
4
6
perimeter △D'E'F'
36
3
area △D'E'F'
54
9
____________
= __
=_
= scale factor; ________
= __
=_
= scale factor squared
perimeter △DEF
12
area △DEF
1
6
1
© Houghton Mifflin Harcourt Publishing Company
20. Draw △WXY with vertices (4, 0), (4, 8), and (-2, 8).
a. Dilate △WXY using a factor of _14 and the origin as the center. Then
dilate its image using a scale factor of 2 and the origin as the center.
Draw the final image.
b. Use the scale factors given in part (a) to determine the scale factor you
could use to dilate △WXY with the origin as the center to the final
image in one step.
1
1
×2=_
= the scale factor you
Multiply the scale factors: _
4
2
multiply the pre-image by to draw the final image in one step.
c.
8
y
Y
X
8
6
Y"
X"
4
Y'2
X'
x
-2
0 W' 2W" 4W
Do you get the same final image if you switch the order of the dilations
in part (a)? Explain your reasoning.
1
1
=_
it will give you the same scale factor
Yes. If you multiply 2 × _
4
2
because multiplication is commutative.
Module 16
IN2_MNLESE389847_U7M16L1 835
835
Lesson 16.1
835
Lesson 1
18/04/14 7:53 PM
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 A rectangle measuring 3 inches by 6 inches is
You’ve hung a sheet on a wall and lit a candle. Now you move your hands into position
between the candle and the sheet and, to the great amusement of your audience, create an
image of an animal on the sheet.
Compare and contrast what you’re doing with what happens when you draw a dilation of a
triangle on a coordinate plane. Point out ways that dilations and hand puppets are alike and
ways they are different. Discuss measures that are preserved in hand-puppet projections and
those that are not. Some terms you might like to discuss:
projected on a wall. The image of the rectangle
on the wall has an area of 200 square inches. What
are the dimensions of the wall rectangle?
Explain. 10 in. x 20 in. ; sample answer: Since the
length of the smaller rectangle is twice its width, the
length of the larger rectangle must also be twice its
width. The problem becomes one of finding two
numbers in the ratio 2:1 with a product of 200. The
solutions 10 in. and 20 in. can be found using the
guess-and-check problem solving strategy.
• pre-image
• image
• center of dilation
• scale factor
• transformation
• input
• output
Points that students may make:
• A coordinate-plane dilation takes place in two dimensions, a shadow
puppet in three dimensions.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 A small rectangle is suspended halfway
• When a shadow puppet is projected on a surface, angle measures are
preserved. Lengths are not preserved. The depth of the hands is not
preserved but is transformed into a two-dimensional image.
Vision/Getty Images
• The pre-image is the hands. The image is the shadow. The center of
dilation is the light source.
• The scale factor for a shadow puppet projection is the ratio of a given
measurement on the wall to the corresponding hand measurement in a
plane parallel to the wall.
Module 16
836
between a flashlight and a wall. It is parallel to the
wall. Viewed from the side, what figure and special
segment does this situation resemble? How could you
use your knowledge of the segment to find unknown
dimensions? Sample answer: The light from the
flashlight, a side of the rectangle, and a side of the
shadow form a triangle with a midsegment drawn.
By the Midsegment Theorem, the side of the
rectangle has half the length of the side of the
Lesson 1
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M16L1 836
Teams of three or four will need a flashlight, an index card, and a way of
supporting the card at a fixed distance from the wall and parallel to it (one team
member can hold the card, keeping it as parallel as possible). One student holds
the flashlight, another measures the flashlight’s distance from the rectangle, and a
third measures the dimensions of the projected rectangle. Teams should take
measurements with the flashlight at various distances from the rectangle. They
can then study their data and hypothesize about the relationship between an
object’s dimensions and the dimensions of its image when it is projected on a wall.
18/04/14 7:53 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Dilations 836
LESSON
16.2
Name
Proving Figures
are Similar Using
Transformations
Essential Question: How can similarity transformations be used to show two figures are
similar?
A similarity transformation is a transformation in which an image has the same shape as its pre-image. Similarity
transformations include reflections, translations, rotations, and dilations. Two plane figures are similar if and only if
one figure can be mapped to the other through one or more similarity transformations.
G-SRT.A.2
Given two figures, use the definition of similarity in terms of similarity
transformations to decide if they are similar … . Also G-C.A.1
A grid shows a map of the city park. Use tracing paper to confirm that the park elements are similar.
A
Mathematical Practices
Trace patio EFHG. Turn the paper so that patio EFHG is mapped onto patio LMON.
Describe the transformation. What does this confirm about the patios?
y
MP.6 Precision
4
Work with a partner to list the essential components needed to prove a
figure is similar to another.
-8
View the Engage section online. Discuss the photo.
Ask students to identify the game that’s being played
and the game board on which it is being played. Then
-6
-4N -2
H
0
2
x
G
4
6
8
-2
L
-4
O
© Houghton Mifflin Harcourt Publishing Company
PREVIEW: LESSON
E
2
ENGAGE
If a sequence of similarity transformations can be
shown to map one figure to another, the figures are
similar.
F
6
Language Objective
Essential Question: How can similarity
transformations be used to show two
figures are similar?
Resource
Locker
Confirming Similarity
Explore
The student is expected to:
COMMON
CORE
Date
16.2 Proving Figures are Similar
Using Transformations
Common Core Math Standards
COMMON
CORE
Class
-6
M
A rotation of 180° around the origin; The fountains are similar.
B
Trace statues ABCDEF and JKLMNO. Fold the paper so that statue ABCDEF is mapped onto
statue JKLMNO. Describe the transformation. What does this confirm about the statues?
y
K
M
-6
-4
N
Module 16
J 2
L
B
4
A
C
D
O
-2
0
-2
2
x
4
6
F
A reflection across the y-axis; The figures are
E
similar.
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 2
837
gh "File info"
Date
Class
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16.2 Pro Transformations
Using
Name
IN2_MNLESE389847_U7M16L2 837
HARDCOVER PAGES 837850
Resource
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G-SRT.A.2
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A grid shows
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Watch for the hardcover
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F
6
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6
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N
Lesson 2
O
837
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47_U7M1
ESE3898
IN2_MNL
837
Lesson 16.2
18/04/14
7:58 PM
18/04/14 7:58 PM

Describe the transformation you can use to map vertices of garden RST to corresponding
vertices of garden DEF. What does this confirm about the gardens?
EXPLORE
y
4 S
Confirming Similarity
2 E
x
-8
-6
-4
-2
D
R
0
2
-2
4
F
6
8
T
1
A dilation with scale factor _
; The
2
INTEGRATE TECHNOLOGY
gardens are similar.
Students have the option of doing the transformation
activity either in the book or online.
Reflect
1.
Look back at all the steps. Were any of the images congruent to the pre-images? If so, what types
of similarity transformations were performed with these figures? What does this tell you about the
relationship between similar and congruent figures?
The two figures in Steps A and B are congruent to each other. The types of transformations
QUESTIONING STRATEGIES
What kind of transformations result in similar
figures? similarity transformations.
that were performed with these figures were rotations and reflections, which are rigid
motions. So congruent figures are also similar figures.
2.
Of these transformations, what kind give
congruent figures? Which transformation
could result in figures that are similar but not
congruent? rigid motions; dilation
If two figures are similar, can you conclude that corresponding angles are congruent? Why or why not?
Yes, the corresponding angles are congruent because rigid motions and dilations preserve
angle measures.
Determining If Figures are Similar
Explain 1
You can represent dilations using the coordinate notation (x, y) → (kx, ky), where k is the scale factor and the center
of dilation is the origin. If 0 < k < 1, the dilation is a reduction. If k > 1, the dilation is an enlargement.

△RST and △XYZ
X
To map △RST onto △XYZ, there must be some factor k that dilates
△RST.
y
R
x
-5
0
T
Z
Pre-image
Image
R(0, 1)
X(0, 3)
5
S
Y
Module 16
S(1, -1)
T(-1, -1)
Y(3, -3)
Z(-3, -3)
838
Determining If Figures Are Similar
© Houghton Mifflin Harcourt Publishing Company
Determine whether the two figures are similar using similarity
transformations. Explain.
Example 1
EXPLAIN 1
QUESTIONING STRATEGIES
Are dilations the only transformations that
result in an image that is similar to the
pre-image? Explain. No, rigid motions like
translations result in figures that are congruent, and
therefore similar.
Lesson 2
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M16L2 838
18/04/14 7:58 PM
Math Background
Two conditions must be met in order to state that two polygons are similar. Two
polygons are similar if and only if their corresponding angles are congruent and
their corresponding sides are proportional. Similarity is an equivalence relation;
that is, similarity is reflexive, symmetric, and transitive. For figures A, B, and C,
A ~ A (reflexivity); if A ~ B, then B ~ A (symmetry); and if A ~ B and B ~ C,
then A ~ C (transitivity).
Proving Figures are Similar Using Transformations
838
You can see that each coordinate of the pre-image is multiplied by 3 to get the image, so this is a
dilation with scale factor 3. Therefore, △RST can be mapped onto △XYZ by a dilation with center at
the origin, which is represented by the coordinate notation (x, y) → (3x, 3y). A dilation is a similarity
transformation, so △RST is similar to △XYZ.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Have students discuss the various approaches
B
PQRS and WXYZ
to defining similarity, including the intuitive
approach (same shape, may be different size);
meeting the conditions of congruent angles and
proportional sides; and the transformational
approach–if one can be obtained from the other by
similarity transformations.
10
To map PQRS onto WXYZ, there must
be some factor k that enlarges PQRS.
y
X(5, 9)
Y(12, 9)
8
6
W(5, 5)
Pre-image
Image
P(2, 2)
W(5, 5)
R(6, 4)
Y(12, 9)
Z(12, 5)
4 Q(2, 4)
R(6, 4)
2 P(2, 2)
S(6, 2)
Q(2, 4)
S(6, 2)
x
0
2
4
6
8
10
12
X(5, 9)
Z(12, 5)
14
Find each distance: PQ = 2, QR = 4 , WX = 4 , and XY = 7
If kPQ = WX, then k = 2. However. 2QR = / ≠ XY.
No value of k can be determine that will map PQRS to WXYZ.
So, the figures are/are not similar.
Determine whether the two figures are similar using similarity transformations. Explain.
3.
LMNO and GHJK
y
© Houghton Mifflin Harcourt Publishing Company
14
K(2, 12)
12
J(8, 12)
10
8
6
O(1, 6)
4
2
N(4, 6)
G(2, 4)
L(1, 2)
0
2
H(8, 4)
M(4, 2)
4
x
6
8
10
Yes, you can use a dilation with scale factor 2 and center at the origin to map LMNO onto GHJK.
The figures are similar because a dilation is a similarity transformation.
Module 16
839
Lesson 2
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M16L2 839
Small Group Activity
Have students work in small groups to design a logo for an imaginary company;
the logo should repeat a single design as several similar figures. Have them write a
description accompanying the logo that describes the transformations they used.
839
Lesson 16.2
18/04/14 7:58 PM
4.
△JKL and △MNP
CDEF and TUVF
5.
J y
C
K
-5
L
0
M
T
N
F
Finding a Sequence of Similarity
Transformations
x
x
5
U 0
-5
P
D
No, the angles are different. △JKL and
△MNP are not similar figures.
Explain 2
EXPLAIN 2
y
V
5
E
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Help students see that scale factors are the
Yes, there is a dilation centered at point F with
a scale factor of 2. CDEF is similar to TUVF.
Finding a Sequence of Similarity Transformations
same as similarity ratios.
In order for two figures to be similar, there has to be some sequence of similarity transformations that maps one
figure to the other. Sometimes there will be a single similarity transformation in the sequence. Sometimes you must
identify more than one transformation to describe a mapping.
Example 2

Find a sequence of similarity transformations that maps the first figure to
the second figure. Write the coordinate notation for each transformation.
ABDC to EFHG
y A
5
B
D
C
F x
E
-5
0
5
G
H
Original Coordinates
A(1, 6)
B(5, 6)
C(-2, 2)
D(2, 2)
Coordinates after
1
dilation k = __
2
1
A′ __, 3
(2 )
5
B′ __, 3
(2 )
C′(-1, 1)
D′(1, 1)
© Houghton Mifflin Harcourt Publishing Company
Since EFHG is_
smaller than ABDC, the scale
_ factor k of the dilation must be between 0 and 1.
The length of AB is 4 and the length of EF is 2; therefore, the scale factor is __12 . Write the new coordinates
after the dilation:
A translation right 2 units and down 3 units completes the mapping.
( )
( )
Coordinates after dilation
1, 3
A′ _
2
5, 3
B′ _
2
C′(-1, 1)
D′(1, 1)
Coordinates after
translation (x + 2, y - 3)
5
E __
,0
2
( )
9
F __
,0
2
( )
G(1, -2)
H(3, -2)
The coordinates after translation are the same as the coordinates of EFGH, so you can map ABDC to
EFHG by the dilation (x, y) → __12 x, __12 y followed by a translation (x, y) → (x + 2, y - 3).
(
Module 16
)
840
Lesson 2
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M16L2 840
18/04/14 7:57 PM
Modeling
To help students understand the concepts in this lesson, have them draw figures
and their transformed images on a coordinate plane. Then have them cut out the
figures and see when they can be made to coincide, and when they cannot.
Proving Figures are Similar Using Transformations
840
△JKL to △PQR
B
QUESTIONING STRATEGIES
y
How do you use the scale factor of a dilation
to find the coordinates of points on the image
of a figure? Multiply the coordinates of points on
the pre-image by the scale factor.
R
8
6
4
-8
AVOID COMMON ERRORS
Students may think that all rectangles are similar.
Challenge them to draw two rectangles that do not
have proportional sides.
-6
-4
J
0
-2
Q
K
2
x
L
2
P4
6
8
You can map △JKL to △PQR with a reflection across the x-axis followed by a dilation followed by a
90 ° counterclockwise rotation about the origin.
°
Reflection: (x, y) → (x, -y) Dilation : (x, y) → ( 3x, 3y ) 90 counterclockwise rotation:
(
)
-y,
x
x,
y
→
( )
Reflect
6.
Using the figure in Example 3A, describe a single dilation that maps ABDC to EFHG.
By connecting the corresponding vertices, you can identify (4, -6) as the center of dilation
for a dilation with scale factor __12 that maps ABDC to EFHG.
7.
Using the figure in Example 3B, describe a different sequence of transformations that will map
△JKL to △PQR.
Reflect △JKL across the y-axis. Then dilate with center at the origin and scale factor 3. Then
© Houghton Mifflin Harcourt Publishing Company
rotate 90° clockwise around the origin.
For each pair of similar figures, find a sequence of similarity transformations that
maps one figure to the other. Use coordinate notation to describe the transformations.
8.
PQRS to TUVW
y
Q
6
R
4
U
V
-8
-6
-4
W
-2
T
0
Reflection: (x, y) → (-x, y)
S
P
2
Followed by…
x
2
Module 16
4
6
You can map PQRS to TUVW by a reflection
followed by a dilation.
8
841
Dilation: (x, y) →
(__13 x, __13 y)
Lesson 2
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M16L2 841
Connect Vocabulary
Make sure students are not confused by the meanings of similarity in everyday
English and in math. In everyday life, things may be called similar based only on
how they appear. But, in math, we must look at the scale factors in corresponding
sides and the angle measures in two similar figures before we call the figures
similar.
841
Lesson 16.2
18/04/14 7:57 PM
9.
△ABC to △DEF
EXPLAIN 3
y
F
8
You can map △ABC to △DEF by a rotation
about the origin 180° followed by a dilation
followed by a translation.
6
Rotation: (x, y) → (-x, -y)
4
2
D
E
Followed by…
x
-8
-6
B
-4
-2 A 0
2
4
6
Proving All Circles Are Similar
8
(
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Help students realize that every dilation has
)
3 _
Dilation: (x, y) → _
x, 3 y
2 2
Followed by…
Translation: (x, y) (x - 3, y + 1.5)
-2
-4
an inverse, and the scale factors are reciprocals.
C
10. Describe a sequence of similarity transformations that maps JKLMN to VWXYZ.
QUESTIONING STRATEGIES
L
How can you perform the dilation in the proof
that two circles are similar without knowing
the lengths of their radii? Using the ratio of the
lengths of the radii as the scale factor ensures that
the dilation will map the circles onto one another.
This is true of all circles of any size, so you do not
need to know specific lengths.
Translate JKLMN right 7 units so that J maps to V.
‹ ›
−
Reflect JKLMN across JN .
M
K
1
Dilate JKLMN with center J and scale factor _
.
2
J
V
N
Z
W
X
Explain 3
Y
Proving All Circles Are Similar
You can use the definition of similarity to prove theorems about figures.
© Houghton Mifflin Harcourt Publishing Company
Circle Similarity Theorem
All circles are similar.
Example 3
Prove the Circle Similarity Theorem.
Given: Circle C with center C and raduis r.
Circle D with center D and raduis s.
s
r
D
C
Prove: Circle C is similar to circle D.
To prove similarity, you must show that there is a sequence of similarity transformations that maps circle C
to circle D.
Module 16
IN2_MNLESE389847_U7M16L2 842
842
Lesson 2
18/04/14 7:57 PM
Proving Figures are Similar Using Transformations
842
→
‾ .
Start by transforming circle C with a translation along the vector CD
A
ELABORATE
s
QUESTIONING STRATEGIES
D
r
C
What scale factor is needed to dilate a circle
r
⇀
CD
Circle C'
translation , the image of point C is
Through this
point D
.
Let the image of circle C be circle Cʹ. The center of circle Cʹ coincides with point
SUMMARIZE THE LESSON
.
Transform circle Cʹ with the dilation with center of dilation D and scale factor _rs .
B
How do you use transformations to determine
whether two polygons are similar? Determine
whether one figure can be mapped to the other
by a similarity transformation.
D
Circle Cʹ is made up of all the points at distance
r
from point D .
_s
After the dilation, the image of circle Cʹ will consist of all the points at distance r × r = s
from point D.
translation followed by the dilation
These are the same points that form circle D . Therefore, the
similarity transformations
translations
D
maps circle C to circle
. Because
and dilations are
,
you can conclude that circle C is similar to circle D .
Reflect
11. Can you show that circle C and circle D are similar through another sequence of similarity
transformations? Explain.
_
Yes, can reflect circle C across the perpendicular bisector of CD, mapping point C to point D.
© Houghton Mifflin Harcourt Publishing Company
Then, follow the same steps for dilation.
12. Discussion Is it possible that circle C and circle D are congruent? If so, does the proof of the similarity of
the circles still work? Explain.
Yes, if the ratio of each circle is the same positive value. The proof still works, because the
ratio of s to r is 1, so the dilation does not affect the size of the circle.
Elaborate
13. Translations, reflections, and rotations are rigid motions. What unique characteristic keeps dilations from
being considered a rigid motion?
Unlike the other transformations, dilations don't preserve distance, meaning the length
of the sides will not stay the same between the pre-image and its image. The lengths of
corresponding sides will be proportional according to the scale factor used.
14. Essential Question Check-In Two squares in the coordinate plane have horizontal and vertical sides.
Explain how they are similar using similarity transformations.
Possible answer: translate the bottom left vertex of one square to the other square. Then
dilate the first square by the ratio of the side lengths.
Module 16
IN2_MNLESE389847_U7M16L2 843
843
Lesson 16.2
843
Lesson 2
18/04/14 7:57 PM
Evaluate: Homework and Practice
EVALUATE
In Exercises 1–4, determine if the two figures are similar using similarity
transformations. Explain.
1.
Looking at the two figures, you can see that EFGH has to be
EFGH and ABCD
5
enlarged by some factor k to be mapped to ABCD.
y
A
E
-5
D
F
H
0
Pre-Image
Image
E(0, 1)
A(0, 4)
G(0, -1)
C(0, -4)
x
B
F(1, 0)
5
G
-5
• Online Homework
• Hints and Help
• Extra Practice
H(-1, 0)
C
ASSIGNMENT GUIDE
B(4, 0)
Concepts and Skills
Practice
D(-4, 0)
Explore
Confirming Similarity
Exercises 11–12
Example 1
Determining If Figures Are Similar
Exercises 1–4
Example 2
Finding a Sequence of Similarity
Transformations
Exercises 5–10
Example 3
Proving All Circles Are Similar
Lesson
Performance
From the table, you can determine that k = 4. Therefore,
EFGH can be mapped onto ABCD with a dilation of 4 with
center at the origin, which is represented by the coordinate
notation (x, y) → (4x, 4y).
2.
△PQR and △STU.
y
No value of k will enlarge △PQR onto
△STU. Therefore, there is no similarity
transformation that maps △PQR onto
△STU, and △PQR is not similar to
△STU.
S
x
-5
0
5
R
Q
T
U
© Houghton Mifflin Harcourt Publishing Company
3.
R
JKLMN and JPQRS
In order to map JKLMN to JPQRS, there
must be a scale factor k that reduces
JKLMN. In this situation, the center of
the dilation is not the origin but point J.
3
The scale factor is _ for the dilation.
5
Since dilations are similarity
transformations, JKLMN is similar to
JPQRS.
y L
K
-8
-6
-4
Exercise
IN2_MNLESE389847_U7M16L2 844
M
4
Q
2
P
-2
J
Module 16
6
R
J
0
x
2
4
-4
S
N
Lesson 2
COMMON
CORE
Mathematical Practices
1–4
1 Recall of Information
MP.1 Problem Solving
5–10
2 Skills/Concepts
MP.4 Modeling
11–13
2 Skills/Concepts
MP.3 Logic
14–17
2 Skills/Concepts
MP.4 Modeling
18–19
2 Skills/Concepts
MP.2 Reasoning
3 Strategic Thinking
MP.2 Reasoning
20
8
-2
844
Depth of Knowledge (D.O.K.)
6
18/04/14 7:57 PM
Proving Figures are Similar Using Transformations
844
4.
AVOID COMMON ERRORS
△UVW and △GHI
y
Some students may confuse congruence and
similarity. Have them draw two similar triangles that
are not congruent, and two similar triangles that are
congruent, and label each set of figures with the
appropriate term. Then have them repeat with pairs
of circles, to make clear that all circles are similar, but
all circles are not congruent.
U
W
V
-5
0
No, the angles are different. Similarity
transformations preserve angle measures
in similar figures, so △UVW and △GHI are
not similar figures.
x
5
H
I
G
For the pair of similar figures in each of Exercises 5–10, find a sequence of similarity transformations
that maps one figure to the other. Provide the coordinate notation for each transformation.
5.
Map △ABC to △PQR.
A
-6
-4
B
-2
6.
y
C y
4
B
2
A
J
0 R
2
4
5
G
F
-5
P
Q
You can map △ABC to △PQR by a reflection
followed by a translation.
You can map ABCD to EFGH by a reflection
followed by a dilation.
Reflection: (x, y)→(-x, y)
Reflection: (x, y)→(x,-y)
Dilation: (x, y)→(2x, 2y)
Map △CED to △CBA.
8.
Map ABCDE to JKLMN.
y
D
A
0C
A
B
C
D
5
J
-2
K
-5
You can map △CED to △CBA by a reflection
followed by a dilation.
Reflection: (x, y)→(-x, y)
1 __
Dilation: (x, y)→ __
, 1
2x 2y
(
B
5
x
-5 E
y
E
2
)
N
x
L 0 M
5
You can map ABCDE to JKLMN by a
reflection followed by a dilation centered
at the origin followed by a translation.
Reflection: (x, y)→(-x, y)
(
)
1 _
Dilation: (x, y)→ _
x, 1 y
2 2
(
Translation: (x, y)→ x +
Module 16
Exercise
IN2_MNLESE389847_U7M16L2 845
845
Lesson 16.2
x
H
6
-2
Translation: (x, y)→(x, y-6 )
© Houghton Mifflin Harcourt Publishing Company
C
D
0 E
-5
x
-4
7.
Map ABCD to EFGH.
2
Lesson 2
845
Depth of Knowledge (D.O.K.)
_1 , y − 2)
COMMON
CORE
Mathematical Practices
21
3 Strategic Thinking
MP.3 Logic
22
3 Strategic Thinking
MP.3 Logic
23
3 Strategic Thinking
MP.3 Logic
18/04/14 7:57 PM
9.
10. Map △JKL to △PQR
Map ABCD to JKLM.
y
K
-8
-6
-4
L
K
6
4
C
D
-2
B
y
4
2
x
A
0
2
2
4
6
8
-2
J
-4
L
-6
-4
-2
Q
J
R
P
0
x
2
4
6
-2
-4
M
You can map ABCD to JKLM by a reflection
followed by a dilation centered at the origin
followed by a translation.
You can map △JKL to △PQR by a reflection
followed by a dilation followed by a 90°
Reflection: (x, y)→(x, y)
Reflection: (x, y)→(x,-y)
(
)
1 _
Dilation: (x, y)→ _
x, 1 y
3 3
Rotation: (x, y)→(y,-x)
Dilation: (x, y)→(3x, 3y)
Translation: (x, y)→(x + 1, y − 2)
Complete the proof.
11. Given: Square ABCD with side length x.
Square EFGH with side length y.
E
A
B
y
C
H
F
x
D
G
Step A: Dilate ABCD with center of dilation A and scale factor _x , producing square A′B′C′D′.
Square ABCD has four sides of length x. Square A′B′C′D′ will have four sides of
y
length _x (x) = y. These are the same side lengths as EFGH. The angles are all 90° in
each square, so A′B′C′D′ is congruent to EFGH.
→
‾ , producing A″B″C″D″ .
Step B: Translate A′B′C′D′ with a translation along the vector A′E
Through this translation, A″ is mapped to E. It may be true that B″ is mapped to F, C″
is mapped to G, and D″ is mapped to H. If not, rotate A″B″C″D″ about E so that B‴ is
mapped to F. Then, C‴ lands on G and D‴ lands on H.
y
Module 16
IN2_MNLESE389847_U7M16L2 846
846
© Houghton Mifflin Harcourt Publishing Company
Prove: Square ABCD is similar to square EFGH.
Lesson 2
18/04/14 7:57 PM
Proving Figures are Similar Using Transformations
846
12. Given: Equilateral △JKL with side length j.
Equilateral △PQR with side length p
Multiple Representations
Ask students to rewrite dilation rules that use
fractions without using fractions. For example:
3 3
(x, y) → __4 x, __4 y . (x, y) → (0.75x, 0.75y)
(
J
P
Prove: △JKL is similar to △PQR.
)
p
j
R
L
Q
K
Step A: Dilate △JKL with center of dilation J and scale factor __j , producing △J′K′L′.
p
△JKL has three sides of length j. After the dilation △J′K′L′ will have three sides
p
of length __j (j) = p. These are the same side lengths as △PQR. By SSS Triangle
Congruence, △J′K′L′ is congruent to △PQR.
→
‾ , producing △J″K″L″. Through this translation,
Step B: Translate △J′K′L′ along the vector J′P
J″ is mapped to P. It may be true that K″ and L″ are mapped to Q and R. If not, rotate
△J″K″L″ about P so that K‴ is mapped to Q. Then, L‴ is mapped to R.
13. Given: △ABC with AB = c, BC = a, AC = b
cx
bx
__
△XYZ with YZ = x, XY = __
a , XZ = a
Y
B
Prove: △ABC is similar to △XYZ.
cx
a
c
Hustace/Corbis
A
x
a
b
C
X
bx
a
Z
x
Step A: Dilate △ABC with center of dilation B and scale factor k _
a , producing △A′B′C′. After
x
the dilation △A′B′C′ will have sides of length B′C′ = ka = _
a (a) = x,
bx
cx
x ( ) __
_x (c) = __
b
=
,
and
A′B′
=
kc
=
.
These
are
the same side lengths of
A′C′ = kb = _
a
a
a
a
△XYZ. By SSS Triangle Congruence, △A′B′C′ is congruent to △XYZ.
→
‾ , producing △A″B″C″. Through this translation,
Step B: Translate △A′B′C′ along the vector B′Y
B″ is mapped to Y. It may be true that A″ and C″ are mapped to X and Z. If not, rotate
△A″B″C″ about Y so that A‴ is mapped to X. Then, L‴ is mapped to R.
14. The dimensions of a standard tennis court are 36 feet. × 78 feet. with a net that is 3 feet. high in the center.
The court is modified for players aged 10 and under such that the dimensions are 27 feet × 60 feet., and
the same net is used. Use similarity to determine if the modified court is similar to the standard court.
4
Dilation factor for the width is _
and
3
13
dilation factor for the length is __
.
10
Therefore the courts are not similar
because the dilation factor is not
constant between the width
and length.
Module 16
IN2_MNLESE389847_U7M16L2 847
847
Lesson 16.2
847
Lesson 2
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15. Represent Real-World Problems A scuba flag is used
to indicate there is a diver below. In North America, scuba
flags are red with a white stripe from the upper left corner
to the lower right corner. Justify the triangles formed on the
scuba flag are similar triangles.
Since the base and the height of the triangles are
equal, no dilation has occurred.
The triangles are 180° reflections of each other. The
triangles are similar to each other.
16. The most common picture size is 4 inches. ×6 inches.
Other common pictures sizes
(in inches) are 5 × 7, 8 × 10, 9 × 12, 11 × 14, 14 × 18, and 16 × 20.
a. Are any of these picture sizes similar? Explain using similarity transformations.
In order for any of the pictures to be similar, there needs to be a scale factor k which maps
one picture to another. The ratio of the sides of each picture are 2:3, 5:7, 4:5, 3:4, 11:14, 7:9,
and 4:5, respectively. Therefore, the only picture sizes that are similar are the 8” × 10” and
16” × 20”, with a scale factor of k = 2.
When resizing pictures one of two things will occur: the picture will be distorted or part
of the picture will need to be cropped out.
Wagner/Shutterstock; (b) ©Dave Reede/Design Pics Inc./Alamy
17. Nicole wants to know the height of the snow sculpture but it is too
tall to measure. Nicole measured the shadow of the snow sculpture's
highest point to be 10 feet long. At the same time of day Nicole's
shadow was 40 inches long. If Nicole is 5 feet tall, what is the height of
the snow sculpture?
Since the shadows were measured at the same time of day
the angle measures are equal and the triangles are similar.
Therefore a proportion can be used to determine the height.
10 ft. = 120 inches
5 ft. = 60 inches
Let h represent the height of the snow sculpture.
60
h
=
; h = 180 in., or 15 ft
120
40
18. Which of the following is a dilation?
A. (x, y) → (x, 3y)
_ _
19. What is not preserved under dilation? Select all that
apply.
B. (x, y) → (3x, -y)
A. Angle measure
D. (x, y) → (x, y - 3)
C. Collinearity
C. (x, y) → (3x, 3y)
B. Betweenness
E. (x, y) → (x - 3, y - 3)
D. Distance
Module 16
IN2_MNLESE389847_U7M16L2 848
E. Proportionality
848
Lesson 2
18/04/14 7:57 PM
Proving Figures are Similar Using Transformations
848
H.O.T. Focus on Higher Order Thinking
JOURNAL
20. Analyze Relationships Consider the transformations below.
Have students define similarity in their own words,
and then justify the definition using properties of
similar figures.
I.
Translation
II.
Reflection
III.
Rotation
IV. Dilation
a. Which transformations preserve distance?
I, II, III
b. Which transformations preserve angle measure?
I, II, III, IV
c.
Use your knowledge of rigid transformations to compare and contrast congruency and similarity.
Rigid transformations preserve angle measure and distance. Therefore, rigid
transformations (translations, reflections, rotations) guarantee congruency. Similarity
transformations (dilations) only preserve angle measures.
Justify Reasoning For Exercises 21–23, use the figure shown. Determine whether the given
assumptions are enough to prove that the two triangles are similar. Write the correct correspondence
of the vertices. If the two triangles must be similar, describe a sequence of similarity transformations
that maps one triangle to the other. If the triangles are not necessarily similar, explain why.
AX
DX
21. The lengths AX, BX, CX, and DX satisfy the equation ___
= ___
.
BX
CX
BX
AX __
AX __
AX
= DX
into __
= CX
. Let k = __
.
Step A: Rearrange __
BX
DX
DX
CX
Step B: Rotate △DXC 180° around point X so that ∠DXC
coincides with ∠AXB.
Step C: Dilate ∠DXC by a factor of k about the center X. This
dilation moves the point D to A, since k(DX) = AX,
and moves C to B, since k(CX) = BX. Since the dilation
is through point X and dilations take line segments to
line segments, △DXC is mapped to △AXB. So △DXC is
similar to △AXB.
A
C
X
D
B
© Houghton Mifflin Harcourt Publishing Company
22. Lines AB and CD are parallel.
¯ is parallel to
Step A: Rotate △DXC so that △DXC coincides with △AXB. Then the image C′D′
¯
¯
¯
the pre-image of CD. So the new side C′D′ is still parallel to side AB.
¯ to a
Step B: Dilate △D′X′C′ with center at point X. This moves vertex C′ to point B and C′D′
‹ ›
‹ ›
−
−
‹ ›
‹ ›
−
−
line through B parallel to CD . Since AB is parallel to C′D′, the dilation moves C′D′
‹ ›
‹ ›
→
−
−
‾ and on AB , D′ must move to A. Therefore,
onto AB . Since D′ moves to a point on XA
the rotation and dilation map △DXC to △AXB. So △DXC is similar to △AXB.
23. ∠XAB is congruent to ∠XCD.
Step A: Draw the bisector of ∠AXC.
¯ onto XA
¯. Since reflections
Step B: Reflect △CXD across the angle bisector. This maps XC
¯ is
¯ onto XB
¯. Since △XCD ≅ △XAB, the image of CD
preserve angles, it also maps XD
¯.
parallel to AB
‹ ›
−
Step C: Dilate △XCD about point X. This moves the new point C to A. Since AB is parallel
‹ ›
‹ ›
‹ ›
−
−
−
to CD , the new CD moves onto AB . Therefore, the new point D is mapped to B and
△XCD is mapped to △XAB. So △XCD is similar to △XAB.
Module 16
IN2_MNLESE389847_U7M16L2 849
849
Lesson 16.2
849
Lesson 2
18/04/14 7:57 PM
AVOID COMMON ERRORS
Students may correctly calculate the areas of circles
A, B, C, and D, but may forget that in order to
calculate the required probabilities, they need to
find the areas of rings. So, for example,
Area Ring C = Area Circle C – Area Circle B.
1. Are the circles similar? Explain, using the concept of a dilation in your explanation.
2. You throw a dart and it sticks in a random location on the board. What is the probability that it sticks
in Circle A? Circle B? Circle C? Circle D? Explain how you found your answers.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Study the pattern in the probabilities you
2′′ 2′′ 2′′ 2′′
A
B
C
Then, predict the probabilities of landing darts in
each ring if a fifth ring of radius 10 inches were
5 ___
9
1 ___
, 3 , ___
, 7 , ___
added outside the 8-inch circle. ___
25 25 25 25 25
D
1. Yes. Each circle can be regarded as a pre-image that can be transformed
through a dilation with its center at the center of the dartboard into
any of the other circles.
2. Area Circle A = πr 2 = π(2) = 4π
2
Area Circle B = πr = π(4) = 16π
2
2
Area Circle C = πr = π(6) = 36π
2
2
Area Circle D = πr 2 = π(8) = 64π
2
Area Ring B = Area Circle B - Area Circle A = 16π - 4π = 12π
Area Ring C = Area Circle C - Area Circle B = 36π - 16π = 20π
Area Ring D = Area Circle D - Area Circle C = 64π - 36π = 28π
© Houghton Mifflin Harcourt Publishing Company
Area of dartboard = Area Circle D = 64π
4π
area Circle A
1
P(A) = __ = _ = _
area of dartboard 64π 16
area Ring B
12π
3
P(B) = __ = _ = _
area of dartboard 64π 16
area Ring C
20π
5
P(C) = __ = _ = _
area of dartboard 64π 16
area Ring D
28π
7
P(D) = __ = _ = _
area of dartboard 64π 16
Module 16
850
Lesson 2
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M16L2 850
Have students design dartboards in basic shapes different from the circular board
(for example, squares or equilateral triangles) but with the same basic properties:
each should have four sections, and the sections should differ in dimensions from
adjacent sections by the same amount. Have students answer the same questions
and equilateral triangles, they should obtain the same results: the figures are
similar and the probabilities of landing a dart in each are (smallest to largest)
5
7
1 __
__
, 3 , __
, and __
.
16 16 16
16
18/04/14 7:57 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Proving Figures are Similar Using Transformations
850
LESSON
16.3
Name
Corresponding Parts
of Similar Figures
Class
Date
16.3 Corresponding Parts
of Similar Figures
Essential Question: If you know two figures are similar, what can you determine about
measures of corresponding angles and lengths?
Common Core Math Standards
Resource
Locker
The student is expected to:
COMMON
CORE
G-SRT.A.2
Explore
… explain using similarity transformations the meaning of similarity for
triangles … . Also G-CO.A.2, G-CO.A.5
Mathematical Practices
COMMON
CORE
A
MP.4 Modeling
ENGAGE
View the Engage section online. Discuss the floor
plan and ask students to explain the relationship of
the floor plan to the house. Then preview the Lesson
N
4
D
m∠A = 80.5°
m∠K = 80.5°
m∠B = 63.4°
m∠L = 63.4°
m∠C = 116.6°
m∠M = 116.6°
m∠D = 99.5°
m∠N = 99.5°
M
C
x
-8
-4
0
4
-4
B
A
K
8
-8
L
Accept reasonable measures; opposite angles should be supplementary
B
© Houghton Mifflin Harcourt Publishing Company
PREVIEW: LESSON
y
8
Measure the angles.
Play “similar or not” with a partner given pairs of figures on coordinate
planes.
Similar figures have corresponding angles that are
congruent and have corresponding sides that are
proportional.
Consider the graph of ABCD and KLMN.
Are corresponding angles congruent? Yes/No
Language Objective
Essential Question: If two figures are
similar, what can you determine about
measures of corresponding angles and
lengths?
Connecting Angles and Sides of Figures
You know that if figures are similar, the side lengths are proportional and the angle measures
are equal. If you have two figures with proportional side lengths and congruent angles, can you
conclude that they are similar?
Are the ratios of corresponding side lengths equal? Yes/No
1
AB = _
_
KL
2
C
1
BC = _
_
LM
2
1
_
KN
2
1
CD = _
_
MN
2
Are the figures similar? Describe how you know using similarity
transformations.
y
8
Yes, there is a dilation centered at the origin that you can
N
4
D
find by connecting the corresponding vertices.
x
-8
-4
0
-4
A
K
Module 16
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EDIT--Chan
DO NOT Key=NL-A;CA-A
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Lesson 3
851
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Parts
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16.3 Corre ilar Figures
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HARDCOVER PAGES 851860
Resource
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know two
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explain using G-CO.A.5
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Quest
Essential
IN2_MNLESE389847_U7M16L3 851
.
and KLMN
of ABCD
the graph
o
Consider
uent? Yes/N
angles congr
ponding
Are corres

Measure
s.
the angle
80.5°
m∠A =
m∠B =
m∠K =
nable
Accept reaso
Are the ratios

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1
AB = _
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KL
2
m∠M =
m∠N =
measures;
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-8
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suppleme
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s equal? Yes/N
side length
1
CD = _
_
MN
2
1
_
KN
2
y
similarity
know using

x
-4
63.4°
opposite
1
BC = _
_
LM
2
M
4
D
-4
-8
80.5°
m∠L =
63.4°
116.6°
m∠C =
99.5°
m∠D =
Watch for the hardcover
student edition page
numbers for this lesson.
y
8
N
how you
Describe
s similar?
you can
Are the figure
ions.
origin that
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on cente
es.
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8
N
-8
M
4
C
D
0
-4
-4
A
K
-8
x
8
4
B
L
Lesson 3
851
Module 16
6L3 851
47_U7M1
ESE3898
IN2_MNL
851
Lesson 16.3
18/04/14
9:04 PM
18/04/14 9:05 PM
D
y
Consider the graph of ABCD and EFGH.
E
The figures are both rectangles, so the measure of each
-8
B
0
-4
D
H
4
Connecting Angles and Sides of
Figures
-4
8
C
G
-8
INTEGRATE TECHNOLOGY
Are the ratios of corresponding side lengths equal? Yes/No
4
AB = _
_
EF
7
F
F
4
A
x
angle is 90°.
E
EXPLORE
8
Are corresponding angles congruent? Yes/No Explain.
3
BC = _
_
FG
5
4
CD = _
_
GH
7
Students have the option of doing the angle measure
activity either in the book or online.
3
_
EH
5
QUESTIONING STRATEGIES
Are the figures similar? Describe how you know using similarity transformations.
EF
, D′ will
No. If you translate ABCD so that A maps to E, and dilate A′B′C′D′ by the ratio __
Given a figure that appears to be a dilation
of another, how could you check whether a
center of dilation exists? Draw lines through
corresponding vertices of the figures. If they
intersect at a point, it is the center of dilation.
If they do not intersect at the point, no center of
dilation exists.
map to H but B′ and C′ will not map to F and G.
G
Consider the graph of PQRS and WXYZ.
y
Are corresponding angles congruent? Yes/No
8
Measure the angles.
Z
m∠W = 76°
m∠Q = 153.4°
m∠X = 104°
m∠R = 26.6°
m∠Y = 76°
m∠S = 153.4°
m∠Z = 104°
4
Y
R
S
-8
-4
P
W
0
-4
Q
4
x
8
X
-8
© Houghton Mifflin Harcourt Publishing Company
m∠P = 26.6°
Accept reasonable measures; opposite angles should be supplementary

Are the ratios of corresponding side lengths equal? Yes/No
1
PQ
_
=_
WX
2

1
QR _
_
=
XY
2
1
RS = _
_
YZ
2
1
PS = _
_
WZ
2
Are the figures similar? Describe how you know using similarity transformations.
No. Dilations map line segments to parallel line segments, so there is no dilation that will
¯ to ZW
¯. The other similarity transformations are all rigid motions, which preserve
map SP
length and angle, so no sequence of similarity transformations will map PQRS to WXYZ.
Module 16
852
Lesson 3
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M16L3 852
Math Background
18/04/14 9:05 PM
Students use similarity to solve a variety of real-world problems. The general
process is this: First, show that two figures are similar; then, use the fact that
corresponding sides are proportional to find an unknown side length.
Corresponding Parts of Similar Figures 852
Reflect
EXPLAIN 1
If two figures have the same number of sides and the corresponding angles are congruent,
does this mean that a pair of corresponding sides are either congruent or proportional?
No, congruence of corresponding angles does not mean there is any relationship between
1.
Justifying Properties of Similar
Figures Using Transformations
a pair of corresponding sides. For example, consider a square and a rectangle.
If two figures have a center of dilation, is a corresponding pair of sides necessarily proportional?
Yes. If two figures have a center of dilation, the figures are similar and any pair of
2.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Use drawings to discuss how similar triangles
corresponding sides is proportional.
If two figures have a correspondence of proportional sides, do the figures necessarily
have a center of dilation?
No. It’s possible to establish a correspondence of proportional sides between two figures
3.
that are not similar (for example, if they don’t have a center of dilation).
can be used to find the heights of buildings or trees
that are difficult to measure directly. The objects’
shadows, measured at the same time of day, are
proportional to the heights of the objects.
Explain 1
Justifying Properties of Similar
Figures Using Transformations
Two figures that can be mapped to each other by similarity transformations (dilations and rigid
motions) are similar. Similar figures have certain properties.
Properties of Similar Figures
QUESTIONING STRATEGIES
Corresponding angles of similar figures are congruent.
How do you know when two rectangles are
similar? They are similar if corresponding
sides are proportional.
Y
B
Corresponding sides of similar figures are proportional.
If △ABC ∼ △XYZ, then
∠A ≅ ∠X
∠B ≅ ∠Y
∠C ≅ ∠Z
© Houghton Mifflin Harcourt Publishing Company
BC = _
AC
AB = _
_
XY
YZ
XZ
A
C
X
Z
To show that two figures with all pairs of corresponding sides having equal ratio k and all pairs
of corresponding angles congruent are similar, you can use similarity transformations.
Dilate one figure using k. The dilated figure is congruent to the second figure by the definition
of congruence. So, there is a sequence of rigid motions (which are also similarity transformations)
that maps one to the other.
Example 1

Identify properties of similar figures.
Figure EFGH maps to figure RSTU by a similarity transformation. Write a proportion that
contains EF and RU. List any angles that must be congruent to ∠G or congruent to ∠U.
EF = _
EH
_
RU
RS

∠T is congruent to ∠G, and ∠H is congruent to ∠U.
Figure JKLMN maps to figure TUVWX by a similarity transformation. Write a proportion
that contains TX and LM. List any angles that must be congruent to ∠V or congruent to ∠K.
JN
LM
_=_
TX
VW
∠ L is congruent to ∠V, and ∠ U is congruent to ∠K.
Module 16
853
Lesson 3
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M16L3 853
Whole Class Activity
Have students bring in magazines and newspapers, and challenge them to find
sets of similar figures. Discuss the use of similar figures in art and design, and the
visual effects that such designs create.
853
Lesson 16.3
18/04/14 9:05 PM
Reflect
4.
EXPLAIN 2
If you know two figures are similar, what angle or side measurements must you know
to find the dilation used in the transformations mapping one figure to another?
You must know measurements of one pair of corresponding sides to find the scale factor.
Applying Properties of Similar
Figures
5.
Triangles △PQR and △LMN are similar. If QR = 6 and MN = 9, what similarity
transformation (in coordinate notation) maps △PQR to △LMN?
QUESTIONING STRATEGIES
MN
The ratio ___
= 1.5, therefore the similarity transformation is (x, y) → (1.5x, 1.5y).
QR
6.
How can you use similar figures to solve
problems? Use the fact that corresponding
sides are proportional to find an unknown side
length.
VW Is the statement true?
DE = _
Error Analysis Triangles △DEF and △UVW are similar. _
UV
EF
DE
EF
= ___
.
No. The proportion must compare corresponding sides. One possibility is ___
UV
VW
Explain 2
Applying Properties of Similar Figures
The properties of similar figures can be used to find the measures of corresponding parts.
Example 2
AVOID COMMON ERRORS
Given that the figures are similar, find the values of x and y.

Find the value of x.
Find the value of y.
∠C ≅ ∠R, so m∠C = m∠R
AB = _
_
PS
PQ
3y - 5
4y _
_
=
5
10
3y
5
4y
_ ⋅ 10 = _
⋅ 10
5
10
4y = 6y - 10
4x + 27 = 95
4x = 68
x = 17
Students may sometimes write proportions that fail
to use corresponding sides. While there are multiple
ways to write a correct proportion, encourage
students to use a pattern that they can remember. For
example, so that the numerators always give side
lengths from one triangle and the denominators
always give side lengths from the other triangle.
A
4y ft
(3y - 5) ft
B
D
(4x + 27)°
P 5 ft Q
10 ft
95°
S
C
R
y=5
(2y - 8) cm
Find the value of x.
m∠LMN = m∠XYZ
5(x - 5) =
5x - 25 =
x
=
4x
4x
25
Z
L
1 cm
Y
Find the value of y.
JK
MN
_
=_
VW
YZ
5(x - 5)°
J
1.5
2x - 8 = _
_
M
V
4x°
X
4 cm
1.5 cm
N
W
1
4
2x - 8 = 1.5(4)
2x - 8 = 6
© Houghton Mifflin Harcourt Publishing Company
K

COMMUNICATE MATH
Give student pairs different pictures of two figures
drawn on coordinate planes. The first student decides
whether a pair of figures is similar or not, and states
three reasons why. The other student must agree or
provide an alternative explanation. Students switch
roles and repeat the process with another pair of
figures.
2x = 14
x= 7
Module 16
854
Lesson 3
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M16L3 854
Modeling
18/04/14 9:05 PM
Have students work with pattern blocks to explore the concept of similarity.
Students can work in pairs, with one student creating a block design and the other
creating a figure that is similar, but not congruent, to the first one. For more sets
of pattern blocks, search for websites that feature “virtual pattern blocks.” These
sites allow students to select, combine, and rearrange blocks on a computer screen.
Corresponding Parts of Similar Figures
854
Reflect
ELABORATE
7.
QUESTIONING STRATEGIES
Discussion What are some things you need to be careful about when solving
problems involving finding the values of variables in similar figures?
Possible Answer: When finding side lengths, you need to be sure to set up the proportion
correctly, and you need to remember that the side lengths are proportional rather than
Is an image always similar to its pre-image?
Explain. No; for example, stretching and
shrinking a figure disproportionally can produce an
image that is not similar.
equal. Once you have the algebraic equation set up, you need to be careful not to make
errors in the calculations.
Use the diagram, in which △ABE ∼ △ACD.
Can two congruent angles have sides that are
of different lengths? Explain. Yes, the
measure of the angle refers to the amount of
opening between the two sides; the length of the
sides of the angle does not affect its size.
5.6 cm
Find the value of x.
9.
5 cm
50°
C
Find the value of y.
CD
_
=
AE
50 = 3x + 14
BE
5.6 + y _
5
_
=
4
5.6
5
5.6 + y = _ ⋅ 5.6
36 = 3x
When two figures are given as similar, what
can you conclude about their sides and
angles? In similar figures, the corresponding sides
are proportional and the corresponding angles are
congruent.
4 cm
B
m∠C = m∠ABE
SUMMARIZE THE LESSON
E
(3x + 14)°
A
8.
D
y cm
12 = x
4
y = 1.4
© Houghton Mifflin Harcourt Publishing Company
Elaborate
10. Consider two similar triangles △ABC and △A'B'C'. If both m∠A' = m∠C and
m∠B' = m∠A, what can you conclude about triangle △ABC ? Explain your reasoning.
Since the two triangles are similar, we know m∠A' = m∠A and m∠B' = m∠B .
Along with the given information, this tells us m∠A = m∠C and m∠B = m∠A.
Therefore, ΔABC is equilateral.
11. Rectangle JKLM maps to rectangle RSTU by the transformation (x, y) → (4x, 4y).
If the perimeter of RSTU is x, what is the perimeter of JKLM in terms of x?
The ratio of corresponding sides is 1:4, and therefore the ratio of the perimeters is 1:4.
x
The perimeter of JKLM is _
.
4
12. Essential Question Check-In If two figures are similar, what can we conclude
Similar figures have corresponding angles that are congruent and have corresponding
sides that are proportional.
Module 16
855
Lesson 3
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M16L3 855
Connect Vocabulary
Students may be very familiar with the connotation of the word translation as it
refers to interpretation between languages. In fact, this use of the word comes
from its mathematically rigorous sense: translating between languages “moves”
the meaning from one to another, just as translating a figure moves it from one
place to another. Make the connection that mathematical translation does not
change the essence of the original figure, much as a linguistic translation does
not significantly alter the original meaning of the message as it enters another
language.
855
Lesson 16.3
18/04/14 9:05 PM
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
In the figures, are corresponding angles congruent? Are corresponding sides
proportional? Are the figures similar? Describe how you know using similarity
transformations.
1.
2.
ASSIGNMENT GUIDE
Yes; no; no. Rotate the smaller square
90° around its upper right corner,
then translate it down and left until
the lower left vertices coincide. If you
dilate it with a scale factor of 3 to 2.5,
the short edges will be congruent but
the long edges will not be.
Yes; yes; yes. Translate the smaller square
down and right so its upper left vertex
coincides with that of the larger square.
Then dilate with a scale factor of 2 about
this vertex.
3.
4.
5.
Figure ABCD is similar to figure MNKL.
Write a proportion that contains BC and KL.
CD
BC
=
NK
KL
6.
△XYZ is similar to △XVW. Write the
congruence statements that must be true.
8.
_ _
_ _
7.
∠X ≅ ∠X, ∠Y ≅ ∠V, and ∠Z ≅ ∠W.
9.
CDEF maps to JKLM with the transformations
(x, y) → (5x, 5y) → (x - 4, y - 4). What is the
1
EF ? __
value of ___
LM
5
Module 16
Exercise
IN2_MNLESE389847_U7M16L3 856
△DEF is similar to △STU. Write a proportion that
contains ST and SU.
ST
SU
=
DE
DF
△MNP is similar to △HJK, and both triangles are
isosceles. If m∠P > 90°, name all angles that are
congruent to ∠H.
If m∠P > 90°, then ∠M ≅ ∠N. So,
∠M, ∠J and ∠N are all ≅ to ∠H.
Practice
Explore
Connecting Angles and Sides of
Figures
Exercises 1–4
Example 1
Justifying Properties of Similar
Figures Using Transformations
Exercises 9–12
Example 2
Applying Properties of Similar
Figures
Exercises 5–8
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Compare the symbols for equality, similarity,
© Houghton Mifflin Harcourt Publishing Company
Yes; yes; yes. Rotate the smaller
figure 90°, then translate it so that
two corresponding vertices coincide.
Then dilate the smaller figure the
centers by a factor of 2 about that
vertex.
No; no; no. You can rotate and
translate the triangle on the right
so the right angles coincide, but no
similarity transformation will make
the acute angles congruent.
Concepts and Skills
and congruence. Point out that combining the
equality and similarity symbols gives the symbol for
congruence. Connect the composition of the symbol
to its meaning: congruence combines equality and
similarity, because congruent figures have the same
size and shape.
10. △PQR maps to △VWX with the transformation
(x, y) →(x + 3, y - 1) → (2x, 2y). If WX = 12,
what does QR equal? QR = 1 (12) = 6
2
__
Lesson 3
856
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–4
2 Skills/Concepts
MP.3 Logic
5–17
2 Skills/Concepts
MP.5 Using Tools
18–20
2 Skills/Concepts
MP.2 Reasoning
21–22
2 Skills/Concepts
MP.4 Modeling
23–24
2 Skills/Concepts
MP.4 Modeling
25
3 Strategic Thinking
MP.3 Logic
26
3 Strategic Thinking
MP.3 Logic
27
3 Strategic Thinking
MP.3 Logic
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Corresponding Parts of Similar Figures
856
11. △QRS maps to △XYZ with the transformation
(x, y) → (6x, 6y). If QS = 7, what is the length
of XZ?
AVOID COMMON ERRORS
When solving a proportion about a scale model,
some students may make mistakes because they
failed to read the units carefully. Remind them that
the lengths of corresponding sides do not necessarily
appear in the same units, and they may need to
convert one of the measures.
12. Algebra Two similar figures are similar based on
the transformation (x, y) → (12x, 3a 2y). What is/
are the value(s) of a?
XZ = 42
If the figures are similar, then 12 = 3a 2,
and a = 2 or a = -2.
13. Algebra △PQR is similar to △XYZ. If PQ = n + 2, QR = n - 2, and XY = n 2 - 4,
what is the value of YZ, in terms of n?
Since n 2 - 4 = (n + 2)(n - 2), the ratio of a pair corresponding sides is
(n - 2) :1. Then the value of YZ is (n - 2)(n - 2) = n 2 - 4n + 4.
14. Which transformations will not produce similar figures? Select all that apply and
Choices B and E will not produce similar
A. (x, y) → (x - 4,y) → (-x, -y) → (8x, 8y)
B. (x, y) → (x + 1, y + 1) → (3x, 2y) → (-x, -y)
C. (x, y) → (5x, 5y) → (x, -y) → (x + 3, y - 3)
D. (x, y) → (x, 2y) → (x + 6, y - 2) → (2x, y)
E. (x, y) → (x, 3y) → (2x, y) → (x - 3, y - 2)
figures, because each sequence contains
transformations that are not dilations or rigid
motions, and are not balanced elsewhere
in the sequence. Choice D contains two
transformations that are not dilations or rigid
motions, but together create a dilation.
15. The figures in the picture are similar to each other.
Find the value of x.
The longer bases of the trapezoids are 6 units and 3 units, so
the scale factor is 2:1. Set up the equation x + 1 = 2(x - 3)
and solve to get x = 7.
x-3
6
3
x+1
16. In the diagram, △NPQ ∼ △NLM and PL = 5.
M
a. Find the value of x.
© Houghton Mifflin Harcourt Publishing Company
3x + 18 = 60
3x = 42
4 cm
x = 14
P
b. Find the lengths NP and NL.
NP
NL
=
NM
NQ
y
5-y
=
4
3.2
3.2(5 - y) = 4y
3.2 cm
(3x + 18)°
L
16 = 7.2y
Module 16
IN2_MNLESE389847_U7M16L3 857
Lesson 16.3
y
N
20
_
=y
9
25
20
So NP = _ and NL = _.
9
857
60°
_ _
_ _
Q
9
857
Lesson 3
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17. △CDE maps to △STU with the transformations
(x, y) → (x - 2, y -2) → (3x, 3y) → (x, -y).
If CD = a + 1, DE = 2a - 1, ST = 2b + 3, and TU = b + 6, find the values of a and b.
Lengths of sides of ∆STU are 3 times the lengths of corresponding
sides of △CDE. This produces the equations 3(a + 1) = 2b + 3 and
3(2a - 1) = b + 6. Solve this system of equations to get a = 2 and b = 3.
18. If a sequence of transformations contains the transformation (ax, by), with a ≠ b, could
the pre-image and image represent congruent figures? Could they represent similar,
Yes; if ab = ±1, then the pre-image and image are congruent (and therefore also similar).
The pre-image and image could represent similar, non-congruent figures if the sequence of
transformations also contains (bx, ay). The transformation (ax, by) followed by (bx, ay) is
equivalent to the dilation (abx, aby).
19. Is any pair of equilateral triangles similar to each other? Why or why not?
Yes, the triangles will be similar. All the angles of each triangle are equal to 60°, and
therefore corresponding angles are congruent. The sides of each triangle are congruent,
and therefore the ratio of corresponding sides is constant.
20. Figure CDEF is similar to figure KLMN. Which statements are false? Select all that apply
and explain why.
CD = _
EF
A. _
KL
MN
CF = _
EF
B. _
KN
MN
CF
DE = _
C. _
LM
KN
LM = _
KL
D. _
DE
CD
KN
LM = _
E. _
DE
CD
Choice E is false. The proportion doesn’t match corresponding sides with each other.
Dalton/Alamy
Consider this model of a train locomotive when answering
the next two questions.
21. If the model is 18 inches long and the actual locomotive is 72 feet long, what is the
similarity transformation to map from the model to the actual locomotive? Express the
answer using the notation x → ax, where x is a measurement on the model and ax is the
corresponding measurement on the actual locomotive.
The length of the model is 1.5 feet and so the similarity transformation is x → 48x.
22. If the diameter of the front wheels on the locomotive is 4 feet, what is the diameter of the
front wheels on the model? Express the answer in inches.
The diameter of the front wheels on the model is 1 inch.
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Lesson 3
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Corresponding Parts of Similar Figures
858
Use the following graph to answer the next two problems.
JOURNAL
y
Have students write a journal entry in which they
make up their own problems using an unknown
length that must be found using similar triangles.
Remind students to include the solutions to their
problems.
8
B
C
K
4
A
-8 D -4 L
x
0
4
J
-4
-8
8
M
23. Specify a sequence of two transformations that will map ABCD onto JKLM.
Possible answer: (x, y) → (2x, 2y) → (x + 14, y - 8) or (x, y) → (x + 7, y - 4) → (2x, 2y)
AC + BD
24. Find the value of _______
.
JL + KM
AC
The two figures are similar, so all corresponding sides/diagonals have the same ratio. ___
JL
BD
1
1
1
= __
and ___
= __
, and so _______
= __
.
2
2
2
KM
JL + KM
AC + BD
H.O.T. Focus on Higher Order Thinking
25. Counterexamples Consider the statement “All rectangles are similar.” Is this
statement true or false? If true, explain why. If false, provide a counterexample.
The statement is false. For example, a rectangle measuring 5 units by 2 units is not similar
to a rectangle measuring 4 units by 3 units.
© Houghton Mifflin Harcourt Publishing Company
26. Justify Reasoning If ABCD is similar to KLMN and MNKL, what special type of
Looking at the corresponding angles, ∠A ≅ ∠K and ∠A ≅ ∠M, which means ∠K ≅ ∠M
by the Transitive Property of Congruence. Also, ∠B ≅ ∠L and ∠B ≅ ∠N, which means
∠L ≅ ∠N by the Transitive Property of Congruence. If both pairs of opposite angles of a
quadrilateral are congruent, then the quadrilateral is a parallelogram. Therefore, KLMN is a
parallelogram. It could also be a rhombus, rectangle, or square, but more information would
be needed to justify that conclusion.
27. Critique Reasoning Consider the statement “If △PQR is similar to △QPR, then
△PQR is similar to △RPQ.” Explain whether or not this statement is true.
The statement is false. If △PQR is similar to △QPR, then ∠P ≅ ∠Q and the triangle is
isosceles. This does not prove any correspondence between ∠P and ∠R, or between ∠Q and
∠R. For △PQR to be similar to △RPQ, the triangle would have to be equilateral.
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Lesson 16.3
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Lesson 3
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You’ve hired an architect to design your dream house and now the house
has been built. Before moving in, you’ve decided to wander through the house
with a tape measure to see how well the builders have followed the architect’s
floor plan. Describe in as much detail as you can how you could accomplish
your goal. Then discuss how you can decide whether the room shapes and
other features of the house are similar to the corresponding shapes on the floor
plan.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Rectangle 1 is a units in length and b units in
BEDROOM
KITCHEN
BEDROOM
LIVING RM
17.5’ x 18.7’
width. Rectangle 2 is obtained by multiplying the
sides of Rectangle 1 by 10.
FOYER
• Is Rectangle 2 similar to Rectangle 1?
Explain. Yes; multiplying the sides by the same
scale factor changes the size of the figure but
not the shape, so the rectangles are similar.
SCALE
1“
2 = 10 ft
Among things students should discuss:
•the scale of the floor plan and how they could use it to check the dimensions of the rooms;
• a method they could use to check whether the shapes of the rooms and other features on
the floor plan could be mapped to the corresponding shapes in the house by a series of
transformations, including dilations;
• How does the area of Rectangle 2 compare to
that of Rectangle 1? Explain. The area of
Rectangle 1 is ab, and the area of Rectangle 2 is
100ab, so the area of Rectangle 2 is 100 times as
great as the area of Rectangle 1.
• the measurements that should be preserved in the transformation from floor plan to house
(e.g. angles) and those that would not be (e.g. lengths).
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Three streets meet to form an equilateral
© Houghton Mifflin Harcourt Publishing Company
triangle 100 yards on each side. In a photo of the
triangle taken directly overhead, each side measures
4 inches. How are the actual triangle of streets and
the triangle in the photo alike? How are they
different? Explain. Their sizes are different but their
shapes are the same; the sides of the triangle in the
photograph are shorter than the actual triangle,
but because they are both equilateral triangles,
their sides are proportional and each has three
60° angles.
Module 16
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Lesson 3
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M16L3 860
Supply yardsticks or meter sticks to students and have them make a floor plan of
the classroom. Students should concentrate on features of the room that are
permanent, such as dimensions and shape, omitting those that are not—desks and
tables, for example. You may wish to specify the approximate size of the floor plan
(for example, an 8.5 in. × 11 in. sheet of paper), thereby forcing students to
calculate an appropriate scale. Students may exchange floor plans with a partner
and check the partner’s scale, shapes, and calculations.
18/04/14 9:05 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Corresponding Parts of Similar Figures
860
LESSON
16.4
Name
AA Similarity of
Triangles
Class
16.4 AA Similarity of Triangles
Essential Question: How can you show that two triangles are similar?
Common Core Math Standards
Explore
The student is expected to:
COMMON
CORE
G-SRT.A.3
Exploring Angle-Angle Similarity
for Triangles
Resource
Locker
Two triangles are similar when their corresponding sides are proportional and their
corresponding angles are congruent. There are several shortcuts for proving triangles
are similar.
Use the properties of similarity transformations to establish the AA
criterion for two triangles to be similar. Also G-SRT.B.5
A
Mathematical Practices
COMMON
CORE
Date
Draw a triangle
_ and label it △ABC. Elsewhere on your page, draw a segment
longer than AB and label the endpoints D and E.
MP.5 Using Tools
C
Language Objective
F
Explain to a partner how to use the Angle-Angle criterion to show
similarity in triangles.
B
A
D
D
B
Copy ∠CAB and ∠ABC to points D and E, respectively. Extend the rays of your copied
angles, if necessary, and label their intersection point F. You have constructed △DEF.
Essential Question: How can you
show that two triangles are similar?
C
You constructed angles D and E to be congruent to angles A and B, respectively. Therefore,
We can use the AA, SSS, or SAS similarity criteria to
prove that triangles are similar.
PREVIEW: LESSON
View the Engage section online. Discuss the
illustration and ask students to speculate on what it
depicts. Then preview the Lesson Performance Task.
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
angles C and F must also be congruent because of the Third Angle Theorem.
D
Check the proportionality of the corresponding sides. Possible answer (ratios should be equal):
4
AB = _ = 0.4
_
DE
10
3
AC = _ =
_
0.4
DF
7.5
2
BC = _
_
= 0.4
EF
5
Since the ratios are equal, the sides of the triangles are proportional .
Reflect
1.
triangles that have two corresponding congruent angles?
If two triangles have two corresponding congruent angles, the triangles must be similar.
Module 16
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 4
861
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Lesson 4
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Explain 1
Proving Angle-Angle Triangle Similarity
EXPLORE
The Explore suggests the following theorem for determining whether two triangles are similar.
Angle-Angle (AA) Triangle Similarity Theorem
Exploring Angle-Angle Similarity
for Triangles
If two angles of one triangle are congruent to two angles of another triangle, then the
two triangles are similar.
Example 1
Prove the Angle-Angle Triangle Similarity Theorem.
Given: ∠A ≅ ∠X and ∠B ≅ ∠Y
B
INTEGRATE TECHNOLOGY
Y
Prove: △ABC ∼ △XYZ
Students have the option of doing the similar
triangles activity either in the book or online.
C
A
Z
X

QUESTIONING STRATEGIES
XY . Let the image of △ABC be △A´B´C.
Apply a dilation to △ABC with scale factor k = _
AB
B’
What does Angle-Angle similarity claim about
triangles? According to the Angle-Angle
similarity criterion, triangles with two pairs of
congruent angles are similar.
Y
B
dilation
A
C
C’
A’
Z
X
∠A and ∠B´ ≅ ∠B
corresponding
angles
of
similar
triangles are congruent .
because
XY
∙ AB = XY
AB
Also, A´B´ = k ∙ AB =
.
EXPLAIN 1
△A´B´C is similar to △ABC, and ∠A´ ≅
_
It is given that ∠A ≅ ∠X and ∠B ≅ ∠Y
By the Transitive Property of Congruence, ∠A´ ≅
∠X and ∠B´ ≅ ∠Y .
the
ASA
Triangle
Congruence
Theorem .
So, △A´B´C´ ≅ △XYZ by
This means there is a sequence of rigid motions that maps △A´B´C´ to △YYZ.
The dilation followed by this sequence of rigid motions shows that there is a sequence of
similarity transformations that maps △ABC to △XYZ. Therefore, △ABC ∼ △XYZ.
Reflect
2.
Discussion Compare and contrast the AA Similarity Postulate with the ASA Congruence Postulate.
Both postulates require that two pairs of angles be congruent, but for congruence you also
© Houghton Mifflin Harcourt Publishing Company

Proving Angle-Angle Triangle
Similarity
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Compare proving two triangles similar with
proving two triangles congruent.
QUESTIONING STRATEGIES
need to know that the included sides are congruent, so that the figures are the same size.
How do you use AA similarity to show two
triangles are similar? Show that two angles
of one triangle are congruent to two angles of the
other triangle. This lets you conclude that the two
triangles are similar.
The AA Similarity Postulate shows only that the two triangles are the same shape.
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Lesson 4
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M16L4.indd 862
Learning Progression
18/04/14 9:01 PM
Students have already proved triangles congruent using SAS, SSS, and ASA. The
same kind of reasoning is used here to explore AA similarity, and to use this
similarity to solve problems.
AA Similarity of Triangles
862
3.
EXPLAIN 2
In △JKL, m∠J = 40° and m∠K = 55°. In △MNP, m∠M = 40° and m∠P = 85°. A student concludes
that the triangles are not similar. Do you agree or disagree? Why?
Disagree; by the Triangle Sum Theorem, m ∠ N = 55°, so the triangles are similar by the
AA Similarity Criterion.
Applying Angle-Angle Similarity
Explain 2
Applying Angle-Angle Similarity
Architects and contractors use the properties of similar figures to find any
unknown dimensions, like the proper height of a triangular roof. They can use
a bevel angle tool to check that the angles of construction are congruent
to the angles in their plans.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Remind students that, in triangle similarity,
they should identify sides that are proportional, rather
than congruent.
Example 2

Find the indicated length, if possible.
BE
First, determine whether △ABC ∼ △DBE.
By the Alternate Interior Angles Theorem, ∠A ≅ ∠D and ∠C ≅ ∠E,
so △ABC ∼ △DBE by the AA Triangle Similarity Theorem.
E
Find BE by solving a proportion.
Lund/Drew Kelly/Blend Images/Corbis
BD = _
BE
_
BA
BC
A
36
54 = _
BE
_
54
36
54
54
54 ∙ 54 = _
BE ∙ 54
_
54
36
BE = 81

B
D
C
RT
Check whether △RSV ∼ △RTU:
It is given in the diagram that ∠ RSV ≅ ∠ T . ∠R is shared by both
triangles,
so ∠R ≅ ∠R by the
Reflexive Property of Congruence.
AA Triangle Similarity Theorem , △RST ∼ △RTU .
So, by the
R
S
10
T
8
12
Find RT by solving a proportion.
V
TU
RT = _
_
RS
SV
U
12
RT = _ ∠RT = 15
_
10
8
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Lesson 4
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M16L4.indd 863
Small Group Activity
Have students work in small groups and draw diagrams to illustrate each of these
statements: all squares are similar; all rectangles are not similar; if two polygons
are congruent, they are also similar; all right triangles are not similar.
863
Lesson 16.4
18/04/14 9:01 PM
Reflect
4.
In Example 2A, is there another way you can set up the proportion to solve for BE?
BD
BA
Yes;
=
would also give the correct result for BE.
BE
BC
5.
Discussion When asked to solve for y, a student sets up the proportion as shown. Explain why the
proportion is wrong. How should you adjust the proportion so that it will give the correct result?
QUESTIONING STRATEGIES
_ _
A
y
y _
14
_
=
8 10
The variable y does not refer to the side of a triangle but just a
14
B
10
C
How can you use the AA similarity postulate
to find unknown dimensions? You can use
AA similarity to establish that two triangles with
two congruent pairs of angles are similar, and then
write a proportion to find unknown lengths of sides.
D
segment of a side.
y____
+8
14
= __
would be a correct way to solve for y.
8
10
8
AVOID COMMON ERRORS
E
Some students may use an incorrect sequence of
points when writing a similarity statement. Compare
the process to writing a congruence statement and
remind them to list corresponding vertices in the
same order.
6.
A builder was given a design plan for a triangular roof as shown. Explain how he knows that
△AED ∼ △ACB. Then find AB.
A
9 ft.
E
D
15 ft.
C
7.
6 ft.
B
By the Corresponding Angles Theorem, ∠AED ≅ ∠C
and ∠ADE ≅ ∠B (or ∠A ≅ ∠A by the Reflexive Property
of Congruence), so △AED ∼ △ACB by the AA Triangle
Similarity Theorem.
AB
_6 = _
→ AB = 10 feet
9
Find PQ, if possible.
_
_
∠TSQ is a right angle because PS and TR are perpendicular.
So △PRS ∼ △TQS by the AA Triangle Similarity Theorem.
Let x = PQ.
P
Q
9
15
Explain 3
S
12
R
x+9 _
12
_
→ x = 11
=
15
9
© Houghton Mifflin Harcourt Publishing Company
T
15
So, PQ = 11.
Applying SSS and SAS Triangle Similarity
In addition to Angle-Angle Triangle Similarity, there are two additional shortcuts for proving
two triangles are similar.
Side-Side-Side (SSS) Triangle Similarity Theorem
If the three sides of one triangle are proportional to the corresponding sides of
another triangle, then the triangles are similar.
Side-Angle-Side (SAS) Triangle Similarity Theorem
If two sides of one triangle are proportional to the corresponding sides of another
triangle and their included angles are congruent, then the triangles are similar.
Module 16
864
Lesson 4
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M16L4.indd 864
Cognitive Strategies
18/04/14 9:01 PM
Have students write their own AAA Similarity Postulates. Then ask them to
explain why this postulate is not necessary. An AAA Similarity Postulate is not
required because of this theorem: Given two triangles, if two pairs of
corresponding angles are congruent, then the remaining pair of corresponding
angles must also be congruent.
AA Similarity of Triangles 864
Example 3
EXPLAIN 3

N
4
Applying SSS and SAS for Triangle
Similarity
4
M
8
P
Q
6
R
You are given two pairs of corresponding side lengths and one pair of congruent
corresponding angles, so try using SAS.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Remind students that similarity statements
Check that the ratios of corresponding sides are equal.
8 =_
MN = _
8 =_
4=_
MP = _
2
2
_
_
8+ 4
3
12
3
6
MR
MQ
Check that the included angles are congruent: ∠NMP ≅ ∠QMR is given in the diagram.
Therefore △NMP ∼ △RMQ by the SAS Triangle Similarity Theorem.
indicate corresponding parts in the same way
congruence statements do.

M
G
4
8
H
15
12
10
J
N
L
6
You are given three pairs of corresponding side lengths and zero congruent
corresponding angles, so try using the SSS Triangle Similarity Theorem .
© Houghton Mifflin Harcourt Publishing Company
Check that the ratios of corresponding sides are equal.
3
12
LM = _
_
=_
GH
8
2
Therefore △
GHJ
3
15
MN = _
_
=_
HJ
10
2
∼△
LMN
3
6
GJ
_
=_=_
LN
2
4
by the SSS Triangle Similarity Theorem .
Since you are given all three pairs of sides, you don’t need to check for congruent angles.
Reflect
8.
Are all isosceles right triangles are similar? Explain why or why not.
Let the legs of one isoc. triangle be x and the legs of another be y. Then the ratios of
the sides would be _xy and _xy . The included angle in each triangle is the right angle, so two
isosceles right triangles are similar by SAS.
9.
Why isn't Angle-Side-Angle (ASA) used to prove two triangles similar?
ASA implies two pairs of angles are congruent, which is sufficient to show the triangles
similar, so checking the included sides is unnecessary.
Module 16
865
Lesson 4
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M16L4.indd 865
Connect Vocabulary
Relate the idea of proof to justifying ideas in math. You use established rules and
conventions to draw some conclusions. In real life, proof means showing something
by gathering evidence by established rules and conventions.
865
Lesson 16.4
18/04/14 9:01 PM
QUESTIONING STRATEGIES
If possible, determine whether the given triangles are similar. Justify your answer.
10.
C
5
10
6
F
B
11.
H
The two triangles cannot be proven similar. Although
3
the two given sides are in proportion, there is not a
G
pair of included congruent angles.
A
AVOID COMMON ERRORS
By the Pythagorean Theorem, NO = 6 and GH = 4,
M
8
Why are ASA and AAS not similarity
theorems? Both of these contain two pairs
of corresponding congruent angles, so ASA and AAS
triangles are already similar by the AA similarity
theorem.
Some students may have difficulty identifying
corresponding sides in similar triangles because of
the orientation of the figures. Show these students
how they can copy one of the triangles onto a piece
of paper, then cut it out and rotate it, so that the two
triangles have the same orientation.
HJ
GJ
GH
1
so ___
= ___
= ___
=_
. △MNO ∼ △GHI by the
MN
2
NO
MO
SSS Triangle Similarity Theorem.
10
J
5
N
O
G
3
H
ELABORATE
Elaborate
12. Is triangle similarity transitive? If you know △ABC ∼ △DEF and △DEF ∼ △GHJ, is △ABC ∼ △GHJ?
Explain.
Yes. If the first two triangles have three pairs of congruent angles and the second two
QUESTIONING STRATEGIES
Two isosceles triangles have congruent vertex
angles. Explain why the two triangles must be
similar. Let the measure of the vertex angles be x°.
Then, by the Isosceles Triangle Theorem, the base
angle in each of the triangles must measure half
of (180 - x)°. So, the triangles are similar by AA
similarity.
triangles do as well, then the first and third triangles will also have those three pairs of
13. The AA Similarity Postulate applies to triangles. Is there an AAA Similarity Postulate for quadrilaterals?
Use your geometry software to test your conjecture or create a counterexample.
No; a square and a rectangle each have three pairs of right angles, but they won’t be
similar because the sides aren’t proportional.
14. Essential Question Check-In How can you prove triangles are similar?
Triangles are similar when their corresponding angles are congruent and their
© Houghton Mifflin Harcourt Publishing Company
congruent angles.
SUMMARIZE THE LESSON
corresponding sides are in proportion, but it is sufficient to use AA Similarity (show two
pairs of congruent angles), SSS Similarity (show three pairs of sides in proportion), or
SAS Similarity (show two pairs of sides are in proportion and their included angles are
congruent).
Module 16
IN2_MNLESE389847_U7M16L4.indd 866
866
Lesson 4
18/04/14 9:01 PM
Which postulates allow you to conclude that triangles
are similar without using transformations to map one
to the other? What do you need to know before you
can apply them? The AA similarity postulate, the
SSS similarity postulate, and the SAS similarity
postulate; you need to know than the triangles have
two pairs of congruent angles, or that all three pairs
of sides are proportional, or that two pairs of sides
are proportional and the included angles are
congruent.
AA Similarity of Triangles 866
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
Show that the triangles are similar by measuring the lengths of their sides and
comparing the ratios of the corresponding sides.
1.
2.
F
A
B
D
E
ASSIGNMENT GUIDE
Concepts and Skills
D
B
C
A
C
E
F
Practice
4.5
DE = _
_
=
AB
3
By the Pythagorean Theorem, NO = 6 and GH = 4,
Explore
Exercises 1–2
HJ
GJ
GH
1
so ___
= ___
= ___
=_
. △MNO ∼ △GHI by the
MN
2
NO
MO
Exploring
Angle-Angle
Similarity in
SSSTriangles
Triangle Similarity Theorem.
Example 1
Proving Angle-Angle Triangle
Similarity
Exercises 3–6
Example 2
Applying Angle-Angle Similarity
Exercises 7–10
Example 3
Applying SSS and SAS for Triangle
Similarity
Exercises 11–14
3
__
or 1.5
2
2.1
DF = _
_
=
AC
1.4
_3 or 1.5
3.9
EF = _
_
=
BC
2.6
_3 or 1.5
2
2
3
AB = _
_
=
DE
6
_1
2
AC = _
_
=
DF
4
_1
2.5
BC = _
_
=
EF
5
_1
2
2
2
Determine whether the two triangles are similar. If they are similar, write the
similarity statement.
3.
4.
D
B
A
65°
16°
48°
© Houghton Mifflin Harcourt Publishing Company
E
A
65°
C
67°
F
Exercise
IN2_MNLESE389847_U7M16L4.indd 867
Lesson 16.4
B
By the AA Triangle Similarity
Postulate, △ABC ∼ △DEF.
Module 16
867
78° D
By the Triangle Angle Sum Theorem,
m∠C = 67°. So ∠A ≅ ∠D and ∠C ≅ ∠F.
C
△ABC and △ACD are isosceles triangles,
so ∠B ≅ ∠ACB and ∠D ≅ ∠ACD. 16 + 2 ∙
m ∠B = 180, so m ∠B = 82. However,
m ∠D = m ∠ACD = 78. None of the
angles in △ABC is congruent to ∠D,
so the triangles are not similar.
Lesson 4
867
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–2
2 Skills/Concepts
MP.5 Using Tools
3–17
2 Skills/Concepts
MP.2 Reasoning
18
2 Skills/Concepts
MP.4 Modeling
19
3 Strategic Thinking
MP.3 Logic
18/04/14 9:01 PM
Determine whether the two triangles are similar. If they are similar, write the similarity
statement.
5.
6.
D
A
D
C
A
AVOID COMMON ERRORS
B
Because they need to know only that two angles of
two triangles are congruent in order to prove
similarity, students might think they need to know
only three angles of two quadrilaterals to do the
same, and so on for any n-gon. Use counter examples
to show that this is incorrect. Stress that triangles are
a special case because they are rigid structures.
C
∠D ≅ ∠C, and ∠BAC ≅ ∠DCA by the
Alternate Interior Angles Theorem.
Therefore △ADC ∼ △BCA by the AA
Triangle Similarity.
E
B
∠A ≅ ∠E, and ∠ACB ≅ ∠ECD by the Vertical
Angles Theorem. Therefore △ABC ∼ △EDC
by the AA Triangle Similarity Theorem.
Explain how you know whether the triangles are similar. If possible,
find the indicated length.
7.
AC
8.
A
F
C
5.0
10.2
9.0
B
A
B D
7.5
15.0
E
D
The triangles are similar by the AA Triangle
Similarity Theorem. It is not possible to find
the indicated length because, the length of
the corresponding side, DF, is not known.
9.
QR
12.0
C
E
The triangles are similar by AA Similarity.
5
__
=_
12
9
10. Find BD.
A
P
A
2.0
B
1.6
B
C
R
Not possible. Only one congruent angle is
identified between △ABC and △DBC, so
similarity cannot be established.
The triangles are similar by AA Similarity.
QR
1.2
___
= ___
→ QR = 0.96
1.6
2
Module 16
Exercise
IN2_MNLESE389847_U7M16L4.indd 868
C
D
Q
1.2
17.0
Lesson 4
868
Depth of Knowledge (D.O.K.)
© Houghton Mifflin Harcourt Publishing Company
8.3
COMMON
CORE
Mathematical Practices
20
3 Strategic Thinking
MP.2 Reasoning
21–22
3 Strategic Thinking
MP.3 Logic
18/04/14 9:01 PM
AA Similarity of Triangles 868
Show whether or not each pair of triangles are similar, if possible. Justify your answer, and
write a similarity statement when the triangles are similar.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 When using the SSS and SAS Similarity
11.
12.
A
D
B
5.6
3.5
C
4.0
16
Theorems, some students have difficulty matching up
the corresponding sides. Tell these students to match
up the smallest side to the smallest side, the longest
side to the longest side, and match the sides that are
neither longest nor shortest.
16
A
E
9
D
6.4
E
3
9
F
∠ACB ≅ ∠ECD by the Vertical Angle
equal, so the two triangles are not similar.
3.5
BC
Theorem. __
= ___
= 0.625;
5.6
CD
AC (
= 4.0/6.4 ) = 0.625. Therefore
CE
△ABC ∼ △DEC by SAS Similarity.
13.
14.
B
C
5
_
16 __
5
AC
AB
__
= __
= __
; BC = _
. The ratios are not
DE
5 EF
DF
3
B
H
6
15
D
8
A
C
20
3.2
6
BD
G
© Houghton Mifflin Harcourt Publishing Company
Therefore △ABC ∼ △BDC by SSS Similarity.
R
The triangles cannot be proven similar using
the given information, because the congruent
angle is not an included angle.
BC
5.5
AB
___
= ___
= 1.15, and ___
= __6 = 1
B
5.5
6.0
EF
F
4.8
A
6.0
E
4.8
DF
6
Because the two ratios are not equal, the two triangles are
not similar.
4.4
7.5
C
D
¯ is the
The student did not compare corresponding sides of the two triangles. AB
_
shortest side of △ABC, so its corresponding side is DE the shortest side of △DEF.
AC
AB ___
The ratios __
, BC and __
are equal, so the triangles are similar by SSS.
DE EF
DF
IN2_MNLESE389847_U7M16L4.indd 869
Lesson 16.4
60°
J
15. Explain the Error A student analyzes the two triangles shown below.
Explain the error that the student makes.
Module 16
869
9.6
60°
12
8
BC
P
10
8
15
20
BC
AC
AB
__
= ___
= 2.5; __
= __
= 2.5; __
= __
= 2.5.
DC
Q
8
3.2
869
Lesson 4
18/04/14 9:01 PM
16. Algebra Find all possible values of x for which these two triangles are similar.
(x + 10)°
(2x - 50)°
x°
70°
The possible values of x are the solutions of
x = 70, x = 2x - 50, x + 10 = 70, x + 10 =
2x - 50, x + x + 10 + 70 = 180, and
x + x + 10 + 2x - 50 = 180. These result
in 50, 55, 60, or 70, of which only 50 results
in similar triangles. So x = 50° is the only
possible value.
17. Multi-Step Identify two similar triangles in the figure, and explain
why they are similar. Then find AB.
B
∠A ≅ ∠A, and ∠ABD ≅ ∠C , so △ABD ≅ △ACB by the AA
A
AB
AB
4
= __
→ __
= __
→ AB = 8
Triangle Similarity Theorem. __
16
AB
AB
AC
4 D
C
12
18. The picture shows a person taking a pinhole photograph of himself. Light entering
the opening reflects his image on the wall, forming similar triangles. What is the
height of the image to the nearest inch?
15 in.
4 ft 6 in.
5 ft 5 in.
5'5"
65
h
_
= _ = _ → h = 18 inches
15
4'6"
54
H.O.T. Focus on Higher Order Thinking
Y
B
XZ and ∠A ≅ ∠X
XY = _
Given: _
AB AC
Prove: △ABC ∼ △XYZ
A
C
ge07sec07l03004a
AB
Z
X
XY
Apply a dilation to △ABC with scale factor k = __
and let the image of △ABC be △A'B'C'.
AB
Then ∠A' ≅ ∠A. It is given that ∠A ≅ ∠X, so by transitivity ∠A' ≅ ∠X.
XY
XY
XZ
Also A'B' = k ∙ AB = __
∙ AB = XY and A'C' = k ∙ AC = __
∙ AC = __
∙ AC = XZ. Therefore,
AB
AB
AC
© Houghton Mifflin Harcourt Publishing Company
19. Analyze Relationships Prove the SAS Triangle Similarity Theorem.
△A'B'C ≅ △XYZ by SAS Congruence. So a sequence of rigid motions maps △A'B'C to △XYZ.
The dilation followed by this sequence of rigid motions shows that there is a sequence of
similarity transformations that maps △ABC to △XYZ. So △ABC ∼ △XYZ.
Module 16
IN2_MNLESE389847_U7M16L4.indd 870
870
Lesson 4
18/04/14 9:01 PM
AA Similarity of Triangles 870
20. Analyze Relationships Prove the SSS Triangle Similarity Theorem.
JOURNAL
Have students write a journal entry to explain what a
scale on a map means, how it is used, and how it is
related to the concept of similarity.
XZ = _
XY = _
YZ
Given: _
AB AC
BC
Y
B
Prove: △ABC ∼ △XYZ
(Hint: The main steps of the proof are similar to
those of the proof of the AA Triangle
Similarity Theorem.)
A
C
Z
X
XY
Apply a dilation to △ABC with scale factor k = __
and let the image of
AB
△ABC be △A'B'C'. Then:
XY
A'B' = k ⋅ AB = __
⋅ AB = XY
AB
XY
XZ
A'C' = k ∙ AC = __
∙ AC = __
∙ AC = XZ
AB
AC
XY
YZ
B'C' = k ∙ BC = __
∙ BC = __
∙ BC = YZ. Therefore, △A'B'C ≅ △XYZ by the
AB
BC
SSS Congruence Theorem. This means there is a sequence of similarity
transformations that maps △ABC to △XYZ. So △ABC ∼ △XYZ.
‹ ›
−
21. Communicate Mathematical Ideas A student is asked to find point X on BC such that △ABC ∼
‹ ›
−
△XBA and XB is as small as possible. The student does so by constructing a perpendicular line to AC at
‹ ›
−
point A, and then labeling X as the intersection of the perpendicular line with BC . Explain why this
procedure generates the similar triangle that the student was requested to construct.
A
10
6
© Houghton Mifflin Harcourt Publishing Company
X
B
8
C
Possible Answer: For XB to be as small as possible, it should correspond to
―
the shortest side of △ABC, which is AB. Thus, X corresponds to A.
22. Make a Conjecture Builders and architects use scale models to help them design and build new
buildings. An architecture student builds a model of an office building in which the height of the model
1
1
is ___
of the height of the actual building, while the width and length of the model are each ___
of the
400
200
corresponding dimensions of the actual building. The model includes several triangles. Describe how a
triangle in this model could be similar to the corresponding triangle in the actual building, then describe
how a triangle in this model might not be similar to the corresponding triangle in the actual building. Use
a similarity theorem to support each answer.
A triangle that lies entirely in a plane parallel to the ground will have side lengths that are
1
of the side lengths of the corresponding triangle in the actual building. So the two
each ___
200
triangles are similar by the SSS Triangle Similarity Theorem. A triangle that has vertices at
several heights will not have side lengths that form a constant ratio with the side lengths of
the corresponding triangle in the actual building, and so would not be similar.
Module 16
IN2_MNLESE389847_U7M16L4.indd 871
871
Lesson 16.4
871
Lesson 4
18/04/14 9:01 PM
CONNECT VOCABULARY
The figure shows a camera obscura and the object being
figure:
1. Explain how the image of the object would be
affected if the camera were moved closer to the
object. How would that limit the height of objects
that could be photographed?
2. How do you know that △ADC is similar to
△GDE?
The name camera obscura comes from the Latin
words camera, meaning room, and obscura, meaning
dark. The plural of Latin words ending in the letter a
is formed by changing a to ae. So, more than one of
these instruments would be referred to as camerae
obscurae.
A
B
D
E
F
G
C
3. Write a proportion you could use to find the height of the
pine tree.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 A camera obscura is square. This means that
4. DF = 12 in., EG = 8 in., BD = 96 ft. How tall is the pine tree?
1. As the object gets closer the vertical angle at D will get larger, and since
△ADC remains similar to △GDE, the height of the image on the back
surface will increase. So the camera must be placed far enough from the
object that the image is no taller than the height of the back surface, or
else the object will not be able to be photographed.
in the figure in the Lesson Performance Task, DF =
EG. Suppose you want to photograph an object that is
n feet from the front of the camera. What is the
maximum height of such an object if you want to
photograph its entire height? Explain your
reasoning. The maximum height is n feet.
AC
EG
Sample answer: Because DF = EG, and , ___
= ___
,
BD
DF
AC = BD = 1.
2. Possible answer: ∠A ≅ ∠G because the angles are alternate interior angles
―
―
for parallel lines AC and EG . ∠ADC ≅ ∠GDE because vertical angles are
congruent. So, △ADC ∼ △GDE by the AA Similarity Theorem.
AC
EG
= __
3. __
BD
DF
AC __
= 8 ⇒ AC = 64 ft
4. __
96 12
© Houghton Mifflin Harcourt Publishing Company
Module 16
872
Lesson 4
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M16L4.indd 872
Have students research methods of making simple models of a camera obscura.
For some models, no more than a cereal box or shoe box, tape, scissors, and a pin
are required. A more elaborate model can be made by blocking a window of the
classroom and projecting an outside scene onto a sheet in the classroom. Either
method will give reasonable results when students compare the dimensions and
angles of the external object with those of the projected image.
18/04/14 9:01 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
AA Similarity of Triangles 872
MODULE
16
MODULE
STUDY GUIDE REVIEW
16
Similarity and
Transformations
Study Guide Review
Essential Question: How can you use similarity and
transformations to solve real-world problems?
ASSESSMENT AND INTERVENTION
KEY EXAMPLE
(Lesson 16.1)
Determine the center of dilation and the scale factor of the
dilation.
8
B′
y
B
6
Assign or customize module reviews.
4
O
A′
A
2
0
C
2
4
C′
x
6
Key Vocabulary
center of dilation (centro de
dilatación)
dilation (dilatación)
scale factor (factor de escala)
Side-Side-Side Similarity
Side-Angle-Side Similarity
similar (semejantes)
similarity transformation
(transformación de
semejanza)
8
Draw a line through A and A′. Draw a line through B and B′. Draw a line through C and C′.
The three lines intersect at point O(2, 3).
Find the distance from point O to points A and A'.
___
_
d A = √(2 - 3) + (3 - 4) = √ 2
2
2
___
_
d A′ = √ (2 - 4) + (3 - 5) = 2√ 2
2
2
Find the distance to point A.
Find the distance to point A'.
© Houghton Mifflin Harcourt Publishing Company
The distance from point O to point A′ is twice the distance from point O to point A. The scale factor
of dilation is 2 to 1.
KEY EXAMPLE
Corresponding sides of similar
_figures
_are proportional.
_ _
_ _
BC
_ = ____
_.
BC = FG and EH = AD, so ___
EH
FG
Corresponding angles of similar figures are congruent.
∠D ≌ ∠H and ∠E ≌ ∠A.
Module 16
IN2_MNLESE389847_U7M16MC 873
873
Module 16
(Lesson 16.3)
△ABCD maps to △EFGH by a similarity transformation. Write a proportion that
BC and ¯
EH. Then list any angles that are congruent to ∠D or ∠E.
contains ¯
873
Study Guide Review
18/04/14 9:35 PM
KEY EXAMPLE
(Lesson 16.4)
Determine whether △ABC and △DEF are similar. If so, justify by SSS or SAS.
A
3
E
B
6
7.5
8
4
C D
10
F
Check that the ratios of corresponding sides are equal.
3
__
4
6 = __
3
__
8 4
7.5 = __
3
_
4
10
Since the ratios of all corresponding sides are equal, the triangles are similar by SSS.
EXERCISES
Determine the following for the dilation. (Lesson 16.1)
8
6
y
A
B
4
D
2
A′
D′
0
center
Module 16
IN2_MNLESE389847_U7M16MC 874
(3, 0)
2 O 4
C
B′
x
6
2.
874
8
scale factor
© Houghton Mifflin Harcourt Publishing Company
1.
C′
1 to 3
Study Guide Review
18/04/14 9:35 PM
Study Guide Review 874
Determine whether the two figures are similar using similarity transformations.
(Lesson 16.2)
MODULE
3.
similar
△ABC to △DEF
8
B
COMMON
CORE
Mathematical Practices: MP.1, MP.2, MP.4, MP.6
G-MG.A.1, G-MG.A.3
-12
-8
D
A
-4
E
SUPPORTING STUDENT REASONING
Students should begin this problem by focusing on
what information they will need. Here are some
issues they might bring up.
B
y
0
A
x
C
C
D
F 0
-4
2
x
G
-2
4
F
E
-8
H
△ABC maps to △DEF by a similarity transformation. (Lesson 16.3)
5.
• What should be the maximum height of the
model: Students can either work backwards from
the desired height or find a percent of the actual
height and use it to determine the scale.
6.
_
_
Write a proportion that contains BC and DF.
7.
List any angles that are congruent to ∠A or ∠E.
• Which dimensions are not needed for a scale
model: Students can decide which dimensions
might not be needed, such as the height at which
the stone color changes.
8.
¯ ___
¯
AB
__
= BC
¯
DE
¯
EF
¯ ___
¯
BC
__
= AC
¯
EF
¯
DF
∠A ≅ ∠D; ∠E ≅ ∠B
Determine whether △ABC and △DEF are similar. If so, justify by SSS or SAS.
(Lesson 16.4)
not similar
6
A
9
13
SAS
9.
B
C D
2
E
4
E
3
B
F
9
6
A
76°
C D
8
76°
12
F
© Houghton Mifflin Harcourt Publishing Company
• What materials should be used: Students can
research the appropriate materials to use or you
can make suggestions.
not similar
y
2
4
_
_
Write a proportion that contains AB and EF.
• How to determine the dimensions of the
model: Students should use proportions to
convert the actual dimensions to the scaled
dimensions. Students may first want to convert
the actual dimensions to decimals.
△ABCD to △EFGH
4.
Module 16
875
Study Guide Review
SCAFFOLDING SUPPORT
IN2_MNLESE389847_U7M16MC 875
• Students should recognize that they will need to convert units of the given
measurements so that all units are the same. Possible units are feet, inches,
meters, or centimeters.
• If students choose to solve the problem by first deciding on a scale factor, they
will need to make sure the scale factor they use results in a model which will
fit inside a classroom.
875
Module 16
18/04/14 9:35 PM
MODULE
SAMPLE SOLUTION
Designing a Model of the
Washington Monument
Use the scale 1 in. = 10 ft.
Your challenge is to design a scale model of the Washington
Monument that would be small enough to fit inside your classroom.
Here are some key dimensions of the Washington Monument for
you to consider in determining the scale factor for your model.
(Note that the color of the stone changes part way up the monument
because of a halt in construction between 1854 and 1877.)
Total height
Height to top of trapezoidal side
First, convert units so that all measurements are in
1 ft
feet, by multiplying inches by the ratio
and
12 in.
adding that to the number of feet. The results are:
_
Total height: 555.42 ft
555 ft. 5 in.
Width at base: 55.08 ft
500 ft
Width at base
55 ft. 1 in.
Width at top of trapezoidal side
34 ft. 5 in.
Height at which stone color changes
16
Width at top of trapezoidal side: 34.42 ft
To find the model’s dimensions, solve the
proportion:
151 ft
What scale factor will you use for your model? What are the key
model height
1 in. = _____________
_____
10 ft
Begin by making some notes in the space below about your strategy for designing the model. Then
actual height
Using the proportion, check the total height of the
model to see if it will fit inside the classroom.
1 in. = _________
x
_____
→ x ≈ 55.5 in.
10 ft
Schwartz/Digital Vision/Getty Images
Module 16
876
555.42 ft
1 ft
≈ 4.6 ft tall, so the scale factor is
(____
12 in. )
This is 55.54
reasonable.
The remaining dimensions are as follows.
Height to top of trapezoidal side: 50 in.
Width at base: 5.5 in.
Width at top of trapezoidal side: 3.4 in.
Height at which stone color changes: 15.1 in.
Study Guide Review
DISCUSSION OPPORTUNITIES
IN2_MNLESE389847_U7M16MC 876
18/04/14 9:35 PM
• If students choose to work backwards from a desired height of exactly six feet,
how will this change the calculations and scale factor used?
• What are some advantages to choosing a scale that makes the model no more
than a foot tall? Sample answer: The amount of material needed to construct
the model will be much less, which in turn will lower both the cost and the
weight and make the model more portable.
Assessment Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain.
0 points: Student does not demonstrate understanding of the problem.
Study Guide Review 876
16.1–16.4 Similarity and Transformations
ASSESS MASTERY
students have mastered the concepts and standards
covered in this module.
• Online Homework
• Hints and Help
• Extra Practice
1.
ASSESSMENT AND INTERVENTION
Yes
Are the two shapes similar?
16
3
y
B′
2.
Find the scale factor k.
3.
Find the center of dilation.
(4, 2)
12
4.
area △A′B′C′D′ .
Compare k to the ratio __
area △ABCD
8
A′
4
A
C
The ratio equals 9, which is the square of the scale factor.
D
0
Determine which of the following transformations are dilations.
(Lesson 16.1)
5.
Access Ready to Go On? assessment online, and
receive instant scoring, feedback, and customized
intervention or enrichment.
(x, y) → (4x, 4y)
6.
(x, y) → (x - 2, y - 2)
8.
Response to Intervention Resources
9.
• Success for English Learners
• Challenge Worksheets
Assessment Resources
© Houghton Mifflin Harcourt Publishing Company
Find the missing length. (Lesson 16.3)
D′ x
16
)
△XYZ maps to △MNO with the transformation (x, y) → (7x, 7y). If XY = 3, what is the length
of MN?
Find the appropriate statements about the triangles. (Lesson 16.4)
10. △ABC is similar to △RTS. Write a proportion that contains AC and RT. Also write the congruence
statements that must be true.
AC _
__
= AB , ∠A ≅ ∠R, ∠B ≅ ∠T, ∠C ≅ ∠S
RS
RT
ESSENTIAL QUESTION
11. How can you determine whether a shape is similar to another shape?
Answers may vary. Sample: Two shapes are only similar if one shape can be mapped
to the other through similarity transformations. These transformations are the rigid
motions, meaning reflections, translations, and rotations, as well as dilations.
Module 16
COMMON
CORE
IN2_MNLESE389847_U7M16MC 877
Module 16
12
21
• Leveled Module Quizzes
877
(
1 x, _
1y
(x, y) → _
3 3
Dilation
Differentiated Instruction Resources
8
(x, y) → (-x, 3y)
Not a dilation
• Reteach Worksheets
4
Not a dilation
Dilation
7.
C′
B
Study Guide Review
877
Common Core Standards
Lesson
Items
16.2
1
16.1
18/04/14 9:35 PM
Content Standards Mathematical Practices
G-SRT.A.2, G-SRT.A.1,
G-CO.A.2
MP.7
2–4
G-SRT.A.1
MP.7
16.2
5–8
G-SRT.A.1, G-CO.A.2
MP.2
16.2
9
G-SRT.A.1, G-CO.A.2
MP.4
16.3
10
G-SRT.A.2
MP.4
MODULE
MODULE 16
MIXED REVIEW
MIXED REVIEW
1. Consider each transformation. Does the transformation preserve distance?
Select Yes or No for A–C.
A. Dilations
Yes
No
B. Reflections
C. Rotations
Yes
Yes
ASSESSMENT AND INTERVENTION
No
No
(
)
1 _
x, 1 y . Choose True or False
2. △MNO maps to △RST with the transformation (x, y) → _
3 3
for each statement.
A. If RT = 3, MO = 9.
True
False
B. If RT = 12, MO = 4.
True
False
C. If RT = 9, MO = 27.
True
False
3. Determine if the following pair of triangles
_ _are
similar. If so, explain how. Note that AC || BD.
A
70°
Possible Answer: ∠ACB and ∠CBD are
congruent because they are alternate interior
angles. ∠ABC and ∠CDB are congruent
because it is given they have the same
measure. So the two triangles are similar by
Angle–Angle Similarity.
prepare students for high-stakes tests.
B
Assessment Resources
70°
C
• Leveled Module Quizzes: Modified, B
D
4. If △ABC is similar to △XYZ and △YZX, what special type of triangle is △ABC? Justify your
reasoning.
COMMON
CORE
AVOID COMMON ERRORS
Item 2 Some students have trouble following a
dilation rule backward. Remind students that if
moving from the image to the original figure,
multiply by the reciprocal of the original dilation.
© Houghton Mifflin Harcourt Publishing Company
Answers may vary. Sample: Examine the corresponding angles. Since
△ABC is similar to △XYZ, this means ∠A ≅ ∠X. Likewise, since △ABC is
similar to △YZX, this means ∠A ≅ ∠Y. Thus, by the Transitive Property
of Congruence, it can be said that ∠X ≅ ∠Y. Also, ∠B ≅ ∠Y and ∠B ≅ ∠Z,
which means ∠Y ≅ ∠Z by the Transitive Property of Congruence. Putting
these facts together, it can be shown that ∠X ≅ ∠Y ≅ ∠Z. Therefore,
△XYZ, △YZX, and by similarity, △ABC are all equilateral triangles.
Module 16
16
Study Guide Review
878
Common Core Standards
IN2_MNLESE389847_U7M16MC 878
18/04/14 9:35 PM
Content Standards Mathematical Practices
Lesson
Items
IM1 15.3,
IM2 16.1
1*
G-SRT.A.2
MP.7
16.3
2
G-SRT.A.1
MP.7
16.4
3
G-SRT.A.3
MP.7
16.3, 15.2
4*
G-SRT.A.2
MP.3
* Item integrates mixed review concepts from previous modules or a previous course.
Study Guide Review 878
MODULE
17
Using Similar
Triangles
Using Similar Triangles
ESSENTIAL QUESTION:
Answer: For one example, similar triangles can be
used to find the height of a tall object by measuring
17
MODULE
Essential Question: How can you use similar
triangles to solve real-world problems?
LESSON 17.1
Triangle
Proportionality
Theorem
LESSON 17.2
Subdividing a Segment
in a Given Ratio
This version is for
Algebra 1 and
Geometry only
PROFESSIONAL DEVELOPMENT
VIDEO
LESSON 17.3
Using Proportional
Relationships
Professional Development Video
Author Juli Dixon models successful
teaching practices in an actual
high-school classroom.
LESSON 17.4
Similarity in Right
Triangles
Professional
Development
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
my.hrw.com
REAL WORLD VIDEO
Check out how properties of similar
triangles can be used to determine
real-world areas of geographic
regions like the Bermuda Triangle.
How Large Is the Bermuda Triangle?
In this module, you will be asked to determine the area of the Bermuda Triangle from a
map. How can indirect measurement and the properties of similar triangles help you find the
answer? Let’s get started on solving this “mystery” of the Bermuda Triangle!
Module 17
DIGITAL TEACHER EDITION
IN2_MNLESE389847_U7M17MO 879
Access a full suite of teaching resources when and
where you need them:
• Access content online or offline
• Customize lessons to share with your class
• Communicate with your students in real-time
• View student grades and data instantly to target
your instruction where it is needed most
879
Module 17
879
PERSONAL MATH TRAINER
Assessment and Intervention
tests, and intervention activities. Prepare your
students with updated, Common Core-aligned
practice tests.
18/04/14 10:00 PM
Complete these exercises to review skills you will need for this module.
Scale Factor and Scale Drawings
Example 1
Determine the length of the
segment A′B′ given AB = 4,
BC = 3, and B′C′ = 6.
A′
A
The image of △ABC is created
as a result of a scale drawing,
so the transformation is a
dilation.
C
students need strategic or intensive intervention for
the module’s prerequisite skills.
• Online Homework
• Hints and Help
• Extra Practice
B
2BC = B′C′, so 2AB = A′B′.
ASSESSMENT AND INTERVENTION
B′
C′
A′B′ = 2(4) = 8
Give the side length.
1.
6
BC, given AC = 5, A′C′ = 15, and B′C′ = 18
Similar Figures
Example 2
The figures PQRS and KLMN are similar. Determine the angle in figure
KLMN that is congruent to ∠Q and find its measure if m∠Q = 45°.
∠Q ≅ ∠L
Corresponding angles of similar
figures are congruent.
m∠Q = m∠L = 45°
Definition of congruency of angle.
3
2
1
Personal Math Trainer will automatically create a
standards-based, personalized intervention
assignment for your students, targeting each student’s
individual needs!
Give each angle measure.
2.
3.
67°
m∠A, given △ABC ≅ △DEF and m∠D = 67°
m∠E, given △PQR ≅ △DEF, m∠R = 13°, and m∠D = 67°
``
100°
Example 3
A right triangle △ABC has side lengths AB = 3 and BC = 4. Find the length
of the hypotenuse AC.
△ABC is a right triangle, so the Pythagorean Theorem can be used.
__
ac = √ ab 2 + bc 2
__
_
AC = √ 3 2 + 4 2 = √ 9 + 16
AC = 5
Write the Pythagorean Theorem.
Substitute and simplify.
Simplify.
Find the side length for each right triangle.
4.
5.
DE, given AB = 15, AC = 17, and AC is the hypotenuse
IN2_MNLESE389847_U7M17MO 880
See the table below for a full list of intervention
resources available for this module.
Response to Intervention Resources also includes:
• Tier 2 Skill Pre-Tests for each Module
• Tier 2 Skill Post-Tests for each skill
13
DE, given DF = 5, EF = 12, and DE is the hypotenuse
Module 17
© Houghton Mifflin Harcourt Publishing Company
The Pythagorean Theorem
TIER 1, TIER 2, TIER 3 SKILLS
8
880
Response to Intervention
Tier 1
Lesson
Intervention
Worksheets
Tier 2
Strategic Intervention Skills
Intervention Worksheets
Reteach 17.1
Reteach 17.2
Reteach 17.3
Reteach 17.4
40 Scale Factor and Scale
Drawings
41 Similar Figures
43 The Pythagorean Theorem
46 Proportional Relationships
Differentiated
Instruction
18/04/14 9:59 PM
Tier 3
Intensive Intervention
Worksheets available
online
Building Block Skills
36, 38, 46, 48, 50, 63,
80, 82, 86, 90, 95, 100
Challenge
worksheets
Extend the Math
Lesson Activities
in TE
Module 17
880
LESSON
17.1
Name
Triangle
Proportionality
Theorem
Essential Question: When a line parallel to one side of a triangle intersects the other two
sides, how does it divide those sides?
Resource
Locker
Constructing Similar Triangles
Explore
The student is expected to:
In the following activity you will see one way to construct a triangle similar to a given triangle.
G-SRT.B.4
A
Prove theorems about triangles. Also G-CO.C.10, G-CO.D.12, G-SRT.B.5
Mathematical Practices
COMMON
CORE
Date
17.1 Triangle Proportionality
Theorem
Common Core Math Standards
COMMON
CORE
Class
Do your work for Steps A–C in the space provided. Draw a triangle.
Label it ABC as shown.
A
MP.5 Using Tools
Check students' constructions.
Language Objective
B
Work with a partner to describe the triangle proportionality theorem and
its converse.
B
C
_
Select a point on AB. Label it E.
A
E
ENGAGE
The line cuts two sides of the triangle into
segments, and the ratios between the two segments
of each side are equal.
PREVIEW: LESSON
View the Engage section online. Discuss the
photograph and ask students to identify the
geometrical shapes seen in the sail. Then preview the
C
© Houghton Mifflin Harcourt Publishing Company
Essential Question: When a line
parallel to one side of a triangle
intersects the other two sides, how
does it divide those sides?
B
C
Construct an angle with vertex E that is congruent
to ∠B. Label the point where the
_
side of the angle you constructed intersects AC as F.
A
E
B
D
E
F
C
_
‹ ›
−
Why are EF and BC parallel?
‹ ›
‹ ›
‹ ›
−
−
−
Coplanar lines EF and BC are cut by transversal AC so that ∠AEF ≅ ∠B. By the Converse
_
‹ ›
−
of the Corresponding Angles Theorem, EF ∥ BC.
_ _ _
_
Graphics\02192014\From Ramesh\Econ 2016\W
Use a ruler to measure AE, EB\\192.168.9.251\07Macdata\07Vol1Data\From
, AF, and FB. Then compare the
AE
AF
ratios ___
and ___
.
EB
FB
Measurements will vary, but, if the constructions are accurate, the ratios should be
approximately equal.
Module 17
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 1
881
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Class
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.10, G-CO.D
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CORE
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Sim
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a
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Explore
to
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ing activit
ed. Draw
In the follow
space provid
Quest
Essential
IN2_MNLESE389847_U7M17L1 881
Steps A–C
work for
Do your
.
as shown
Label it ABC
A

HARDCOVER PAGES 881890
Watch for the hardcover
student edition page
numbers for this lesson.
in the
nts'
Check stude
ns.
constructio
C
B

Select a point
_
it E.
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A
E
point where
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‹ › and BC parallel?
−
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EF
‹ › are cut by trans _
‹ › and −
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881
Module 17
ESE3898
7L1 881
47_U7M1
IN2_MNL
881
Lesson 17.1
18/04/14
10:03 PM
18/04/14 10:04 PM
Reflect
1.
EXPLORE
Discussion How can you show that △AEF ∼ △ABC? Explain.
You can show that there are three pairs of congruent angles. ∠A ≅ ∠A (Reflexive Property
‹ › _
−
of Equality) and ∠AEF ≅ ∠B (by construction). Also, because EF || BC, ∠AFE ≅ ∠C. c. Then
Constructing Similar Triangles
you can use the AA Similarity Criterion because you can show that there are two pairs of
congruent angles. For instance,
2.
INTEGRATE TECHNOLOGY
AE
AF
What do you know about the ratios ___
and ___
? Explain.
EB
FB
Because △ABC ∼ △AEF and corresponding sides of similar triangles are
Students have the option of doing the Explore activity
either in the book or online.
AE
AF
proportional, __
and ___
? Explain.
AB
AB
3.
segments produced when a line parallel to one side of a triangle intersects the other
two sides.
The parallel line divides the other two sides so the lengths of the segments are
QUESTIONING STRATEGIES
How do you use the AA Similarity Criterion
to show two triangles are similar? Show that
two angles of one triangle are congruent to two
angles of the other triangle. This lets you conclude
that the two triangles are similar.
proportional.
Explain 1
Proving the Triangle Proportionality
Theorem
As you saw in the Explore, when a line parallel to one side of a triangle intersects the other
two sides of the triangle, the lengths of the segments are proportional.
Triangle Proportionality Theorem
Theorem
Hypothesis
A
E
B
EXPLAIN 1
Conclusion
AE = _
AF
_
EB FC
F
C
EF ∥ BC
Example 1

Prove the Triangle Proportionality Theorem
‹ › _
−
Given: EF ∥ BC
A
AE
AF
= ___
Prove: ___
EB
FB
Step 1 Show that △AEF ∼ △ABC.
‹ › _
−
Because EF ∥ BC, you can conclude that ∠1 ≅ ∠2 and
by ∠3 ≅ ∠4 the Corresponding Angles Theorem.
E 1
Proving the Triangle
Proportionality Theorem
© Houghton Mifflin Harcourt Publishing Company
If a line is parallel to a side
of a triangle intersects
the other two sides, then
it divides those sides
proportionally.
3 F
4 C
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Review with students which geometric
statements can be used to provide justification for
each step of a proof.
2
B
QUESTIONING STRATEGIES
So, △AEF ∼ △ABC by the AA Similarity Criterion .
Module 17
882
x+y
Is it always true that the fraction _____
z can be
y
x + __? Yes, provided that z ≠ 0.
written as __
z z
Lesson 1
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M17L1 882
Math Background
The Triangle Proportionality Theorem and its converse may be stated as a single
biconditional: a line that intersects two sides of a triangle is parallel to the third
side of the triangle if and only if it divides the intersected sides proportionally.
18/04/14 10:04 PM
How do you take the reciprocal of both sides
of an equation such as x = y? The reciprocal
of an expression is 1 divided by the expression. One
1
way is to multiply both sides by ___
xy (xy ≠ 0).
Triangle Proportionality Theorem 882
Step 2 Use the fact that corresponding sides of similar triangles are proportional to
AE
AF
prove that ___
= ___
.
EB
FC
EXPLAIN 2
AC
___
AB =
_
AE
Applying the Triangle
Proportionality Theorem
AF+ FC
______
AE + EB
_
=
AE
AF
EB =
1+_
AB
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Some students may misunderstand the
Corresponding sides are proportional.
AF
1+
EB =
_
AE
FC
_
a+b
a
b
__
__
Use the property that ____
c = c + c.
AF
FC
___
Subtract 1 from both sides.
AF
___
Take the reciprocal of both sides.
AF
AE =
_
EB
FC
Reflect
Triangle Proportionality Theorem to mean that a line
parallel to one side of a triangle divides the other two
sides into congruent segments rather than
proportional ones. Show these students that
congruent segments result only in the special case of
the line intersecting the two sides at their midpoints.
4.
Explain how you conclude that ▵AEF ∼▵ABC without using ∠3 and ∠4.
∠A ≅ ∠A by the Reflexive Property of Congruence, and ∠1 ≅ ∠2 since they
are corresponding angles; △AEF ∼ △ABC by the AA Similarity Criterion.
Explain 2
Example 2

Applying the Triangle Proportionality
Theorem
Find the length of each segment.
_
CY
_ _ AX ___
AY by the Triangle Proportionality
It is given that XY ∥ BC so ___
= YC
BY
Theorem.
B
4 X
9
Substitute 9 for AX, 4 for XB, and 10 for AY.
© Houghton Mifflin Harcourt Publishing Company
Then solve for CY.
C
9=_
10
_
4
CY
Take the reciprocal of both sides.
4=_
CY
_
9
10
Next, multiply both sides by 10.
() ( )
CY 10
4 = _
10 _
9
10

NP
___
PL
by the
L
Triangle Proportionality Theorem.
for N Q, 2
Multiply both sides by
3
A
()
_ _
NQ
It is given that PQ ∥ LM, so ___
=
QM
5=_
NP
_
2
3
10
40 = CY, or 4 _
4 = CY
_
9
9
→
Find PN.
Substitute 5
Y
3
5
for QM, and 3 for PL .
: 3
Module 17
()
5 = 3
_
2
NP →
(_
3 )
883
M
Q 2
P
N
15
2
_
or 7_
1
2
= NP
Lesson 1
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M17L1 883
Small Group Activity
Give small groups of students each a sheet with three parallel lines cut by a
transversal. Have students measure the segments of the transversal and verify the
ratio formed. Then have them draw another transversal across the parallel lines
and measure the resulting segments. Have students try to form a conjecture based
on their results.
883
Lesson 17.1
18/04/14 10:04 PM
Find the length of each segment.
_
5. DG
E
32
24
F
6.
QUESTIONING STRATEGIES
_
RN
40
C
8
D
M
G
How can you use the Triangle Proportionality
Theorem to find unknown dimensions? If a
line cutting two sides of a triangle is parallel to the
third, you know that it cuts the two sides
proportionally. Knowing this allows you to set up
and solve a proportion for the missing side length.
5 P
Q
10
N
R
EC
DG
ED 32
24
40 24
10 _
8 RN
5
5
_
_ = MQ
_; _
= _; _ =_; _ = _; 40(_) = DG MR
= ; _ = _; RN = (_)10
DG 24
CF
DG 32
40
32
RN
8
8
4
4
8
Explain 3
5 10
QP RN
1
50
25
RN = _ = _ or 6_
Proving the Converse of the Triangle
Proportionality Theorem
AVOID COMMON ERRORS
Some students may write proportions that do not
compare corresponding parts of the figure. Have
them use different colors to show the
proportional parts.
The converse of the Triangle Proportionality Theorem is also true.
Converse of the Triangle Proportionality Theorem
Theorem
Hypothesis
If a line divides two sides of
a triangle proportionally,
then it is parallel to the
third side.
Example 3

A
Conclusion
AE
AF
=
EB
FC
F
E
B
‹ › _
−
EF || BC
EXPLAIN 3
C
Proving the Converse of the Triangle
Proportionality Theorem
Prove the Converse of the Triangle Proportionality Theorem
AE = _
AF
Given: _
EB
FC
_
‹ ›
−
Prove: EF ∥ BC
A
F
E
AE = _
AF , and taking the reciprocal
It is given that _
EB
FC EB
= FC
AF . Now add 1 to
of both sides shows that AE
___ ___
C
B
AF to the right side.
AE to the left side and _
AE AF
AF
AE
FC
EB
+ AE = AF + AF
This gives AE
.
AB __
__
= AC
___ ___ ___ ___
Since ∠A ≅ ∠A, △AEF ∼ △ABC by the SAS Similarity
Criterion.
‹ › _
−
Step 2 Use corresponding angles of similar triangles to show that EF ∥ BC.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Remind students that the converse of a
© Houghton Mifflin Harcourt Publishing Company
Step 1 Show that △AEF ∼ △ABC.
theorem switches the condition and the conclusion.
Review converses from previous lessons, such as
converses of triangle congruence theorems, and
discuss why they are useful.
∠AEF ≅ ∠ ABC and are corresponding angles, and ∠AEF ≅ ABC .
‹ › _
−
So, EF ∥ BC by the Converse of the Corresponding Angles Theorem.
Module 17
884
QUESTIONING STRATEGIES
What are the main steps of the proof? First,
show that the two triangles are similar by
SAS. Then, use congruent corresponding angles to
show that the lines are parallel.
Lesson 1
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M17L1 884
Modeling
Use transparencies to explain the Triangle Proportionality Theorem. Put the
parallel line on a separate transparency from the diagram of the triangle. Slide the
line up and down relative to the triangle to show that regardless of the line’s
position, the sides will be divided proportionally.
18/04/14 10:04 PM
How is this proof similar to that of the
Triangle Proportionality Theorem? Both
proofs use properties of fractions and the Segment
Addition Postulate to work with proportions. Both
proofs also use the fact that you can take the
reciprocal of both sides of a proportion.
Triangle Proportionality Theorem 884
R
Reflect
EXPLAIN 4
_
Critique Reasoning
A student states that UV must
_
be parallel to ST. Do you agree? Why or why not?
7.
RU
RV
Yes; because RU = US and RV = VS, ___
= ___
= 1.
US
VT
Applying the Converse of the Triangle
Proportionality Theorem
S
T
― ―
So UV ∥ ST by the Converse of the Triangle Proportionality Theorem.
Applying the Converse of the Triangle
Proportionality Theorem
Explain 4
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Have students discuss the special case in
You can use the Converse of the Triangle Proportionality Theorem to verify that a line is
parallel to a side of a triangle.
Example 4
Verify that the line segments are parallel.
_
_
MN and KL
JN
JM
30 = 2
42 = 2
_
_
=_
=_
21
15
MK
NL
JM
JN _ _
Since _ = _, MN || KL by the Converse of the
MK
NL

which the parallel line intersects the two sides at their
midpoints.
42
21 M
K
L
15
J
30
N
Triangle Proportionality Theorem.
QUESTIONING STRATEGIES
_
_
DE and AB (Given that AC = 36 cm, and BC = 27 cm)

A
D
20 cm
AD = AC - DC = 36 - 20 = 16
B
BE = BC - EC = 27 - 15 = 12
15 cm
E
C
20
5
CD = _ = _
_
DA
4
16
15
5
CE = _ = _
_
4
EB
12
CE , DE || AB by the Converse of the Triangle Proportionality Theorem.
CD = _
Since _
DA
EB
© Houghton Mifflin Harcourt Publishing Company
How is using the Converse of the Triangle
Proportionality Theorem different from using
the theorem itself? You use the TPT when you are
given that a line cutting two sides of a triangle is
parallel to the third side, and want to prove that it
cuts the sides proportionally. You use the converse
when you are given that the line cuts two sides
proportionally, and want to prove that it is parallel
to the third side.
V
U
Reflect
8.
Communicate Mathematical Ideas In △ABC, in the example, what is the value
AB
of ___
? Explain how you know.
DE
_ _
5
CD
AB
___
= __
=_
; Possible answer: because DE || AB, corresponding angles A and CDE are
4
DA
DE
CD
CE
AB
congruent, as are corresponding angles B and CDE. So, △ABC ̴ △DEC and __
= __
= __
.
DE
DA
EB
9.
_
_
Verify that TU and RS are parallel.
R
72
TR
U
4
54
108
4
_ _
90
67.5
72
US
_
_
VU
VT
=
, so RS || TU.
TR
US
T
V
90 _
5
VU
67.5
135
5
VT _
_
=
= , _=_=_=_
54
S
Module 17
885
Lesson 1
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M17L1 885
Connect Vocabulary
Some students may have difficulty identifying a transversal. Remind them that the
prefix trans- means across, as in transportation, transfer, and transatlantic.
Transverse means situated or lying across. Therefore, the transversal is the line that
lies across other lines.
885
Lesson 17.1
18/04/14 10:04 PM
Elaborate
_ _
10. In △ABC, XY || BC. Use what you know about similarity
and proportionality to identify as many different proportions
as possible.
AC AX
YC B
AY AB
AY XB
AX
=
;
=
;
=
;
=
AY XB
AY
AB
AC AX
YC AX
_ __ __ __ _
ELABORATE
A
X
Y
C
QUESTIONING STRATEGIES
What is the difference between triangles
having congruent sides and triangles having
proportional sides? Congruent sides have equal
measures, but proportional sides have equal ratios
to one another.
11. Discussion What theorems, properties, or strategies are common to the proof of
the Triangle Proportionality Theorem and the proof of Converse of the Triangle
Proportionality Theorem?
taking reciprocals of both sides of a proportion, and use of Similarity Criteria
(AA for the Triangle Proportionality Theorem and SAS for its converse)
―
12. Essential Question Check-In Suppose a line parallel to side BC of
―
―
AX
▵ABC intersects sides AB and AC at points X and Y, respectively, and ___
= 1. What
XB
do you know about X and Y? Explain. _
_ __
AY
X and Y are the midpoints of sides AB and AC. If AX
= 1, then __
= 1.
XB
YC
_
Then AX = XB, so X is the midpoint of AB. Similarly, Y is the midpoint
_
of AC.
SUMMARIZE THE LESSON
When does a line cut two sides of a triangle
proportionally? What does it mean for the
sides to be cut proportionally? When it is parallel to
the third side; the ratios of the corresponding parts
are equal.
Evaluate: Homework and Practice
1.
• Online Homework
• Hints and Help
• Extra Practice
Check students’ constructions.
_ −
‹ ›
XN
XM
ZY || MN . Write a paragraph proof to show that ___
= ___
.
NY
MZ
Y
X
N
_ −
‹ ›
Since ZY|| MN , ∠XNM ≅ ∠XYZ and ∠X MN ≅ ∠XZY by
the Corresponding Angles Theorem.
Z
M
So △XYZ ~ △XNM by the AA Similarity Criterion. Since
XZ
XY
corresponding sides of similar triangles are proportional, ___
= __
. Use the Segment
XM
XN
Addition Postulate to rewrite XZ as XM + MZ, and rewrite XY as XN + NY. So ______
XM
XM + MZ
© Houghton Mifflin Harcourt Publishing Company
2.
Copy the triangle ABC that you drew for the Explore activity. Construct a line
_
‹ ›
−
FG parallel to AB using the same method you used in the Explore activity.
XN + NY
XM
MZ
XN
NY
= ______
. This can be rewritten as ___
+ ___
= __
+ __
. Since any number divided by itself
XN
XM
XM
XN
XN
MZ
NY
MZ
___
__
is 1, you can rewrite the expression as 1 + XM = 1 + XN . Subtracting 1 from both sides, ___
XM
NY
XM
XN
= __
. Taking the reciprocals of both sides, ___
= __
.
XN
MZ
NY
Module 17
IN2_MNLESE389847_U7M17L1 886
886
Lesson 1
18/04/14 10:04 PM
Triangle Proportionality Theorem 886
Find the length of each segment.
_
3. KL
EVALUATE
H
G
6
4.
X
m
M
n
K
Concepts and Skills
18
Explore
Constructing Similar Triangles
Exercise 1
Example 1
Proving the Triangle Proportionality
Theorem
Exercise 2
Example 2
Applying the Triangle
Proportionality Theorem
Exercises 3–5
Example 3
Proving the Converse of the Triangle
Proportionality Theorem
Exercise 13
Example 4
Applying the Converse of the
Triangle Proportionality Theorem
Exercises 6–9
6
8
_
_
AB and CD
6.
7.
D
14
B
© Houghton Mifflin Harcourt Publishing Company
9.
18
30
_
_
MN and QR
8.
9
10
N
3
R
2.7 M
Q
PM = 9 - 2.7 = 6.3
PN = 10 - 3 = 7
6.3
PM
7
PN
=
= =
2.7
3
N
MQ
_ _
So, MN || QR.
_ _ _ _
CA
3.5
2.5
E
1.5
FW
___
= ___
= 0.6;
2.5
WD
2.1
FX
__
= ___
= 0.6
XE
3.5
Since
16
20
16
5 AM
CN _
12 _
4 CL
4 AL
_
=
= ; _ = _ = _; _ = _ = _; _ = _ = 2
15
5 LA
5 LC
4 MB
16
20
8
NB
_ _
CL
CN _
_
=
and LN||AB.
So,
LA
NB
_
_
AM
AL
So, _ ≠ _ and LM is not parallel to BC.
FX _ _
FW _
_
=
, WX || DE.
XE
WD
M
16
A
8
B
15
20
MB
LC
N
L
16
12
C
10. On the map, 1st Street and 2nd Street are parallel. What is the distance from City Hall
to 2nd Street along Cedar Road?
2.8
2.1
mi
mi
2.4 mi
1st St.
As p
en
Rd.
Library
Cedar Rd.
2nd St.
Let x be the distance in miles from City Hall to 1st Street.
Exercise
IN2_MNLESE389847_U7M17L1 887
( )
2.1
x
___
= ___
2.4
2.1
x = ___
2.4 = 1.8
2.8
2.8
1.8 + 2.4 = 4.2 miles
The distance is 4.2 miles.
Module 17
Lesson 17.1
14
_
_
WX and DE
F
1.5
2.1
W
X
18
Use the Converse of the Triangle Proportionality Theorem to
identify parallel lines in the figure.
City
Hall
887
49
V
T
N 14
D
3
ED _
14
_
= 14 ∙ _ = 3
=
14
2
DB
4_
3
EC _
12
_
=
=3
4
CA
_ _
EC
ED
So, _ = _. So, AB || CD.
DB
U
V
P
2
43
8
35 _
XY
XZ
VN _
VM _
VM
_
XY = 12; _ = _;
=
;
=
;
8
YU
ZV
NT
MU 14
35
XZ
12 _
12
_
=
; XZ = (_)30 = 20 VM = (_)8 = 20
3
A
C 4
12
E
30
U
KL
4
1
_4 = _
; KL = (_)8 = 5_
Practice
Z
Y
30
L
6
_
VM
5.
8
J
4
ASSIGNMENT GUIDE
_
XZ
Lesson 1
887
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–9
2 Skills/Concepts
MP.4 Modeling
10–12
2 Skills/Concepts
MP.4 ge07se_c07l04007a
Modeling
13–15
2 Skills/Concepts
MP.2 Reasoning
16
3 Strategic Thinking
MP.2 Reasoning
17
3 Strategic Thinking
MP.3 Logic
18
3 Strategic Thinking
MP.1 Problem Solving
18/04/14 10:04 PM
11. On the map, 5th Avenue, 6th Avenue, and 7th Avenue are parallel. What is the length
of Main Street between 5th Avenue and 6th Avenue?
5th
Ave.
6th
Ave.
Main
St.
t.
gS
S
0.3
5th Avenue and 6th Ave.
0.4
0.5
___
= ___
m
0.4 km
in
pr
0.3
0.3
m
___
= ___
0.4
0.5
In exercises that give a length with an algebraic
expression, a common error is to give the value of the
variable as the answer. Remind students they need to
use the value of the variable to evaluate the
expressions and find the segment lengths.
( )
0.3
m = ___
0.4 = 0.24
0.5
The length is 0.24 kilometer.
0.5
AVOID COMMON ERRORS
Let m be the length in kilometers of Main Street between
7th
Ave.
km
km
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 For problems with two triangles, encourage
G
12. Multi-Step The storage unit has horizontal siding that is parallel to the base.
a. Find LM.
2.6
GL
GH ___
10.4 ___
2.6
__
= __
; 11.3 = ___
; LM = ___
; LM = 11.3(___
) = 2.825 ft
LM
HJ
LM
2.6 11.3
11.3 ft
10.4
10.4
b. Find GM.
L
GM = GL + LM = 11.3 + 2.825 = 14.125
M
N
H
D 2.6 ft
E
F
J 2.2 ft
K
Find MN to the nearest tenth of a foot.
c.
10.4 ft
students to draw (or color) the two triangles
separately to identify corresponding parts.
GJ _____
2.2
GM
13 _____
2.2
___
= __
; 14.125 = ___
; MN = ___
; MN = 14.125(___
) ≈ 2.4ft
MN
JK
MN
2.2 14.125
13
13
HJ
HJ
LM
___
≈ 1.18 and __
= 1.18.
LM
and __
as
d. Make a Conjecture Write the ratios ___
MN
JK
decimals to the nearest hundredth and compare them.
Make a conjecture about the relationship between parallel
‹ ›
‹ ›
‹ › −
‹ ›
‹ ›
−
−
−
−
lines LD , ME , and NF and transversals GN and GK.
MN
JK
The parallel lines divide the
transversals proportionally.
E
Statements
‹ › −
‹ › −
‹ › −
‹ ›
−
1. AB ∥ CD , CD ∥ AF
‹ ›
‹ ›
−
−
2. Draw EB intersecting CD at X.
F
Reasons
1. Given
2. Two points determine a line.
AC
BX
3. ___
= __
XE
CE
3. Triangle Proportionality Theorem
BD
BX
4. __
= ___
XE
DF
4. Triangle Proportionality Theorem
AC
BD
5. ___
= ___
DF
CE
5. Transitive Property of Equality
Module 17
IN2_MNLESE389847_U7M17L1 888
© Houghton Mifflin Harcourt Publishing Company
13. A corollary to the Converse of the Triangle Proportionality Theorem states that
if three or more parallel lines intersect two transversals, then they divide the
transversals proportionally. Complete the proof of the corollary.
‹ › −
‹ ›−
‹ › −
‹ ›
−
Given: Parallel lines AB ∥ CD, CD ∥ EF
B
A
AC
AC
BX
BX
BD
BD
= ___
, ___ = ___
, ___ = ___
Prove: ___
XE XE
DF CE
DF
CE
X
D
C
888
Lesson 1
18/04/14 10:03 PM
Triangle Proportionality Theorem 888
14. Suppose that LM = 18.75. Use the Triangle Proportionality Theorem
to find PM.
JOURNAL
L
15 ______
3
PM
___
= __
; PM = _
10 24 - PM
2
LP
3(
PM
3
16
16
)
PM = _
24
PM
;
PM
= 36 - _
PM ___
= __
PM = (__
22 = 14.08
2
22
2
25
25 )
3
_
PM = 14 or 14.4
K
Have students write a journal entry in which they
explain in their own words the Triangle
Proportionality Theorem and its converse. Ask
students to include figures with their explanations.
P
10
2
15.
― ―
Which of the given measures allow you to conclude that UV ∥ ST? Select
all that apply.
A. SR = 12, TR = 9
B. SR = 16, TR = 20
C. SR = 35, TR = 28
D. SR = 50, TR = 48
E. SR = 25, TR = 20
20
5 __
16
__
=_
; 20; ≠ __
12
3 12
9
20
5 __
16
4 __
__
=_
; 16 = _
; 20 ≠ __
4 20
5 16
16
20
20
16
20
16
4 __
4 __
__
_
_
__
=
;
=
;
=
7 28
7 35
35
28
20
16
20
16
__
_2; __
_1; __
__
=
=
≠
5 48
50
3 50
48
20
16
4 __
4 __
__
=_
; 16 = _
; 20 = __
5 20
25
5 25
15
N
M
U
20
S
T
V
16
R
20
― ―
Answer: Only C and E allow you to conclude that UV ∥ ST.
H.O.T. Focus on Higher Order Thinking
16. Algebra For what value of x is GF ∥ HJ?
G
30
F
45
― ―
H
EJ
+1
EH
____
GF ∥ HJ if __
= __
, that is, if _____
= 5x
; 360
45
40
JF
HG
4x + 4
J 5x + 1
4x + 4
36x + 36 = 40x; x = 7
―
―.
E Then, for x = 7, GF ∥ HJ
17. Communicate Mathematical Ideas John used △ABC to write a proof
―
―
of the Centroid Theorem. He began by drawing medians AK and CL,
―
―
intersecting at Z. Next he drew midsegments LM and NP, both
―
parallel to median AK.
© Houghton Mifflin Harcourt Publishing Company
+4
5x + 1
= 360( _____
;
(4x_____
45 )
40 )
―
―
―
Given: △ABC with medians AK and CL, and midsegments LM
―
and NP
A
L
B
Prove: Z is located _23 of the distance from each vertex of △ABC to
the midpoint of the opposite site.
M
L
a. Complete each statement to justify the first part of John’s proof.
By the definition of midsegment , MK = _12BK. By the definition
M
KC
=
2.
of median , BK = _12KC. So, by substitution , MK = _12KC, or ___
MC
― ― ZC = ___
KC
by the Triangle Proportionality
Consider △LMC. LM∥AK, so ___
LZ
Z
K
N
P
C
Z
K
P
C
MK
ZC
2LZ
= ___
= _23 , and Z is located _23
Theorem, and ZC = 2LZ. Because LC = 3LZ, ___
LC
3LZ
of the distance from vertex C of △ABC to the midpoint of the opposite side.
b. Explain how John can complete his proof.
2
of the distance
He can repeat the same process twice to show that Z is located _
3
from vertices A and B of △ABC to the midpoints of the opposite side.
Module 17
IN2_MNLESE389847_U7M17L1 889
889
Lesson 17.1
889
Lesson 1
18/04/14 10:03 PM
18. Persevere in Problem Solving Given △ABC with FC = 5, you want to find BF.
First, find the value that y must have for the Triangle Proportionality to apply. Then
describe more than one way to find BF, and find BF.
AVOID COMMON ERRORS
When students attempt to find the area of △AEF,
they may mistake the triangle for a right triangle with
1 bh to find the area.
right angle ∠AEF, then use A = __
2
This gives an incorrect area of 31.6875 ft 2, slightly
greater than the actual area, which can be found by
adding the areas of two different right triangles,
△AGE and △EGF.
For the triangle Proportionality Theorem to apply, y
y
―
9
B (5.5, y)
8
7
F (7, 6)
6
5
4
3
2
C (10, 2)
1 A (1, 2) E (4, 2)
x
0
1 2 3 4 5 6 7 8 9
―
must be such that AB and EF are parallel.
― y-2 y-2
―
-2
4
Slope of EF = 6____
=_
, and slope of AB = _____ = ____.
7-7
y-2
5.5 - 1
4
()
4.5
4
4
Let ____
=_
and solve for y: y - 2 = 4.5 _
= 6; y = 8.
4.5
3
3
To find BF, you can use the Pythagorean Theorem, the
Distance Formula, or the Triangle Proportionality Theorem:
()
3
3
BF
AE __
AE = 3, EC = 6, FC = 5; __
= __
; BF = _
; BF = 5 _
= 2.5
6
6
FC
EC 5
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 A theorem for rectangles similar to the
Shown here is a triangular striped sail, together with some of its dimensions.
Find each of the following:
A
1. HG
2.5 ft
2. BC
B
3. AJ
C
1.5 ft
4. GF
6 ft
7. the number of sails you could make for \$10,000 if the sail
material costs \$30 per square yard
HG
4.
1.5(HG) = 42
1.8
BC
___
= ___
1.5
1.2
1.2(BC) = 2.7
BC = 2.25
2.5
AJ
___
= ___
3.
2.25
2
2
6.
2
(GF) = 6.25
1
∙ base ∙ height
Area = _
2
= 30.9
2
5.
F
6.5 ft
1(
=_
10.3)(6)
2
36 + (GF) = 42.25
HG = 2.8
2.
6 2 + (GF) = (6.5)
E
GF = 2.5
Perimeter = 2.5 + 2.25 +
1.5 + 3.5 + 6.5 +
2.5 + 2.8 + 1.2 +
1.8 + 2 = 26.55
7.
30.9 ft
_____
= 3.43 yd 2
9 ft
\$30
3.43 yd 2 ∙ ___ = \$103 per sail
2
2
yd 2
\$10,
000
total
_________ = 97 sails
\$103 per sail
© Houghton Mifflin Harcourt Publishing Company
3.5
G
3.5 ft
6. the area of △AEF
1.5
1.2
___
= ___
Triangle Proportionality Theorem might go like this:
If a line segment parallel to one side of a rectangle
intersects the two sides that are perpendicular to the
segment, then the segment divides the perpendicular
sides proportionally. Is the theorem true? Give an
¯ is drawn
answer: In rectangle ABCD, segment EF
¯. AF = 3 and
¯ and BC
parallel to AB
FD = 5. Since AB = FE, ABEF is a rectangle, so BE =
3. Since FE = DC, FECD is a rectangle, so EC = 5.
3 and the segment divides the
BE = __
AF = ___
So, ___
FD
EC 5
perpendicular sides proportionally.
1.8 ft
I
1.2 ft
H
D
5. the perimeter of △AEF
1.
J
1.8
4.5 = 2.25(AJ)
2 = AJ
Module 17
890
Lesson 1
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M17L1 890
Refer students to the sail in the Lesson Performance Task. Then give these
directions:
_ _
_
1. Use your knowledge of similar triangles to find the lengths of BJ, CI, and DH.
Round to the nearest hundredth. 1.54 ft; 2.92 ft; 3.85 ft
2.
Find the perimeters of △ABJ, △ACI, △ADH, and △AEG. 6.04 ft; 11.47 ft;
15.1 ft; 23.55 ft
3.
Are the perimeters of the triangles in the same ratio as the side lengths?
Explain. The perimeters are in the same ratio as the side lengths. Students
perimeter of ▵ACI
AC = 1.9 = _________________
.
should give sample results such as ___
AB
perimeter of ▵ABJ
18/04/14 10:03 PM
Scoring Rubric
2 Points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Triangle Proportionality Theorem 890
LESSON
17.2
Name
Subdividing a Segment
in a Given Ratio
Class
Date
17.2 Subdividing a Segment in
a Given Ratio
Essential Question: How do you find the point on a directed line segment that partitions the
given segment in a given ratio?
Common Core Math Standards
Resource
Locker
The student is expected to:
COMMON
CORE
G-GPE.B.6
It takes just one number to specify an exact location on a number line. For this
reason, a number line is sometimes called a one-dimensional coordinate system.
The mile markers on a straight stretch of a highway turn that part of the highway
into a one-dimensional coordinate system.
Mathematical Practices
COMMON
CORE
MP.5 Using Tools
Language Objective
On a straight highway, the exit for Arthur Avenue is at mile marker 14. The exit for
Collingwood Road is at mile marker 44. The state highway administration plans to
put an exit for Briar Street at a point that is _23 of the distance from Arthur Avenue
to Collingwood Road. Follow these steps to determine where the new exit should
be placed.
Work in groups to find ratios of subdivided segments.
ENGAGE
If the segment lies on a number line, subtract the
coordinates to find the distance between the
endpoints. Then multiply the length by the ratio to
find the coordinate of the point that divides the
segment in that ratio. If there are no coordinates,
use a compass and straightedge to divide the given
segment into equal parts, and identify the point
that divides the segment in the given ratio.
A
Mark Arthur Avenue (point A) and Collingwood Road (point C) on the
number line.
A
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
Essential Question: How do you find
the point on a directed line segment
that partitions the given segment in a
given ratio?
Partitioning a Segment in a
One-Dimensional Coordinate System
Explore
Find the point on a directed line segment between two given points that
partitions the segment in a given ratio. Also G-CO.D.12
10
B
D
B
20
C
30
40
50
What is the distance from Arthur Avenue to Collingwood Road? Explain.
30 miles; find the absolute value of the difference of the coordinates: ⎜44 - 14⎟ = 30.
C
How far will the Briar Street exit be from Arthur Avenue? Explain.
2
20 miles; · 30 miles = 20 miles
3
D
What is the mile marker number for the Briar Street exit? Why?
_
mile marker 34; 14 + 20 = 34
E
Plot and label the Briar Street exit (point B) on the number line.
Check students’ number lines.
PREVIEW: LESSON
Module 17
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 2
891
gh “File info”
Date
Class
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ividing a
17.2 Subd en Ratio
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Find the point
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segment in
Resource
Locker
Quest
Essential
COMMON
CORE
HARDCOVER PAGES 891902
a
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System
ing a Seg
Coordinate
Partition
ensional
line. For this .
One-Dim
a number
Explore
Watch for the hardcover
student edition page
numbers for this lesson.
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an exact locatio mensional coordi
ay
the highw
r to specify
called a one-diay turn that part of
one numbe
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reason, a
a
.
The exit for
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The mile
marker 14.
plans to
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e is at mile
istration
into a one-di
Arthur Avenu
r Avenue
the exit for 44. The state highw
ce from Arthu should
r
t highway,
On a straigh Road is at mile marke that is _23 of the distan
new exit
the
where
Collingwood Briar Street at a pointsteps to determine
for
put an exit
to Collingwo
C) on the
be placed.

r Avenue
Mark Arthu
number line.
A
(point A)
and Collin
(point
C
B
D
50
40
30
Credits:
20
Explain.
14⎟ = 30.
s: ⎜44 e to Collin
coordinate
Arthur Avenu
ence of the
distance from
the differ
What is the
value of
absolute
; find the
Explain.
30 miles
r Avenue?
be from Arthu
Street exit
Briar
the
How far will
= 20 miles
2
; · 30 miles
20 miles 3
Why?
Street exit?
for the Briar
r number
mile marke
What is the
+ 20 = 34
er 34; 14
er line.
mile mark
on the numb
(point B)
Street exit
Briar
label the
Plot and
er lines.
nts’ numb
stude
Check
10
y • Image
g Compan
IN2_MNLESE389847_U7M17L2 891

Publishin
View the Engage section online. Discuss the
photograph and point out that the Golden Ratio is
found throughout nature—even, some say, in the
ratio of the length of a person’s forearm to the length
of the person’s hand. Then preview the Lesson
Harcour t
n Mifflin
erstock
36/Shutt

_


Lesson 2
891
Module 17
7L2 891
47_U7M1
ESE3898
IN2_MNL
891
Lesson 17.2
18/04/14
10:08 PM
18/04/14 10:08 PM

The highway administration also plans to put an exit for Dakota Lane at a point that divides
the highway from Arthur Avenue to Collingwood Road in a ratio of 2 to 3. What is the
mile marker number for Dakota Lane? Why? (Hint: Let the distance from Arthur Avenue to
Dakota Lane be 2x and let the distance from Dakota Lane to Collingwood Road be 3x.)
EXPLORE
Partitioning a Segment in a OneDimensional Coordinate System
mile marker 26; since the distance from Arthur Avenue to Collingwood Road is 30 miles,
2x + 3x = 30, 5x = 30, and x = 6. Therefore, the distance from Arthur Avenue to Dakota
Lane is 2x = 12 miles.

INTEGRATE TECHNOLOGY
Plot and label the Dakota Lane exit (point D) on the number line.
Show how to use geometry software to partition a
segment.
Check students’ number lines.
Reflect
1.
How can you tell that the location at which you plotted point B is reasonable?
2
of the way (or about 67% of the way) from point A to
Point B appears to be about _
3
QUESTIONING STRATEGIES
point C.
2.
What does it mean to say that point P
_
divides AB in the ratio 3 to 1? Point P divides
the segment into two segments, and the length of
one is three times the length of the other.
Would your answer in Step F be different if the exit for Dakota Avenue divided the highway from Arthur
Avenue to Collingwood Road in a ratio of 3 to 2? Explain.
Yes; in this case, the exit would be closer to Collingwood Avenue. It would be at mile
marker 32.
Explain 1
Partitioning a Segment in a Two-Dimensional
Coordinate System
EXPLAIN 1
A directed line segment is a segment between two points A and B with a specified direction, from A to B or
from B to A. To partition a directed line segment is to divide it into two segments with a given ratio.
Example 1
A (-8, -7), B(8, 5); 3 to 1
Step 1 Write a ratio that expresses the distance of point P along the segment from A to B.
3 =_
3 of the distance from A to B.
Point P is _
4
3+1
Step 2 Find the run and the rise of the directed
line segment.
8
y
B(8, 5)
4
x
-8
-4
0
-4
A(-8, -7) -8
run = 8 - (-8) = 16
4
Partitioning a Segment in a TwoDimensional Coordinate System
© Houghton Mifflin Harcourt Publishing Company

Find the coordinates of the point P that divides the directed line segment
from A to B in the given ratio.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Have students draw a segment and find the
midpoint to divide it into two segments with a ratio
of 1 to 1. Then have them find the midpoint of one of
those segments to divide the segment with a ratio of 3
to 1. Ask students what other ratios they could divide
the segments into by using midpoints.
Rise
Run (8, -7)
rise = 5 - (-7) = 12
Module 17
892
Lesson 2
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M17L2 892
Learning Progressions
18/04/14 10:08 PM
Students have previously studied slope and learned to find the slope of a line.
They also used slope to write the equation of a line, either in slope-intercept form
(y = mx + b) or in point-slope form (y – y 1 = m (x – x 1)). Here students use the
definition of slope (the ratio of the rise to the run) to help them find a point that
divides a given line segment in a given ratio.
Subdividing a Segment in a Given Ratio
892
3 of the distance from point A to point B, so find _
3 of both the rise and the run.
Step 3 Point P is _
4
4
3 of run = _
3 (16) = 12
3 of rise = _
3 (12) = 9
_
_
4
4
4
4
y
Step 4 To find the coordinates of point P, add the
8
values from Step 3 to the coordinates
B(8, 5)
of point A.
4
P
x
x-coordinate of point P = -8 + 12 = 4
-8 -4 0
4
y-coordinate of point P = -7 + 9 = 2
Rise
-4
QUESTIONING STRATEGIES
How can you predict whether the point to
divide a line in a given ratio will be closer to
one end or the other? If the ratio is less than onehalf, the point will be closer to the first endpoint,
and if it is greater than one-half, the point will be
closer to the second endpoint.
FPO
The coordinates of point P are (4, 2).
Can a proportion compare measurements that
have different units? Explain. Yes; if the
numerators are both in one unit and the
denominators in another, or if one fraction is in one
unit and the other fraction is in another
A(-8, -7) -8
Run (8, -7)
A(-4, 4), B(2, 1); 1 to 2
B
Step 1 Write a ratio that expresses the distance of point P along the segment from A to B.
1
1
Point P is ___________ = _ of the distance from A to B.
1 + 2
3
A
P 4
Step 2 Graph the directed line segment. Find the rise and the run
y
2
B
of the directed line segment.
-4
run = 2 - (-4) = 6
rise = 1 - 4 = -3
-2
0
2
x
4
-2
-4
1
Step 3 Point P is _____ of the distance from point A to point B.
3
( )
© Houghton Mifflin Harcourt Publishing Company
1
1
_ of run = _ (6) = 2
3
3
1
1
_ of run = _ -3 = -1
3
3
Step 4 To find the coordinates of point P, add the values from Step 3 to the coordinates of point A.
x-coordinate of point P = -4 + 2 = -2
(
The coordinates of point P are -2 , 3
)
y-coordinate of point P = 4 + -1 = 3
. Plot point P on the above graph.
Reflect
3.
In Part A, show how you can use the Distance Formula to check that point P partitions the directed line
segment in the correct ratio.
2
2
2
2
AP = (4 - (-8)) + (2 - (-7)) = √225 = 15; PB = √(8 - 4) + (5 - 2) = √25 = 5.
―――――――――
√
――
―――――――
―
The ratio of AP to BP is 15 to 5 or 3 to 1, which is the correct ratio.
4.
Discussion What can you conclude about a point that partitions a segment in the ratio 1 to 1? How can
you find the coordinates of such a point?
The point is the midpoint of the segment. Use the Midpoint Formula.
Module 17
893
Lesson 2
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M17L2 893
Peer-to-Peer Activity
Have students work in pairs to draw and label a line, a line segment, and a
directed line segment. Have them discuss their similarities and differences.
893
Lesson 17.2
18/04/14 10:08 PM
EXPLAIN 2
Find the coordinates of the point P that divides the directed line segment from A to B
in the given ratio.
5.
A(-6, 5), B(2, -3); 5 to 3
5
5
Point P is
= of the distance
5+3
8
from A to B.
_ _
6.
Constructing a Partition of a Segment
_ _
run = 2 - (-6) = 8; rise = -3 - 5 = -8
5
5
of run = 5; of rise = -5
8
8
x-coordinate of point P = -6 + 5 = −1;
run = -6 - 4 = −10; rise = −13 - 2 = −15
QUESTIONING STRATEGIES
5
5
x-coordinate of point P = 4 - 6 = −2;
The coordinates of point P are (−1, 0).
The coordinates of point P are (−2, -7).
Can you partition a line (as opposed to a line
segment)? Explain. No; because a line
continues infinitely in both directions, it cannot be
divided into segments with a definite ratio.
_
_
_3 of run = -6; _3 of rise = -9
y-coordinate of point P = 2 + (-9) = -7
y-coordinate of point P = 5 + (-5) = 0
Explain 2
Example 2

A(4, 2), B(-6, -13); 3 to 2
3
3
= of the distance
Point P is
5
3+2
from A to B.
Constructing a Partition of a Segment
Given the directed line segment from A to B, construct the point P that
divides the segment in the given ratio from A to B.
2 to 1
A
B
→
‾ . The exact measure of the angle
Step 1 Use a straightedge to draw AC
is not important, but the construction is easiest for angles from
C
A
C
F
E
D
A
Step 3 Use the straightedge to connect points B and F. Construct an angle
congruent to ∠AFB with D as its vertex. Construct an angle
congruent to ∠AFB with E as its vertex.
_
Step 4 The construction partitions AB into 3 equal parts. Label point P at
the point that divides the segment in the ratio 2 to 1 from A to B.
A
Module 17
894
B
B
C
F
E
D
P
B
© Houghton Mifflin Harcourt Publishing Company
→
‾ . Label
Step 2 Place the compass point on A and draw an arc through AC
the intersection D. Using the same compass setting, draw an arc
centered on D and label the intersection E. Using the same
compass setting, draw an arc centered on E and label the
intersection F.
Lesson 2
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M17L2 894
Multiple Representations
18/04/14 10:08 PM
¯
Discuss whether it matters if the segment with endpoints C and D is named as DC
_
or as CD. Both are correct. The order does not matter as long as C and D are
endpoints. If the segment is a directed line segment, the order does matter.
Subdividing a Segment in a Given Ratio
894
B
AVOID COMMON ERRORS
1 to 3
C
G
Remind students to pay attention to the order of the
letters in the problem in order to ensure they find the
correct point for the given direction.
F
E
D
A
B
P
_
Step 1 Use a straightedge to draw AC.
_
Step 2 Place the compass point on A and draw an arc through AC. Label the intersection D. Using the
same compass setting, draw an arc centered on D and label the intersection E. Using the same
compass setting, draw an arc centered on E and label the intersection F. Using the same compass
setting, draw an arc centered on F and label the intersection G.
Step 3 Use the straightedge to connect points B and G. Construct angles congruent to ∠AGB with D, E,
and F as the vertices.
_
Step 4 The construction partitions AB into 4 equal parts. Label point P at the point that divides the
segment in the ratio 1 to 3 from A to B.
Reflect
7.
© Houghton Mifflin Harcourt Publishing Company
8.
_
_
In Part A, why is EP is parallel to FB?
∠AEP was constructed to be congruent to ∠AFB. These are congruent corresponding
_ _
angles, so EP∥FB.
How can you use the Triangle Proportionality Theorem to explain why this construction method works?
The construction ensures that the segments of the ray you constructed are congruent.
Also, the segments that are drawn in Step 3 are all parallel. By the Triangle Proportionality
_
Theorem, you can conclude that the segments along AB are congruent to each other.
Given the directed line segment from A to B, construct the point P that divides the
segment in the given ratio from A to B.
9.
1 to 2
10. 3 to 2
C
F
A
E
D
E
F
G
H
C
D
A
P
B
P
B
Module 17
895
Lesson 2
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M17L2 895
Connect Vocabulary
Help students develop an understanding of the meaning of dimension by
considering one-dimensional figures, such as a line, whose only dimension is
length, and two-dimensional figures, such as a square. The two dimensions of a
square are length and width.
895
Lesson 17.2
18/04/14 10:08 PM
Elaborate
ELABORATE
11. How is a one-dimensional coordinate system similar to a two-dimensional coordinate
system? How is it different?
In both types of coordinate systems, numbers are used to specify the locations of points.
QUESTIONING STRATEGIES
In a one-dimensional coordinate system, a single number is used to specify the location of
Can a proportion compare measurements that
have different units? Explain. Yes, but each
ratio must compare measurements with the same
unit; for example, you can write a proportion with a
ratio of feet to feet equal to a ratio of meters to
meters.
a point on a line. In a two-dimensional coordinate system, two numbers are used to specify
the location of a point on a plane.
4
_
12. Is finding a point that
_ is 5 of the distance from point A to point B the same as finding a
point that divides AB in the ratio 4 to 5? Explain.
4
No; the point that is _
of the distance from point A to point B is 80% of the distance from
5
_
4
point A to point B; the point that divides AB in the ratio 4 to 5 is _
or approximately 44% of
9
the distance from point A to point B.
SUMMARIZE THE LESSON
13. Essential Question Check-In What are some different ways to divide a segment in
the ratio 2 to 1?
If the segment lies on a number line, subtract the coordinates to find the distance between
How can you divide a directed line segment in
a given ratio? Directed line segments are
always read in the order in which the points are
given. A segment partition divides the segment
from a given point A to another point B in a ratio of
a : b. In a coordinate plane, the rise and run can be
used to find the point dividing the segment in the
ratio a : b. Segment partitions can also be
constructed using a compass and straightedge.
2
the endpoints. Then find the point that is _
of the distance from one endpoint to the other.
3
If the segment is on a coordinate plane, use the run and the rise to find the point that is
_2 of the distance from one endpoint to the other. If there are no coordinates, use a
3
compass and straightedge to divide the given segment into 3 equal parts. Then identify
the point that divides the segment in the ratio 2 to 1.
A choreographer uses a number line to position dancers for a ballet.
Dancers A and B have coordinates 5 and 23, respectively. In Exercises 1–4,
find the coordinate for each of the following dancers based on the given locations.
1.
Dancer C stands at a point that is _56 of the
distance from Dancer A to Dancer B.
AB = ⎜23 - 5⎟ = 18
2.
3
6
Dancer C is 15 units from Dancer A, so the
coordinate for Dancer C is 5 + 15 = 20.
IN2_MNLESE389847_U7M17L2 896
Dancer D stands at a point that is _13 of the
distance from Dancer A to Dancer B.
AB = ⎜23 - 5⎟ = 18
_1 ⋅ 18 = 6
_5 ⋅ 18 = 15
Module 17
• Online Homework
• Hints and Help
• Extra Practice
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
Evaluate: Homework and Practice
Dancer D is 6 units from Dancer A, so the
coordinate for Dancer D is 5 + 6 = 11.
896
Lesson 2
18/04/14 10:08 PM
Subdividing a Segment in a Given Ratio
896
3.
EVALUATE
Dancer E stands at a point that divides the line segment from Dancer A to Dancer B in a ratio of 2 to 1.
AB = ⎜23 - 5⎟ = 18
Let the distance from Dancer A to Dancer E be 2x and let the distance from Dancer E to
Dancer B be x. Then 2x + x = 18, 3x = 18, and x = 6. So the distance from Dancer A to
Dancer E is 2x = 2(6) = 12.
The coordinate for Dancer E is 5 + 12 = 17.
4.
Dancer F stands at a point that divides the line segment from Dancer A to Dancer B in a ratio of 1 to 5.
AB = ⎜23 - 5⎟ = 18
Let the distance from Dancer A to Dancer F be x and let the distance from Dancer F to
Dancer B be 5x. Then x + 5x = 18, 6x = 18, and x = 3. So the distance from Dancer A to
Dancer F is x = 3.
ASSIGNMENT GUIDE
Concepts and Skills
Practice
Explore
Partitioning a Segment in a OneDimensional Coordinate System
Exercises 1–4
Example 1
Partitioning a Segment in a TwoDimensional Coordinate System
Exercises 5–8
Example 2
Constructing a Partition of a
Segment
Exercises 9–12
The coordinate for Dancer F is 5 + 3 = 8.
Find the coordinates of the point P that divides the directed line
segment from A to B in the given ratio.
5.
A( −3, −2 ), B(12, 3); 3 to 2
3
3
P is ____
=_
of the distance from A to B.
5
3+2
A(−1, 5), B(7, −3); 7 to 1
6.
7
7
P is ____
=_
of the distance from A to B.
7+1
8
run = 12 - (−3) = 15; rise = 3 - (−2) = 5
The coordinates of point P are (6, 1).
The coordinates of point P are (6, −2).
5
© Houghton Mifflin Harcourt Publishing Company
constructions and labels in the portioning of
segments.
_7 of run = _7(8) = 7; _7 of rise = _7(-8) = -7
5
5
5
8
x-coordinate of point P = −3 + 9 = 6;
y-coordinate of point P = −2 + 3 = 1
7.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Discuss the importance of careful
run = 7 - (−1) = 8; rise = −3 - 5 = -8
_3 of run = _3(15) = 9; _3 of rise = _3(5) = 3
A(−1, 4), (B-9, 0); 1 to 3
1
1
P is ____
=_
of the distance from A to B.
4
1+3
8
run = -9 - (−1) = -8; rise = 0 - 4 = -4
_1 of run = _1(-8) = -2; _1 of rise = _1(-4) = -1
4
4
4
4
x-coordinate of point P = −1 + (−2) = −3;
y-coordinate of point P = 4 + (−1) = 3
The coordinates of point P are (−3, 3).
3
3
=_
of the distance from A to B.
P is ____
7
3+4
run = -7 - 7 = −14; rise = 4 - (−3) = 7
_3 of run = _3(-14) = -6; _3 of rise = _3 (7) = 3
7
7
7
7
x-coordinate of point P = 7 + (-6) = 1;
y-coordinate of point P = −3 + 3 = 0
The coordinates of point P are (1, 0).
Given the directed line segment from A to B, construct the point P that divides the
segment in the given ratio from A to B.
9. 3 to 1
10. 2 to 3
C
H
C
G
G
F
F
E
D
A
D
P
B
Module 17
Exercise
IN2_MNLESE389847_U7M17L2 897
Lesson 17.2
8
A(7, −3), B(-7, 4); 3 to 4
8.
E
897
8
x-coordinate of point P = −1 + 7 = 6;
y-coordinate of point P = 5 + (-7) = −2
B
P
A
Lesson 2
897
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–4
2 Skills/Concepts
MP.4 Modeling
5–8
2 Skills/Concepts
MP.2 Reasoning
9–12
2 Skills/Concepts
MP.5 Using Tools
13–16
2 Skills/Concepts
MP.4 Modeling
17–20
2 Skills/Concepts
MP.4 Modeling
21
2 Skills/Concepts
MP.2 Reasoning
22
3 Strategic Thinking
MP.3 Logic
23
3 Strategic Thinking
MP.3 Logic
18/04/14 10:08 PM
Given the directed line segment from A to B, construct the point P
that divides the segment in the given ratio from A to B.
11. 1 to 4
12. 4 to 1
D
A
F
E
G
H
C
AVOID COMMON ERRORS
Some students may confuse various ratios. Remind
them that the similarity ratio refers only to the ratio
of the lengths of the sides. It is equal to the perimeter
ratio and the square root of the area ratio.
A
D
E
P
F
G
H
B
C
P
B
Find the coordinate of the point P that divides each directed line segment in the
given ratio.
J
K
-10
-20
13. from J to M; 1 to 9
JM = ⎜10 - (-15)⎟ = 25
L
M
0
N
10
20
14. from K to L; 1 to 1
KL = ⎜5 - (-6)⎟ = 11
Let JP = x and let PM = 9x.
Let KP = x and let PL = x.
The coordinate of point P is −15 + 2.5 = 12.5
The coordinate of point P is -6 + 5.5 = −0.5
Then x + 9x = 25, 10x = 25, and x = 2.5.
15. from N to K; 3 to 5
Then x + x = 11, 2x = 11, and x = 5.5.
16. from K to J; 7 to 11
KJ = ⎜-15 - (-6)⎟ = 9
NK = ⎜-6 - 18⎟ = 24
Let KP = 7x and let PJ = 11x.
The coordinate of point P is 18 - 9 = 9.
The coordinate of point P is -6 - 3.5 = -9.5.
Then 7x + 11x = 9, 18x = 9, x = 0.5.
So KP = 7(0.5) = 3.5.
Then 3x + 5x = 24, 8x = 24, and x = 3.
So NP = 3(3) = 9.
17. Communicate Mathematical Ideas Leon constructed a point P that divides the directed segment from
A to B in the ratio 2 to 1. Chelsea constructed a point Q that divides the directed segment from B to A in
the ratio 1 to 2. How are points P and Q related? Explain.
2
Points P and Q are the same point. Sample explanation: Point P is _
of the
3
1
of
the
distance
from
B
to
A.
This
means
the
distance from A to B. Point Q is _
3
points lie at the same location along the line segment.
Module 17
Exercise
IN2_MNLESE389847_U7M17L2 898
Lesson 2
898
Depth of Knowledge (D.O.K.)
© Houghton Mifflin Harcourt Publishing Company
Let NP = 3x and let PK = 5x.
COMMON
CORE
Mathematical Practices
24
3 Strategic Thinking
MP.2 Reasoning
25
3 Strategic Thinking
MP.2 Reasoning
18/04/14 10:08 PM
Subdividing a Segment in a Given Ratio
898
18. City planners use a number line to place landmarks along a new street. Each unit of
the number line represents 100 feet. A fountain F is located at coordinate −3 and
a plaza P is located at coordinate 21. The city planners place two benches along the
street at points that divide the segment from F to P in the ratios 1 to 2 and 3 to 1.
What is the distance between the benches?
FP = ⎜21 - (-3)⎟ = 24; Then x + 2x = 24, 3x = 24, and x = 8.
Let the distance from F to the first bench be x and let the distance from the first bench to
P be 2x.
The coordinate for the first bench is −3 + 8 = 5.
Let the distance from F to the second bench be 3x and let the distance from the second
bench to P be x.
Then 3x + x = 24, 4x = 24, x = 6. So the distance from F to the second bench is
3x = 3(6) = 18 units.
The coordinate for the second bench is −3 + 18 = 15.
The distance between the benches is 15 - 5 = 10 units or 1000 feet.
19. The course for a marathon includes a straight segment from
city hall to the main library. The planning committee wants
to put water stations along this part of the course so that
the stations divide the segment into three equal parts. Find
the coordinates of the points at the which the water stations
should be placed.
4
2
y
Main Library
M
x
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
-4
0
-2
2
4
-2
C
City Hall
-4
From C to M, run = 3 - (−3) = 6; rise = 3 - (−2) = 5
1
Let the water stations be at points P and Q. Point P is _
of the distance from C to M.
3
_1 of run = _1(6) = 2; _1 of rise = _1(5) = 1_2
3
3
3
3
3
2
1
x-coordinate of P = −3 + 2 = −1; y-coordinate of P = -2 + 1_
= -_
3
3
(
)
1
The coordinates of point P are -1, -_
.
3
2
Point Q is _
of the distance from C to M.
3
_2 of run = _2(6) = 4; _2 of rise = _2(5) = 3_1
3
3
3
3
3
1
1
x-coordinate of Q = −3 + 4 = 1; y-coordinate of Q = -2 + 3_
= 1_
3
3
(
)
1
The coordinates of point Q are 1, 1_
.
3
1
1
The water stations should be placed at -1, -_
and 1, 1_
.
3
3
Module 17
IN2_MNLESE389847_U7M17L2 899
899
Lesson 17.2
(
)
899
(
)
Lesson 2
18/04/14 10:08 PM
20. Multi-Step Carlos is driving on a straight section of highway from Ashford to
Lincoln. Ashford is at mile marker 433 and Lincoln is at mile marker 553. A rest stop
is located along the highway _23 of the distance from Ashford to Lincoln. Assuming
Carlos drives at a constant rate of 60 miles per hour, how long will it take him to drive
from Ashford to the rest stop?
The distance from Ashford to Lincoln is ⎜553 - 433⎟ = 120 miles.
AVOID COMMON ERRORS
Students may forget to use the correct direction to
partition line segments. Remind students to doublecheck direction in order to choose the correct
partition point.
_2 ∙ 120 = 80, so Carlos must drive 80 miles from Ashford to the rest stop.
3
1
d = rt, where d is the distance, r is the rate, and t is the time, so 80 = 60t, and t = 1_
.
3
1
So it will take 1_
hours (1 hour and 20 minutes) to drive from Ashford to the rest stop.
3
21. The directed segment from J to K is shown in the figure.
y
4
J
Points divide the segment from J to K in the each of the following
ratios. Which points have integer coordinates? Select all that apply
2
x
A. 1 to 1
-4
0
-2
2
4
B. 2 to 1
-4
C. 2 to 3
K
D. 1 to 3
E. 1 to 2
From J to K, run = 0 - (−3) = 3; rise = −3 - 3 = -6
1
1
1
1( )
1 _
A. The point is ____
=_
of the distance from J to K. _
of run = _
3 = 1_
;1
1+1
2
2
2
2 2
1(
1
1
of rise = _
-6) = -3. x-coordinate of the point is -3 + 1_
= -1_
; y-coordinate of the
2
2
2
point is 3 + (−3) = 0. The point does not have integer coordinates.
2
2
2
2( )
2
B. The point is ____
=_
of the distance from J to K. _
of run = _
3 = 2; _
2+1
3
3
3
3
2
2
2
2( )
1 _
C. The point is ____
=_
of the distance from J to K. _
of run = _
3 = 1_
;2
5
5
5
5 5
2+3
2(
2
1
4
of rise = _
-6) = -2_
. x-coordinate of the point is -3 + 1_
= -1_
; y-coordinate of the
5
5
5
5
(
)
3
2
point is 3 + -2_
=_
. The point does not have integer coordinates.
5
5
1
1
1
1( ) _
1
____
_
D. The point is 1 + 3 = 4 of the distance from J to K. _
of run = _
3 = 34 ; _
4
4
4
3
1(
1
1
of rise = _
-6) = -1_
. x-coordinate of the point is -3 + _
= -2_
; y-coordinate of the
4
4
4
2
1
_3. The point does not have integer coordinates.
point is 3 + -1_
=
1
4
4
(
)
1
1
1
1
1( )
E. The point is ____
=_
of the distance from J to K. _
of run = _
3 = 1; _
1+2
3
3
3
3
1(
of rise = _
-6) = -2. x-coordinate of the point is -3 + 1 = -2; y-coordinate of the
3
© Houghton Mifflin Harcourt Publishing Company
2(
of rise = _
-6) = -4. x-coordinate of the point is -3 + 2 = -1; y-coordinate of the
3
point is 3 + (-4) = −1. The point has integer coordinates.
point is 3 + (−2) = 1. The point has integer coordinates.
Module 17
IN2_MNLESE389847_U7M17L2 900
900
Lesson 2
18/04/14 10:08 PM
Subdividing a Segment in a Given Ratio
900
H.O.T. Focus on Higher Order Thinking
JOURNAL
22. Critique Reasoning Jeffrey was given a directed line segment and was asked to
use a compass and straightedge to construct the point that divides the segment in
the ratio 4 to 2. He said he would have to draw a ray and then construct 6 congruent
segments along the ray. Tamara said it is not necessary to construct 6 congruent
segments along the ray. Do you agree? If so, explain Tamara’s shortcut. If not, explain
why not.
Yes; the ratio 4 to 2 is equivalent to the ratio 2 to 1. To construct a point that divides a
segment in the ratio 2 to 1, it is only necessary to construct 3 congruent segments along
the ray.
Have students summarize the process for finding the
point on a directed line segment that partitions the
segment in a given ratio.
23. Explain the Error Point A has coordinate -9 and point B has coordinate 9. A
student was asked to find the coordinate of the point P that is _23 of the distance from
A to B. The student said the coordinate of point P is −3.
a. Without doing any calculations, how can you tell that the student made an error?
Point P must be closer to point B than to point A, so the coordinate of point P should be positive.
b. What error do you think the student made?
2
of the distance from B to A.
Sample answer: The student found the coordinate of the point that is _
3
24. Analyze Relationships Point P divides the directed segment from A to B in the
ratio 3 to 2. The coordinates of point A are (-4, -2) and the coordinates of point P
are (2, 1). Find the coordinates of point B.
3
3
=_
of the distance from A to B.
Point P is ____
5
3+2
3
The run from A to P is 2 - (-4) = 6. Let the run from A to B be x. Then 6 = _
x and x = 10.
5
3
The rise from A to P is 1 - (-2) = 3. Let the rise from A to B be y. Then 3 = _
y and y = 5.
5
© Houghton Mifflin Harcourt Publishing Company
x-coordinate of point B = -4 + 10 = 6; y-coordinate of point B = -2 + 5 = 3
The coordinates of point B are (6, 3)
_
_
25. Critical Thinking RS passes through R(−3, 1) and S(4, 3). Find a point P on RS
such that the ratio of RP to SP is 5 to 4. Is there more than one possibility? Explain.
_
5
5
If point P is on RS, then point P is ____
=_
of the distance from R to S.
5+4
9
run = 4 - (-3) = 7; rise = 3 - 1 = 2
_5 of run = _5(7) = 3_8; _5 of rise = _5(2) = 1_1
9
9
9 9
9
9
8
_8; y-coordinate of point P = 1 + 1_1 = 2_1
x-coordinate of point P = -3 + 3_
=
9
9
9
9
8 _
In this case, the coordinates of point P are (_
, 2 1).
9
9
¯, that lies beyond point S. Let P have coordinates (x, y).
There is also a point P, not on RS
¯
x - (-3)
of RP
_______
_5 ______ _5
)
(
Then rise
¯ = 4 so x - 4 = 4 , 4(x + 3) = 5 x - 4 , 4x + 12 = 5x - 20, 12 = x - 20,
rise of SP
and x =32.
¯
y-1
of RP
_______
_5 ____ _5
Also, rise
¯ = 4 , so y - 3 = 4 , 4(y - 1) = 5(y - 3), 4y - 4 = 5y - 15, -4 = y - 15,
rise of SP
and y = 11.
In this case, the coordinates of point P are (32, 11).
Module 17
IN2_MNLESE389847_U7M17L2 901
901
Lesson 17.2
901
Lesson 2
18/04/14 10:08 PM
In this lesson you will subdivide line segments in given ratios. The
diagram shows a line segment divided into two parts in such a way
that the longer part divided by the shorter part equals the entire
length divided by the longer part:
a+b
a=_
_
a
b
Each of these ratios is called the Golden Ratio. To find the point on
a line segment that divides the segment this way, study this figure:
a
L
S
M
R
Q
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 The opening of the Lesson Performance Task
N
states that for the divided line segment a + b, “the
longer part divided by the shorter part equals the
entire length divided by the longer part.”
a+b
a = _____
So, __
a . Use that same relationship to
b
complete this equation
_ LM
LN
= ?. __
relating to line segment LN∶ ____
LM
MN
P
b
a+ b
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 The Lesson Performance Task shows a line
LN
In the figure, LMQS is a square. ___
equals the Golden Ratio (the entire segment length divided
LM
by the longer part).
_
1. Describe how, starting with line segment LM, you can find the location of point N.
LN
LN
2. Letting LM equal 1, find ___
= ___
= LN, the Golden Ratio. Describe your
1
LM
method.
_
_
1. Using LM as one side, construct square
_ LMQS. Construct R, the midpoint of SQ. Place
the compass point on R _
and use RM as a radius to construct an arc intersecting line
SQ. Mark point P where RM intersects SQ. Construct a perpendicular through P
intersecting line RM. Mark point N where the perpendicular intersects LM.
Module 17
902
© Houghton Mifflin Harcourt Publishing Company
2. RM equals the hypotenuse of a right triangle with sides 1 and 0.5. By
the Pythagorean Theorem, RM ≈ 1.118. So, RP ≈ 1.118. SR = 0.5, so
SP ≈ 1.118 + 0.5 ≈ 1.618. LN = SP, so LN, the Golden Ratio, equals
approximately 1.618.
segment divided into two parts a and b such
a+b
a+1
a = _____
a _____
__
that __
a . If b = 1, then 1 = a . Solve the
b
equation for a. Compare your results with your
results in the Lesson Performance Task and make a
―
1 + √5
value of a. a = ________ ≈ 1.618; a appears to
2
equal the Golden Ratio.
Lesson 2
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M17L2 902
The first two numbers in the Fibonacci sequence of natural numbers
are 1 and 1. To find each new term in the sequence, add the two previous terms.
So, the first few terms are 1, 1, 2, 3, 5, and 8. Have students use calculators to
write the ratios of each of the first ten numbers in the sequence to the previous
number, rounding to the nearest ten-thousandth. Here are the first four
3 = 1.5; __
5 ≈ 1.333. After students complete their
2 = 2; __
1 = 1; __
ratios: __
1
1
2
3
calculations, have them compare their results with their results in the Lesson
calculated. The ratios get closer and closer to the Golden Ratio the farther the
sequence of ratios continues.
18/04/14 10:08 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points:The student’s answer contains no attributes of an appropriate response.
Subdividing a Segment in a Given Ratio
902
LESSON
17.3
Name
Using Proportional
Relationships
Class
Date
17.3 Using Proportional
Relationships
Essential Question: How can you use similar triangles to solve problems?
Common Core Math Standards
The student is expected to:
COMMON
CORE
Explore
G-SRT.B.5
Exploring Indirect Measurement
In this Explore, you will consider how to find heights, lengths, or distances that are too great to
be measured directly, that is, with measuring tools like rulers. Indirect measurement involves
using the properties of similar triangles to measure such heights or distances.
Use congruence and similarity criteria for triangles to solve problems and
to prove relationships in geometric figures.
Mathematical Practices
COMMON
CORE
Resource
Locker
A
MP.5 Using Tools
During the day sunlight creates shadows, as shown in the figure below.
The dashed segment represents the ray of sunlight. What kind of
triangle is formed by the flagpole, its shadow, and the ray of sunlight?
Language Objective
Explain the difference between direct and indirect measurement to a
partner.
A right triangle
ENGAGE
You can use similar triangles to measure things
indirectly, using the fact that similar triangles have
side lengths that are proportional.
PREVIEW: LESSON
View the Engage section online. Discuss the
photograph and ask if students can cite evidence that
suggests the Earth is round. Then preview the Lesson
B
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
Essential Question: How can you use
similar triangles to solve problems?
Suppose the sun is shining, and you are standing near a flagpole, but out of its shadow. You
will cast a shadow as well. You can assume that the rays of the sun are parallel. What do you
The triangles are similar. The rays of the sun are parallel and the line representing the
ground is a transversal. So the acute angles formed by the bases and hypotenuses of the
right triangles are congruent, as are the right angles. So the triangles are similar by the AA
Similarity Criterion.
C
In the diagram, what heights or lengths do you already know?
You probably know your own height.
D
What heights or lengths can be measured directly?
Module 17
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 3
903
gh "File info"
Date
Class
al
Proportion
17.3 Usingionships
Relat
Name
IN2_MNLESE389847_U7M17L3.indd 903
problems?
to solve
and to prove
r triangles
problems
you use simila
es to solve
How can
for triangl
Question:
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Essential
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ment
COMMON
G-SRT.B.5 Use geometric figures.
CORE
in
Measure
to
relationships
too great
Indirect
ces that are involves
Exploring
s, or distan
rement
heights, length Indirect measu
Explore
find
to
how
rulers.
consider
tools like
distances.
e, you will
heights or
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In this Explordirectly, that is, with les to measure such
red
triang
be measu
below.
of similar
properties
in the figure
using the
of
ws, as shown
Resource
Locker
HARDCOVER PAGES 903912
Watch for the hardcover
student edition page
numbers for this lesson.
What kind
ht creates
ht?
sunlight.
day sunlig
the ray of
ray of sunlig
During the
represents
w, and the
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The dashe
the flagpo
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What height
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

903
Module 17
ESE3898
7L3.indd
47_U7M1
IN2_MNL
903
Lesson 17.3
903
18/04/14
10:12 PM
18/04/14 10:13 PM
Reflect
1.
EXPLORE
How could you use similar triangles to measure the height of the flagpole indirectly?
Exploring Indirect Measurement
and solve a proportion in which the unknown is the height of the flagpole. Use the fact
that corresponding sides of similar triangles are proportional.
Explain 1
Example 1

INTEGRATE TECHNOLOGY
Finding an Unknown Height
After you have the known measurements, use
geometry software to calculate the missing
measurements.
Find the indicated dimension using the measurements shown in the figure
and the properties of similar triangles.
A
In order to find the height of a palm tree, you measure
the tree’s shadow and, at the same time of day, you
measure the shadow cast by a meter stick that you hold at
X
a right angle to the ground. Find the height h of the tree.
_ _
Because ZX ǁ CA, ∠Z ≅ ∠C. All right angles are
congruent, so ∠Y ≅ ∠B. So △XYZ ≅ △ABC.
Set up proportion.
Substitute.
Multiply each by 7.2.
Simplify.
QUESTIONING STRATEGIES
1m
Z
1.6 m
Y
C
What assumptions are made, in this kind of
indirect measurement, about the rays of the
Sun and the time of day when the shadows are
measured? The Sun’s rays are parallel and the
same time of day.
B
7.2 m
BC
AB = _
_
XY
YZ
h =_
1
_
7.2
1.6
1
h = 7.2 _
1.6
h = 4.5
( )
The tree is 4.5 meters high.

EXPLAIN 1
h
72 in.
The triangles are similar by the AA Similarity Criterion.
Set up proportion.
48 in.
128 in.
flagpole’s height __
__
=
person’s height
128
h =_
_
72
48
Substitute.
( )
Multiply both sides by 6.
128
h = 72 _
128
Simplify.
x = 192
Finding an Unknown Height
© Houghton Mifflin Harcourt Publishing Company
Sid is 72 inches tall. To measure a flagpole, Sid stands near the flag. Sid’s
friend Miranda measures the lengths of Sid’s shadow and the flagpole’s
shadow. Find the height h of the flagpole.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Have students compare the process of using
similar triangles to the process of using congruent
triangles.
The flagpole is 192 inches tall.
Module 17
904
Lesson 3
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M17L3.indd 904
Learning Progressions
18/04/14 10:13 PM
Students have already used congruence to solve real-world problems. For example,
to find the distance across a pond, students showed that two triangles were
congruent and then used CPCTC to find an unknown side length that
corresponded to the distance across the pond. This lesson presents similar
problems (problems in which unknown lengths must be determined), but now
students will use similarity and the proportionality of corresponding sides to find
the unknown length.
Using Proportional Relationships 904
Reflect
QUESTIONING STRATEGIES
2.
Which line in the figure is the transversal that
is cutting two parallel lines? the ground
In the tree example, how can you check that your answer is reasonable?
The length of the meter stick’s shadow is a little more than 1.5 times the length of the
meter stick. So the length of the tree’s shadow, should be a little more than 1.5 times the
height of the tree. Since 1.5(4.5) = 6.75, an answer of 4.5 is reasonable
EXPLAIN 2
Finding an Unknown Distance
Liam is 6 feet tall. To find the height of a tree, he measures
two shadows are shown. Find the height h of the tree.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Have students name the corresponding parts
The triangles are similar by the AA Similarity Criterion.
tree’s height
=
Liam’s height
28
28
h
h=6
=
6
8
8
168
x = 21
h=
8
The tree is 28 feet tall.
3.
__ __
_ _
(_)
_
before they begin to solve for any lengths.
Explain 2
h
Liam
6 ft
28 ft
8 ft
Finding an Unknown Distance
In real-world situations, you may not be able to measure an object directly because there is a physical barrier
separating you from the object. You can use similar triangles in these situations as well.
QUESTIONING STRATEGIES
Example 2

© Houghton Mifflin Harcourt Publishing Company
How could you predict if an unknown length
is longer or shorter than the corresponding
given length? If the triangle with the unknown
length looks smaller than the triangle with the
given length, you can predict that the unknown
length will be smaller. If the triangle with the
unknown length looks greater than the triangle
with the given length, you can predict that the
unknown length will be greater.
Tree
Explain how to use the information in the figure to find the
indicated distance.
A hiker wants to find the distance d across a canyon. She locates points as described.
1. She identifies a landmark at X. She places a marker (Y) directly across the canyon from X.
2. At Y, she turns 90° away from X and walks 400 feet in a straight line. She places a
marker (Z) at this location.
3. She continues walking another 600 feet, and places a marker (W) at this location.
4. She turns 90° away from the canyon and walks until the
marker Z aligns with X._
She places a marker (V) at this
location and measures WV.
∠VWZ ≅ ∠XYZ (All right angles are congruent) and
∠VZW ≅ ∠XZY (Vertical angles are congruent). So,
△VWZ ≅ △XYZ by the AA Similarity Criterion.
d =_
400 , or _
d =_
XY = _
YZ , So _
2
_
327
327
3
600
VW
WZ
2 = 218.
Then d = 327 _
3
X
W
d
Z
600 ft
400 ft
Y
327 ft
V
()
The distance across the canyon is 218 feet.
Module 17
905
Lesson 3
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M17L3.indd 905
Small Group Activity
Have students work in a small group to estimate the length of something in the
school building that would be difficult to measure. Examples include the height of
a stairwell or a flagpole. Then have the group write an explanation of how to
measure the structure, including diagrams.
905
Lesson 17.3
18/04/14 10:13 PM
B
To find the distance d across the gorge, a student identifies points as
shown in the figure. Find d.
J
AVOID COMMON ERRORS
d
42 m
∠JKL ≅ ∠NML by the AA Similarity Criterion.
K
24 m
Remind students to pay attention to the units in their
final answers, and to make sure that the lengths they
find for real-world problems are reasonable ones.
M
L
JK = _
KL
_
NM
ML
35 m
d =_
24
_
35
24
N
24
4
d = 35 ⋅ _ = 35 ⋅ _
7
42
ELABORATE
140
d=_
7
QUESTIONING STRATEGIES
d = 20
What is the difference between direct and
indirect measurement? If you can use a
measurement tool like a ruler to measure a
dimension, the measurement is direct. If you use the
measurement tool to measure a corresponding
dimension and relate it to the desired dimension
using a proportion, the measurement is indirect.
The distance across the gorge is 20 meters .
Reflect
4.
In the example, why is ∠EGF ≅ ∠IGH?
∠EGF ≅ ∠IGH are vertical angles and vertical angles are congruent.
5.
To find the distance d across a stream, Levi located points as shown in the
figure. Use the given information to find d.
A
SUMMARIZE THE LESSON
△JKL ≈ △NML by the AA Similarity Criterion.
d
BC _
AB _
12
12
_
=
;
= _; d = 12(_) = 24
DE
EC 12
6
d = 24 meters
6m
12 m C
E
Elaborate
D
Discussion Suppose you want to help a friend prepare for solving indirect measurement problems.
What topics would you suggest that your friend review?
as the Vertical Angles Theorem), writing and solving proportions, and properties of
similar triangles, including the relationships between corresponding angles and between
© Houghton Mifflin Harcourt Publishing Company
B
12 m
6.
How can you use indirect measurement and
similar triangles to solve problems? You can
show that two triangles are similar using the AA or
SAS Similarity Criterion. Then, you can use the fact
that corresponding sides are proportional to find an
unknown side length.
6
corresponding sides.
7.
Essential Question Check-In You are given a figure including triangles that represent a real-world
situation. What is the first step you should take to find an unknown measurement?
You must first be sure that the triangles are similar.
Module 17
906
Lesson 3
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M17L3.indd 906
Modeling
18/04/14 10:13 PM
If possible, encourage students to go outdoors to construct and solve indirect
measurement problems with shadows. Find a distance that may be difficult to
measure directly with a measuring tape and use similar triangles to find the
distance. Direct students to draw a diagram of the problem, make measurements,
write distance measurements on the diagram, and use these measurements and
the appropriate steps to find the distance.
Using Proportional Relationships 906
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
Finding distances using similar triangles is called Indirect measurement .
1.
Use similar triangles △ABC and △XYZ to find the missing height h.
2. B
3.
h
h
ASSIGNMENT GUIDE
A
Concepts and Skills
Practice
Explore
Exploring Indirect Measurement
Exercise 1
Example 1
Finding an Unknown Height
Exercises 2–5
Example 2
Finding an Unknown Distance
Exercises 6–9
C
60 ft
6 ft
60
60
AC _
AB _
h
_
=
; = _; h = 6(_) = 24 ft
XZ 6
XY
15
Y
X 15 ft
Z
15
4.
A
C
156 ft
XZ 156
16.5
5.
X
14 ft
Z
4 ft
A
16.5
14
C
208 ft
X
15.2 ft
Y
3.8 ft
Z
XY
XZ 208
h = 52 feet
14
15.2
15.2
Use similar triangles △EFG and △IHG to find the missing distance d.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Discuss how you could use a yardstick or a
6.
7.
E
8.
E
E
d
© Houghton Mifflin Harcourt Publishing Company
d
60 m
F
d
H
48 m G
F
78 m G
80 m
I
d
48
FG _
EF _
_
=
;
= _;
60
IH
HG 80
48
d = 80(_) = 64
60
d = 34 meters
9.
d
E
F
45 m
H
18 m
H
180 m
140.4 m
I
I
78
FG h
EF
_
= _; _ = _;
45
IH
HG 180
78
EF = 180(_) = 312
45
d = 312 meters
388.8 m
211.2 m
F
27 m G
d
EF _
27
FG _
_
;
=
= _;
18
IH
HG 140.4
27
EF = 140.4(_) = 210.6
18
d = 210.6 meters
H
64.8 m
I
G
d
64.8
64.8
FG _
EF = _
_
= _; d = 211.2(_) = 35.2 ; d = 35.2 meters
;
IH
HG 211.2
388.8
388.8
Module 17
Exercise
IN2_MNLESE389847_U7M17L3.indd 907
907
Lesson 17.3
Z
3.8
3.8
AC _
AB _
h
_
=
;
= _; h = 208(_) = 52
AC _
AB _
h
4
4
_
=
;
= _; h = 108.5(_) = 31
XY
XZ 108.5
h = 31 feet
16.5 ft
h
C
108.5 ft
Y
X
B
h
A
5.5 ft
5.5
5.5
AC _
AB _
h
_
=
;
= _; m = (_) 156 = 52 ft
XY
B
Y
ruler to take an indirect measurement of objects in
the classroom that are difficult to measure directly,
such as a chalkboard or a door.
B
907
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–5
1 Recall of Information
MP.4 Modeling
6–9
2 Skills/Concepts
MP.2 Reasoning
10–16
2 Skills/Concepts
MP.4 Modeling
17–19
2 Skills/Concepts
MP.2 Reasoning
20
3 Strategic Thinking
MP.3 Logic
21
3 Strategic Thinking
MP.2 Reasoning
Lesson 3
18/04/14 10:13 PM
10. To find the height h of a dinosaur in a museum, Amir placed
a mirror on the ground 40 feet from its base. Then he stepped
back 4 feet so that he could see the top of the dinosaur in the
mirror. Amir’s eyes were approximately 5 feet 6 inches above
the ground. What is the height of the dinosaur?
The two triangles are similar by the AA Similarity
Criterion. All the dimensions have to be in the same units,
so write 5 feet 6 inches as 5.5 feet.
40
40
h
=
; h = 5.5
= 55; The dinosaur is 55 feet tall.
5.5
4
4
5 ft 6 in.
4 ft
(_)
_ _
11. Jenny is 5 feet 2 inches tall. To find the height h
of a light pole, she measured her shadow and the
pole’s shadow. What is the height of the pole?
40 ft
ge07sec07l05005aa
AB
The two triangles are similar by the AA
5 ft 2 in.
Similarity Criterion. All the dimensions
15.5 ft
2
1
have to be in the same units, so write 5 feet 2 as 5___
= 5__
feet. Write 7 feet 9 inches as
6
12
3
9
1
___
__
7 = 7 feet. To avoid mixing decimals and fractions, write 15.5 feet as 15__
feet.
12
4
_1
3
7_
2
56 31 31
31
31 __
31 _
62
h
2
___
= __
= __(__ ÷ __
= 20 _
)= __
(31)(__4 ) = __
(4) = __
1
15_
2
4
2
6
4
2
3
31
2 3
3
2
feet, or 20 feet 8 inches tall.
The light pole is 20 __
7 ft 9 in.
3
3
D
12. A student wanted to find the height h of a statue of a pineapple in Nambour,
Australia. She measured the pineapple’s shadow and her own shadow. The student’s
height is 5 feet 4 inches. What is the height of the pineapple?
△ABC ≈ △DEP by the AA Similarity Criterion.
AC = 5 ft 4 in. = 64 in.
BC = 2 ft = 24 in.
A
105 64
24
24
EF h
DF
The height h of the pineapple is 280 inches or 23 feet 4 inches.
B
2
1
2 ft C E
8 ft 9 in.
13. To find the height h of a flagpole, Casey measured her own shadow
and the flagpole’s shadow. Given that Casey’s height is 5 feet 4 inches,
what is the height of the flagpole?
ge07se_ c07105002aa
The triangles are similar by the AA Similarity Criterion.
AB
Casey's height: 5 ft 4 in. = 64 in.; Casey's shadow: 3 ft = 36 in.
flagpole's shadow: 14 ft 3 in. = 171 in.
64
36
x
5 ft 4 in.
3 ft
flagpole's height __
h
171
171
__
=
;
= _; h = 64(_) = 304
Casey's height
F
© Houghton Mifflin Harcourt Publishing Company
EF = 8 ft 9 in. = 105 in.
64 _
105
105
AC _
BC _
24 _
h
_
=
;
=
;
= _; h = 64(_) = 280
14 ft 3 in.
36
The height h of the flagpole is 304 inches or 25 feet 4 inches.
Module 17
IN2_MNLESE389847_U7M17L3.indd 908
908
Lesson 3
18/04/14 10:13 PM
Using Proportional Relationships 908
A city is planning an outdoor concert for an Independence Day
celebration. To hold speakers and lights, a crew of technicians sets
up a scaffold with two platforms by the stage. The first platform
is 8 feet 2 inches off the ground. The second platform is 7 feet
6 inches above the first platform. The shadow of the first platform
stretches 6 feet 3 inches across the ground.
AVOID COMMON ERRORS
Some students may not think they can find the
similarity ratio of two figures without knowing the
lengths of the sides. Remind them that the perimeters
have the same similarity ratio as the corresponding
sides, and the areas have the same ratio as the squares
of the corresponding sides.
E
7 ft 6 in.
C
8 ft 2 in.
A 6 ft 3 in. B
D
14. Explain why △ABC is similar to △ADE. (Hint: rays of light are parallel.)
Communicate Mathematical Ideas Possible answer: Because rays of light are parallel
‹ ›
−
and AD is a transversal, ∠ABC and ∠ADE are corresponding angles, so they are congruent.
∠A is common to both triangles. So △ABC ~ △ADE by the AA Similarity Postulate.
15. Find the length of the shadow of the second platform in feet and inches to the
nearest inch.
First, convert all lengths to inches:
AC = 8 ft 2 in. = 96 in.
CE = 7 ft 6 in. = 90 in.
© Houghton Mifflin Harcourt Publishing Company
AB = 6 ft 3 in. = 75 in.
AE = AC + CE = 98 + 90 = 188
188
AE
=
;
=
= 144
75
98
AB 98
AC
_
The shadow of the second platform is represented by BD.
_ __ _
BD = AD - AB = 144 - 75 = 69
So the shadow of the second platform is 69 inches or 5 feet 9 inches.
16. A technician is 5 feet 8 inches tall. The technician is standing on top of the second
platform. Find the length s of the shadow that is cast by the scaffold and the
technician to the nearest inch.
Height of technician: 5 ft 8 in. = 68 in.
Height of scaffold with technician: 188 + 68 = 256 in.
256
256
s
=
; s = 75
= 196
75
98
98
The shadow cast by the scaffold and the technician is 196 inches or 16 feet 4 inches.
_ _
Module 17
IN2_MNLESE389847_U7M17L3.indd 909
909
Lesson 17.3
(_)
(_)
909
Lesson 3
18/04/14 10:13 PM
17. To find the distance XY across a lake, you locate points as
shown in the figure. Explain how to use this information to
find XY.
300 ft
△XYZ ~ △VUZ by the SAS Similarity Criterion,
XY
XY
XY
800
=
. Then
=
, so XY = 1,000 ft.
so
500
VU
VZ
400
_ _
500 ft
U
V
400 ft
Z
_ _
600 ft
800 ft
Y
X
18. In order to find the height of a cliff, you stand at the bottom of the cliff, walk 60 feet
from the base, and place a mirror on the ground. Then you face the cliff and step back
5 feet so that can see the top of the cliff in the mirror. Assuming your eyes are 6 feet
above the ground, explain how to use this information to find the height of the cliff.
(The angles marked congruent are congruent because of the nature of the reflection of
light in a mirror.)
J
P
M
6 ft
Q 5 ft
60 ft
K
Mirror
△JKM ~ △PQM by the AA Similarity Criterion, so
so JK = 72, and the height of the cliff is 72 feet.
JK
JK
60
MK
_
= _. Then _ = _,
PQ
5
6
MQ
19. To find the height of a tree, Adrian measures the tree’s shadow and then his shadow.
Which proportion could Adrian use to find the height of the tree? Select all that apply.
D
© Houghton Mifflin Harcourt Publishing Company
BC
AC = _
A. _
DF
EF
DF = _
EF
B. _
AC
BC
BC
AB = _
C. _
DF
EF
DF = _
EF
D. _
BC
AC
AC
BC = _
E. _
EF
DF
Tree
A
5.6 ft
B
4.2 ft
C E
42.3 ft
F
―
―
―
―
―
―
―
―
DF and AC are corresponding sides, as are EF and BC.
―
―
AB and DF are not corresponding sides.
―
―
―
―
DF and BC are not corresponding sides, nor are EF and AC.
―
―
―
―
BC and EF are corresponding sides, as are AC and DF.
A. AC and DF are corresponding sides, as are BC and EF.
B.
C.
D.
E.
Module 17
IN2_MNLESE389847_U7M17L3.indd 910
910
Lesson 3
18/04/14 10:13 PM
Using Proportional Relationships 910
JOURNAL
H.O.T. Focus on Higher Order Thinking
Have students make up their own problem in which
an unknown length must be found using similar
triangles. Remind students to include the solutions to
their problems.
20. Critique Reasoning Jesse and Kyle are hiking. Jesse is carrying a walking stick.
They spot a tall tree and use the walking stick as a vertical marker to create similar
triangles and measure the tree indirectly. Later in the day they come upon a rock
formation. They measure the rock formation’s shadow and again want to use similar
triangles to measure its height indirectly. Kyle wants to use the shadow length they
measured earlier for the stick. Jesse says they should measure it again. Who do you
think is right?
Jesse is right. The length of a shadow is dependent not only on the height of an object,
but on the position of the sun in the sky. To create similar triangles, the shadows of the
two objects must be measured at the same time of day.
21. Error Analysis Andy wants to find the distance d across a river. He located points
as shown in the figure, then use similar triangles to find that d = 220.5 feet. How can
you tell without calculating that he must be wrong? Tell what you think he did wrong
and correct his error.
C
A
300 ft
E
200 ft D
147 ft
B
―
―
AB is the shortest side of right △ABE, so corresponding side DC of △DCE must be shorter
―
than DE, that is, DE < 200. the triangles are similar, Andy must have used the wrong
( )
d
200
200
proportion. The correct proportion is ____
= ____
, so d = 147 ____
= 98. The distance across
147
300
300
© Houghton Mifflin Harcourt Publishing Company
the river is 98 ft.
Module 17
IN2_MNLESE389847_U7M17L3.indd 911
911
Lesson 17.3
911
Lesson 3
18/04/14 10:13 PM
AVOID COMMON ERRORS
Around 240 B.C., the Greek astronomer Eratosthenes
was residing in Alexandria, Egypt. He believed that the
Earth was spherical and conceived of an experiment
to measure its circumference. At noon in the town of
Syene, the sun was directly overhead. A stick stuck
vertically in the ground cast no shadow. At the same
moment in Alexandria, 490 miles from Syene, a vertical
stick cast a shadow that veered 7.2° from the vertical.
Percent error compares the error in a measurement
to the correct measurement. So, to find the percent
error in Eratosthenes’ calculations, students should
compare the 401-mile error to the correct
circumference, not to the incorrect circumference.
401
401
, not _____
.
The correct ratio is _____
24,901
24,500
Alexandria
7.2°
490 mi
7.2°
Syene
1. Refer to the diagram. Explain why
Eratosthenes reasoned that the angle at the
center of the Earth that intercepted a
490-mile arc measured 7.2 degrees.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 The ancient Greeks used several different
2. Calculate the circumference of the Earth using Eratosthenes’s
3. Calculate the radius of the Earth using Eratosthenes’s figures.
which, by one modern estimate, equaled 185.4
meters. How far was Alexandria from Syene, in
4. The accepted circumference of the Earth today is 24,901 miles.
Calculate the percent error in Eratosthenes’s calculations.
1. The line from the center of the Earth to Syene and the line from the top of the Alexandria
stick to the north end of the Alexandria shadow are parallel. By the Alternate Interior
Angles Theorem, the angle at the center of the Earth is congruent to the 7.2° angle at
Alexandria.
2. In a complete rotation of 360° at the Earth’s center, each 7.2° angle will intercept a
360
= 50 such 490-mile arcs in the entire circumference. So, by
490-mile arc. There are ____
7.2
Eratosthenes’ figures, the circumference of the Earth measures 50 × 490 = 24, 500 miles.
C = 2πr
24, 500 ≈ 2(3.14)r
24, 500
_
≈r
© Houghton Mifflin Harcourt Publishing Company
3.
2(3.14)
3901.3 ≈ r
C = 2πr
24, 500 ≈ 2(3.14)r
_
24, 500
≈r
2(3.14)
3901.3 ≈ r
401
4. 24,901 - 24,500 = 401 miles; ______
≈ 1.6%
24, 901
Module 17
912
Lesson 3
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M17L3.indd 912
Give these directions to students:
1.
2.
3.
Choose one of the planets of the Solar System other than Earth and conduct
research to find its radius or diameter.
Calculate the circumference of the planet (use 3.14 for π).
Suppose that you placed two sticks in the ground 490 miles apart on your
planet. At the moment the Sun was directly over one of the sticks, what angle
from the vertical would the shadow of the second stick cast? Explain.
Example: Mars: diameter 4200 mi; circumference: 13,888 mi;
490 = _____
x ; x = 12.7°
_______
13,888
360°
18/04/14 10:13 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Using Proportional Relationships 912
LESSON
17.4
Name
Similarity in Right
Triangles
Date
17.4 Similarity in Right Triangles
Essential Question: How does the altitude to the hypotenuse of a right triangle help you
use similar right triangles to solve problems?
Resource
Locker
Common Core Math Standards
Explore
The student is expected to:
COMMON
CORE
Class
G-SRT.B.4
Identifying Similar Triangles
A
Make two copies of the right triangle on a piece of paper and cut them out.
B
Choose one of the triangles. Fold the paper to find the altitude to the hypotenuse.
C
Cut the second triangle along the altitude. Label the triangles as shown.
Prove theorems about triangles. Also G-SRT.B.5
Mathematical Practices
COMMON
CORE
MP.8 Patterns
Language Objective
Explain to a partner how to use the Angle/Angle criterion to show
similarity in triangles.
Essential Question: How does the
altitude to the hypotenuse of a right
triangles to solve problems?
You can use the geometric means theorems to find
the missing measurement using indirect
measurement.
PREVIEW: LESSON
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
2
1
3
View the Engage section online. Discuss the
photograph, asking students if they can describe what
is happening and where. Then preview the Lesson
Module 17
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 4
913
gh "File info"
Date
Class
Name
ilarity
17.4 Sim
angles
in Right Tri
a right triang
enuse of
to the hypot
altitude
problems?
does the
les to solve
ion: How
r right triang
B.5
use simila
Also G-SRT.
triangles.
Resource
Locker
Quest
Essential
Explore
IN2_MNLESE389847_U7M17L4 913

s
theore
G-SRT.B.4 Prove
COMMON
CORE
Identifying
copies
Make two
y
g Compan

Choose one
triangle on
of the triang
Cut the second
a piece
les. Fold the
triangle along
cut them
of paper and
paper to find
HARDCOVER PAGES 913924
out.
Watch for the hardcover
student edition page
numbers for this lesson.
enuse.
e to the hypot
the altitud
e. Label the
the altitud
triangles
Publishin

of the right
ngle
Similar Tria
.
as shown
n Mifflin
Harcour t
2
1
3
Lesson 4
913
Module 17
7L4 913
47_U7M1
ESE3898
IN2_MNL
913
Lesson 17.4
18/04/14
10:18 PM
18/04/14 10:19 PM
D
Place triangle 2 on top of triangle 1. What do you notice about the
angles?
EXPLORE
The corresponding angles are congruent.
E
Identifying Similar Triangles
What is true of triangles 1 and 2? How do you know?
They are congruent by the AA Similarity Criterion.
F
INTEGRATE TECHNOLOGY
Use geometry software to explore drawing a right
angle and an altitude to the hypotenuse.
Repeat Steps 1 and 2 for triangles 1 and 3. Does the same relationship
hold true for triangles 1 and 3?
Yes; the corresponding angles are congruent, and the triangles are similar by the AA
Similarity Criterion.
QUESTIONING STRATEGIES
Reflect
1.
How are the hypotenuses of the triangles 2 and 3 related to triangle 1?
The hypotenuses of the smaller triangles are the legs of the original triangle.
2.
What is the relationship between triangles 2 and 3? Explain.
They are similar because triangle similarity is transitive.
When you draw the altitude to the hypotenuse of a right triangle, what kinds of
figures are produced?
Two triangles that are similar to the original triangle and to each other.
4.
Suppose you draw △ABC such that
_∠B is a right angle and the altitude to the
hypotenuse intersects hypotenuse AC at point P. Match each triangle to a similar
B
A. △ABC
△PAB
B. △PBC
C
△CAB
C. △BAP
A
△BPC
What does it mean when you say that two
triangles are similar? The triangles have the
same shape; that is, corresponding angles are
congruent and corresponding sides are
proportional.
© Houghton Mifflin Harcourt Publishing Company
3.
What kind of angle is formed by the altitude
to the hypotenuse? The altitude to the
hypotenuse forms a right angle with the
hypotenuse.
A. Angles B and P are corresponding right angles. Angle C corresponds to itself.
B. ∠BPC corresponds to ∠APB and ∠BCP corresponds to ∠ABP.
C. Angles B and P are corresponding right angles. Angle A corresponds to itself.
Module 17
914
Lesson 4
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M17L4 914
Math Background
18/04/14 10:19 PM
In mathematics, the word mean describes a relationship between numbers. The
arithmetic mean of two numbers is the average of the numbers. This is different
from the geometric mean, which, for positive numbers, is the positive square root
of their product. The geometric mean of a and b is the positive number x such
―
that x = √ab . In later math courses, students may also study harmonic means.
Similarity in Right Triangles 914
Explain 1
EXPLAIN 1
Finding Geometric Means of Pairs of Numbers
Consider the proportion __ax = __bx where two of the numbers in the proportion are the same. The number x is the
geometric mean of a and b. The geometric mean of two positive numbers is the positive square root of their product.
So the geometric mean of a and b is the positive number x such that x = √ab or x 2 = ab.
―
Finding Geometric Means of
Pairs of Numbers
Example 1

Find the geometric mean x of the numbers.
4 and 25
x
4 =_
_
x
25
x
4 = 25x · _
25x · _
x
25
2
100x = _
25x
_
x
25
100 = x 2
Write proportion.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Point out that only a positive number can
Multiply both sides by the product of the denominators.
Multiply.
Simplify.
_
√ 100
Take the square root of both sides.
represent the length of a segment in a triangle.
10 = x
Simplify.

9 and 20
QUESTIONING STRATEGIES
9
x
_=_
x
20
Write proportion.
In a proportion involving a and b, where is the
geometric mean between the two
numbers? The geometric mean is the denominator
of one fraction and the numerator of the other.
_
= √x 2
9
x
_
20x · _
x = 20x · 20
Multiply both sides by the product of the denominators.
2
180 x _
_
= 20x
x
20
Multiply.
180 = x 2
Simplify.
_
√ 180 = √_x
5 =x
6 √―
2
Take the square root of both sides.
© Houghton Mifflin Harcourt Publishing Company
Simplify.
Reflect
5.
―
How can you show that if positive numbers a and b are such that __ax = __bx , then x = √ab ?
Multiply both sides of the proportion by xb, then take the square root of both sides:
()
()
―
― ―
a
_x
2
2
xb _
x = xb b ; ab = x ; √ab = √x ; √ab = x.
Find the geometric mean of the numbers. If necessary, give the answer in simplest radical form.
6.
6 and 24
√――
6 · 24 = √――
144 = 12
Module 17
7.
5 and 12
―
――
―
√60 = √4 · 15 = 2√15
915
Lesson 4
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M17L4 915
Small Group Activity
Have small groups of students work together to research and present to the class
other proofs of the Pythagorean Theorem.
915
Lesson 17.4
18/04/14 10:19 PM
Explain 2
Proving the Geometric Means Theorems
EXPLAIN 2
In the Explore activity, you discovered a theorem about right triangles and similarity.
The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to
the original triangle.
Proving the Geometric
Means Theorems
That theorem lead to two additional theorems about right triangles. Both of the theorems involve geometric means.
Geometric Means Theorems
Theorem
Example
The length of the altitude to the
hypotenuse of a right triangle is the
geometric mean of the lengths of the
segments of the hypotenuse.
h = xy or
h = √xy
The length of a leg of a right triangle is
the geometric mean of the lengths of
the hypotenuse and the segment of the
a 2 = xc or
a = √xc
b 2 = yc or
b = √yc
2
Diagram
―
x
―
―
Prove the first Geometric Means
Theorem.
_
Given: Right triangle ABC with altitude BD
A
CD = _
BD
Prove: _
BD
B
Example 2
Statements
a
c
h
QUESTIONING STRATEGIES
Reasons
2. The altitude to the hypotenuse of a right
triangle forms two triangles that are similar to
the original triangle and to each other.
3. Corresponding sides of similar triangles are
proportional.
Reflect
8.
Discussion How can you prove the second Geometric Means Theorem?
Use the triangle shown in the Example. Show that △ABC ∼ △ADB because the altitude
to the hypotenuse of a right triangle forms two triangles that are similar to the original
AB
triangle and to each other. Then show that __
= __
because corresponding sides of similar
AB
AC
How could you express this relationship as a
proportion? The ratio of one segment to the
altitude is equal to the ratio of the altitude to the
other segment.
© Houghton Mifflin Harcourt Publishing Company
What is the relationship between the altitude
of a hypotenuse and the segments on the
hypotenuse? It is the geometric mean of the lengths
of the segments.
C
BD
angle and the hypotenuse, as well as the altitude to
the hypotenuse.
D
1. Given
CD _
BD
_
=
y
b
_
1. △ABC with altitude BD
3.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Make sure that students can identify the right
DC
BC
triangles are proportional. Then prove that __
= __
in the same way.
BC
AC
Module 17
916
Lesson 4
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M17L4 916
Modeling
18/04/14 10:19 PM
Have students cut a 15–20–25-unit right triangle out of a sheet of graph paper.
Ask them to draw the altitude to the hypotenuse, forming a 12–16–20-unit right
triangle and a 9–12–15-unit right triangle. Then have students use the side lengths
in these triangles to verify the relationships in this lesson.
Similarity in Right Triangles 916
Using the Geometric Means Theorems
Explain 3
EXPLAIN 3
You can use the Geometric Means Theorems to find unknown segment lengths in a right triangle.
Example 3
Using the Geometric Means Theorems
Find the indicated value.
y
2
10
QUESTIONING STRATEGIES

x
If you know the lengths of the segments that
the altitude divides the hypotenuse into, how
can you find the length of the altitude? You can write
a proportion with the lengths of the segments and
length of the hypotenuse and solve for the
unknown.
x
2 =_
_
x
10
x
2 = 10x · _
10x · _
x
10
2
20x = _
10x
_
x
10
20 = x 2
Write proportion.
Multiply both sides by the product of the denominators.
Multiply.
Simplify.
― ―
2√5― = x
√20 = √x 2
Take the square root of both sides.
Simplify.

AVOID COMMON ERRORS
y
y
10 = _
_
y
12
Write proportion.
Some students may find it puzzling that a leg in
one triangle can become the hypotenuse in a related
triangle. Have them use colored pencils to distinguish
the various triangles, marking the hypotenuse in each
with a double or darker line.
y
10 =
12y _
12y _
y
12
Multiply both sides by the product of the denominators.
12y 2
120y
_=_
y
12
Multiply.
120 = y 2
Simplify.
© Houghton Mifflin Harcourt Publishing Company
z
x
――
―――
√ 120 = √ y
Take the square root of both sides.
―
=y
2√30
Simplify.
2
Reflect
9.
Possible answer: Use the Pythagorean Theorem.
10. Find x.
y
5
x
z
_5 = _x ; 7x(_5 ) = 7x(_x ); 35 = x ; √―
35 = x 16 cm
x
7
x
7
2
7
Module 17
917
Lesson 4
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M17L4 917
Connect Vocabulary
Differentiate the geometric mean from previous uses of mean such as the
arithmetic mean, in which we add a set of numbers and divide by the number of
values. The geometric mean is the nth root of the product of n numbers. That
means, we multiply the set of numbers, then take the nth root, where n is the
number of values that were multiplied.
917
Lesson 17.4
18/04/14 10:19 PM
Explain 4
Proving the Pythagorean Theorem
using Similarity
EXPLAIN 4
You have used the Pythagorean Theorem in earlier courses as well as in this one.
There are many, many proofs of the Pythagorean Theorem. You will prove it now
using similar right triangles.
Proving the Pythagorean Theorem
using Similarity
The Pythagorean Theorem
In a right triangle, the square of the sum of the lengths of the legs is equal to the square of the length of
the hypotenuse.
Complete the proof of the Pythagorean Theorem.
_
Given: Right △ABC with altitude CD
A
Example 4
c
b
Prove: a 2 + b 2 = c 2
Draw the altitude to the hypotenuse.
Label the point of intersection X.
A
.
∠B ≅ ∠B by the Reflexive Property of Congruence .
B
a
C
Part 1
∠BXC ≅ ∠BCA because all right angles are congruent
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 The Pythagorean Theorem was known in
many ancient civilizations, including those of Egypt,
India, and China. The oldest known axiomatic proof
of the theorem, however, dates to Euclid’s Elements
(circa 300 bce). Since then, hundreds of proofs have
been given, including proofs that rely on area, proofs
that use calculus, and the proof that is given in this
lesson, which is based on similar triangles.
x
c
b
B
a
C
So, △BXC ∼ △BCA by the AA Similarity Criterion .
∠AXC ≅ ∠ACB because all right angles are congruent .
∠A ≅ ∠A by the Reflexive Property of Congruence .
So, △AXC ∼ △ACB by the AA Similarity Criterion .
Part 2
Let the lengths of the segments of the hypotenuse
be d and e, as shown in the figure.
Use the fact that corresponding sides of similar triangles are proportional
to write two proportions.
e
a = _.
Proportion 1: △BXC ∼ △BCA,so _
a
c
d x
c
b
C
e
a
B
d
b = _____
Proportion 2: △AXC ∼ △ACB, so _
.
c
b
Module 17
IN2_MNLESE389847_U7M17L4 918
918
© Houghton Mifflin Harcourt Publishing Company
A
Lesson 4
18/04/14 10:19 PM
Similarity in Right Triangles 918
Part 3
QUESTIONING STRATEGIES
Now perform some algebra to complete the proof as follows.
To which triangles does the Pythagorean
Theorem apply? Why is it useful? To right
triangles only; it allows you to relate the lengths of
the sides of right triangles.
Multiply both sides of Proportion1 by ac. Write the resulting equation.
Multiply both sides of Proportion12 by bc. Write the resulting equation.
a2 = ce
b2 = cd
2
2
Adding the two resulting equations give this: a + b = ce + cd
2
2
Factor the right side of the equation: a + b = c (e + d)
Finally, use the fact that e + d = c
by the Segment Addition Postulate to rewrite the equation
as a + b = c .
2
2
2
Reflect
11. Error Analysis A student used the figure in Part 2 of the example, and wrote the following
incorrect proof of the Pythagorean Theorem. Critique the student’s proof. △BXC ∼ △BCA and
△BCA ∼ △CXA, so △BXC ∼ △CXA by transitivity of similarity. Since corresponding
f
sides of similar triangles are proportional, _e = _ and f 2 = ed. Because △BXC ∼ and △CXA
f
d
are right triangles, a2 = e2 + f 2 and b2 = f 2 + d 2.
a2 + b2 = e2 + 2f 2 + d 2
Substitute.
= e2 + 2ed + d 2
Factor.
= (e + d)
= c2
2
The proof is incorrect because the student assumes the result that is to be proved.
The student assumes that the Pythagorean Theorem is true in order to write that
© Houghton Mifflin Harcourt Publishing Company
a 2 = e 2 + f 2 and b 2 = f 2 + d 2.
Module 17
IN2_MNLESE389847_U7M17L4 919
919
Lesson 17.4
919
Lesson 4
18/04/14 10:19 PM
Elaborate
ELABORATE
12. How would you explain to a friend how to find the geometric mean of two numbers?
Possible answers: Write a proportion and multiply both sides by the product of the
denominators, then simplify and take the square root of both sides; multiply the two
QUESTIONING STRATEGIES
numbers and then take the square root.
What is the relationship between the original
triangle and the two smaller triangles formed
by the altitude to the hypotenuse? All three triangles
are similar to one another.
_
13. △XYZ is an
_isosceles right triangle and the right angle is ∠Y. Suppose the altitude to hypotenuse XZ
intersects XZ at point P. Describe the relationships among triangles △XYZ, △YPZ and △XPY.
Possible answer: All three triangles are similar. In addition, △YPZ ≅ △XPY.
How can you identify the hypotenuse in a
right triangle? It is opposite the right angle,
and it is the longest side.
14. Can two different pairs of numbers have the same geometric mean? If so, give an example. If not, explain
why not.
Yes; possible answer: the geometric mean of 4 and 16 is 8, and the geometric mean of
2 and 32 is also 8.
15. Essential Question Check-In How is the altitude to the hypotenuse of a right
triangle related to the segments of the hypotenuse it creates?
The length of the altitude is the geometric mean of the lengths of the segments of the
SUMMARIZE THE LESSON
How are geometric means related to the
Pythagorean Theorem? Geometric means
can be used to prove the Pythagorean Theorem.
hypotenuse.
© Houghton Mifflin Harcourt Publishing Company
Module 17
IN2_MNLESE389847_U7M17L4 920
920
Lesson 4
18/04/14 10:19 PM
Similarity in Right Triangles 920
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
Write a similarity statement comparing the three triangles to each diagram.
1.
2.
P
B
3.
C
X
W
R
ASSIGNMENT GUIDE
E
Q
S
△PQR ∼ △SPR ∼ △SQP
D
Y
△BDE ∼ △DEC ∼ △BEC
Z
△XYZ ∼ △XWY ∼ △YWZ
Concepts and Skills
Practice
Explore
Identifying Similar Right Triangles
Exercises 1–3
Find the geometric mean x of each pair of numbers. If necessary, give the answer in
Example 1
Finding Geometric Means of
Pairs of Numbers
Exercises 4–9
4.
Example 2
Proving the Geometric
Means Theorems
Exercises 13–18
Example 3
Using the Geometric
Means Theorems
Exercises 10–12
Example 4
Proving the Pythagorean Theorem
using Similarity
Exercise 22
5 and 20
x ;x
_5 = _
x
6.
2
x
7.
―
_ _
1.5 and 84
1.5
x ;x
_
=_
84
2
9.
―
2
= 36; x = 6
3.5 and 20
3.5 _
_
= x ;x
20
2
27
2 and _
_
40
3
2
3 √5
x
3
9
2
x = 27 ; x = 20 ; x = 10
40
_
_ _
_
= 126; x = 3 √14
―
= 70; x = √70
_―
_
Find x, y, and z.
10.
11.
25
65
x
y
x
y
z
z
―――
――
30
y = √652 - 252 = √3600 = 60
2
2
2
Exercise
40
30 _
x
_
√―
x = 40 ; x = 1200; x = 20 3
y
70 _
_
√―
y = 30 ; y = 2100; y = 10 21
70 _
z
_
=
7
; z = 2800; z = 20√―
25
25 _
60
_
_y _
y = x ; 60 = x ; 25x = 3600; x = 144
25 + x _
z
z 169 _
_
= x; _
z
z = 144 ; z = 24,366;
―――
z = √24,336 = 156
IN2_MNLESE389847_U7M17L4 921
Lesson 17.4
12
x
Module 17
921
3 and 12
x ;x
_3 = _
= 100; x = 10
8 and 13
x
© Houghton Mifflin Harcourt Publishing Company
with the altitude drawn to the hypotenuse. Have
them identify three similar triangles. Ask them to
write three different proportions involving a
geometric mean.
20
8
x 2
√
x = 13 ; x = 104; x = 2 26
8.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Have students draw and label a right triangle
5.
z
40
2
Lesson 4
921
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–9
2 Skills/Concepts
MP.2 Reasoning
10–12
2 Skills/Concepts
MP.4 Modeling
13–18
2 Skills/Concepts
MP.1 Problem Solving
19–22
2 Skills/Concepts
MP.5 Using Tools
23
3 Strategic Thinking
MP.1 Problem Solving
24
3 Strategic Thinking
MP.3 Logic
18/04/14 10:19 PM
――
12.
Some students may have difficulty visualizing the
corresponding angles in the overlapping triangles.
Suggest that they redraw the diagram to separate the
three triangles.
12.8
9.6
_
= _; 12.8z = 92.16; z = 7.2
y
9.6
AVOID COMMON ERRORS
2
2
y2 = (12.8) + (9.6) = 256; y = √256 = 16
12.8
z
9.6
z
12.8 + 7.2 _
x
_
=
;x
x
x
7.2
= 144; x = 12
2
Use the diagram to complete each equation.
c
a
d
e
b
e
13. _ec = _
d
c _
a
14. _
a=
c + d
c+d
b
15. _ = _
b
d
17. c(c + d) = a
d = _e
16. _
c
e
2
2
18.
e
= ab
© Houghton Mifflin Harcourt Publishing Company
Find the length of the altitude to the hypotenuse under the given
conditions.
B
D
A
19. BC = 5
AC = 4
_ _
AB = 3; AD = 4 ;
5
3
Module 17
IN2_MNLESE389847_U7M17L4 922
C
20. BC = 17
AC = 15
_ _
_
15
;
17
8
120
≈ 7.06
17
922
21. BC = 13
AC = 12
_ _
_
12
;
5
13
60
≈ 4.62
13
Lesson 4
18/04/14 10:19 PM
Similarity in Right Triangles 922
22. Communicate Mathematical Ideas The area of a rectangle with a length of ℓ and
a width of w has the same area as a square. Show that the side length of the square is
the geometric mean of the length and width of the rectangle.
―
Area of the square = s2 = Area of rectangle = ℓw; s2 = ℓw; s = √ℓw
JOURNAL
Ask students to demonstrate, with examples and
reasons, that the geometric mean of two positive
numbers is always smaller than their arithmetic
mean.
H.O.T. Focus on Higher Order Thinking
23. Algebra An 8-inch-long altitude of a right triangle divides the hypotenuse into two
segments. One segment is 4 times as long as the other. What are the lengths of the
segments of the hypotenuse?
x in. = shorter segment; 4x in. = longer segment; The length of the hypotenuse is the
8
x
geometric mean of se the lengths, so =
. Then 4x2 = 64, x2 = 16, and x = 4.
4x
8
The segments of the hypotenuse measure 4 inches and 16 inches.
_ _
24. Error Analysis Cecile and Amelia both found a value for EF in △DEF. Both
students work are shown. Which student’s solution is correct? What mistake did the
other student make?
12
EF
= ___
Cecile: ___
8
EF
E
So EF 2 = 12(8) = 96.
_
8
_
© Houghton Mifflin Harcourt Publishing Company
Then EF = √96 = 4√6 .
_
F
_
Then EF = √32 = 4√2 .
Cecile’s solution is correct. Amelia wrote the wrong proportion.
IN2_MNLESE389847_U7M17L4 923
Lesson 17.4
D
So EF 2 = 8(4) = 32.
Module 17
923
G
4
8
EF
Amelia: ___
= ___
4
EF
923
Lesson 4
18/04/14 10:19 PM
INTEGRATE MATHEMATICAL
PRACTICES
MP.8 Focus on Patterns
In the example at the beginning of the lesson, a \$100 investment grew for one year at the rate
of 50%, to \$150, then fell for one year at the rate of 50%, to \$75. The arithmetic mean of
+50% and -50%, which is 0%, was not a good predictor of the change, for it predicted the
investment would still be worth \$100 after two years, not \$75.
• Find the geometric mean of 1, 1, 2, 4, and 4. 2
1. Find the geometric mean of 1 + 50% and 1 - 50%. (Each 1 represents the fact that
at the beginning of each year, an investment is worth 100% of itself.) Round to the
nearest thousandth.
• Write an expression representing the
geometric mean of the numbers a, b, c, d,
5 ――
5 ―
a · b · c · d · e or abcde
and e. 
2. It is the geometric mean, not the arithmetic mean, that tells you what the interest rate
would have had to have been over an entire investment period to achieve the end
result. You can use your answer to Exercise 1 to check this claim. Find the value of a
\$100 investment after it increased or decreased at the rate you found in Exercise 1 for
AVOID COMMON ERRORS
In Question 4 of the Lesson Performance Task, some
students may calculate the geometric mean and then
assume that it represents the interest rate they are
looking for. Remind students that a geometric mean
of 1 represents 0% change, and they must subtract the
mean they found from 1 to find the average percent
change per year.
3. Copy the right triangle shown here. Write the terms “Year 1 Rate”, “Year 2 Rate”, and
“Average Rate” to show geometrically how the three investment rates relate to each
other.
Average
Rate.
Year 1 Rate
Year 2 Rate
4. The geometric mean of n numbers is the nth root of the product of the numbers. Find
what the interest rate would have had to have been over 4 years to achieve the result
of a \$100 investment that grew 20% in Year 1 and 30% in Year 2, then lost 20% in
Year 3 and 30% in Year 4. Show your work. Round your answer to the nearest tenth
of a percent.
――――――
= √――
.75
―――
√(1 + .5)(1 - .5) = √(1.5)(.5)
© Houghton Mifflin Harcourt Publishing Company
1.
= .866
2. After 1 year: 100 × 0.866 = 86.6
After 2 years: 86.6 × 0.866 = 74.9956, or \$75
3. See diagram.
4 ―――――――
4 ―――
4. √(1.2)(1.3)(0.8)(0.7) = √0.8736
≈ 0.967
Since 1 represents 0% change, 0.967 represents (1 - 0.967) = 0.033,
or a 3.3% average loss per year.
Module 17
924
Lesson 4
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M17L4 924
The Pythagorean Theorem is only one of many contributions the Greek
mathematician and philosopher Pythagoras made to mathematics. He was the
first to identify three means for weighting numbers: the arithmetic mean (or
average), the geometric mean, and the harmonic mean. Have students research
and explain the harmonic mean. Then they should show how to find all three
Pythagorean means for the numbers 4 and 9. arithmetic: 6.5; geometric: 6;
72 ≈ 5.54
harmonic: ___
13
18/04/14 10:19 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Similarity in Right Triangles 924
MODULE
17
MODULE
STUDY GUIDE REVIEW
17
Using Similar Triangles
Study Guide Review
Essential Question: How can you use similar triangles to solve
real-world problems?
ASSESSMENT AND INTERVENTION
KEY EXAMPLE
Key Vocabulary
indirect measurement
(medición indirecta)
geometric mean (media
geométrica)
(Lesson 17.1)
Find the missing length x.
ZU = _
ZV
_
UX
VY
16
x =_
___
10
9
16 (10)
x = ___
9
( )
Assign or customize module reviews.
x ≈ 17.8
KEY EXAMPLE
MODULE
Mathematical Practices: MP.1, MP.2, MP.4, MP.6
G-SRT.B.5, G-MG.A.1
• The best way to work with the GPS
coordinates: Students may want to express the
coordinates as (latitude, longitude) so they can
use the distance formula to find the lengths of
the sides of the triangle (see Module 1
Performance Task). This gives the following
coordinates: Miami: (–80.226529, 25.789106);
San Juan: (–66.1057427, 18.4663188); Hamilton:
(–64.781380, 32.294887)
• How to find the area from the lengths of the
sides: Students can experiment with ways to find
the area or you can make a suggestion.
925
Module 17
© Houghton Mifflin Harcourt Publishing Company
10 U
x
Z
16
Multiply both sides by 10.
Y
Simplify.
9
V
(Lesson 17.2)
Use a straightedge to connect points B and H.
SUPPORTING STUDENT REASONING
• The GPS coordinates of the vertices of the
Bermuda Triangle: Students can research
these. The GPS coordinates for Miami are
(25.789106, –80.226529), for San Juan are
(18.4663188, –66.1057427), and for Hamilton
are (32.294887, –64.781380).
X
Substitute.
Given the directed line segment from A to B, construct
the point P that divides the segment in the ratio 2 to 3
from A to B.
→
‾ .
Use a straightedge to draw AC
→
‾ .
Place a compass on point A and draw an arc through AC
Label the intersection D. Continue this for intersections D
through H.
COMMON
CORE
Students should begin this problem by focusing on
what information they will need. Here are some
issues they might bring up.
Write a proportion.
C
H
G
F
E
D
A
B
P
Construct angles congruent to ∠AHB with points D through G.
_
AB is partitioned into five equal parts. Label point P at the point that divides the
segment in the ratio 2 to 3 from A to B.
KEY EXAMPLE
(Lesson 17.3)
A 5.8-foot-tall man is standing next to a basketball hoop that casts an 11.2-foot shadow. The man’s
Let x be the height of the basketball hoop.
x =_
5.8
_
Write a proportion.
11.2
6.5
5.8 (11.2)
x = ___
Multiply both sides by 11.2.
6.5
x ≈ 10
( )
Module 17
925
Study Guide Review
SCAFFOLDING SUPPORT
IN2_MNLESE389847_U7M17MC 925
• For this region of the globe, assume that each unit of latitude or longitude
is equal to 100 km.
• One way to find the area is to enclose the triangle shown in the figure in a
rectangle then find the area of the rectangle, and subtract the areas in the
rectangle that are outside of the triangle.
• Students can also use the area formula for a triangle with side lengths a, b
a+b+c
and c given by A = √s(s - a)(s - b)(s - c) , where s = _________.
2
――――――――
18/04/14 10:23 PM
EXERCISES
Find the missing lengths. (Lesson 17.1)
4.5
A
1.
_
BG =
SAMPLE SOLUTION
B 3.5
E
8
F
G
5D
15
7
Find the unknown length. (Lesson 17.3)
A 5.9-foot-tall-man stands near a 12-foot
statue. The man places a mirror on the ground
a certain distance from the base of the statue,
and then stands another 7 feet from the mirror
to see the top of the statue in it. How far is the
mirror from the base of the statue?
14.2 feet
4.
C
3.
G
F
E
D
A
Find the lengths. (Lesson 17.4)
9
x
6.
x = 16.2
7.
―――――――
A 45-foot flagpole casts a 22-foot shadow. At
the same time of day, a woman casts a 2.7-foot
shadow. How tall is the woman?
5.5 feet
5.
B
P
Find the GPS coordinates of the three locations
and write them as (x, y) = (latitude, longitude).
Then use the distance formula,
―――
d = (x 2 - x 1) 2 + (y 2 - y 1) 2 , to find the lengths of
the three sides, a, b, and c, of the triangle. To find
the distances, multiply the results by 100 km.
Finally, use the area formula for a triangle,
a+b+c
A = √s(s - a)(s - b)(s - c) where s = _________,
2
to determine the area.
C
_1
_
CE = 10 2
2.
Given the directed line segment from A to B,
construct the point P that divides the segment
in the ratio 3 to 1 from A to B. (Lesson 17.2)
Methodology:
7
Coordinates:
Miami: (-80.226529, 25.789106);
San Juan: (-66.1057427, 18.4663188);
Hamilton: (-64.781380, 32.294887)
20
y
z
y = 13.4
8.
Using the distance formula and multiplying the
results by 100 km gives the following distances.
z = 24.1
Miami to San Juan: 1590 km
San Juan to Hamilton: 1389 km
Hamilton to Miami: 1676 km
The boundaries of the Bermuda Triangle are not well defined, but the
region is often represented as a triangle with vertices at Miami, Florida; San
Juan, Puerto Rico; and Hamilton, Bermuda. The distance between Miami
and San Juan is about 1,034 miles. What is the approximate area of this
region? One tool that you may find helpful in solving this problem is the
similar triangle shown here with angle measures labeled.
and assumptions. Then use graphs, diagrams, words, or numbers to explain
H
61°
M
50°
69°
S
© Houghton Mifflin Harcourt Publishing Company
How Large Is the Bermuda Triangle?
Calculate s, s - a, s - b, and s - c.
s=
1590 + 1389 + 1676
__
= 2327.5
2
s - a = 2327.5 - 1590 = 737.5
s - b = 2327.5 - 1389 = 938.5
s - c = 2327.5 - 1676 = 651.5
Find the area:
―――――――
――――――――――
= √2327.5(737.5)(938.5)(651.5)
A=
Module 17
926
≈ 1, 024, 472 km 2
Study Guide Review
DISCUSSION OPPORTUNITIES
IN2_MNLESE389847_U7M17MC 926
√s(s - a)(s - b)(s - c)
18/04/14 10:23 PM
• If students choose to add a height of the triangle to the drawing they are
provided, and then use the familiar area of a triangle formula, how does this
affect the accuracy of their calculations?
• How does the fact that the Bermuda Triangle is not actually a triangle in the
plane, but a triangle on a sphere, affect the accuracy of the calculations? Is the
actual area more or less than the one calculated?
Assessment Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain.
0 points: Student does not demonstrate understanding of the problem.
Study Guide Review 926
17.1–17.4 Using Similar Triangles
ASSESS MASTERY
students have mastered the concepts and standards
covered in this module.
• Online Homework
• Hints and Help
• Extra Practice
Find the missing lengths. (Lesson 17.1)
B
1.
5
ASSESSMENT AND INTERVENTION
A
3
2.
7
D
E
15
y
x
20
12
z
G
x = 4.2
z = 16 ; y= 9
Given the directed line segment from A to B, construct the point P that divides the segment in the
given ratio from A to B. (Lesson 17.2)
C
3. 3 to 4
G
Access Ready to Go On? assessment online, and
receive instant scoring, feedback, and customized
intervention or enrichment.
D
A
• Success for English Learners
• Challenge Worksheets
Assessment Resources
B
P
6.25 feet
© Houghton Mifflin Harcourt Publishing Company
Differentiated Instruction Resources
F
The height of a street light is 25 feet. It casts a 12-foot shadow. At the same time, a man standing next
to the street light casts a 3-foot shadow. How tall is the man?
• Reteach Worksheets
I
Find the missing height. (Lesson 17.3)
4.
Response to Intervention Resources
E
H
ESSENTIAL QUESTION
How can you use similar triangles to find the missing parts of a triangle?
You can use proportions or corresponding parts of a triangle
to find the missing parts of a triangle that is similar to another.
5.
• Leveled Module Quizzes
Module 17
COMMON
CORE
IN2_MNLESE389847_U7M17MC 927
927
Module 17
Study Guide Review
927
Common Core Standards
18/04/14 10:23 PM
Content Standards Mathematical Practices
Lesson
Items
17.1, 17.3
1
G-SRT.B.4, G-SRT.B.5,
G-CO.C.10
MP.7
17.1, 17.3,
17.4
2
G-SRT.B.4, G-SRT.B.5,
G-CO.C.10
MP.7
17.2
3
G-CO.D.12, G-GPE.B.6
MP.5
17.1, 17.3,
17.4
4
G-SRT.B.4, G-SRT.B.5,
G-CO.C.10
MP.4
MODULE
MODULE 17
MIXED REVIEW
MIXED REVIEW
SELECTED RESPONSE
1. △XYZ is given by the points X(-1, −1), Y(3, 5), and Z(5, 1). Consider each of the points
below. Is each point a vertex of the image under the transformation
(x, y) → (x + 3, y - 2) →
Select Yes or No for A–C.
(
ASSESSMENT AND INTERVENTION
)
1
_
x, y → (y, -x)?
2
A. X‴ (-3, -1)
B. Y‴ (3, -3)
C. Z‴ (-1, -4)
Yes
Yes
Yes
No
No
No
2. Which of the following statements are true about the triangle at the
right? Choose True or False for each statement.
A. The value of x is 15.
True
False
B. The value of y is 12.
True
False
C. The value of y is 16.
True
False
prepare students for high-stakes tests.
28
30
Y
7
X
3. △ABC is given by the points A(-1, 2), B(2, 5), and C(4, -1). What is
(2 2)
Assessment Resources
1 _
the point _
, 5 ? Explain what this means.
• Leveled Module Quizzes: Modified, B
Possible Answer: The orthocenter; When the altitude is drawn
from each of the three vertices of the triangle, the point they
1 _
,5 .
intersect at is _
2 2
( )
COMMON
CORE
AVOID COMMON ERRORS
Item 1 Some students will stop working on this
problem too early, not completing all three
transformations. Encourage students to double-check
the problem to make sure they have completely
finished before making their answer choices.
© Houghton Mifflin Harcourt Publishing Company
4. Given the directed segment from A to B, construct
I C
H
the point P that divides the segment in the ratio 1
G
F
to 5 from A to B. Explain your process and how it
E
D
relates to similar triangles.
→
A
B
‾ . Use a compass to draw 6 arcs
Draw AC
P
→
‾ and label
(equidistant spacing) through AC
each intersection D through I. Connect the points B and I. Construct angles
¯ is partitioned into 6
congruent to ∠AIB for the remaining intersections. AB
equal parts. Label the point P that divides the segment in the ratio 1 to 5
from A to B. Each triangle created is a similar triangle to △AIB.
Module 17
17
Study Guide Review
928
Common Core Standards
IN2_MNLESE389847_U7M17MC 928
18/04/14 10:23 PM
Content Standards Mathematical Practices
Lesson
Items
IM1 17.1
1*
G-SRT.B.5, G-CO.A.5
MP.6
17.4
2
G-SRT.C.8
MP.6
15.3
3*
G-CO.D.12, G-CO.C.10,
G-GPE.B.5
MP.6
17.2
4
G-GPE.B.6, G-CO.D.12
MP.5
* Item integrates mixed review concepts from previous modules or a previous course.
Study Guide Review 928
MODULE
18
18
MODULE
Trigonometry with
Right Triangles
Trigonometry with
Right Triangles
Essential Question: How can you use
ESSENTIAL QUESTION:
trigonometry with right triangles to solve
real-world problems?
Answer: Trigonometric ratios allow you to find the
side length of a right triangle given an angle
measure, or vice versa. This can be useful whenever
a triangular shape appears in the real world, such as
in a metal bracket or a sculpture.
LESSON 18.1
Tangent Ratio
LESSON 18.2
Sine and Cosine Ratios
LESSON 18.3
Special Right Triangles
LESSON 18.4
Problem Solving with
Trigonometry
This version is for
Algebra 1 and
PROFESSIONAL
DEVELOPMENT
Geometry only
VIDEO
LESSON 18.5
Using a Pythagorean
Identity
Author Juli Dixon models successful
teaching practices in an actual
high-school classroom.
Professional
Development
my.hrw.com
Shutterstock
Professional Development Video
REAL WORLD VIDEO
Check out how right triangle
trigonometry is used in real-world
warehouses to minimize the space
needed for items being shipped or stored.
How Much Shorter Are Staggered Pipe Stacks?
In this module, you will investigate how much space can be saved by stacking pipes in a
staggered pattern rather than directly on top of each other. How can trigonometry help you
find the answer to this problem? Get prepared to discover the “staggering” results!
Module 18
DIGITAL TEACHER EDITION
IN2_MNLESE389847_U7M18MO 929
Access a full suite of teaching resources when and
where you need them:
• Access content online or offline
• Customize lessons to share with your class
• Communicate with your students in real-time
• View student grades and data instantly to target
your instruction where it is needed most
929
Module 18
929
PERSONAL MATH TRAINER
Assessment and Intervention
tests, and intervention activities. Prepare your
students with updated, Common Core-aligned
practice tests.
18/04/14 11:08 PM
Complete these exercises to review skills you will need
for this module.
Angle Relationships
Example 1
Find the angle complementary to the given angle. 75°
x + 75° = 90°
students need strategic or intensive intervention for
the module’s prerequisite skills.
• Online Homework
• Hints and Help
• Extra Practice
Write as an equation.
x = 90° − 75°
Solve for x.
x = 15°
Find the complementary angle.
ASSESSMENT AND INTERVENTION
70°
1.
20°
2.
35°
55°
3.
67°
23°
Find the supplementary angle.
4.
80°
5.
65°
6.
34°
100°
115°
3
2
146°
1
Find the remaining angle or angles for △ABC.
m∠A = 50°, m∠B = 40°
m∠C = 90°
8.
m∠A = 60°, m∠C = 20°
m∠B = 100°
9.
m∠B = 70° and ∠A ≅ ∠C
Personal Math Trainer will automatically create a
standards-based, personalized intervention
assignment for your students, targeting each student’s
individual needs!
m∠A = 55°, m∠C = 55°
m∠A = 60°, m∠B = 60°, m∠C = 60°
10. ∠A ≅ ∠B ≅ ∠C
1 m∠C
11. m∠B = 30° and m∠A = _
2
m∠A = 50°, m∠C = 100°
12. △ABC is similar to △DEF and m∠D = 70° and m∠F = 50° m∠A = 70°, m∠B = 60°, m∠C = 50°
13. △ABC is similar to △PQR and m∠R = 50° and ∠P ≅ ∠Q
14. m∠A = 45° and m∠B = m∠C
15. m∠B = 105° and m∠A = 2 · m∠C
16. m∠A = 5° and m∠B = 9 · m∠C
Module 18
IN2_MNLESE389847_U7M18MO 930
Tier 1
Lesson Intervention
Worksheets
Reteach 18.1
Reteach 18.2
Reteach 18.3
Reteach 18.4
Reteach 18.5
m∠A = 65°, m∠B = 65°, m∠C = 50°
m∠B = 67.5°, m∠C = 67.5°
m∠A = 50°, m∠C = 25°
© Houghton Mifflin Harcourt Publishing Company
7.
TIER 1, TIER 2, TIER 3 SKILLS
See the table below for a full list of intervention
resources available for this module.
Response to Intervention Resources also includes:
• Tier 2 Skill Pre-Tests for each Module
• Tier 2 Skill Post-Tests for each skill
m∠B = 157.5°, m∠C = 17.5°
930
Response to Intervention
Tier 2
Strategic Intervention
Skills Intervention
Worksheets
Tier 3
Intensive Intervention
Worksheets available
online
38 Angle Relationships
43 The Pythagorean
Theorem
46 Proportional
Relationships
Building Block Skills 7,
15, 16, 38, 46, 53, 56,
63, 66, 82, 90, 95, 98,
100, 102
Differentiated
Instruction
18/04/14 11:08 PM
Challenge
worksheets
Extend the Math
Lesson Activities
in TE
Module 18
930
LESSON
18.1
Name
Tangent Ratio
Class
Date
18.1 Tangent Ratio
Essential Question: How do you find the tangent ratio for an acute angle?
Common Core Math Standards
The student is expected to:
COMMON
CORE
Resource
Locker
G-SRT.C.6
Explore
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. Also G-SRT.C.8
Mathematical Practices
COMMON
CORE
Investigating a Ratio in a Right Triangle
In a given a right triangle, △ABC, with a right angle at vertex C, there are three sides. The side
adjacent to ∠A is the leg that forms one side of ∠A. The side opposite ∠A is the leg that does
not form a side of ∠A. The side that connects the adjacent and opposite legs is the hypotenuse.
B
Hypotenuse
MP.4 Modeling
Language Objective
Explain to a partner how to find the tangent of an angle given a diagram
of a right triangle with given angle measure and leg lengths.
A

Measurements may vary slightly.
opposite
2.5 cm
40°
3.1 cm
F
© Houghton Mifflin Harcourt Publishing Company
View the Engage section online. Discuss the
photograph, asking students to describe the forces
that might create the shape of a sand dune. Then
C
E
Essential Question: How do you find
the tangent ratio for an acute angle?
PREVIEW: LESSON
to ∠A
In △DEF, label the legs opposite and adjacent to ∠D. Then measure the lengths of
the legs in centimeters and record their values in the rectangles provided.
ENGAGE
For an acute angle in a right triangle, the tangent
ratio is the ratio of the length of the opposite leg to
the length of the adjacent leg.
Leg
opposite
to ∠A

D
What is the ratio of the opposite leg length to the adjacent leg length,
rounded to the nearest hundredth?
EF ≈ 0.81 Ratios may vary slightly depending on measurements.
_
DF

Using a protractor and ruler, draw right triangle △JKL with a right angle at vertex L
and ∠J = 40° so that △JKL ~ △DEF. Label the opposite and adjacent legs to ∠J and
include their measurements.
Drawings will vary but should be similar to the triangle in Step A. For all
triangles, it is true that opposite length < adjacent length.

What is the ratio of the opposite leg length to the adjacent leg length,
rounded to the nearest hundredth?
KL ≈
_
0.81
JL
Module 18
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 1
931
gh “File info”
Date
Class
ent Ratio
18.1 Tang
Name
angle?
an acute
nt ratio for
the tange
angles in
ties of the
do you find
ion: How
es are proper G-SRT.C.8
in right triangl
. Also
ity, side ratios ratios for acute angles
by similar
metric
tand that
of trigono
G-SRT.C.6 Unders
ngle
to definitions
a Right Tria
the triangl
a Ratio in are three sides. The side
g
atin
does
C, there
Investig
the leg that
angle at vertex
ite ∠A is
nuse.
Explore
, with a right ∠A. The side oppos ite legs is the hypote
le, △ABC
of
a right triang that forms one side the adjacent and oppos
In a given
leg
cts
to ∠A is the The side that conne
B
a side of ∠A.
not form
Resource
Locker
Quest
Essential
COMMON
CORE
IN2_MNLESE389847_U7M18L1 931
nt

Watch for the hardcover
student edition page
numbers for this lesson.
Leg
opposite
to ∠A
Hypotenuse
to ∠A
HARDCOVER PAGES 931940
C
s of
A
re the length
Then measu
nt to ∠D.
provided.
ite and adjace in the rectangles
legs oppos
their values
, label the
In △DEF
and record
ly.
centimeters
vary slight
the legs in
ents may
E
Measurem
y
g Compan
opposite
2.5 cm

n Mifflin
Harcour t
Publishin

40°
D
F
3.1 cm
,
nt leg length
leg length
opposite
ratio of the
urements.
the
is
edth?
on meas
What
t hundr
ly depending
to the neares
rounded
vary slight
L
Ratios may
at vertex
EF ≈ 0.81
a right angle
_
and
le △JKL with adjacent legs to ∠J
DF
right triang
and
ite
draw
oppos
ruler,
ctor and
. Label the
Using a protra that △JKL ~ △DEF
A. For all
40° so
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ts.
and ∠J =
the triang
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similar to
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but
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Drawings
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opposite
ratio of the
What is the
t hundredth?
to the neares
rounded

KL ≈ 0.81
_
JL
Module 18
8L1 931
47_U7M1
ESE3898
IN2_MNL
931
Lesson 18.1
Lesson 1
931
18/04/14
11:19 PM
18/04/14 11:20 PM
Reflect
1.
EXPLORE
Discussion Compare your work with that of other students. Do all the triangles
have the same angles? Do they all have the same side lengths? Do they all have the
same leg ratios? Summarize your findings.
The triangles are all similar, so they have angles that have the same
Investigating a Ratio in a
Right Triangle
measure but not the same side lengths. Yet, for a given angle, the ratio of
the length of the opposite leg to the length of the adjacent leg is constant.
2.
INTEGRATE TECHNOLOGY
If you repeated Steps A–D with a right triangle having a 30° angle, how would your
results be similar? How would they be different?
The triangles would all be similar, with the ratio of the length of the leg
Students have the option of doing the Explore activity
either in the book or online.
opposite the 30° angle to the length of the leg adjacent to the 30° angle
constant, but the actual value of the ratio would be different from 0.81.
CONNECT VOCABULARY
Finding the Tangent of an Angle
Explain 1
Draw and label a variety of right triangles with
different orientations. For each of the triangles, have
students name the hypotenuse, the leg adjacent to a
given angle, and the leg opposite the angle.
The ratio you calculated in the Explore section is called the tangent of an angle. The tangent
of acute angle A, written tan ∠A, is defined as follows:
length of leg opposite ∠A
tan A = ___
length of leg adjacent to ∠A
You can use what you know about similarity to show why the tangent of an angle is constant.
By the AA Similarity Postulate, given ∠ D ≅ ∠J and also ∠ F ≅ ∠L, then △DEF ~ △JKL.
This means the lengths of the sides of △JKL are each the same multiple, k, of the lengths of the
corresponding sides of △DEF. Substituting into the tangent equation shows that the ratio of the
length of the opposite leg to the length of the adjacent leg is always the same value for a given
acute angle.
E
40°
F
tangent defined for specified angle △DEF
D
J
Find the tangent of each specified angle. Write each ratio as a C
fraction and as a decimal rounded to the nearest hundredth.
∠A
18
∠B
B
length of leg opposite ∠ A
18 = _
3 = 0.75
tan A = ___ = _
4
24
length of leg adjacent to ∠ A
4
24
length of leg opposite ∠ B
tan B = ___ = _ = _ ≈ 1.33
3
length of leg adjacent to ∠ B
18
Module 18
932
24
A
© Houghton Mifflin Harcourt Publishing Company

⎧

⎪
Example 1
40°
⎪
k∙DF
⎨
L
The ratio of the length of the opposite leg to
the adjacent leg of an angle of a right triangle
is constant for any right triangle with the same angle
measures. What does this imply about right triangles
with that given angle measure? Explain. The
triangles must be similar because the angle
measures are constant and the ratios of the sides
are proportional.
△JKL
leg opposite ∠ 40°
k·EF = _
KL = _
EF
EF = _
tan 40° = __ = _
JL k·DF
DF
DF
k∙DE
k∙EF
⎪
⎪
⎩
K
QUESTIONING STRATEGIES
AVOID COMMON ERRORS
If students use calculators to compute the tangent
ratio of an angle, remind them to check that their
calculators are in degree mode. Otherwise, they will
not find the correct value.
30
Lesson 1
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M18L1 932
Learning Progressions
Students work with three trigonometric ratios in this course: tangent, sine, and
cosine. The tangent is treated separately for two reasons. First, working with a
single ratio gives students a chance to focus on the conceptual underpinnings of
trigonometry. That is, right triangles with a given acute angle, ∠A, are all similar,
so the ratio of the length of the leg opposite ∠A to the length of the leg adjacent to
∠A is constant for all such triangles. Second, students sometimes have difficulty
deciding which trigonometric ratio to use when solving a problem. By working
with tangents first, students can instead focus on the general process for solving a
problem using trigonometry.
18/04/14 11:20 PM
EXPLAIN 1
Finding the Tangent of an Angle
CONNECT VOCABULARY
The abbreviation tan is read tangent. For example,
tan 30° is read as tangent of 30 degrees.
Tangent Ratio 932
Reflect
QUESTIONING STRATEGIES
3.
If you know the tangent ratio for an acute
angle of a right triangle, and the length of one
of the legs, can you reconstruct the triangle?
Explain. Yes; you can use the ratio to find the length
of the other leg, and then use the Pythagorean
Theorem to find the length of the hypotenuse.
What is the relationship between the ratios for tan A and tan B? Do you believe this
relationship will be true for acute angles in other right triangles? Explain.
The ratios are reciprocals of each other. This will always be true because the opposite side
of one acute angle is the adjacent side of the other acute angle, and vice versa.
4.
Why does it not make sense to ask for the value of tan L?
The tangent ratio is defined only for the acute angles of a right triangle,
not for the right angle.
You can use a table of values to find the
tangent of a given acute angle. How do you
think the tables were compiled? by constructing
right triangles with the given angle measure and its
complement, measuring the opposite and adjacent
side lengths, and then computing the tangent ratio
Find the tangent of each specified angle. Write each ratio as a fraction
and as a decimal rounded to the nearest hundredth.
5.
∠Q
6.
15
5
tan ∠ Q = _ = _ ≈ 0.42
36 12
39
S
15
R
36 12
tan ∠ R = _ = _
= 2.4
5
15
When you know the length of a leg of a right triangle and the measure of one of the acute
angles, you can use the tangent to find the length of the other leg. This is especially useful in
real-world problems.
Apply the tangent ratio to find unknown lengths.
Example 2
Finding a Side Length using Tangent
In order to meet safety guidelines, a roof contractor determines that she must place the base of
her ladder 6 feet away from the house, making an angle of 76° with the ground. To the nearest
tenth of a foot, how far above the ground is the eave of the roof?
C
© Houghton Mifflin Harcourt Publishing Company
QUESTIONING STRATEGIES
What if you want to find the tangent of the
other acute angle in the triangle? How would
you find the angle? When might this be
necessary? The measure of the angle is the
complement of the given angle. You might want to
do this if the missing side is in the denominator of
the tangent ratio.
∠R
36
Finding a Side Length using Tangent
Explain 2
EXPLAIN 2
Why is it important to draw and label a
diagram when solving a real-world problem
about relationships in a right triangle? The diagram
makes it possible to identify values and
relationships that you can use to solve the problem.
Q
A
76°
6 ft
B
Step 1 Write a tangent ratio that involves the unknown length.
length of leg opposite ∠A
BC
tan A = ___ = _
BA
length of leg adjacent to ∠A
Step 2 Identify the given values and substitute into the tangent equation.
Given: BA = 6 ft and m∠A = 76°
BC
Substitute: tan 76° = _
6
Module 18
933
Lesson 1
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M18L1 933
Peer-to-Peer Activity
Provide pairs of students with graph paper and ask each student to draw two
different right triangles, with unit side lengths for the legs. Have them use a
protractor to estimate one of the acute angles in one triangle, then have pairs work
together to find the tangent. Then, have them choose an acute angle in the
remaining triangle and find the inverse tangent of the angle. For additional work,
have students exchange triangles with other pairs.
933
Lesson 18.1
18/04/14 11:20 PM
Step 3 Solve for the unknown leg length. Be sure the calculator is in degree mode
and do not round until the final step of the solution.
Multiply each side by 6.
BC
6 ·_
6 · tan 76° = _
1 6
Use a calculator to find tan 76°.
6 · tan 76° = BC
Substitute this value in for tan 76°.
6(4.010780934) = BC
Multiply. Round to the nearest tenth.
24.1 ≈ BC
AVOID COMMON ERRORS
x,
When solving an equation of the form tan 35° = ___
12
students may forget to multiply by 12 to solve for x
after they find tan 35°. These students may benefit
from rewriting the equation as 12 ∙ tan 35° = x
before evaluating tan 35°.
So, the eave of the roof is about 24.1 feet above the ground.
B
For right triangle △STU, what is the length of the leg adjacent to ∠S?
S
54°
Step 1 Write a tangent ratio that involves the unknown length.
EXPLAIN 3
TU
length of leg opposite ∠ S
tan S = ___ = _
length of leg adjacent to ∠ S
SU
T
87
U
Finding an Angle Measure
using Tangent
Step 2 Identify the given values and substitute into the tangent equation.
Given: TU = 87 and m∠S = 54 °
87
Substitute: tan 54 ° = _
SU
Step 3 Solve for the unknown leg length.
QUESTIONING STRATEGIES
In the equation y = tan -1x, explain what x
and y represent. In the equation, x represents
the tangent of the angle with measure y°.
87
Multiply both sides by SU, then divide both sides by 54°. SU = __
tan 54 °
87
Use a calculator to find 54° and substitute.
SU ≈ __
Divide. Round to the nearest tenth.
SU ≈ 63.2
© Houghton Mifflin Harcourt Publishing Company
7.
A ladder needs to reach the second story window, which is 10 feet above the ground,
and make an angle with the ground of 70°. How far out from the building does the
base of the ladder need to be positioned?
10
10
10
_____
__
3.6 ft; tan 70° = __
x , so x = tan 70° ≈ 2.747477419 ≈ 3.6
Explain 3
Finding an Angle Measure using
Tangent
In the previous section you used a given angle measure and leg measure with the tangent
ratio to solve for an unknown leg. What if you are given the leg measures and want to find the
measures of the acute angles? If you know the tan A, read as “tangent of ∠A,” then you can use
the tan -1 A, read as “inverse tangent of ∠A,” to find m∠A. So, given an acute angle ∠A, if
tan A = x, then tan -1 x = m∠A.
Module 18
If tan x = 1, what is the measure of angle x?
Explain. The measure of angle x must be 45°
because the legs of the triangle must be the same
length to make the tangent ratio 1.
1.37638192
934
How could you evaluate the inverse tangent of
an angle without using a calculator’s inverse
tangent key or a table of values? Use the
measurements of the tangent ratio as the legs to
draw the corresponding right triangle. Then use a
protractor to measure the angle formed by the
hypotenuse and the adjacent side to estimate the
angle measure.
Lesson 1
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M18L1 934
Kinesthetic Experience
18/04/14 11:20 PM
To help students identify the opposite and adjacent legs, model a right triangle on
the floor with masking tape. Have a student stand at one acute angle and walk to
the leg next to, or touching, that angle. That is the adjacent leg. Then have the
student walk from the acute angle to the leg opposite, or across from, the angle.
This is the opposite leg.
Tangent Ratio 934
Example 3
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Discuss why students will get more accurate
19 in.

What is m∠A?
A
()
1
results if they evaluate an expression like tan -1 __
3
rather than finding a decimal approximation for the
1 rounding, and evaluating tan -1(0.33).
fraction __
3,
ELABORATE

SUMMARIZE THE LESSON
How do you find the tangent ratio for an acute
angle of a right triangle? Find the ratio of the
length of the leg opposite the angle to the length of
the leg adjacent to the angle.
935
Lesson 18.1
Step 2 Write the inverse
tangent equation.
Step 3 Evaluate using
a calculator and
round as indicated.
19
tan A = _
36
19
tan -1 _ = m∠ A
36
m∠ A ≈ 27.82409638 ≈ 28°
Step 2 Write the
inverse tangent
equation.
36
tan -1
tan B = _____
36
_____ = m∠ B
19
19
Step 3 Evaluate using a
calculator and round as
indicated.
°
°
m∠ B ≈ 62.17590362 ≈ 62
AVOID COMMON ERRORS
8.
J
Find m∠J.
46
26°, m∠J = tan -1 __
≈ 26.31808814
93
K
93
46
© Houghton Mifflin Harcourt Publishing Company
How are tangent and inverse tangent
related? The tangent of an acute angle of a
right triangle is the ratio of the lengths of the
opposite side to the adjacent side of the right
triangle. The inverse tangent is the angle with that
tangent ratio.
C
Step 1 Write the tangent
ratio for ∠ A using
the known values.
Step 1 Write the
tangent ratio
for ∠ B using the
known values.
Have students graph y = tan x on their graphing
calculators on the interval [0, 90°]. Discuss how to
interpret the graph, especially as x nears 90°.
QUESTIONING STRATEGIES
36 in.
What is m∠B?
INTEGRATE TECHNOLOGY
Some students may have the misconception that the
opposite leg is always a vertical side of the triangle
and the adjacent leg is always a horizontal side of the
triangle. Remind students that the opposite and
adjacent sides are determined by the location of the
associated angle, not by the orientation of the
triangle.
B
Find the measure of the indicated angle.
Round to the nearest degree.
L
Elaborate
9.
Explain how to identify the opposite and adjacent legs of a given acute angle.
The opposite leg does not form a side of the given angle. The adjacent leg
forms a side of the given angle and is not the hypotenuse.
10. Discussion How does tan A change as m∠A increases? Explain the basis for the
identified relationship.
Tan A increases as m∠A increases. As m∠A increases, the length of the leg
adjacent to ∠A remains the same and the length of the leg opposite ∠A
increases. Since the length of the opposite leg is the numerator of the tangent
ratio, and the denominator of that ratio does not change, the ratio increases.
Module 18
935
Lesson 1
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M18L1 935
Connect Vocabulary
18/04/14 11:20 PM
Have students look up the word inverse in the dictionary. A definition may include
reversed in position. Discuss how this corresponds to the relationship between the
inverse tangent and the tangent of an angle.
11. Essential Question Check-In Compare and contrast the use of the tangent and
inverse tangent ratios for solving problems.
The tangent and inverse tangent both apply only to acute angles in
right triangles. When the leg lengths are known, the tangent can be
used to solve for the constant ratio. Whereas, the inverse tangent can be
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
used to solve for the associated acute angle. When the measures of one
leg and one angle are known, the tangent can be used to solve for the
unknown leg whereas the inverse tangent is not applicable.
ASSIGNMENT GUIDE
Evaluate: Homework and Practice
In each pair of right triangles, measure the lengths of the adjacent
side and the opposite side of the angle specified in the figure. Then
calculate and compare the ratios.
1.
In each triangle, measure the length of the adjacent side and the opposite side of the
22° angle. Then calculate and compare the ratios.
22°
22°
Opposite
Opposite
______
= 0.40; ______ = 0.40; the ratios are the same.
Concepts and Skills
Practice
Explore
Investigating a Ratio in a Right
Triangle
Exercise 1
Example 1
Finding the Tangent of an Angle
Exercises 2–8
Example 2
Finding a Side Length using Tangent
Exercises 9–14,
18–19
Example 3
Finding an Angle Measure using
Tangent
Exercises 15–17,
20–21
In each right triangle, find the tangent of each angle that is not the
right angle.
2.
3.
A
E
5.0
COMMUNICATING MATH
F
10
9.0
C
8
B
10.3
D
8
tan ∠A = __
6
6
tan ∠B = _
8
5
tan ∠D = __
9
tan ∠A = 1.33
tan ∠B = 0.75
tan ∠D = 0.56 tan ∠F = 1.8
4.
B
32
81.5
A
75
5.
C
9
tan ∠F = _
5
19
Q
3
P
R
19.2
32
tan ∠A = ___
75
75
tan ∠B = __
32
19
tan ∠P = ___
3
3
tan ∠R = __
19
tan ∠A = 0.43
tan ∠B = 2.34
tan ∠P = 6.33
tan ∠R = 0.16
Module 18
Exercise
IN2_MNLESE389847_U7M18L1 936
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Discuss rounding errors that occur when
finding the tangent and inverse tangent of an angle.
Have students find the tangent of an angle on their
calculators. Then have them use the value to find the
inverse tangent for that tangent. Compare results.
Repeat with several exercises. Have students also start
with the inverse tangent.
Lesson 1
936
Depth of Knowledge (D.O.K.)
Point out that the hypotenuse of a right triangle will
never be the opposite or the adjacent side for the
tangent ratio. Discuss why this must be so.
© Houghton Mifflin Harcourt Publishing Company
6
COMMON
CORE
Mathematical Practices
1–4
1 Recall of Information
MP.6 Precision
5–17
2 Skills/Concepts
MP.5 Using Tools
18–22
2 Skills/Concepts
MP.2 Reasoning
23–26
2 Skills/Concepts
MP.4 Modeling
18/04/14 11:20 PM
Tangent Ratio 936
Let △ABC be a right triangle, with m∠C = 90°. Given the tangent of one of the
complementary angles of the triangle, find the tangent of the other angle.
AVOID COMMON ERRORS
6.
Students may confuse the opposite and the adjacent
sides of a given angle in a right triangle. Suggest that
students use one color to highlight the opposite side
and another color to highlight the adjacent side of a
given acute angle. Emphasize that the adjacent side
forms the given angle’s ray that is not the hypotenuse
of the triangle.
tan ∠A = 1.25
1
tan ∠B = ___
1.25
7.
tan ∠B = 0.80
tan ∠B = 0.50
1
tan ∠A = ___
0.50
8.
tan ∠A = 2.0
tan ∠B = 1.0
1
tan ∠A = ___
1.0
tan ∠A = 1.0
Use the tangent to find the unknown side length.
9.
Find QR.
10. Find AC.
Q
A
R
C
11. Find PQ.
P
7.0
7.0
tan 60° = ___
QR
7.0
QR = _____
E
13
tan 85°
14. Find PR.
P
45°
8.4
C
R
54°
4.6
Q
AB
tan 45° = ___
8.4
DE
AB = tan 45°(8.4)
PR
tan 54° = ___
4.6
tan 21°
AB = 8.4
PR = 6.3
21°
13
tan 21° = ___
DE =
9
PQ = _____
PQ = 0.79
A
B
D
9
tan 85° = __
PQ
AC = 4.1
13. Find AB.
F
R
85°
B
AC = tan 27°(7.0)
QR = 4.0
12. Find DE.
27°
8.0
AC
tan 27° = ___
8.0
tan 60°
9
Q
P
60°
13
_____
© Houghton Mifflin Harcourt Publishing Company
DE = 33.9
PR = tan 54°(4.6)
Module 18
IN2_MNLESE389847_U7M18L1 937
937
Lesson 18.1
937
Lesson 1
18/04/14 11:20 PM
Find the measure of the angle specified for each triangle. Use the inverse
nearest degree.
15. Find ∠A.
16. Find ∠R.
A
C
B
R
9
B
6.8
For right triangles with unit side lengths for the legs,
when they are finding the inverse tangent, students
may benefit from drawing the right triangle on graph
paper and then using a protractor to estimate the
acute angle measure. They can also do this to check
their work.
17. Find ∠B.
24
Q
3.0
MANIPULATIVES
17
P
6.8
tan -1 ___
= m∠A
3.0
9
tan -1 __
= m∠R
24
m∠A = 66°
m∠R = 21°
A
C
16
16
tan -1 B = __
17
AVOID COMMON ERRORS
m∠B = 43°
Call attention to the notation used to represent the
inverse tangent: tan -1x. The expression tan -1x is a
short way to indicate “the angle whose tangent ratio
is x.” Be sure students understand that the raised –1
is not an exponent and that tan -1x does not
1
mean ____
tanx . Emphasize that when students find the
inverse tangent, the result will be an angle with a
degree measure.
-1
Write an equation using either tan or tan to express the
measure of the angle or side. Then solve the equation.
18. Find BC.
B
40°
10
38
Q
P
A
20. Find ∠A and ∠C.
19. Find PQ.
R
14.5
A
B
75°
14.5
38
PQ = _____
= 10.2
tan 75°
C
C
( )
BC = 10 tan 40° = 8.4
14.5
= 45°
m∠A = m∠C = tan -1 A ___
14.5
21. Multi-Step Find the measure of angle D. Show your work.
B
BC
tan 30° = __
; BC = 12 tan 30°
12
( )
(
)
A
30°
Module 18
IN2_MNLESE389847_U7M18L1 938
12 cm
C 4 cm
© Houghton Mifflin Harcourt Publishing Company
12 tan30°
BC
m∠D = tan -1 __
= tan -1 _______
= 60°
4
CD
D
938
Lesson 1
18/04/14 11:20 PM
Tangent Ratio 938
22. Engineering A client wants to build a ramp that carries
1.4 m
people to a height of 1.4 meters, as shown in the diagram. What
additional information is necessary to identify the measure of
angle a, the angle the ramp forms with the horizontal? After
measure of the angle. Possible answer: Measure the distance x of the ramp across
JOURNAL
Have students summarize what they know about the
tangent ratio. Remind them to include at least one
figure and at least one example in their descriptions.
a
( )
1.4
the horizontal. Then find tan -1 ___
x , which equals the measure of angle a.
23. Explain the Error A student uses the triangle shown to
6.5
2.5
calculate a. Find and explain the student’s error.
a
6.5
a = tan -1 _ = tan -1(2.6) The student’s calculations are correct only if the triangle
2.5
is a right triangle.
a = 69.0°
( )
24. When m∠A + m∠B = 90°, what relationship is formed by tan ∠A and tan ∠B? Select
all that apply.
1
C. (tan∠A)(tan∠B) = 1
A. tan∠A = _
tanB
B. tan∠A + tan∠B = 1
D. (tan∠A)(tan∠B) = -1
A, C; Tan A and tan B are reciprocals of each other so they have a product of 1.
H.O.T. Focus on Higher Order Thinking
25. Analyze Relationships To travel from Pottstown to Cogsville, a man drives his car
83 miles due east on one road, and then 15 miles due north on another road. Describe
the path that a bird could fly in a straight line from Pottstown to Cogsville. What
angle does the line make with the two roads that the man used?
© Houghton Mifflin Harcourt Publishing Company
The bird’s path is the hypotenuse of the right triangle formed by the path of
the man. The bird’s path forms an angle of 10.2° with the first road and an
angle of 79.8° with the second road.
26. Critical Thinking A right triangle has only one 90° angle. Both of its other angles
have measures greater than 0° and less than 90°. Why is it useful to define the tangent
of 90° to equal 1, and the tangent of 0° to equal 0?
Answers will vary. Possible answer: Consider a right triangle in which the
complementary angles are very close to 0° and 90°. Let the length of the
0
hypotenuse be 1. The tangent of the small angle is very close to _
. The
1
1
tangent of the larger angle will be very close to _
.
1
Module 18
IN2_MNLESE389847_U7M18L1 939
939
Lesson 18.1
939
Lesson 1
18/04/14 11:20 PM
AVOID COMMON ERRORS
When they form conical piles, granular materials such
as salt, gravel, and sand settle at different “angles of repose,”
depending on the shapes of the grains. One particular
13-foot tall cone of dry sand has a base diameter of 38.6 feet.
When using the tangent function to find the angle of
repose of a conical pile, students may mistakenly use
the diameter of the base of the pile as the side
adjacent to the angle of repose. The triangle they
should use has the radius of the base as the side
adjacent to the angle of repose and the height of the
pile as the side opposite.
angle of repose
1. To the nearest tenth of a degree, what is the angle
of repose of this type of dry sand?
2. A different conical pile of the same type of sand is 10 feet tall.
What is the diameter of the cone’s base?
3. Henley Landscaping Supply sells a type of sand with a 30°
angle of repose for \$32 per cubic yard. Find the cost of an
11-foot-tall cone of this type of sand. Show your work.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Are values of the tangent function
13
tan (x) = ____
38.6
(___
2 )
13
1. tan (x) = ___
19.3
( )
13
x = tan -1 ___
19.3
x = 34°
proportional? If they are, the tangent of 40°, for
example, should be double the tangent of 20°. Use
tangent tables or a calculator to investigate the
question. Give examples to support your
conclusion. Not proportional; example: tan 20° ≈
0.36, but tan 40° ≈ 0.84 ≠ 2tan 20°
10
tan (34) = ___
(_2x )
20
2. tan (34) = ___
x
x = 29.7
3. Let r = radius of cone. Then:
11
tan 30° = __
r
© Houghton Mifflin Harcourt Publishing Company
11
0.5774 = __
r
r ≈ 19.1 ft
Volume of cone:
1
V=_
Bh
3
1 2
=_
πr h
3
2
1(
= __
3.14)(19.1) (11)
3
= 4200.2 ft 3
1 yd 3 = 27 ft 3 so 4200.2 ft 3 = 4200.2 ÷ 27 ≈ 155.6 yd 3
At \$32 per yd 3 55.6 yd 3 will cost 155.6 × 32 = \$4979.20.
Module 18
940
Lesson 1
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M18L1 940
Have students research the angle of repose of at least four granular substances. For
each one they should draw an equilateral triangle showing a cross-section of a pile
of the substance labeled with a base diameter of 20 feet, base angles measuring the
angle of repose, and the correct height to the nearest tenth.
18/04/14 11:20 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Tangent Ratio 940
LESSON
18.2
Name
Sine and Cosine Ratios
Class
Date
18.2 Sine and Cosine Ratios
Essential Question: How can you use the sine and cosine ratios, and their inverses,
in calculations involving right triangles?
Common Core Math Standards
The student is expected to:
COMMON
CORE
Resource
Locker
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. Also G-SRT.C.7, G-SRT.C.8
Explore
You can use geometry software or an online tool to explore ratios of side lengths in right triangles.
Mathematical Practices
COMMON
CORE
Investigating Ratios in a Right Triangle

Construct three points A, B, and C.
→
→
‾ and AC
‾ . Move C so that ∠A is acute.
Construct raysAB
MP.4 Modeling
_
Construct point D_
on AC. Construct a line through D
perpendicular to AB. Construct
_ point E as the intersection of
the perpendicular line and AB.
Language Objective

Explain to a partner how to find the sine and cosine of an angle given a
diagram of a right triangle with given angle measure and opposite or

Measure ∠A. Measure the side lengths DE, AE, and

DE and _
AE .
Calculate the ratios _
Essential Question: How can you use
the sine and cosine ratios, and their
inverses, in calculations involving right
triangles?
Given the measure of one acute angle and the
length of the hypotenuse, you can use the sine and
cosine ratios to find the lengths of each leg. Given
the length of the hypotenuse and the length of a
leg, you can use their inverses to find the measure
of each acute angle.
PREVIEW: LESSON
View the Engage section online. Discuss the
photograph, asking students to speculate on what the
person in the photo might be doing. Then preview
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
Reflect
1.
2.
3.
→
→
‾ . What happens to m∠A as D moves along AC
‾ ? What postulate or
Drag D along AC
theorem guarantees that the different triangles formed are similar to each other?
m∠A does not change; AA Similarity Postulate, given also that ∠AED remains a right angle.
→
DE and _
AE ?
‾ , what happens to the values of the ratios _
As you move D along AC
Use the properties of similar triangles to explain this result.
DE →_
DE and _
AE → _
AE’ = _
AE .
D’E’ = _
_
Move C. What happens to m∠A? With a new value of m∠A, note the values of
→
‾ ?
the two ratios. What happens to the ratios if you drag D along AC
m∠A changes in value; the new values of the ratios do not change as D is dragged
→
‾ .
along AC
Module 18
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IN2_MNLESE389847_U7M18L2 941
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Module 18
8L2 941
47_U7M1
ESE3898
IN2_MNL
941
Lesson 18.2
18/04/14
11:31 PM
18/04/14 11:33 PM
Explain 1
Finding the Sine and Cosine of an Angle
EXPLORE
Trigonometric Ratios
A trigonometric ratio is a ratio of two sides of a right triangle. You have already seen
one trigonometric ratio, the tangent. There are two additional trigonometric ratios, the
sine and the cosine, that involve the hypotenuse of a right triangle.
The sine of ∠A, written sin A, is defined as follows:
lenngth of leg opposite ∠A
BC
sin A = ___ = _
AB
length of hypotenuse
The cosine of ∠A, written cos A, is defined as follows:
length of leg adjacent to ∠A
AC
cos A = ___ = _
AB
length of hypotenuse
B
INTEGRATE TECHNOLOGY
A
C
You can use these definitions to calculate trigonometric ratios.
Example 1

Investigating Ratios in a Right
Triangle
Students have the option of doing the Explore activity
either in the book or online.
D
Write sine and cosine of each angle as a fraction and as a decimal
rounded to the nearest thousandth.
∠D
15
E
length of leg opposite ∠D EF
8
sin D = ___ = _ = _ ≈ 0.471
DF
17
length of hypotenuse
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Have students examine what happens to the
17
8
F
ratio of the opposite side length to the hypotenuse
length as the acute angle gets closer to 90°. Repeat for
the ratio of the adjacent side length to the hypotenuse
length as the acute angle gets closer to 0°.
length of leg adjacent to ∠D
15 ≈ 0.882
DE = _
cos D = ___ = _
17
DF
length of hypotenuse

∠F
15
length of leg opposite to ∠F
DE = _ ≈
cos F = ___ = _
DF
length of hypotenuse
0.882
17
8
QUESTIONING STRATEGIES
0.471
Reflect
4.
What do you notice about the sines and cosines you found? Do you think this relationship
will be true for any pair of acute angles in a right triangle? Explain.
sin D = cos F and cos D = sin F; this relationship always holds because the leg opposite one
acute angle is adjacent to the other one.
5.
In a right triangle △PQR with hypotenuse 5, m∠Q = 90°, and PQ > QR, what are the
values of sin P and cos P?
sin P =
cos P =
_3
If the measure of the acute angle of the right
triangle does not change but the side lengths
of the triangle change, how do the ratios change?
Explain. The values of the numerator and the
denominators will change but the ratios are equal
to the original ratios of opposite length to
hypotenuse and adjacent length to hypotenuse
because the triangles are similar.
© Houghton Mifflin Harcourt Publishing Company
length of leg adjacent to ∠F
cos F = ___ = _ ≈
length of hypotenuse
17
5
_4
EXPLAIN 1
5
Module 18
942
Finding the Sine and Cosine of
an Angle
Lesson 2
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M18L2 942
Math Background
Trigonometry is the branch of mathematics concerned with angle relationships in
triangles. The ancient Egyptians used trigonometry to reset land boundaries after
the Nile River flooded each year. The Babylonians used trigonometry to measure
distances to nearby stars. Trigonometry is used in modern engineering,
cartography, medical imaging, and many other fields.
18/04/14 11:33 PM
AVOID COMMON ERRORS
Students often use the wrong ratio for sine or cosine.
Help students review these relationships by using
flashcards, mnemonics, or other memory aids.
Encourage students to research mnemonics or
produce their own.
Sine and Cosine Ratios
942
Explain 2
QUESTIONING STRATEGIES
Using Complementary Angles
The acute angles of a right triangle are complementary. Their trigonometric ratios are related to
each other as shown in the following relationship.
If you know only the sine for an acute angle of
a right triangle, how could you find the
cosine? The sine gives the ratio of the opposite side
to the hypotenuse, so you can construct a right
triangle with a hypotenuse and a leg that match the
ratio. Then you can use the Pythagorean Theorem to
find the length of the other leg, and use it to write
the cosine ratio.
Trigonometric Ratios of Complementary Angles
If ∠A and ∠B are the acute angles in a right triangle, then sin A = cos B and cos A = sin B.
Therefore, if θ (“theta”) is the measure of an acute angle, then sinθ° = cos (90 - θ)° and cos θ° = sin (90 - θ)°.
(90 - θ°)
A
Do you think it is possible for the value of a
sine or cosine to be greater than 1? Why or
why not? It is not possible; because the hypotenuse
is the longest side of a right triangle, any ratio that
has the length of the hypotenuse as the
denominator will be less than 1.
θ°
B
C
You can use these relationships to write equivalent expressions.
Example 2

Write each trigonometric expression.
Given that sin 38° ≈ 0.616, write the cosine of a complementary angle in terms of
the sine of 38°. Then find the cosine of the complementary angle.
Use an expression relating trigonometric ratios of complementary angles.
sin θ° = cos(90 - θ)°
EXPLAIN 2
Substitute 38 into both sides.
sin 38° = cos(90 - 38)°
Simplify.
sin 38° = cos 52°
Using Complementary Angles
Substitute for sin 38°.
0.616 ≈ cos 52°
QUESTIONING STRATEGIES
Why do the sine and cosine have a
complementary angle relationship? The ratios
are for a right triangle. Since one angle must be a
right angle, the other two angles must be
complementary because the sum of the measures of
the angles of a triangle is 180°.
How can you write equivalent expressions for
sin x° using cosine and cos y° using sine?
Explain. Use the complement to write
sin x° = cos (90° - x°) and cos y° = sin (90° - y°).
How can the relationship between the sine and
solve equations involving sines and cosines? Apply
the fact that the total measure of the angles is 90° to
set up an equation to solve for unknown values.
943
Lesson 18.2
© Houghton Mifflin Harcourt Publishing Company
So, the cosine of the complementary angle is about 0.616.

Given that cos 60° = 0.5, write the sine of a complementary angle in terms of
the cosine of 60°. Then find the sine of the complementary angle.
Use an expression relating trigonometric ratios of complementary angles.
cos θ° = sin(90 - θ)°
(
Substitute 60 into both sides.
cos 60 ° = sin 90 - 60
Simplify the right side.
cos 60 ° = sin 30 °
Substitute for 60 °.
0.5 = sin 60 °
)°
So, the sine of the complementary angle is 0.5.
Module 18
943
Lesson 2
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M18L2 943
Small Group Activity
18/04/14 11:33 PM
Have students work in small groups to investigate the relationship between the
size of an angle and its sine, using geometry software or graph paper, rulers, and
protractors. They should draw several triangles with unit side lengths and an angle
that increases. Those using graph paper will need to measure with protractor and
ruler. Have them determine how the sine changes as the angle size increases; then
repeat for cosine. Have students share their results and how they drew their
conclusions.
Reflect
COLLABORATIVE LEARNING
6.
What can you conclude about the sine and cosine of 45°? Explain.
sin 45° = cos 45°; 45° is complementary to itself.
7.
Discussion Is it possible for the sine or cosine of an acute angle to equal 1? Explain.
No; the hypotenuse of a right triangle is always longer than its legs, so the side ratios
Have students construct right triangles with unit side
lengths on graph paper, then use a protractor to
measure each of the acute angles to the nearest
degree. Have students verify the complementary
relationship between the sine and cosine for their
triangles for each of the acute angles, and then share
their results with the class.
defining sine and cosine must always be less than 1.
Write each trigonometric expression.
8.
Given that cos 73° ≈ 0.454, write the sine of a complementary angle.
EXPLAIN 3
θ = 73°, so 90 - θ = 27°.
sin 27° ≈ 0.454
9.
Finding Side Lengths using Sine and
Cosine
Given that sin 45° ≈ 0.707, write the cosine of a complementary angle.
θ = 45°, so 90 - θ = 45°.
cos 45° ≈ 0.707
Explain 3
INTEGRATE TECHNOLOGY
Finding Side Lengths using Sine and Cosine
It may be helpful to review with students how to
evaluate expressions of the form asin (x°) and
bcos (y°), with given values for a, b, x, and y, using
their calculators.
You can use sine and cosine to solve real-world problems.
Example 3
A 12-ft ramp is installed alongside some steps to provide wheelchair
access to a library. The ramp makes an angle of 11° with the ground. Find
each dimension, to the nearest tenth of a foot.
C
12 ft
B
y
Find the height x of the wall.
Multiply both sides by 12.
A
length of leg opposite ∠A
AB
sin A = ___ = _
AC
length of hypotenuse
Use the definition of sine.
Substitute 11° for A, x for BC, and 12 for AC.
11°
© Houghton Mifflin Harcourt Publishing Company

wall x
x
sin 11° = _
12
12sin 11° = x
QUESTIONING STRATEGIES
Why is a trigonometric ratio useful in solving
a real-world problem involving right
triangles? Sample answer: It makes it possible to
use known lengths and angle measures to find
unknown lengths that might be difficult to measure.
If you use a trigonometric ratio, such as sine
or cosine, to find the length of one of the legs
of a right triangle, do you have to use a trigonometric
ratio to find the length of the other leg as well?
Explain. No, you could also use the Pythagorean
Theorem to find the other leg since you will know
the lengths of the hypotenuse and one leg.
x ≈ 2.3
Use a calculator to evaluate the expression.
So, the height of the wall is about 2.3 feet.
Module 18
944
Lesson 2
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M18L2 944
Critical Thinking
18/04/14 11:32 PM
sinx
Discuss how to use the trigonometric ratios to prove that tan x = ____
students if they find this surprising and why or why not.
Sine and Cosine Ratios
944
B
EXPLAIN 4
Find the distance y that the ramp extends in front of the wall.
Substitute 11 ° for A, y for AB, and 12 for AC.
Finding Angle Measures using Sine
and Cosine
Multiply both sides by 12 .
y
cos 11 ° = _
12
12 cos 11 ° = y
y ≈ 11.8
Use a calculator to evaluate the expression.
QUESTIONING STRATEGIES
_
length of leg adjacent to ∠A
cos A = ___ = AB
AC
length of hypotenuse
Use the definition of cosine.
So, the ramp extends in front of the wall about 11.8 feet.
In the equation y = sin -1 x, explain what x
and y represent. In the equation, x represents
the sine of the angle with measure y°.
Reflect
_
10. Could you find the height of the wall using the cosine? Explain.
x
and
Since ∠A and ∠B are complementary, m∠C = 79°. Then cos 79° =
12
x = 12cos 79° ≈ 2.3 ft.
11. Suppose a new regulation states that the maximum angle of a ramp for wheelchairs
is 8°. At least how long must the new ramp be? Round to the nearest tenth of a foot.
C
2.3 ft
z
wall
8°
B
A
The ramp must be at least long enough to create an 8° angle at ∠A.
BC
2.3
⇒ sin 8° ≈ _
sin A =
z
AC
z ≈ 16.5
© Houghton Mifflin Harcourt Publishing Company
_
The ramp must be at least 16.5 ft long.
Explain 4
Finding Angle Measures using Sine and Cosine
5 =_
1 . However, you already know that 30° = _
1 . So you can
In the triangle, sinA = _
10
2
2
conclude that m∠A = 30°,
1 = 30°.
and write sin -1 _
2
Extending this idea, the inverse trigonometric ratios for sine and cosine are
A
defined as follows:
( )
10
C
5
B
Given an acute angle, ∠A,
• if sin A = x, then sin -1 x = m∠A, read as “inverse sine of x”
• if cos A = x, then cos -1 x = m∠A, read as “inverse cosine of x”
You can use a calculator to evaluate inverse trigonometric expressions.
Module 18
945
Lesson 2
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M18L2 945
Connect Vocabulary
Distinguishing between sine and cosine may be challenging for some students.
Explain that the prefix co- can mean together, as it does in the word cooperate.
Point out that the cosine ratio for an acute angle of a triangle involves the adjacent
leg. Tell students to remember this by thinking of the adjacent leg as “coming
together” with the hypotenuse from the angle.
945
Lesson 18.2
18/04/14 11:32 PM
Example 4

Find the acute angle measures in △PQR, to the nearest degree.
P
QUESTIONING STRATEGIES
Write a trigonometric ratio for ∠R.
Since the lengths of the hypotenuse and the opposite leg are given,
PQ
use the sine ratio.
sinR = _
PR
7
Substitute 7 for PQ and 13 for PR.
sinR = _
13

7
Q
How could you evaluate the inverse sine or
cosine of an angle without using the
calculator’s inverse trigonometric keys or a table of
values? Use the measurements of the sine or cosine
ratio in the Pythagorean Theorem to find the length
of the missing leg. Then draw a right triangle with
the side lengths and use a protractor to measure the
angle formed by the hypotenuse and the opposite
side or the adjacent side to estimate the angle
measure.
13
R
Write and evaluate an inverse trigonometric ratio to find m∠R and m∠P.
7
__
sinR = 13
7
__
Use the definition of the inverse sine ratio.
m∠R = sin -1 13
Use a calculator to evaluate the inverse sine ratio.
m∠R = 33 °
Write a cosine ratio for ∠P.
cosP =
Substitute 7 for PQ and 13 for PR.
cosP =
Use the definition of the inverse cosine ratio.
m∠P = cos - 1
Use a calculator to evaluate the inverse cosine ratio.
m∠P = 57 °
PQ
_
PR
7
__
13
7
__
AVOID COMMON ERRORS
13
Remind students about the notation used to represent
the inverse sine and cosine: sin -1 x and cos -1 x. As
with the inverse tangent, the –1 is not an exponent.
Rather, it denotes the inverse trigonometric ratio
whose value is an angle with a degree measure.
Reflect
12. How else could you have determined m∠P?
∠P and ∠R are complementary. Therefore, m∠P = 90° - m∠R ≈ 90° - 33° = 57°.
Find the acute angle measures in △XYZ, to the nearest degree.
XY = _
14
Z
X
cos Y = _
13. m∠Y
YZ 23
14
m∠Y = cos -1
23
14
23
14. m∠Z m∠Z = 90° - m∠Y ≈ 90° - 37° = 53°
Y
Elaborate
B
hypotenuse
15. How are the sine and cosine ratios for an acute angle of
a right triangle defined?
sin A =
opposite
BC
AC
__
= _ and cos A = __ = _
hypotenuse
Module 18
IN2_MNLESE389847_U7M18L2 946
AB
hypotenuse
946
AB
A
ELABORATE
© Houghton Mifflin Harcourt Publishing Company
(_ ) ≈ 37°
INTEGRATE TECHNOLOGY
Have students graph y = sin x and y = cos x on their
graphing calculators on the interval [0°, 90°]. Discuss
similarities and differences.
opposite
C
Lesson 2
18/04/14 11:32 PM
Sine and Cosine Ratios
946
16. How are the inverse sine and cosine ratios for an acute angle of a right triangle defined?
Because the sine ratio for a given acute angle is always the same, the measure of that angle
QUESTIONING STRATEGIES
can be defined as an inverse sine ratio:
BC
BC
→ m∠A = sin -1
sin A =
or sin A = x → m∠A = sin -1 x
AB
AB
Similarly,
AC
AC
→ m∠A = cos -1
cos A =
or cos A = y → m∠A = cos -1 y
AB
AB
(_ )
_
How are cosine and the inverse cosine
related? The cosine of an acute angle of a
right triangle is the ratio of the lengths of the
adjacent side to the hypotenuse of the right
triangle. The inverse cosine is the angle with that
cosine ratio.
(_ )
_
17. Essential Question Check-In How do you find an unknown angle measure in a right triangle?
First, use two known side lengths to form a ratio. Then use the appropriate inverse
trigonometric ratio to find the angle measure.
SUMMARIZE THE LESSON
Evaluate: Homework and Practice
How do you decide which trigonometric ratio
to use to find a missing side length or angle
measure in a right triangle? Use sine or inverse sine
for opposite side and hypotenuse relationships and
cosine or inverse cosine for adjacent side and
hypotenuse relationships.
• Online Homework
• Hints and Help
• Extra Practice
Write each trigonometric expression. Round trigonometric ratios to the
nearest thousandth.
1.
Given that sin 60° ≈ 0.866, write the
cosine of a complementary angle.
Given that cos 26° ≈ 0.899, write the
sine of a complementary angle.
2.
sin 60° = cos(90° - 60°)
cos 26° = sin(90° - 26°)
0.866 ≈ cos 30°
0.899 ≈ sin 64°
Write each trigonometric ratio as a fraction and as a decimal,
rounded (if necessary) to the nearest thousandth.
B
© Houghton Mifflin Harcourt Publishing Company
13
A
Module 18
IN2_MNLESE389847_U7M18L2 947
947
Lesson 18.2
4. cos A
5
AC
cos A =
=
=
≈ 0.385
13
AB
hypotenuse
5. cos B
cos B =
AB
hypotenuse
13
__ _ _
__
12
_
≈ 0.9231
13
hypotenuse
C
25
D
sin A =
12
5
opposite
BC
12
__
= _ = _ ≈ 0.923
3. sin A
24
F
6. sin D
cos A =
opposite
EF
7
__
= _ = _ = 0.28
7. cos F
cos F =
EF
7
__
= _ = _ = 0.28
8. sin F
sin F =
opposite
24
__
= _ = 0.96
7
E
947
hypotenuse
hypotenuse
hypotenuse
DF
DF
25
25
25
Lesson 2
18/04/14 11:32 PM
Find the unknown length x in each right triangle, to the nearest tenth.
9.
10.
C
EVALUATE
D
37°
15
27
B
F
A
x
EF
cos E = _
DE
x
x
cos 53° = _ ⇒ 16.2 ≈ x
AB
_
AC
x
sin 37° = _ ⇒ 9.0 ≈ x
sin C =
ASSIGNMENT GUIDE
53°
27
E
15
11.
12.
J
K
14
L
R
19
x
75°
JL
_
KL
14
cos 75° = _ ⇒ ≈ 54.1
24°
P
cos L =
Q
x
PR
_
PQ
19
sin 24° = _ ⇒ ≈ 46.7
sin P =
x
x
Find each acute angle measure, to the nearest degree.
11
V
P
W
18
8
15
Q
U
13. m∠P
8
PR
cos P = _ = _
18
PQ
8
m∠P = cos (_) ≈ 64°
-1
VW
11
sin U = _ = _
15
UW
11
_
m∠U = sin ( ) ≈ 47°
Exercise
16. m∠W
m∠W = 90° - m∠U ≈ 90° - 47° = 43°
Explore
Investigating Ratios in a Right
Triangle
Exercise 17
Example 1
Finding the Sine and Cosine of an
Angle
Exercises 3–8
Example 2
Using Complementary Angles
Exercises 1–2
Example 3
Finding Side Lengths using Sine and
Cosine
Exercises 9–12
Example 4
Finding Angle Measures using Sine
and Cosine
Exercises 13–16
Remind students that their calculators must be set on
degrees, not radians, to get the correct values for the
trigonometric ratios.
Communicating Math
Discuss the importance of being able to draw and
interpret right triangle diagrams to use trigonometric
ratios to solve real-world problems.
Lesson 2
948
Depth of Knowledge (D.O.K.)
Practice
AVOID COMMON ERRORS
15
Module 18
IN2_MNLESE389847_U7M18L2 948
m∠Q = 90° - m∠P ≈ 90° - 64° = 26°
18
15. m∠U
-1
14. m∠Q
© Houghton Mifflin Harcourt Publishing Company
R
Concepts and Skills
COMMON
CORE
Mathematical Practices
1–16
1 Recall of Information
MP.4 Modeling
17–24
2 Skills/Concepts
MP.4 Modeling
25
3 Strategic Thinking
MP.3 Logic
26
3 Strategic Thinking
MP.2 Reasoning
27
3 Strategic Thinking
MP.3 Logic
18/04/14 11:32 PM
Sine and Cosine Ratios
948
AVOID COMMON ERRORS
Students may have difficulty solving equations such
a when the variable is in the denominator.
as sin x = __
c
Review with students how to use inverse operations
to write an equivalent equation with the variable in
a
the numerator, such as c = _____
sin x.
17. Use the property that corresponding sides of similar triangles are
proportional to explain why the trigonometric ratio sin A is the same
when calculated in △ADE as in △ABC.
C
By AA ~, △ABC~△ADE, so corresponding sides are proportional:
(BC)(AE)
AC
BC
AC
AE
=
⇒ AC =
⇒
=
A
AE
DE
DE
DE
BC
B
E
∠A ≅ ∠A and ∠ABC ≅ ∠ADE, since both are right angles.
_ _
_ _ _
D
These ratios determine sin A, and since they are equal, sin A is the same
when calculated in either right triangle.
18. Technology The specifications for a laptop computer describe its
screen as measuring 15.6 in. However, this is actually the length of
a diagonal of the rectangular screen, as represented in the figure.
How wide is the screen horizontally, to the nearest tenth of an inch?
INTEGRATE TECHNOLOGY
In addition to calculators, students can use geometry
trigonometric ratios.
cos P =
R
15.6 in
PQ
PQ
_
⇒ cos29° = _ ⇒ 13.6 in. = PQ
P
15.6
PR
19. Building Sharla’s bedroom is directly under the roof of her
house. Given the dimensions shown, how high is the ceiling at its
highest point, to the nearest tenth of a foot?
sin(m∠AEB) =
AB
AB
_
⇒ sin 22° = _ ⇒ AB ≈ 4.0
10.6
AE
Maximum height: AC = AB + BC ≈ 4.0 + 5.5 = 9.5 ft
29°
Q
A
10.6 ft
roof
B
E
22°
5.5 ft
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
C
20. Zoology You can sometimes see an eagle gliding with its wings flexed in
a characteristic double-vee shape. Each wing can be modeled as two right
triangles as shown in the figure. Find the measure of the angle in the middle
of the wing, ∠DHG to the nearest degree.
cos(m∠DHE) =
32
EH
_
=_
49
32
m∠DHE = cos -1
≈ 49.2°
49
38
FG
sin(m∠FHG) =
=
43
GH
38
≈ 62.1°
m∠FHG = sin -1
43
m∠DHG = m∠DHE + m∠FHG ≈
DH
(_)
_ _
(_)
H
43 cm
32 cm
49.2 + 62.1 = 111.3°
Module 18
IN2_MNLESE389847_U7M18L2 949
949
Lesson 18.2
D
D
949
49 cm
F
38 cm
G
E
Lesson 2
18/04/14 11:32 PM
21. Algebra Find a pair of acute angles that satisfy the equation sin(3x + 9) = cos(x +
COGNITIVE STRATEGIES
The expressions must be the measures of two complementary angles:
Discuss how students can use the facts that sin 0° = 0
and the sine increases as the angle increases (between
0° and 90°), and cos 0° = 1 and the cosine decreases
as the angle increases (between 0° and 90°), as a quick
check when they evaluate sine and cosine
trigonometric ratios.
(3x + 9) + (x + 5) = 90 ⇒ x = 19
(3x + 9)° = (3(19) + 9)° = 66° and (x + 5)° = ((19) + 5)° = 24°
Check: 66° + 24° = 90°
22. Multi-Step Reginald is planning to fence his back yard. Every side
of the yard except for the side along the house is to be fenced, and
fencing costs \$3.50/yd. How much will the fencing cost?
(23 - 13)
LM
sin(m∠LKM) =
⇒ sin 32° =
⇒ LK ≈ 18.9
LK
LK
distance to fence: JK + KL + KN ≈ 13 + 18.9 + 23 = 54.9 yd
_
_
J
K
32°
yard
house
cost of fencing: about 54.9(3.50) = \$192.15
13 yd
N
M
23 yd
L
23. Architecture The sides of One World Trade Center in New York City form
eight isosceles triangles, four of which are 200 ft long at their base BC. The
length AC of each sloping side is approximately 1185 ft.
A
B
200 ft
Utech/Alamy
1185 ft
C
Find the measure of the apex angle BAC of each isosceles
_triangle, to the
nearest tenth of a degree. (Hint: Use the midpoint D of BC to create two
right triangles.)
1
In △ABD, AB = 1185 ft and BD = (200 ft) = 100 ft. Therefore,
2
-1 100
100
BD
=
≈ 4.8°
1185
1185
_ _
Module 18
IN2_MNLESE389847_U7M18L2 950
_
(_)
950
Lesson 2
18/04/14 11:32 PM
Sine and Cosine Ratios
950
PEERTOPEER DISCUSSION
H.O.T. Focus on Higher Order Thinking
Have students investigate the following identities by
evaluating them for different angles:
sin -1(sin x) = x and cos -1(cos x) = x.
_
24. Explain the Error Melissa has calculated the length of XZ in △XYZ. Explain why
Melissa’s answer must be incorrect, and identify and correct her error.
XZ
cos X = _
XY
XY
XZ = _
cos X
XZ = 27 cos 42° ≈ 20.1
JOURNAL
Have students describe how they remember the
difference between the sine and cosine ratios, and the
relationship between the ratios. Remind them to
include at least one figure and at least one example in
their descriptions.
20.1 < 27, but the hypotenuse should
be the longest side. Her definition of
cosine was inverted.
27
XY
⇒ XZ =
≈ 36.3
cos X =
XZ
cos 42°
Z
Melissa’s solution:
_
42°
X
_
Y
27
25. Communicate Mathematical Ideas Explain why the sine and
cosine of an acute angle are always between 0 and 1.
0
BC
AB
<
<
⇒ 0 < sinA < 1
0 < BC < AB ⇒ AB
AB
AB
The same argument shows that 0 < cos A < 1.
B
_ _ _
A
_
26. Look for a Pattern In △ABC, the hypotenuse AB has a length of 1.
Use the Pythagorean Theorem to explore the relationship between the
squares of the sine and cosine of ∠A, written sin 2 A and cos 2A. Could
you derive this relationship using a right triangle without any lengths
specified? Explain.
BC
BC
AC
AC
sin A = ___
= ___
= BC and cos A = ___
= __
= AC
1
AB
AB
1
C
B
1
A
C
By the Pythagorean Theorem, AC 2 + BC 2 = AB 2
© Houghton Mifflin Harcourt Publishing Company
(sin A) 2 + (cos A) 2 = 1 2
sin 2 A + cos 2 A = 1
BC
Yes: sin 2 A + cos 2 A =
AB
AC
(_) + (_
AB )
2
2
=
BC + AC
AB
_
=_=1
2
2
AB 2
2
AB 2
27. Justify Reasoning Use the Triangle Proportionality Theorem
to explain why the trigonometric ratio cos A is the same when
calculated in △ADE as in △ABC.
E
C
Two segments ⊥ to the same line are ∥ to each other,
¯ ∥ DE
¯
¯. By the Triangle Proportionality Theorem, BC
so BC
¯ and AE
A
B
D
AB + BD
AC + CE
CE
CE
BD
BD
AB
AE
__
= __
⇒ 1 + __
= 1 + __
⇒ ______ = ______ ⇒ __
= __
⇒ __
= __
AB
AC
AB
AC
AB
AC
AB
AC
AE
AC
These ratios determine cos A, and since they are equal, cos A is the same when
calculated in either ⇒ triangle.
Module 18
IN2_MNLESE389847_U7M18L2 951
951
Lesson 18.2
951
Lesson 2
18/04/14 11:32 PM
As light passes from a vacuum into another medium, it is refracted—that is, its
direction changes. The ratio of the sine of the angle of the incoming incident ray,
I, to the sine of the angle of the outgoing refracted ray, r, is called the index of
refraction:
AVOID COMMON ERRORS
incident ray
Question 4 in the Lesson Performance Task identifies
an error commonly made by students in working
with the trigonometric functions. Here is a similar
error:
gem
refracted ray
sin I
n = ___
. where n is the index of refraction.
sin r
This relationship is important in many fields, including gemology, the study
of precious stones. A gemologist can place an unidentified gem into an instrument called a
refractometer, direct an incident ray of light at a particular angle into the stone, measure the
angle of the refracted ray, and calculate the index of refraction. Because the indices of refraction
of thousands of gems are known,
the gemologist can then identify the gem.
Gem
1.
Identify the gem, given these angles obtained from a refractometer:
a. I = 71°, r = 29°
b. I = 51°, r = 34°
I = 45°, r = 17°
c.
2.
3.
4.
An incident ray of light struck a slice of serpentine. The resulting
angle of refraction measured 21°. Find the angle of incidence to the
nearest degree.
2.94
Diamond
2.42
Zircon
1.95
Azurite
1.85
Sapphire
1.77
Tourmaline
1.62
Serpentine
1.56
Coral
1.49
Describe the error(s) in a student’s solution and explain why they
Opal
1.39
were error(s):
sin
I
sin(56°)
n=_
2. 1.77 = ______
sin r
sin(x°)
sin 51°
=_
sin(56°)
______
sin 34°
sin(x°) = 1.77
51°
=_
34°
sin(56°)
x = sin -1 ______
= 28°
1.77
= 1.5 → coral
a.
sin(71°)
______
= 1.95
sin(29°)
3.
zircon
c.
sin(45°)
______
= 2.42
sin(17°)
sin(x°)
1.56 = ______
sin(21°)
)
x = sin -1(1.56 ⋅ sin(21°)) =34°
sin(51°)
b. ______ = 1.39 opal
sin(34°)
4.
Index of
Refraction
Hematite
(
sin 10°
10°
Explain to students that the function and the angle
cannot be separated. The correct solution is:
diamond
Possible answer: The student divided out the word “sin” from the
sin 50° ≈ _______
0. 7660 ≈ 4.41
____
sin 10°
0.1736
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Using a refractometer, a gemologist found
that the index of refraction of a substance was 2.0.
Using the same angle of incidence, the gemologist
found that the angle of the refracted ray for a second
substance was greater than the angle of the refracted
ray for the first substance. Was the index of refraction
of the second substance greater or less than 2.0?
Explain. Sample answer: Let i = angle of incidence,
r 1 = refraction angle 1, and r 2 = refraction angle 2.
sin i . Since the sine function increases as
Then, 2 = _____
sin r 1
an angle increases from 0° to 90°, sin r 2 > sin r 1
sin i , the index of
sin i < _____
sin i = 2. So, _____
and _____
sin r 2
sin r 2
sin r 1
refraction of the second substance, was less than 2.
© Houghton Mifflin Harcourt Publishing Company
1.
A thin slice of sapphire is placed in a refractometer. The angle of the
incident ray is 56°. Find the angle of the refracted ray to the nearest
degree.
sin 50° = sin ____
50° = sin 5° ≈ 0.0872
______
numerator and the denominator in Step 2. This was incorrect because
“sin 51°” and “sin 34°” are indivisible units, with each representing a
single real number.
Module 18
952
Lesson 2
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M18L2 952
Physicists define the index of refraction of a material in terms of the speed of light.
Research the formula used by physicists that relates the index of refraction of a
substance, n, the speed of light in a vacuum, c, and the speed of light through the
c
substance, v. n = __
v
The speed of light in azurite is approximately 100,693 miles per second. Use that
fact and the table in the Lesson Performance Task to find the speed of light in a
vacuum. Express your answer in miles per second and miles per hour.
186,282 mi/sec; 670,615,200 mi/hr
The Sun is about 93 million miles from Earth. How long does it take light from
the Sun to reach Earth? about 8.3 min
18/04/14 11:32 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Sine and Cosine Ratios
952
LESSON
18.3
Name
Special Right Triangles
Class
Date
18.3 Special Right Triangles
Essential Question: What do you know about the side lengths and the trigonometric ratios in
special right triangles?
Common Core Math Standards
The student is expected to:
COMMON
CORE
Resource
Locker
G-SRT.C.8
Explore 1
Use trigonometric ratios and the Pythagorean Theorem to solve right
triangles in applied problems.
Discover relationships that always apply in an isosceles right triangle.
A
Mathematical Practices
COMMON
CORE
The figure shows an isosceles right triangle. Identify the base angles, and
use the fact that they are complementary to write an equation relating their
x
∠A, ∠B; m∠A + m∠B = 90°
Language Objective
Explain to a partner how to find the sine, cosine, and tangent of a
30° -60° -90° triangle or a 45° -45° -90° triangle.
B
Use the Isosceles Triangle Theorem to write a different equation relating the
base angle measures.
C
Use the Pythagorean Theorem to find the length of the hypotenuse in terms of the length of
each leg, x.
AB = x 2 + x 2
―
√3
tan 30° =
,
3
_
PREVIEW: LESSON
© Houghton Mifflin Harcourt Publishing Company
2
_
2
What must the measures of the base angles be? Why?
D
The side lengths of a 45° -45° -90° triangle are
―
always in the ratio 1: 1: √2 . The trigonometric
―
√2
ratios associated with this triangle are sin 45° =
2
and tan 45° = 1. The side lengths of a 30° -60°
―
-90° triangle are always in the ratio 1: √3 : 2. The
trigonometric ratios associated with this triangle
are
_1
2
―
√3
sin 60° = cos 30° = _,
Reflect
1.
Is it true that if you know one side length of an isosceles right triangle, then you know all the side lengths?
Explain.
Yes; If you know either leg length x, then the other leg length is also equal to x, and the
―
length, then each leg length is this value divided by √―
2.
length of the hypotenuse is this value multiplied by √2 . If you know the hypotenuse
2.
_
What if? Suppose_
you draw the perpendicular from C to AB. Explain how to
find the length of CD.
Since △ABC is isosceles, m∠A = 45°, and since △ADC is a right
triangle, m∠ACD = 90° - m∠A = 45°. Therefore △ADC is an
_
√―
2
1
isosceles right triangle, so CD = ___
.
― or ___
2
√2
Module 18
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
D
A
1
C
1
Lesson 3
953
gh "File info"
B
Date
Class
s
ht Triangle
Name
al Rig
18.3 Speci
in
ic ratios
trigonometr
s and the
side length
es in
do you know
right triangl
ion: What
m to solve
triangles?
orean Theore
special right
the Pythag
ratios and
Quest
Essential
1
Explore

apply
es right triang
in an isoscel
le.
that always
HARDCOVER PAGES 953966
B
x
A
Watch for the hardcover
student edition page
numbers for this lesson.
C
x


be? Why?
base angles
m∠A = 45°.
res of the
= 90°, so
the measu
n, 2m∠A
What must
substitutio
length of
= 45°; using
terms of the
enuse in
m∠A = m∠B
of the hypot
the length
em to find
gorean Theor
√2
Use the Pytha
AB = x
2
x.
leg,
2
each
AB = 2x
2
2
2
+x
s?
AB = x
the side length
know all
le, then you
les right triang
x, and the
of an isosce
Reflect
equal to
one side length
h is also
e
if you know
leg lengt
true that
hypotenus
the other
1. Is it
know the
h x, then
√2 . If you
r leg lengt
Explain.
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know eithe
multi
you
If
value
Yes;
this
by √2 .
tenuse is
divided
the hypo
this value
length of
_
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length is
AB. Explai
each leg
from C to
length, then
ndicular
D
the perpe
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C is a right
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2. What length of CD.
= 45°, and
an
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find the
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Therefore
A
1
Since △ABC
m∠A = 45°. ―2
√
= 90° - _
D
___
.
1
m∠AC
___
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= ― or 2
le, so CD √2
right triang
les
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―
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IN2_MNLESE389847_U7M18L3 953
g an
Investigatin
Right Tria
Isosceles
, and
the base angles g their
le. Identify
relatin
right triang
equation
isosceles
to write an
shows an
ementary
The figure
are compl
that they
use the fact
90°
+ m∠B =
measures.
∠A, ∠B; m∠A
relating the
equation
nt
differe
to write a
le Theorem
les Triang
Use the Isosce res.
measu
m∠A = m∠B
base angle relationships
Discover
Resource
Locker
ngle
trigonometric
G-SRT.C.8 Use
ms.
applied proble
View the Engage section online. Discuss the
photograph, asking students to speculate on whether
the Lesson Performance Task will involve the dog or
the flying disc, and why. Then preview the Lesson
―
AB = x √2
AB 2 = 2x 2
COMMON
CORE
B
1
C
Lesson 3
953
Module 18
8L3 953
47_U7M1
ESE3898
IN2_MNL
Lesson 18.3
C
x
m∠A = m∠B = 45°; using substitution, 2m∠A = 90°, so m∠A = 45°.
Essential Question: What do you
know about the side lengths and the
trigonometric ratios in special right
triangles?
sin 45° = cos 60° = ,
A
m∠A = m∠B
ENGAGE
953
B
measures.
MP.2 Reasoning
―
and tan 60° = √3 .
Investigating an Isosceles Right Triangle
18/04/14
11:49 PM
18/04/14 11:50 PM
Explore 2
Investigating Another Special Right Triangle
EXPLORE 1
Discover relationships that always apply in a right triangle formed as half of an equilateral triangle.
A
_
_
△ABD is an equilateral triangle and BC is a perpendicular from B to AD.
Determine all three angle measures in △ABC.
m∠C = 90°; each angle in an equilateral triangle measures 60°,
B
Investigating an Isosceles Right
Triangle
so m∠A = 60°, and therefore m∠ABC = 90° - m∠A = 30°.
B
A
Explain why △ABC ≅ △DBC.
INTEGRATE TECHNOLOGY
D
C
∠A ≅ ∠D (Equilateral Triangle Theorem), ∠ACB ≅ ∠DCB (all right angles are congruent),
_ _
and BC ≅ BC (Reflexive Property of Congruence), so △ABC ≅ △DBC by AAS Congruence.
_ _
_ _
Or, AB ≅ DB (△ABD equilateral), BC ≅ BC (Reflexive Property of Congruence), and since
Students have the option of doing the Explore activity
either in the book or online.
△ABC and △DBC are right triangles, △ABC ≅ △DBC by HL Congruence.
C
_
_
Let the length of AC be x. What is the length of AB, and why?
_ _
From Step B, △ABC ≅ △DBC, so AC ≅ DC (CPCTC) and therefore
CD = AC = x. In △ABD, AD = AC + DC = x + x = 2x (Seg. Add. Post.);
_ _
D
_
Using the Pythagorean Theorem, find the length of BC.
(2x) 2 = x 2 + BC 2
QUESTIONING STRATEGIES
B
A
x
How can you use the relationship between the
angle measure and the length of the sides to
right triangle? The legs are opposite the 45° angles
and measure 1 unit while the hypotenuse is
―
opposite the right angle and measures √2 units.
C
―
3x 2 = BC 2
x3 = BC
Reflect
3.
―
―
x : x3 : 2x, or 1 : 3 : 2.
4.
Explain the Error A student has drawn a right triangle with a 60° angle and a
hypotenuse of 6. He has labeled the other side lengths as shown. Explain how you
can tell at a glance that he has made an error and how to correct it.
―
Since 63 > 6, leg JK is longer than the hypotenuse, which is
―
impossible. The side lengths must be in the ratio 1 :
―
For this to be true, the length of JK should be 3
Module 18
―3 .
―3 : 2.


954
J
6
L
60°
3
6√3
K
EXPLORE 2
© Houghton Mifflin Harcourt Publishing Company
What is the numerical ratio of the side lengths in a right triangle with acute angles that measure 30° and
60°? Explain.
The right triangle is similar to △ABC (AA Similarity), so its side lengths are in the ratio
Investigating Another Special Right
Triangle
QUESTIONING STRATEGIES
How can you use the relationship between the
angle measure and the length of the side
simplest form of a 30° -60° -90° triangle? The
shortest side is opposite the 30° angle and the
longest side is the hypotenuse. So label the side
opposite the 30° angle 1 unit, the side opposite the
―
60° angle √3 units, and the hypotenuse 2 units.
Lesson 3
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M18L3 954
18/04/14 11:50 PM
Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.2,
which calls for students to “reason abstractly and quantitatively.” Students
investigate the relationships among the side lengths and angles of the special right
triangles, and use them to find the trigonometric ratios and angle measures
associated with these relationships. This recognition can often provide a quicker
solution to a problem involving a special triangle.
Special Right Triangles 954
Applying Relationships in Special Right Triangles
Explain 1
AVOID COMMON ERRORS
The right triangles you explored are sometimes called 45°-45°-90° and 30°-60°-90° triangles. In a 45°-45°-90°
triangle, the hypotenuse is 2 times as long as each leg. In a 30°-60°-90° triangle, the hypotenuse is twice as long as
the shorter leg and the longer leg is √3 times as long as the shorter leg. You can use these relationships to find side
lengths in these special types of right triangles.
―
Students may confuse the side lengths when labeling
the sides of a standard 30° -60° -90° triangle.
―
Review with students why √3 > 1 and why this side
length or its multiple must be opposite the 60° angle.
30°
45°
x√2
EXPLAIN 1
45°
Applying Relationships in Special
Right Triangles
Example 1

QUESTIONING STRATEGIES
√2
the length without the radical in the denominator
(
)
―
―
√2
√2
1
_
· _
= _ . Remind students this is called
―
―
2
√2
√2
x
Find the unknown side lengths in △ABC.
A
―
―
―2 = BC ―2
The hypotenuse is √2 times
as long as each leg.
AB = AC2 = BC2
Substitute 10 for AB.
10 = AC
―
45°
10

―
―2 = AC = BC
102 = 2AC = 2BC
Divide by 2.
C
45°
B
5
―
In right △DEF, m∠D = 30°and m∠E = 60°. The shorter leg measures 53 . Find the
remaining side lengths.
E
© Houghton Mifflin Harcourt Publishing Company
1
unit. The lengths are _
― . Review how to write
x√3
Find the unknown side lengths in each right triangle.
Multiply by 2 .

legs of an isosceles right triangle with a hypotenuse of
one
60°
x
_
If you know the exact value of the length of
one side of a special right triangle, can you
always find the exact value of the remaining sides?
Explain. Yes, because of the special side-length
relationships that exist in the special right triangles.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Work with students to find the lengths of the
2x
x
60°
5√3
D
30°
F
The hypotenuse is twice as long as the shorter leg.
DE = 2 EF
Substitute 5 √3 for
DE = 2 5 √3
―
EF .
The longer leg is
―
Substitute 5 √3 for
rationalizing the denominator.
―
―3
= EF
= 5 √―
3 3―
DE =
Simplify.
_
√3
―
times as long as the shorter leg.
EF .
DF


DF = 15
Simplify.
Module 18
DF
10 √3
955
Lesson 3
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M18L3 955
Small Group Activity
955
18/04/14 11:50 PM
Have students work in small groups to draw and label right triangles with the
following properties:
• isosceles right triangles with legs whose lengths are whole numbers greater than 1
• isosceles right triangles with a hypotenuse whose length is a whole number
greater than 1
• 30° -60° -90° triangles with a shortest leg whose length is a whole number
greater than 1
• 30° -60° -90° triangles with a hypotenuse whose length is a whole number
greater than 2
Lesson 18.3
EXPLAIN 2
Find the unknown side lengths in each right triangle.
L
5.
6.
30°
Q
Trigonometric Ratios of Special Right
Triangles
45°
4√3
J
KL = 2JK
―
―3 = JK
43 = 2JK
60°
JL = JK
K
P
―3
―3 ) ―3
2√6
―
If a trigonometric ratio includes √2 in the
ratio, which special right triangle does it likely
refer to? Explain. an isosceles right triangle because
that is the length of the hypotenuse
―
If a trigonometric ratio includes √3 in the
ratio, which special right triangle does it likely
refer to? Explain. a 30° -60° -90° triangle because
that is the length of the side opposite the 60° angle
R
―
_ _
PR ≅ QR , so QR = PR = 2 √6

JL = (2
QUESTIONING STRATEGIES
45°

PQ = PR
―2
―6 ) √―2 = 4 ―3

2
JL = 6
Explain 2
Trigonometric Ratios of Special Right Triangles
= (2

You can use the relationships you found in special right triangles to find trigonometric ratios for
the angles 45°, 30°, and 60°.
Example 2

For each triangle, find the unknown side lengths and trigonometric ratios
for the angles.
If you know the exact value of a trigonometric
ratio for a special right triangle, can you find
the measure of the angle that corresponds to the
ratio? Explain. Yes. Draw the corresponding special
right triangle and then find the angle that
corresponds to the trigonometric ratio.
A 45°-45°-90° triangle with a leg length of 1
Step 1
―
―
Since the lengths of the sides _
opposite the 45° angles are congruent, they are both 1. The
length of the hypotenuse is √2 times as long as each leg, so it is 1(2 ), or 2 .
√2
© Houghton Mifflin Harcourt Publishing Company
45°
1
45°
1
Step 2
Use the triangle to find the trigonometric ratios for 45°. Write each ratio as a simplified
fraction.
Angle
45°
Sine =
opp
_
hyp
―
√2
_
2
Module 18
Cosine =
―2
_
_
hyp

2
956
Tangent =
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Point out to students that the side lengths and
the trigonometric ratios for the special right triangles
are usually left in exact answer form. This means they
are often expressed as ratios in simplest form, with
rationalized denominators, when applicable.
opp
_
1
Lesson 3
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M18L3 956
Kinesthetic Experience
18/04/14 11:50 PM
Kinesthetic learners may benefit from verifying the relationships of an isosceles
right triangle through a simple paper-folding activity. Have students take a square
piece of paper and fold it in half diagonally. This creates an isosceles right triangle.
Then ask students to measure one leg of the triangle to the nearest millimeter, and
apply their understanding of the relationships among the legs and the hypotenuse
to predict the length of the hypotenuse. Finally, have students measure the length
of the hypotenuse to check their predictions.
Special Right Triangles 956
B
A 30°-60°-90° triangle with a shorter leg of 1
Step 1
30°
The hypotenuse is twice
√―
3
as long as the shorter leg,
2
so the length of the hypotenuse is
The longer leg is
.
2
√3
times as long as the shorter leg,
―3

so the length of the longer leg is
.
60°
1
Step 2
Use the triangle to complete the table. Write each ratio as a simplified fraction.
Angle
Sine =
opp
_
hyp
Cosine =
√―
3
_
_1
2
√―
3
_
30°
60°
_
hyp
Tangent =
√―
3
_
2
3
―
_1
2
opp
_
√3
2
Reflect
7.
Write any patterns or relationships you see in the tables in Part A and Part B as equations. Why do these
patterns or relationships make sense?
―
―
3
2
1
sin 30° = cos 60° = _
; sin 45° = cos 45° = ___
; sin 60° = cos 30° = ___
; the sine of an angle
2
2
2
equals the cosine of its complement.
© Houghton Mifflin Harcourt Publishing Company
8.
For which acute angle measures θ, is tanθ less than 1? equal to 1? greater than 1?
tan θ < 1 for θ < 45; tan θ = 1 for θ = 45°; tan θ > 1 for θ > 45°.
Find the unknown side lengths and trigonometric ratios for the 45 angles.
9.
10
CB
_
_
― , AB = ―2 = 5 ―2 ;
2
AC = AB, AC = 5 ―
2
10
sin 45° = cos 45° = _
―2 = 5 ―2 ;
C
AB =
45°
45°



10
B

tan 45° = 1


A
Module 18
957
Lesson 3
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M18L3 957
Connect Vocabulary
Point out that what makes special right triangles special are the unique
relationships among the sides. It may be easier for students to focus on the special
triangles by referring to them by their angle measures. Instead of discussing an
isosceles right triangle, refer to the triangle as a 45° -45° -90° triangle.
957
Lesson 18.3
18/04/14 11:50 PM
Explain 3
Investigating Pythagorean Triples
EXPLAIN 3
Pythagorean Triples
A Pythagorean triple is a set of positive integers a, b, and c
that satisfy the equation a 2 + b 2 = c 2. This means that a, b,
and c are the legs and hypotenuse of a right triangle.
Right triangles that have non-integer sides will not form
Pythagorean triples.
B
c
A
b
Examples of Pythagorean triples include 3, 4, and 5; 5, 12,
and 13; 7, 24, and 25; and 8, 15, and 17.
Example 3

Investigating Pythagorean Triples
a
C
QUESTIONING STRATEGIES
How can understanding the relationship
between a Pythagorean triple and its multiples
you can identify the triple and the multiple, you can
multiply the missing triple length by the multiple to
find the missing side length without having to use
the Pythagorean Theorem.
Use Pythagorean triples to find side lengths in right triangles.
Verify that the side lengths 3, 4, and 5; 5, 12, and 13; 7, 24, and 25; and 8, 15, and 17 are
Pythagorean triples.
3 2 + 4 2 = 9 + 16 = 25 = 5 √
5 2 + 12 2 = 25 + 144 = 169 = 13 2 √
7 2 + 24 = 49 + 576 = 625 = 25 2 √
8 2 + 15 2 = 64 + 225 = 289 = 17 2 √
2
2
The numbers in Step A are not the only Pythagorean triples. In the following steps you will discover that
multiples of known Pythagorean triples are also Pythagorean triples.

INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Have students create a list of Pythagorean
In right triangles DEF and JKL, a, b, and c form a Pythagorean triple, and k is a positive
integer greater than 1. Explain how the two triangles are related.
J
D
a
F
c
b
kc
ka
E
L
kb
© Houghton Mifflin Harcourt Publishing Company
△DEF is similar to △JKL by the Side-Side-Side (SSS) Triangle Similarity Theorem
because the corresponding sides are proportional. Complete the ratios to verify
Side-Side-Side (SSS) Triangle Similarity.
a : b : c = ka : kb : kc

You can use the Pythagorean Theorem to compare the lengths of the sides of ▵JKL. What
must be true of the set of numbers ka, kb, and kc?
2 2
(ka 2) + (kb 2) = k a + k 2b 2
( 2 2)
= k 2 a +b
= k 2c 2 =
triples. Students do not need to memorize every
triple on the list, but they can use it as a reference to
help them recognize the triples and to look for
multiples. Students may want to include side columns
with multiples to help them recognize the patterns
between the original triple and its multiples.
K
(kc) 2
The set of numbers ka, kb, and kc form a Pythagorean triple .
Module 18
IN2_MNLESE389847_U7M18L3 958
958
Lesson 3
18/04/14 11:50 PM
Special Right Triangles 958
Reflect
ELABORATE
10. Suppose you are given a right triangle with two side lengths. What would have to be true for you to use a
Pythagorean triple to find the remaining side length?
The given side lengths would have to be two numbers in a Pythagorean triple. Also, if the
AVOID COMMON ERRORS
legs are given, the side lengths would have to be the smaller two numbers in the triple,
Students may confuse Pythagorean triples. Emphasize
that students can always use the Pythagorean
Theorem to verify that a triple or a multiple of a triple
is correct.
whereas if one leg and the hypotenuse are given, the side lengths would have to include
the largest number in the triple.
Use Pythagorean triples to find the unknown side length.
11.
SUMMARIZE THE LESSON
What can you say about the side lengths and
the trigonometric ratios associated with
special right triangles? The sides of an isosceles
―
right triangle are in the ratio 1 - 1 - √2 . The
sides of a 30° -60° -90° triangle are in the ratio
―
1 - √3 - 2. You can find the related trigonometric
ratios by drawing the triangles and finding the sine,
cosine, and tangent of the corresponding 30°, 45°,
and 60° angles from the triangles.
R
_
12. In △XYZ, the hypotenuse
XY has length 68, and
_
the shorter leg XZ has length 32.
32 : 68 = 8 : 17, so the side lengths are
multiples of 8, 15, and 17.
24
P
18
8 : 15 : 17 = 4(8) : 4(15) : 4(17) = 32 : 60 : 68,
so YZ = 60.
Q
18 : 24 = 3 : 4, so the side lengths are
multiples of the Pythagorean triple 3, 4, 5.
3 : 4 : 5 = 6(3) : 6(4) : 6(5) = 18 : 24 : 30,
so PR = 30.
Elaborate
13. Describe the type of problems involving special right triangles you can solve.
Once you identify the side (longer leg, shorter leg, hypotenuse) that is given, you can
find the lengths of the other two sides by applying relationships such as the length of the
© Houghton Mifflin Harcourt Publishing Company
hypotenuse in a 30°-60°-90° triangle being twice as long as the length of the shorter leg.
14. How can you use Pythagorean triples to solve right triangles?
Suppose two side lengths of a right triangle are given and correspond to two numbers in
a Pythagorean triple. If one of the given sides is the hypotenuse and one of the lengths is
the largest number of the triple, then the unknown length is the remaining number in the
triple. Likewise, if neither given side is the hypotenuse and the lengths are the smaller two
numbers of the triple, the unknown side length must be the largest number of the triple.
15. Discussion How many Pythagorean triples are there?
Infinitely many; for example, since 3, 4, and 5 is a Pythagorean triple, so are
2(3) = 6, 2(4) = 8, and 2(5) = 10; 9, 12, and 15; 12, 16, and 20; and so on.
16. Essential Question Check-In What is the ratio of the length of the hypotenuse to the length of the
shorter leg in any 30°-60°-90° triangle?
The ratio is 2 to 1.
Module 18
IN2_MNLESE389847_U7M18L3 959
959
Lesson 18.3
959
Lesson 3
18/04/14 11:50 PM
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
For each triangle, state whether the side lengths shown are possible.
Explain why or why not.
1.
2.
3√3
45°
6√3
6
3√3
60°
3√3
6 √―
3
6
:_
=
No; 3 √―
3 : 6 : 6 √―
3=1:_
―
―
√
√
―
3
3
3
3
2 √3
1 : _ : 2 ≠ 1 : √―
3:2
― ―
_―
―
―
ASSIGNMENT GUIDE
Yes; 3 √3 : 3 √3 : 3 √6 =
3 √6
= 1 : 1 : √2
1:1:
3 √3
3
3.
3√6
―
4.
8√3
12
6
30°
45°
4√3
―
―
Yes; 4 √3 : 12 : 8 √3 = 1 :
―
1 : √3 : 2
2√3
8 √―
3
12
_
:_
=
―
4 √3 4 √―
3
―
1 : 1 : √―
3
―
2√3
No; 2 √3 : 2 √3 : 6 = 1 : 1 :
6
_
≠
2 √―
3
Find the unknown side lengths in each right triangle.
5.
6.
A
L
18
B
9
C
―
―
―
―
18 = AB √2 = BC √2
18 √―
2 = 2AB = 2BC
9 √―
2 = AB = BC
―
―
AC = AB √2 = BC √2
Module 18
Exercise
IN2_MNLESE389847_U7M18L3 960
30°
© Houghton Mifflin Harcourt Publishing Company
45°
JL = JK √3
9 = JK √3
―
√
3 ―
3 = JK
9 √3 = 3JK
J
KL = 2JK
K
―
KL = 2(3 √3 )
KL = 6 √3
―
Practice
Explore 1
Investigating an Isosceles Right
Triangle
Exercises 1–4
Explore 2
Investigating Another Special Right
Triangle
Exercises 5–8
Example 1
Applying Relationships in Special
Right Triangles
Exercises 9–12
Example 2
Trigonometric Ratios of Special
Right Triangles
Exercises 13–16
Example 3
Investigating Pythagorean Triples
Exercises 17–20
INTEGRATE TECHNOLOGY
Although students should write answers in exact
form for special right triangles, they can use a
calculator to check that their answers make sense.
Lesson 3
960
Depth of Knowledge (D.O.K.)
60°
Concepts and Skills
COMMON
CORE
Mathematical Practices
1–20
2 Skills/Concepts
MP.4 Modeling
21
2 Skills/Concepts
MP.4 Modeling
22–24
2 Strategic Thinking
MP.4 Modeling
25
3 Strategic Thinking
MP.3 Logic
26
3 Strategic Thinking
MP.3 Logic
18/04/14 11:50 PM
Special Right Triangles 960
AVOID COMMON ERRORS
7.
Students working with special right triangles can
mislabel diagrams when the side lengths differ from
the standard reference triangles. Remind students
they can refer to those triangles, but that they must
identify the appropriate relationship for each triangle
based on its corresponding problem statement.
Right triangle UVW has acute angles U
measuring 30°
and W measuring 60°.
_
Hypotenuse UW measures 12. (You may
30°
Q
U
45°
P
12 = 2VW
6 = VW
Have students construct several triangles with legs
whose measures are the legs of Pythagorean triples
using centimeter graph paper. Then have them use a
ruler to verify that the length of the hypotenuse
corresponds to the length of the hypotenuse in the
corresponding Pythagorean triple.
45°
―
― ―
= 5 √―
20
= 10 √―
5
R
5√10
PQ = PR √2
=(5 √10 ) √2
UV = VW √―
3
UV = 6 √―
3
UW = 2VW
MANIPULATIVES
V
Use trigonometric ratios to solve each right triangle.
9.
10.
D
A
14
45°
C
AC
_
AB
―
√2
AC
_
=_
14
2
7 √―
2 = AC
© Houghton Mifflin Harcourt Publishing Company
60°
E
B
10√3
_
―
_ _
―
BC
_
AB
―
√2
BC
_
=_
14
2
7 √―
2 = BC
F
_
_ _
―
―
cos 60° = DE
sin 60° = DF
EF
EF
√3
DF
DE
1
=
=
2
2
10 √3
10 √3
15 = DF
5 √3 = DE
cos 45° =
sin 45° =
―
11. Right △KLM with m∠J = 45°, leg JK = 4 √3
12. Right △PQR with m∠Q = 30°, leg QR = 15
Q
L
30°
45°
15
J
45°
4√3
K
P
KL
_
tan 45° = _
JK
―
―
4 √―
3
_ _
1=_
JK
JL = (4 √―
3 ) √―
2 = 4 √―
6
JK= 4 √―
3
sin 45° = JK
JL
√2
4 √3
=
2
JL
Module 18
IN2_MNLESE389847_U7M18L3 961
Lesson 18.3
Right triangle PQR has
P and Q
_ acute angles _
measuring 45°. Leg PR measures 5√10 . (You may
12
W 60°
961
8.
961
PR
_
QR
√―
3 _
PR
_
=
15
3
5 √―
3 = PR
tan 30° =
R
PR
_
PQ
5 √―
3
_1 = _
PQ
2
PQ =2(5 √―
3 ) = 10 √―
3
sin 30° =
Lesson 3
18/04/14 11:49 PM
COGNITIVE STRATEGIES
For each right triangle, find the unknown side length using a Pythagorean triple. If it
is not possible, state why.
13.
14.
K
Discuss the importance of being able to draw and
interpret right triangle diagrams to use trigonometric
ratios to solve real-world problems.
B
25
A
60
8
C
25:60 = 5:12, so the side lengths are
multiples of 5, 12, and 13.
J
5 : 12 : 13 = 5(5) : 5(12) : 5(13) = 25 : 60 : 65 ,
so AB = 65.
L
4
Not possible: 8 - 4 = 64 - 16 = 48;
48 is not a perfect square.
2
2
15. In right △PQR, the legs have lengths PQ = 9 and QR = 21.
Not possible: 9 2 + 21 2 = 81 + 441 = 522 ; 522 is not a perfect square.
_
_
16. In right △XYZ, the hypotenuse XY has length 35, and the shorter leg YZ has
length 21.
21 : 35 = 3 : 5, so the side lengths are multiples of 3, 4, and 5.
3 : 4 : 5 = 7(3) : 7(4) : 7(5) = 21 : 28 : 35 , so XZ = 28.
17.
E
12
12 D
x
F
H
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
G
DEFG is a rhombus, so its diagonals bisect each other, and EH = HG. Since
EH + HG = EG =12 (Seg. Add. Post.), 2EH = 12, and therefore EH = 6. By
the Pythagorean Thm.,
x 2 + 6 2 = 12 2
x 2 + 36 = 144
x 2 = 108
x = √108 = 6 √3 .
――
―
18. Represent Real-World Problems A baseball “diamond” actually forms a
square, each side measuring 30 yards. How far, to the nearest yard, must the third
baseman throw the ball to reach first base?
The line from first base to third base forms the diagonal of the
square, so with home plate they form a 45°-45°-90° triangle with
legs of length 30 yd. Therefore, the third baseman must throw the
ball 30 √2 yd ≈ 42 yd.
―
Module 18
IN2_MNLESE389847_U7M18L3 962
962
Lesson 3
18/04/14 11:49 PM
Special Right Triangles 962
_
19. In a right triangle, the longer leg is exactly √3 times the length of the shorter leg. Use
the inverse tangent trigonometric ratio to prove that the acute angles of the triangle
measure 30° and 60°.
BC
, use an inverse tangent ratio:
Given: BC = AC √3 . Since tan A =
B
AC
√
AC
3
BC
√
tanA =
=
= 3
AC
AC
m∠A = tan -1 √3 = 60°
―
_ _―
―
A
―
_
∠A and ∠B are complementary, so m∠B = 90° - m∠A = 90° - 60° = 30°.
C
Algebra Find the value of x in each right triangle.
20.
21.
C
5x - 6
B
x√3
L
x
x
J
A
―
Since the longer leg is √3 times the length
of the shorter leg, △ABC is a 30°-60°-90°
triangle. Therefore,
(3x - 25)√2
x√2
2
K
―
Since the hypotenuse is √2 times the
length of one of the legs, △JKL is a
45°-45°-90° triangle. Therefore,
―
_
BC = 2AC
JK = KL
x √2
= (3x - 25) √2
2
x = 2(3x - 25)
5x - 6 = 2x
3x = 6
x=2
―
x = 6 x - 50
50 = 5 x
© Houghton Mifflin Harcourt Publishing Company
10 = x
sides of a right triangle
22. Explain the Error Charlene is trying to find the unknown
_
with a 30° acute angle, whose hypotenuse measures 12√2 . Identify, explain, and
correct Charlene’s error.
12√2
P
Module 18
IN2_MNLESE389847_U7M18L3 963
Lesson 18.3
12
R
6√2
―
―
―
Charlene appears to have used the ratio 1 : √2 : 2, whereas the correct
ratio for a 30°-60°-90° triangle is 1 : √3 : 2. PR = 6 √2 is correct, but
QR = PR √3 = (6 √2 ) √3 = 6 √6 .
―
963
Q
30°
― ―
―
963
Lesson 3
18/04/14 11:49 PM
23. Represent Real-World Problems Honeycomb blinds form a string of almostregular hexagons when viewed end-on. Approximately how much material, to the
nearest ten square centimeters, is needed for each 3.2-cm deep cell of a honeycomb
blind that is 125 cm wide? (Hint: Draw a picture. A regular hexagon can be divided
into 6 equilateral triangles.)
...
3.2 cm
...
125 cm
B
3.2 cm A
C
_
1
△ABC is a 30°-60°-90° triangle, and BC = (3.2 cm) = 1.6 cm. Therefore,
2
BC
cos 30° =
AB
√3
1.6 cm
=
2
AB
AB = 3.2 √3 ≈ 5.542 cm
_
―
_ _
―
The amount of material needed is the perimeter of the hexagon × the width of the blind:
24. Which of these pairs of numbers are two out of three integer-valued side lengths of a
right triangle? (Hint: for positive integers a, b, c, and k, ka, kb, and kc are side lengths
of a right triangle if and only if a, b, and c are side lengths of a right triangle.)
A. 15, 18
True
False
B. 15, 30
True
False
C. 15, 51
True
False
D. 16, 20
True
False
E. 16, 24
True
False
A. False; 15 = 3(5) and 18 = 3(6); 5 2 + 6 2 = 25 + 36 = 61 and 6 2 - 5 2 = 36 - 25 = 11,
and neither are perfect squares.
B. False; 15 = 15(1) and 30 = 15(2); 1 2 + 2 2 = 5 and 2 2 - 1 2 = 3, and neither are perfect
squares.
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
6(AB) × width = 6(5.542) × 125 ≈ 4,160 cm 2
C. True: 15 = 3(5) and 39 = 3(13); 5, 12, 13 is a Pythagorean triple.
D. True; 16 = 4(4) and 20 = 4(5); 3, 4, 5 is a Pythagorean triple.
E. False; 16 = 8(2) and 24 = 8(3); 2 2 + 3 2 = 13 and 3 2 - 2 2 = 5, and neither are perfect
squares.
Module 18
IN2_MNLESE389847_U7M18L3 964
964
Lesson 3
18/04/14 11:49 PM
Special Right Triangles 964
PEERTOPEER DISCUSSION
H.O.T. Focus on Higher Order Thinking
25. Communicate Mathematical Ideas Is it possible for the three side lengths of a
right triangle to be odd integers? Explain.
Have students use the special right triangles to make
up several inverse trigonometric problems to
1
. Have them exchange and
evaluate, such as sin -1 _
2
solve problems with another pair of students. Have
students use a calculator to check their work.
No; if the two shorter side lengths are odd, then their squares are odd,
because the square of an odd number is always odd. But the sum of their
squares is even, because the sum of two odd numbers is always even.
Therefore the sum of the squares of the two shorter side lengths cannot
itself be the square of an odd number.
26. Make a Conjecture Use spreadsheet software to investigate this question: are there
sets of positive integers a, b, and c such that a 3 + b 3 = c 3? You may choose to begin
with these formulas:
JOURNAL
Have students summarize what they know about
special right triangles. Remind them to include a
chart that summarizes trigonometric ratios for
relevant angles and diagrams that support the ratios.
© Houghton Mifflin Harcourt Publishing Company
Actually there are no sets of positive integers a, b, and c such
that a 3 + b 3 = c 3, but this is very hard indeed to prove. Students should
be able to extend the spreadsheet example as follows, testing triples
with different smallest numbers by changing the value in cell A1; for
example:
Module 18
IN2_MNLESE389847_U7M18L3 965
965
Lesson 18.3
965
Lesson 3
18/04/14 11:49 PM
AVOID COMMON ERRORS
Kate and her dog are longtime flying disc players. Kate has decided to start a small business
making circles of soft material that dogs can catch without injuring their teeth. Since she also
likes math, she’s decided to see whether she can apply Pythagorean principles to her designs.
She used the Pythagorean triple 3-4-5 for the dimensions of her first three designs.
When students check three measurements to see if
they could represent the sides of a right triangle, they
may add the two smaller measurements before
squaring them to see if the result is the square of the
third measurement. Caution students to square each
smaller measurement first, before finding their sum.
r = 3 in.
small
r = 4 in.
medium
r = 5 in.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 If you look at a list of Pythagorean Triples,
large
1. Is it true that the (small area) + (medium area) = (large area)? Explain.
2. If the circles had radii based on the Pythagorean triple 5-12-13, would the above
equation be true? Explain.
you’ll notice that at least one of the numbers forming
the triple is even. Must this be true for all
Pythagorean Triples? Explain. Yes; sample answer:
The square of an odd number is always odd. The
sum of two odd numbers is always even. So, if a and
b are both odd, a 2 + b 2 must be even, which means
that c must also be even.
3. Three of Kate’s circles have radii of a, b, and c, where a, b, and c form a
Pythagorean triple (a 2 + b 2 = c 2). Show that the sum of the areas of the small and
medium circles equals the area of the large circle.
5. Explain the discrepancy between your results in Exercises 3 and 4.
1. yes; small area + medium area = 9 in. 2 + 16 in. 2 = 25 in. 2 = large area
2. yes; small area + medium area = 25 in. 2 + 144 in. 2 = 169 in. 2 = large area
3. small area + medium area = a 2 + b 2 =( a 2 + b 2) = c 2 = large area
1
1
2
4. no; small volume + medium volume = 36 in. 3 + 85 in. 3 = 121 in. 3 ≠ 166 in. 3
3
3
3
(large area)
_
_
_
5. Possible answer: Numbers are squared in both the Pythagorean Theorem and the
formula for the area of a circle. In the formula for the volume of a sphere, numbers are
cubed. There is no theorem analogous to the Pythagorean Theorem for cubes of numbers.
Module 18
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
4. Kate has decided to go into the beach ball business. Sticking to her Pythagorean
principles, she starts with three spherical beach balls--a small ball with radius 3
in., a medium ball with radius 4 in., and a large ball with radius 5 in. Is it true that
(small volume) + (medium volume) = (large volume)? Show your work.
Lesson 3
966
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M18L3 966
Choose two whole numbers m and n, such that m < n. Make a table like the
following for at least 10 pairs of values of m and n:
M
n
n2 - m2
2mn
n2 + m2
Pythagorean Triple?
1
2
3
4
5
yes
Describe your results, and use algebra to explain them. m and n always generate
a Pythagorean Triple.
(n 2 - m 2) 2 + (2mn) 2 = n 4 - 2n 2m 2 + m 4 + 4m 2n 2
= n 4 + 2n 2 m 2 + m 4
2
= (n 2 + m 2)
18/04/14 11:49 PM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Special Right Triangles 966
LESSON
18.4
Name
Problem Solving with
Trigonometry
Class
Date
18.4 Problem Solving
with Trigonometry
Essential Question: How can you solve a right triangle?
Resource
Locker
Common Core Math Standards
The student is expected to:
COMMON
CORE
Explore
G-SRT.C.8
Use trigonometric ratios and the Pythagorean Theorem to solve right
triangles in applied problems. Also G-SRT.D.9(+), G-GPE.B.7
_
_
Suppose you draw an altitude AD to side BC of △ABC. Then write an equation using a
trigonometric ratio in terms of ∠C, the height h of △ABC, and the length of one of its sides.
A
Mathematical Practices
COMMON
CORE
Deriving an Area Formula
You can use trigonometry to find the area of a triangle without knowing its height.
b
Language Objective
C
Explain to a partner how to solve a right triangle, and how to solve a right
triangle in the coordinate plane.
sin C =
ENGAGE
B
Essential Question: How can you solve
a right triangle?
C
b
c
C
B
a
h
a
D
c
B
h
_
=
AC
b
Solve your equation from Step A for h.
b sin C = h
Complete this formula for the area of △ABC in terms of h and
1 ah
another of its side lengths: Area = _
2
© Houghton Mifflin Harcourt Publishing Company
You can use trigonometric ratios to find side
lengths, or their inverses to find angle measures;
you can use the Pythagorean Theorem to find the
third side length; you can use the fact that the acute
angles are complementary; if the triangle is in the
coordinate plane, you can use the distance formula
to find side lengths.
A
A
MP.2 Reasoning
your expression for h from Step B into your formula from Step C.
D Substitute
_1 ab sin C
2
Reflect
1.
Does the area formula you found work if ∠C is a right angle? Explain.
1
Yes; in this case, sin C = sin 90° = 1, so the formula becomes Area = __
ab, and this is correct
2
since a and b are now the base and height of △ABC.
PREVIEW: LESSON
View the Engage section online. Discuss the
photograph. Ask students if they know why water is
sometimes referred to as H 2O. Then preview the
Module 18
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 4
967
gh “File info”
Date
Class
ng
blem Solvi
18.4 Pro Trigonometry
with
Name
IN2_MNLESE389847_U7M18L4 967
HARDCOVER PAGES 967980
Resource
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right triangl
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Formula
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.
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Deriving
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its sides.
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draw an altitud of ∠C, the height
A
Suppose you
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trigonometr
A
h
b

B
C
Watch for the hardcover
student edition page
numbers for this lesson.
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B
c
b
a
C
D
a
_ _
b
sin C = AC

from Step
equation
Solve your
h
b sin C =
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Lesson 4
967
Module 18
8L4 967
47_U7M1
ESE3898
IN2_MNL
967
Lesson 18.4
19/04/14
12:02 AM
19/04/14 12:03 AM
Suppose you used a trigonometric ratio in terms of ∠B, h, and a different side length. How would this
change your findings? What does this tell you about the choice of sides and included angle?
2.
EXPLORE
Step C would be the same, but the other steps would change:
h
= c
sin B =
AB
c sin B = h
1
Area = ac sin B
2
You can choose different sides and included angle and derive a slightly different
_ _
Deriving an Area Formula
_
INTEGRATE TECHNOLOGY
formula for the area, but in the same form.
Explain 1
Students have the option of doing the Explore activity
either in the book or online.
Using the Area Formula
INTEGRATE TECHNOLOGY
Area Formula for a Triangle in Terms of its Side Lengths
The area of △ABC with sides a, b, and c can be found using the lengths of two
of its sides and the sine of the included angle: Area = __12 bc sin A, Area = __12 ac sin B,
or Area = __12 ab sin C.
Students are familiar with the trigonometric ratios
from 0° to 90°. Ask them to graph y = sin x from
0° to 180° using their graphing calculators to see that
the ratios are defined for angles greater (and less)
than 90°.
C
a
B
b
c
A
You can use any form of the area formula to find the area of a triangle, given two side
lengths and the measure of the included angle.
Example 1
QUESTIONING STRATEGIES
Find the area of each triangle to the nearest tenth.

Trigonometric ratios are defined using right
triangles. Why does the area formula work for
all types of triangles when it uses the sine ratio? The
formula works because when the altitude, or height,
is drawn to apply the usual formula for the area of a
triangle, a right triangle is created.
3.2 m
142°
4.7 m
Let the known angle be ∠C.
1 ab sin C.
Substitute in the formula Area = _
2
Evaluate, rounding to the nearest tenth.
© Houghton Mifflin Harcourt Publishing Company
a = 3.2 m and b = 4.7 m
Let the known side lengths be a and b.
m ∠C = 142°
1 (3.2)(4.7)sin 142°
Area = _
2
Area ≈ 4.6 m 2
If you know the area of a triangle and the
lengths of two sides, how can you find the
measure of the included angle? Find the inverse sine
of twice the area divided by the product of the
two sides.
EXPLAIN 1
Module 18
968
Lesson 4
Using the Area Formula
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M18L4 968
Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.2,
which calls for students to “reason abstractly and quantitatively.” Students derive
the formula for the area of a triangle by recognizing the relationships that occur
within the triangle when the altitude is constructed. They apply this formula to a
variety of triangles. As students solve a right triangle, they must identify
relationships that can be used to find missing measures, and they can often choose
which of the three inverse trigonometric ratios to apply.
19/04/14 12:03 AM
AVOID COMMON ERRORS
Students may have difficulty substituting values into
the sine area formula from a diagram. Emphasize that
the two sides are both adjacent to the angle. Or, the
angle is the included angle formed by the sides of the
triangle.
Problem Solving with Trigonometry 968
In △DEF, DE = 9 in., DF = 13 in., and m∠D = 57°.
B
QUESTIONING STRATEGIES
Sketch △DEF and check that ∠D is the included angle.
Why is the sine area formula useful when
there is already a formula for the area of a
triangle? It provides another method to find the
area of a triangle without having to construct the
height.
E
9 in.
D 57°
Why does the angle have to be the included
angle to use the sine area formula? The
formula is derived by constructing a right triangle
so that the height is opposite the angle and the
hypotenuse is adjacent to the angle to make it
possible to apply the sine ratio.
F
13 in.
( )
( )( )
1 (DE) DF sin D
Area = _
2
1 9
13 sin 57 °
Area = _
2
Write the area formula in terms of △DEF.
Substitute in the area formula.
Area ≈ 49.1 in. 2
Evaluate, rounding to the nearest tenth.
Find the area of each triangle to the nearest tenth.
3.
CONNECT VOCABULARY
12 mm
Remind students that another word for the altitude of
a triangle is its height.
one version of the sine area formula to apply the
formula to find the area of any triangle.
15 mm
Let the known side lengths be a and b.
a = 12 mm and b = 15 mm
Let the known angle be ∠C.
m ∠C = 34°
1
Area = (12)(15)sin 34°
2
Area ≈ 50.3 mm 2
1
Substitute in the formula Area = _ ab sin C.
2
Evaluate, rounding to the nearest tenth.
© Houghton Mifflin Harcourt Publishing Company
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Discuss why students need remember only
34°
4.
_
In △PQR, PQ = 3 cm, QR = 6 cm, and m∠Q = 108°.
R
6 cm
P
108°
Q
3 cm
_1 (PQ)(QR)sin Q
2
1
= _(3)(6)sin 108°
Area =
2
≈ 8.6 cm 2
Module 18
969
Lesson 4
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M18L4 969
Small Group Activity
Have students verify that they can use different trigonometric ratios and different
inverse trigonometric ratios to solve a right triangle. Have students work together
to solve a right triangle. Then have each student verify the measures using
different trigonometric and inverse trigonometric ratios.
969
Lesson 18.4
19/04/14 12:03 AM
Explain 2
Solving a Right Triangle
EXPLAIN 2
Solving a right triangle means finding the lengths of all its sides and the measures of all its angles.
To solve a right triangle you need to know two side lengths or one side length and an acute
angle measure. Based on the given information, choose among trigonometric ratios, inverse
trigonometric ratios, and the Pythagorean Theorem to help you solve the right triangle.
Solving a Right Triangle
A shelf extends perpendicularly 7 in. from a wall. You want to place a 9-in. brace
under the shelf, as shown. To the nearest tenth of an inch, how far below the shelf
will the brace be attached to the wall? To the nearest degree, what angle will the
brace make with the shelf and with the wall?
QUESTIONING STRATEGIES
What information do you need to solve a right
triangle? Explain. You need the lengths of
two sides or the length of one side and one acute
angle measure. With this information you can find
the remaining angle measures and side lengths.
Shelf
C
7 in.
A
9 in.
Wall
A
B
Brace
Find BC.
Substitute 7 for AC and 9 for AB.
7 2 + BC 2 = 9 2
Find the squares.
49 + BC 2 = 81
Be sure students remember how to interpret the
notation that represents an inverse function. Remind
a
them that the –1 in an expression such as tan -1 __
b
represents the angle whose tangent is __ab . It does not
1 . It should generally
represent the reciprocal _______
a
tan __
b
be clear from the context whether the –1 is an inverse
function or a negative exponent.
()
BC 2 = 32
Subtract 49 from both sides.
BC ≈ 5.7
Find the square root and root.
B
AVOID COMMON ERRORS
AC 2 + BC 2 = AB 2
Use the Pythagorean Theorem to find the length
of the third side.
()
Find m∠A and m∠B.
Use an inverse trigonometric ratio to find m∠A. You know the lengths of the
adjacent side and the hypotenuse, so use the cosine ratio.
cos A = _
9
Write an inverse cosine ratio.
m∠A = cos -1
Evaluate the inverse cosine ratio and round.
m∠A ≈ 39 °
© Houghton Mifflin Harcourt Publishing Company
Write a cosine ratio for ∠A.
7
∠ A and ∠B are complementary.
m∠ A
Substitute 39 ° for m∠ A .
(_)
7
9
+ m∠B = 90°
39 ° + m∠B ≈ 90°
Subtract 39 ° from both sides.
m∠B ≈ 51 °
Reflect
5.
(_)
Is it possible to find m∠B before you find m∠A? Explain.
7
7
≈ 51°.
Yes; use an inverse sine ratio; sin B = , so m∠B = sin -1
9
9
Module 18
_
970
Lesson 4
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M18L4 970
Modeling
19/04/14 12:03 AM
Have students draw an acute triangle with two adjacent sides that have whole
number measures. Ask them to use a protractor to measure the included angle.
Then have them find the area of the triangle using the sine area formula. Repeat
with an obtuse triangle.
Auditory Cues
Students can use the mnemonic soh-cah-toa to remember that
opposite leg
opposite leg
sin A = __________, cos A = __________, and tan A = __________.
hypotenuse
hypotenuse
Problem Solving with Trigonometry 970
EXPLAIN 3
Solving a Right Triangle in the
Coordinate Plane
6.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Discuss how to identify the legs and the
7.
hypotenuse of a right triangle in the coordinate plane
if the legs of the right triangle are not formed by
horizontal and vertical line segments. Use the
distance formula to find the length of each side. The
longest side is the hypotenuse.
8.
Use a trigonometric ratio to find the distance EF.
DE
tan F =
EF
33
tan 27° =
EF
33
≈ 65 m
EF =
tan 27°
Use another trigonometric ratio to find the distance DF.
DE
sin F =
DF
33
sin 27° =
DF
33
≈ 73 m
DF =
sin 27°
Use the fact that acute angles of a right triangle are complementary
to find m∠D.
_
_
_
F
Building
D 33 m E
_
_
_
m∠D + m∠F = 90°
m∠D + 27° = 90°
m∠D = 63°
QUESTIONING STRATEGIES
Explain 3
Solving a Right Triangle in the
Coordinate Plane
You can use the distance formula as well as trigonometric tools to solve right triangles
in the coordinate plane.
© Houghton Mifflin Harcourt Publishing Company
How is solving a right triangle in the
coordinate plane different from solving a
labeled right triangle? Neither side lengths nor an
angle measure are given in the coordinate plane.
You must use the distance formula to find side
lengths and then use the inverses of trigonometric
ratios to find the angles.
27°
A building casts a 33-m shadow when the Sun is at an angle of 27° to
the vertical. How tall is the building, to the nearest meter? How far is
it from the top of the building to the tip of the shadow? What angle
does a ray from the Sun along the edge of the shadow make with
the ground?
Example 3

Solve each triangle.
Triangle ABC has vertices A(-3, 3), B(-3, -1), and C(4, -1). Find the side lengths to the
nearest hundredth and the angle measures to the nearest degree.
Plot points A, B, and C, and draw △ABC.
5
Find the side lengths: AB = 4, BC = 7
A
_
Use the distance formula to find the length of AC.
AC =
―――――――――
+ (-1 - 3) = √―
65 ≈ 8.06
√(4 -(-3))
2
y
x
2
_
_
Find the angle measures: AB ⊥ BC , so m∠B = 90°.
-5
B
0
C
5
Use an inverse tangent ratio to find
( )
()
AB = tan -1 _
4 ≈ 30°.
m∠C = tan -1 _
7
BC
∠A and ∠C are complementary, so m∠C ≈ 90° - 30° = 60°.
Module 18
971
-5
Lesson 4
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M18L4 971
Connect Vocabulary
Discuss the meaning of solve in the phrase solve a right triangle. Compare solving
a triangle to solving an equation. To use the non-mathematical meaning of solve,
compare solving a triangle to solving a mystery. A detective uses clues to find
missing information. Similarly, to solve a right triangle means to use given
information and tools such as the trigonometric ratios to find all the missing side
lengths and angle measures.
971
Lesson 18.4
19/04/14 12:03 AM
B
Triangle DEF has vertices D(-4, 3), E(3, 4), and F(0, 0). Find the side lengths
to the nearest hundredth and the angle measures to the nearest degree.
Plot points D, E, and F, and draw △DEF.
5
∠F appears to be a right angle. To check, find the slope
x
-5
_
-4
4
0 - -4
__ = _ = _;
EF :
0 -3
3
-3
Find the side lengths using the distance formula:
―――――――――
――
――
√(3 - -4 ) + ( 4 - 3) = √ 50 = 5 √ 2
――――――――――― ――
DF = √( 0 - -4 ) + ( 0 - 3) = √ 25 = 5 ,
――――――――――― ――
EF = √( 0 - 3 ) + ( 0 - 4) = √ 25 = 5
2
2
2
0 F
5
-5
so m∠F = 90 °.
DE =
E
D
-3
_ 0 -3
3
of DF: _ = _ = -_
;
4
0 - -4
4
slope of
y
≈
7.07
,
2
2
2
( )
Use an inverse sine ratio to find m∠D.
( )
5
EF = sin -1 _ =
m∠D = sin -1 _
45 °
DE
5 √2
―
∠D and ∠ E are complementary, so m∠ E = 90° - 45 ° = 45 °.
9.
How does the given information determine which inverse trigonometric ratio you
should use to determine an acute angle measure?
It doesn’t – you can use any inverse trigonometric ratio, as long as you use the correct
two side lengths. However, if two sides are vertical and horizontal, it makes sense to use
inverse tangent.
Module 18
IN2_MNLESE389847_U7M18L4 972
972
© Houghton Mifflin Harcourt Publishing Company
Reflect
Lesson 4
19/04/14 12:03 AM
Problem Solving with Trigonometry 972
ELABORATE
10. Triangle JKL has vertices J(3, 5), K(-3, 2), and L(5, 1). Find the
side lengths to the nearest hundredth and the angle measures to the
nearest degree.
Plot points J, K, and L, and draw △JKL.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Some students may find it difficult to choose
――――――― ―
―
―――――――
JL = √(5 - 3) + (1 - 5) = √―
20 = 2 √―
5 ≈ 4.47,
―――――――
―
√
KL = √(-3 - 5) + (1 - 2) = 65 ≈ 8.06
JK = √(-3 - 3) + (2 - 5) = √45 = 3 √5 ≈ 6.71,
2
2
5
-5
2
_ _ _
_ _
(_)
―
(_― )
Elaborate
11. Would you use the area formula you determined in this lesson for a right
triangle? Explain.
Possible answer: No. You could use this formula, but since in a right triangle, the lengths of
1
the legs are the base and height, it is simpler to use the formula A = bh.
2
_
12. Discussion How does the process of solving a right triangle change when
its vertices are located in the coordinate plane?
Generally, you need to find all the side lengths and all the angles, checking that one is
© Houghton Mifflin Harcourt Publishing Company
a right angle. The distance formula is generally used to find side lengths rather than
trigonometric ratios or the Pythagorean Theorem (although the distance formula depends
on the Pythagorean Theorem). The acute angles are found in the same way in both cases.
13. Essential Question Check-In How do you find the unknown angle
measures in a right triangle?
You first use two known side lengths to form a ratio and then use the appropriate inverse
trigonometric ratio to find one angle measure. Then subtract that measure from 90° to
find the measure of the other acute angle.
Module 18
IN2_MNLESE389847_U7M18L4 973
Lesson 18.4
2
Lx
0
-5
¯.
∠J appears to be a right angle. To check, find the slope of JK
―
―
-3
2-5
1-5
1
-4
= ; slope of JL:
=
=
= -2; so m∠J = 90°.
Slope of JK:
-6
-3 - 3
2
5-3
2
-1 JK
-1 √45
Use an inverse cosine ratio to find m∠K = cos
= cos
≈ 34°.
KL
√65
∠J and ∠L are complementary, so m∠L ≈ 90° - 34° = 66°.
Students may miss parts when solving a right
triangle. Remind students that three side lengths and
three angle measures are needed. Encourage students
to always draw a diagram and to label it whenever
they determine a measure.
973
J
2
2
AVOID COMMON ERRORS
How do you solve a right triangle? Use the
Pythagorean Theorem, trigonometric ratios, inverse
trigonometric ratios, and complementary angle
relationships to find missing side lengths and angle
measures in a right triangle.
y
K
Find the side lengths using the distance formula:
a trigonometric or inverse trigonometric ratio to
solve a right triangle. Discuss with the class how
students choose a particular relationship. For
example, some students may prefer to always use the
inverse tangent ratio if the lengths of both legs are
known to find an acute angle measure.
SUMMARIZE THE LESSON
5
973
Lesson 4
19/04/14 12:03 AM
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
Find the area of each triangle to the nearest tenth.
1.
In △PQR, PR = 23 mm, QR = 39 mm, and
m∠R = 163°.
2.
62°
4.1 cm
3.2 cm
Q
39 mm
_
R
163°
P
23 mm
ASSIGNMENT GUIDE
_
1
1
Area = (PR)(QR)sin R = (23)(39) sin 163° ≈ 131.1 mm 2.
2
2
a = 4.1 cm, b = 3.2 cm, and m∠C = 62°;
1
1
Area = ab sin C = (4.1)(3.2)sin 62° ≈ 5.8 cm 2
2
2
_
_
Solve each right triangle. Round lengths to the nearest tenth and angles to the nearest degree.
3.
AC 2 + BC 2 = AB 2
A
AC 2 + (2.7) = (3.1)
2
3.1 cm
4.
F
5.
E
――
(_ )
AC = √2.32 ≈ 1.5 cm
2.7
BC
m∠A = sin -1
= sin -1
≈ 61°
3.1
AB
∠A and ∠B are complementary, so m∠B ≈ 90° - 61° = 29°.
(_)
∠D and ∠F are complementary, so m∠F = 90° - 56° = 34°.
DE
EF
cos D =
tan D =
DE
DF
26
EF
cos 56° =
tan 56° =
26
DF
26
26tan 56° = EF
DF =
≈ 46.5 m
cos56°
38.5 m ≈ EF
_
_
_
_
Right △PQR with PQ ⊥ PR , QR = 47 mm, and m∠Q = 52°
∠Q and ∠R are complementary, so m∠R = 90° - 52° = 38°.
PQ
PR
sin P =
cos P =
QR
QR
PQ
PR
cos 52° =
sin 52° =
47
47
R
_
_
47 mm
52°
P
Q
47 cos 52° = PQ
37.0 mm ≈ PR
28.9 mm ≈ PQ
Module 18
Exercise
IN2_MNLESE389847_U7M18L4 974
_
_
47 sin 52° = PR
Explore
Deriving an Area Formula
Exercise 16
Example 1
Using the Area Formula
Exercises 1–2
Example 2
Solving a Right Triangle
Exercises 3–5
Example 3
Solving a Right Triangle in the
Coordinate Plane
Exercises 6–8
Students may forget to include units when solving
area problems. Remind students that the units are an
COMMUNICATING MATH
Encourage students solving right triangles to
communicate their understanding of the
relationships between the trigonometric ratios and
their inverses.
Lesson 4
974
Depth of Knowledge (D.O.K.)
Practice
AVOID COMMON ERRORS
_
_
_
© Houghton Mifflin Harcourt Publishing Company
56°
26 m
D
AC 2 = 2.32
B
2.7 cm
C
2
Concepts and Skills
COMMON
CORE
Mathematical Practices
1–8
1 Recall of Information
MP.4 Modeling
9–15
2 Skills/Concepts
MP.4 Modeling
16
3 Strategic Thinking
MP.3 Logic
17
3 Strategic Thinking
MP.2 Reasoning
18
3 Strategic Thinking
MP.4 Modeling
19
3 Strategic Thinking
MP.3 Logic
19/04/14 12:03 AM
Problem Solving with Trigonometry 974
Solve each triangle. Find the side lengths to the nearest hundredth
and the angle measures to the nearest degree.
AVOID COMMON ERRORS
Students may have difficulty solving right triangles.
They may find it easier when they first identify
whether they are looking for a ratio or an angle
measure. If they are looking for a ratio, they should
use sin, cos, or tan. If they are looking for an angle
measure, they should use sin -1, cos -1, or tan -1.
6.
Triangle ABC with vertices A(-4, 4), B(3, 4), and C(3, -2)
5
A
Plot points A, B, and C, and draw △ABC.
y
¯, so m∠B = 90°
¯ ⊥ BC
AB
B
5
C
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 For greater accuracy, remind students to use
2
( )
x
0
-5
―――――――
―
AB = 7, BC = 6, AC = √(3 - 4) + (-2 - 4) = √85 ≈ 9.2,
6
BC
m∠A = tan -1 _ = tan -1 _ ≈ 41°
7
AB
∠A and ∠C are complementary, so m∠C ≈ 90° - 41° = 49°.
2
()
-5
7.
Triangle JKL with vertices J(-3, 1), K(-1, 4), and L(6, 5)
K
given measurements instead of calculated values in
the later steps of solving a right triangle.
5
Plot points J, K, and L, and draw △JKL.
y
J
x
0
-5
∠J appears to be a right angle. To check,
3
4-1
¯:
= ;
slope of JK
2
-1 - (-3)
¯: -5 - 1 = -6 = - 2 ; so m∠J = 90°.
slope of JL
9
3
6 - (-3)
5
__ _
_ _ _
――――――――
13 ≈ 3.6,
JK = √(-1 - (-3)) + (4 - 1) = √―
――――――――
117 ≈ 10.8,
JL = √(6 - (-3)) + (-5 - 1) = √――
―――――――――
130 ≈ 11.4
KL = √(6 - (-1)) + (5 - (-4)) = √――
―
√
13
JK
≈ 18°
m∠L = sin (_) = sin _
KL
√――
130
2
2
2
-5
L
2
2
2
(
© Houghton Mifflin Harcourt Publishing Company
-1
)
-1
∠K and ∠L are complementary, so m∠K ≈ 90° - 18° = 72°.
8.
Triangle PQR with vertices P(5, 5), Q(-5, 3), and R(-4, -2)
5
y
P
Q
x
0
-5
5
Plot points J, K, and L, and draw △JKL.
∠Q appears to be a right angle. To check,
― 3-5
-2
1
=
= ;
slope of PQ:
-5 - 5
5
-10
― -2 - 3 = -5 = -5 ; so m∠Q = 90°.
slope of PR:
1
-4 - (-5)
_ _ _
__ _
――――――― 104 = 2 √―
26 ≈ 10.2,
PQ = √(-5 - 5) + (3 - 5) = √――
―――――――――
26 ≈ 5.1,
QR = √(-4 - (-5)) + (-2 - 3) = √―
―――――――― 130 ≈ 11.4
PR = √(-4 - 5) + (-2 - 5) = √――
2
2
R
2
-5
2
( )
2
2
()
QR
1 ≈ 27°
m∠P = tan -1 _ = tan -1 _
2
PR
∠P and ∠R are complementary, so m∠R ≈ 90° - 27° = 63°.
Module 18
IN2_MNLESE389847_U7M18L4 975
975
Lesson 18.4
975
Lesson 4
19/04/14 12:03 AM
9.
Surveying A plot of land is in the shape of a triangle, as shown. Find the area of the
plot, to the nearest hundred square yards.
_1 ab sin C
2
1
= _(142)(227) sin 128° ≈ 12,700 yd
142 yd
COLLABORATIVE LEARNING
Have students construct a right triangle and measure
the lengths of two sides. Ask them to find the length
of the third side and the acute angle measures
without using the Pythagorean Theorem or a
protractor. Have students construct a second right
triangle and measure one acute angle and a side.
Have them find the other acute angle measure and
the lengths of the other two sides. Finally, have
students exchange triangles and check each other’s
work.
Area =
128°
2
2
227 yd
10. History A drawbridge at the entrance to an ancient castle is raised and lowered by a
pair of chains. The figure represents the drawbridge when flat. Find the height of the
suspension point of the chain, to the nearest tenth of a meter and the acute angles the
chain makes with the wall and the drawbridge, to the nearest degree.
height of wall: AB 2 + BC 2 = AC 2
C
(3.2)2 + BC 2 = (5.0)2
10.24 + BC 2 = 25
5.0 m
Chain
A
3.2 m
drawbridge
B
COMMUNICATE MATH
BC 2 = 14.76
Wall
Have students explain to each other when they would
use the Pythagorean Theorem, the sin, cos, or tan
ratios, and the sin -1, cos -1, or tan -1 ratios to solve a
right triangle.
BC ≈ 3.8 m
3.2
BC
-1 _
m∠A = cos
= cos -1 _ ≈ 50°
5.0
AC
∠A and ∠C are complementary, so m∠C ≈ 90° - 50° = 40°.
( )
( )
11. Building For safety, the angle a wheelchair ramp makes with the horizontal should
be no more than 3.5°. What is the maximum height of a ramp of length 30 ft? What
distance along the ground would this ramp cover? Round to the nearest tenth of
a foot.
AB
_
AC
AB
sin 3.5° = _
sin C =
30 ft
BC
_
AC
BC
cos 3.5° = _
30
30 cos 3.5° = BC
1.8 ft ≈ AB
29.9 ft ≈ BC
IN2_MNLESE389847_U7M18L4 976
C
cos C =
30
30 sin 3.5° = AB
Module 18
3.5°
© Houghton Mifflin Harcourt Publishing Company
A
B
976
Lesson 4
19/04/14 12:03 AM
Problem Solving with Trigonometry 976
12. Multi-Step The figure shows an origami crane as well as a stage of its construction.
The area of each wing is shown by the shaded part of the figure, which is symmetric
about its vertical center line. Use the information in the figure to find the total wing
area of the crane, to the nearest tenth of a square inch.
2.2 in.
A 45°
F
B
C
E
4.0 in.
3.7 in.
D
22.5°
Area of each wing = Area of △ ABF + Area of △ DBF - Area of △ DCE
1
Area of △ ABF = (2.2)(2.2) sin 45°; ≈ 1.711 in. 2
2
1
Area of △ DBF = (4.0)(4.0) sin 22.5°; ≈ 3.061 in. 2
2
1
Area of △ DCE = (3.7 in.)(3.7 in.) sin 22.5°; ≈ 2.619 in. 2
2
Area of each wing: 1.711 in. 2 + 3.061 in. 2 - 2.619 in. 2 = 2.153 in. 2
_
_
_
Kitano/iStockPhoto.com
Area of both wings: 22.153 in. 2 ≈ 4.3 in. 2
13. Right triangle △XYZ has vertices X(1, 4) and Y(2, -3). The vertex Z has positive
integer coordinates, and XZ = 5. Find the coordinates of Z and solve △XYZ, giving
y
5
X
Z
x
0
-5
5
Y
-5
Based on 3–4–5 right triangles as well as horizontal and vertical displacements of 5, possible
coordinates for Z are (1, 9), (4, 8), (5, 7), (6, 4), and (5, 1) . Z(5, 1) is the only possibility that
appears to create a right triangle.
― 1 - 4 = -3 = - 3 ; slope of YZ
―: 1 -(-3) = 4 ; so m∠Z = 90°.
To check, slope of XZ:
5-1
4
5 -2
3
4
_ _ _
_ _
――――――― 50 = 5 √―2 , XZ = √―――――――
(5 - 1) + (1 - 4) = √―
25 = 5,
XY = √(2 - 1) + (-3 - 4) = √―
――――――――
25 = 5
YZ = √(5 - 2) + (1 - (-3)) = √―
2
2
―
2
2
2
2
―
Since XZ ≅ YZ, △ XYZ is a 45°-45°-90° triangle. Therefore, m∠X = m∠Y = 45°.
Module 18
IN2_MNLESE389847_U7M18L4 977
977
Lesson 18.4
977
Lesson 4
19/04/14 12:03 AM
14. Critique Reasoning Shania and Pedro are discussing whether it is always possible
to solve a right triangle, given enough information, without using the Pythagorean
Theorem. Pedro says that it is always possible, but Shania thinks that when two side
lengths and no angle measures are given, the Pythagorean Theorem is needed. Who is
correct, and why?
Pedro; Possible answer: When one side and one acute angle are given, trigonometric ratios
can be used to find the other sides, and complementary angles give the other acute angle.
When two sides are given, an inverse trigonometric ratio can be used to find one acute
angle, and this angle together with a given side can be used with a different trigonometric
ratio to find the other side. The other acute angle is complementary as before.
15. Design The logo shown is symmetrical about one of its diagonals. Find the angle
measures in △CAE, to the nearest degree. (Hint: First find an angle in △ABC,
△CDE or △AEF) Then, find the area of △CAE, without first finding the areas of the
other triangles.
3
BC
6 mm
= tan -1 _
≈ 26.565°
m∠BAC = tan -1 __
A
6
AB
( )
()
B
∠BAC ≅ ∠EAF, so m∠EAF = m∠BAC ≈ 26.565°
3 mm
m∠BAC + m∠CAE + m∠EAF = 90°
C
26.565° + m∠CAE + 26.565° ≈ 90°
3 mm
m∠CAE ≈ 90° - 2(26.565°)
F
≈ 36.869° ≈ 37°
AC 2 = AB 2 + BC 2
∠AEC ≅ ∠ACE, so m∠AEC = m∠ACE
AC 2 = 6 2 + 3 2 = 45
m∠AEC + m∠ACE + m∠CAE = 90°
2(m∠AEC) + 36.869° ≈ 90°
m∠AEC ≈ 27°
m∠ACE ≈ 27°
D
E
―
―
―
―
AC = √45 = 3 √5
1
Area = (AC)(AE) sin(∠CAE)
2
1 ( √ )( √ )
≈_
3 5 3 5 sin(36.869°) ≈ 13.5 mm 2
2
_
A
1
ac sin B
Area formula: Area = _
2
b
But △ABC has base a and height h = c sin (∠ABD),
1
(a)(c sin (∠ABD)).
so Area = _
2
C
a
c h
B
D
Equating these two expressions for the area, sin B = sin ∠ABD
But since, for any obtuse ∠B, ∠B and ∠ABD are supplementary in this
© Houghton Mifflin Harcourt Publishing Company
16. Use the area formula for obtuse ∠B in the diagram to show that if an acute
angle and an obtuse angle are supplementary, then their sines are equal.
construction, this proves the result.
Module 18
IN2_MNLESE389847_U7M18L4 978
978
Lesson 4
19/04/14 12:02 AM
Problem Solving with Trigonometry 978
JOURNAL
H.O.T. Focus on Higher Order Thinking
Have students create an example that shows how to
solve a right triangle. Students should specify the
known side or angle measures and explain how to
find the unknown side lengths and angle measures.
17. Communicate Mathematical Ideas The HL Congruence
Theorem states that for right triangles
ABC and DEF such that
― _
― ―
∠A and ∠D are right angles, BC ≅ EF, and AB ≅ DE,
△ABC ≅ △DEF.
E
C
A
B
Explain, without formal proof, how solving a right triangle
F
with given leg lengths, or with a given side length and acute
angle measure, shows that right triangles with both legs
congruent, or with corresponding sides and angles congruent,
must be congruent.
¯ ≅ DF
¯ ≅ DE
¯ and AC
¯. Solving either of these right triangles determines the
Suppose AB
D
length of the hypotenuse in the same way, e.g., using the Pythagorean Theorem, so
BC = EF and therefore, by SSS ≅, △ABC ≅ △DEF.
Suppose ∠B ≅ ∠E. The given corresponding side lengths allow the unknown sides to
be calculated in the same way using trigonometric ratios, so that all corresponding
side lengths are equal and therefore all corresponding sides are congruent. Again, by
SSS ≅, △ABC ≅ △DEF.
18. Persevere in Problem Solving Find the perimeter and area of
△ABC, as exact numbers. Then, find the measures of all the angles
to the nearest degree.
――――――――
26 ,
√(3 -(-2)) + (3 - 2) = √―
―――――――――
―
2
AC = √(3 -(-2)) + (-3 -(-2)) = √50 = 5 √―
26 + 5 √―
2
perimeter: BC + AB + AC = 6 + √―
2
BC = 6, AB =
2
_―
_
_
A
x
2
_
―
y
B
2
0
-5
With base b = BC = 6, height h = 5,
1
1
so area = bh = (6)(5) = 15.
2
2
1
1
Area = ab sin C
15 = (5 √2 )( √26 ) sin A
2
2
3 √13
15 = 1 (6)(5 √2 )sin C
= sin A
2
13
√2
= sin C
√13
-1 3
2
sin
= m∠A
13
45° = m∠C
56° ≈ m∠A
_
© Houghton Mifflin Harcourt Publishing Company
5
―
_
―
)
(_
_
―
5
C
-5
―
Angle sum of a triangle: 56° + m∠B + 45° ≈ 180°; m∠B ≈ 79°
19. Analyze Relationships Find the area of the triangle using two different
formulas, and deduce an expression for sin2θ.
1
1
Area = (x)(x)sin 2θ
Area = bh
2
2
1
1
= x 2 sin 2θ
= (2x sin θ)(x cos θ)
2
2
= x 2 sin θ cos θ
1 2
x sin 2θ = x 2 sin θ cos θ
→
sin 2θ = 2sin θ cos θ
2
_
_
_
_
x
θθ
x
_
Module 18
IN2_MNLESE389847_U7M18L4 979
979
Lesson 18.4
979
Lesson 4
19/04/14 12:02 AM
95
.84
p
1. Draw and label a triangle with the dimensions shown.
2. Find the area of the triangle in square centimeters. Show
H
O
104.45°
Students are probably familiar with the metric
prefixes deci-, centi-, and milli-, but may not know the
next three smaller prefixes. They are:
pm
.84
95
Every molecule of water contains two atoms of hydrogen and one atom of oxygen.
The drawing shows how the atoms are arranged in a molecule of water, along with
the incredibly precise dimensions of the molecule that physicists have been able to
determine. (1 pm = 1 picometer = 10 -12m)
m
CONNECT VOCABULARY
micro- = 10 -6 = 0.000 001
H
nano- = 10 -9 = 0.000 000 001
pico- = 10 -12 = 0.000 000 000 001
3. Find the distance between the hydrogen atoms in centimeters.
1.
A
.84
pm
95
104.45°
AVOID COMMON ERRORS
95
.84
pm
B
2.
The conversion from 95.84 picometers to centimeters
may trip students up. Here are the steps and reasons:
C
95.84 pm = 95.84 × 10 -12 m
AB = AC = 95.84 pm = 95.84 × 10 -12 m = 95.84 × 10 -10 cm = 9.584 × 10 -9 cm
1
1
area = AB(AC) sin A = (9.584 × 10 -9 cm)(9.584 × 10 -9 cm) sin 104.45°
2
2
1
= (9.584 × 10 -9 cm)(9.584 × 10 -9 cm)(0.9684)
2
≈ 44.48 × 10 -18 cm 2
_
_
_
(Definition of pm)
= 95.84 × 10 -10 cm
(Multiply by 10 2 cm = 1 m.)
≈ 4.45 × 10 -17 cm 2
3.
―
―_
= 9.584 × 10 -9 cm
two 52.225° angles.
(Divide 95.84 by 10 and multiply 10 -10 by 10)
_
_
-8
BC = 2 × 75.71 = 151.42 pm = 151.42 × 10 -12 m = 151.42 × 10 -10 cm = 1.51 × 10 cm
Module 18
980
© Houghton Mifflin Harcourt Publishing Company
In right triangle ABD,
BD
sin 52.225 =
AB
BD
0.79 ≈
95.84
75.71 ≈ BD
Lesson 4
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M18L4 980
This Lesson Performance Task will give students a glimpse into the complex field
of molecular geometry. While most of the subject lies beyond the range of
high-school geometry students, there are many questions students can research
and report on at a basic level. Among them:
• What is molecular geometry?
• How was the molecular structure of water discovered?
• What determines the angles between the atoms in a molecule?
• What holds a molecule together?
19/04/14 12:02 AM
Scoring Rubric
2 points: The student’s answer is an accurate and complete execution of the task
1 point: The student’s answer contains attributes of an appropriate response but
is flawed.
0 points: The student’s answer contains no attributes of an appropriate response.
Problem Solving with Trigonometry 980
LESSON
18.5
Name
Using a Pythagorean
Identity
Essential Question: How can you use a given value of one of the trigonometric functions to
calculate the values of the other functions?
Explore
The student is expected to:
Proving a Pythagorean Identity
In the previous lesson, you learned that the coordinates of any point (x, y) that lies on the unit circle where the
y
terminal ray of an angle θ intersects the circle are x = cos θ and y = sin θ, and y = sin θ that tan θ = _
x . Combining
F-TF.C.8
Prove the Pythagorean identity sin 2(θ) + cos 2(θ) = 1 and use it to
find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the
sin θ
these facts gives the identity tan θ = ____
, which is true for all values of θ where cos θ ≠ 0. In the following Explore,
cos θ
you will derive another identity based on the Pythagorean theorem, which is why the identity is known as a
Pythagorean identity.
Mathematical Practices
COMMON
CORE
Date
18.5 Using a Pythagorean Identity
Common Core Math Standards
COMMON
CORE
Class

MP.1 Problem Solving
The terminal side of an angle θ intersects the unit circle at the point (a, b) as shown. Write a and b
in terms of trigonometric functions involving θ.
a=
Language Objective
b=
Explain to a partner what the Pythagorean Theorem and the Pythagorean
identity are and how they are related.
cos θ
sin θ
y
(a, b)
1
b
θ
a
Essential Question: How can you use a
given value of one of the trigonometric
functions to calculate the values of the
other functions?
The values of the sine, cosine, and tangent functions
are connected by the identities
sinθ
sin 2(θ) + cos 2(θ) = 1 and tanθ =
. Once you
cosθ
know the value of one of the functions, you can use
the identities to solve for the others, using the
quadrant of the terminal side of θ to assign the
correct sign.
_
PREVIEW: LESSON

Apply the Pythagorean theorem to the right triangle in the diagram. Note that when a trigonometric
function is squared, the exponent is typically written immediately after the name of the function. For
2
instance, (sinθ) = sin 2θ.
Write the Pythagorean Theorem.
a + b2 = c
2
Substitute for a, b, and c.
Square each expression.
Module 18
2
( cos θ ) + ( sin θ ) =
2
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
2
+
2
cos θ
=
2
sin θ
2
1
1
Lesson 5
981
gh "File info"
Date
Class
n Identity
thagorea
Name
a Py
18.5 Using
to
ic functions
trigonometr
one of the
value of
Resource
use a given other functions?
(θ), cos(θ), or
Locker
can you
it to find sin
of the
ion: How
2
1 and use
the values
2
cos (θ) =
calculate
y sin (θ) +
of the angle.
orean identit
the Pythag
(θ) and the
F-TF.C.8 Prove (θ), cos(θ), or tan
tity
Iden
sin
where the
tan(θ) given
agorean
y
the unit circle _
ving a Pyth
. Combining
that lies on
y
=
)
x,
Pro
θ
x
point (
e,
θ that tan
Explore
nates of any
ing Explor
and y = sin
the coordi
the follow
y = sin θ,
θ ≠ 0. In
learned that
cos θ and
cos
=
you
,
x
a
as
are
of θ where
us lesson
known
cts the circle
identity is
In the previo
for all values
angle θ interse
is why the
sin θ , which is true
____
ray of an
m, which
θ = cos θ
terminal
orean theore
identity tan
the Pythag
gives the
and b
y based on
these facts
. Write a
another identit
b) as shown
you will derive
point (a,
circle at the
identity.
cts the unit
Pythagorean
interse
θ
angle
θ.
Quest
Essential
COMMON
CORE
IN2_MNLESE389847_U7M18L5.indd 981
an
involving
al side of
ic functions
The termin
of trigonometr
in terms
cos θ
a=
sin θ
b=

HARDCOVER PAGES 981992
Watch for the hardcover
student edition page
numbers for this lesson.
y
(a, b)
1
θ
y
g Compan
View the online Engage. Discuss the photo and how
light and shadows can help determine where to place
a cell tower. Then preview the Lesson
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
x
n Mifflin
Harcour t
Publishin

x
ometric
when a trigon
m. Note that of the function. For
the diagra
triangle in
after the name
to the right
immediately
theorem
lly written
Pythagorean
ent is typica
Apply the
d, the expon
2
is square
2 .
2
2
2
function
c
θ
a +b =
(sinθ) = sin
2
instance,
2
em.
Theor
2
1
Pythagorean
Write the
sin θ =
cos θ +
for a,
Substitute
b
a
b, and c.
expression.
Square each
(
2
cos θ
) (
+
2
sin θ
)
=
1
Lesson 5
981
Module 18
8L5.indd
47_U7M1
ESE3898
IN2_MNL
981
Lesson 18.5
981
19/04/14
12:24 AM
19/04/14 12:26 AM
Reflect
1.
EXPLORE
The identity is typically written with the sine function first. Write the identity this way, and explain why it
is equivalent to the one in Step B.
2
2
The rewritten identity is sin θ + cos θ = 1 . It is equivalent to the original identity because
Proving a Pythagorean Identity
of the Commutative Property of Addition.
2.
π.
Confirm the Pythagorean identity for θ = _
3
√―
3
π
π
1 ; therefore,
sin _ = ___
and cos _ = _
2
3
3
2
2
√―
3
2
2
2 _
2 _
π
π
1
sin θ + cos θ = sin
+ cos
= ____
+ _
2
3
3
2
( )
3.
INTEGRATE TECHNOLOGY
( )
( )
( ) ( ) ( ) = _34 + _14 = 1.
Students have the option of completing the Explore
activity either in the book or online.
2
3π .
Confirm the Pythagorean identity for θ = _
4
√―
√―
2
2
3π
3π
and cos _ = -___
. Therefore,
sin _ = ___
2
2
4
4
( )
( )
sin θ + cos θ = sin
2
2
Explain 1
2
( )
( )
2 3π
3π
_
+ cos _ =
4
4
√2
)
(_
_
2
2
CONNECT VOCABULARY
(
√2
_
)
_
+ -
2
2
Have the class list expressions that contain the word
identity, such as identity theft, or student Identity card.
Ask them to discuss the meaning of identity in each
expression. Remind them that 0 is the identity
element for addition and 1 is the identity element for
multiplication. Then have them discuss the use of the
word identity in the context presented in the lesson.
1 +_
1 = 1.
=_
2 2
Finding the Value of the Other Trigonometric
Functions Given the Value of sin θ or cos θ
You can rewrite the identity sin θ + cos θ = 1 to express one trigonometric function in terms of the other.
As shown, each alternate version of the identity involves both positive and negative square roots. You can
determine which sign to use based on knowing the quadrant in which the terminal side of θ lies.
2
2
Solve for sin θ
Solve for cos θ
sin2 θ + cos2 θ = 1
sin2 θ = 1 - cos2θ
__
cos θ = ± √1 - sin2θ
© Houghton Mifflin Harcourt Publishing Company
sin θ = ± √1 - cos2θ

How do you know the hypotenuse is equal to
1? The radius of a unit circle is 1.
cos2 θ = 1 - cos2θ
__
Example 1
QUESTIONING STRATEGIES
sin2 θ + cos2 θ = 1
Find the approximate value of each trigonometric function.
π , find cos θ.
Given that sinθ = 0.766 where 0 < θ < _
2
__
Use the identity to solve for cosθ.
cos θ = ±√1 - sin2 θ
__
Substitute for sinθ.
= ±√1 -(0.766)
Use a calculator, then round.
≈ ±0.643
2
Is (cosθ) the same as cos 2θ? Is (sinθ) the
same as sin 2θ? Yes, they are different ways of
writing the same thing, because the angle measure
is not actually squared.
2
2
If cos 2θ has a value of 0.36, what do you know
2
sin θ = 1 - 0.36 = 0.64, so sinθ = √0.64 = 0.8.
――
The terminal side of θ lies in Quadrant I, where cos θ > 0. So, cos θ ≈ 0.643.
EXPLAIN 1
Module 18
982
Lesson 5
PROFESSIONAL DEVELOPMENT
IN2_MNLESE389847_U7M18L5.indd 982
Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.1,
which calls for students to “make sense of problems and persevere in solving
them.” Students analyze given information and constraints and plan a pathway to
the solution. The solution process may involve transforming established identities
in order to make use of the given information. In the course of solving the
problem, students may revisit the given information to make any additional
decisions regarding the solution.
19/04/14 12:26 AM
Finding the Value of the Other
Trigonometric Functions Given the
Value of sin θ or cos θ
QUESTIONING STRATEGIES
Is 1 - sin 2θ ever negative? Why or why
not? No; -1 ≤ sinθ ≤ 1, so 0 ≤ sin 2θ ≤ 1.
Therefore, 1 - sin 2θ can never be less than 0.
Using a Pythagorean Identity 982
B
QUESTIONING STRATEGIES
3π
Given that cos θ = -0.906 where π < θ < ___
, find sinθ.
2
Do you need to know the precise value of θ to
solve these problems? Why or why not? No;
the Pythagorean identity allows you to find the
cosine of θ given the sine of θ without knowing the
value of θ itself.
__
sin θ = ±√1 - cos2θ
Use the identity to solve for sinθ.
___
√ (
Substitute for cos θ.
= ± 1 - -0.906
Use a calculator, then round.
≈ ± 0.423
)
2
The terminal side of θ lies in Quadrant III , where sin θ < 0. So, sin θ ≈ -0.423 .
Reflect
Do you need to know the quadrant where θ
terminates to solve these problems? Why or
why not? Yes; the sign of the answer varies
depending on the quadrant in which θ terminates.
4.
π
π
Suppose that __
< θ < π instead of 0 < θ < __
in part A of this Example. How does
2
2
this affect the value of sin θ?
π
< θ < π, then the terminal side of θ lies in Quadrant II. Since cosine is negative in
If __
2
Quadrant II, the value of cos θ would be negative.
5.
AVOID COMMON ERRORS
3π
3π
Suppose that ___
< θ < 2π instead of π < θ < ___
in part B of this Example. How
2
2
does this affect the value of sin θ ?
< θ < 2π, then the terminal side of θ lies in Quadrant IV. Sine is also negative in
If 3π
2
___
―
Students may erroneously reason that since
sin 2θ + cos 2θ = 1, sinθ + cosθ = √1 = 1. Use a
special angle to demonstrate that this is not true, and
correct the erroneous reasoning by reminding
students that √x 2 + y 2 ≠ x + y.
Quadrant IV, so the value of sin θ would not change.
6.
―――
Explain how you would use the results of part A of this Example to determine the
approximate value for tan θ. Then find it.
sin θ
To determine the approximate value for tanθ, use the identity tan θ =
.
cos θ
0.766
____
tan θ =
≈ 1.191
_
0.643
7.
3π
Given that sinθ = -0.644 where π < θ < ___
, find cosθ.
2
_
© Houghton Mifflin Harcourt Publishing Company
cos θ = ±√1 - sin2θ
___
= ±√1 - (-0.644)
2
≈ ±0.765
Since θ lies in Quadrant III, where cos θ < 0, cos θ ≈ -0.765.
8.
π
Given that cos θ = -0.994 where __
< θ < π, find sin θ. Then find tanθ.
2
__
sin θ = ±√1 - cos2 θ
___
2
= ±√1 - (-0.994)
= ±0.109
Since θ lies in Quadrant II, where sin θ > 0, sin θ ≈ 0.109.
0.109
tan θ = sinθ ≈ _____
≈ -0.110
cosθ -0.994
_
Module 18
983
Lesson 5
COLLABORATIVE LEARNING
IN2_MNLESE389847_U7M18L5.indd 983
Peer-to-Peer Activity
Have students work in pairs. Instruct pairs to write three different identities that
show the relationships among the sine, cosine, and tangent functions. Then have
them verify each of the three identities for θ = 30°, θ = 45°, and θ = 60°. Have
each pair compare their work with that of another pair.
983
Lesson 18.5
19/04/14 12:26 AM
Explain 2
Finding the Value of Other Trigonometric
Functions Given the Value of tanθ
EXPLAIN 2
sinθ
If you multiply both sides of the identity tanθ = ____
by cosθ, you get the identity cosθtanθ = sinθ, or
cosθ
sinθ
sinθ = cosθ tanθ. Also, if you divide both sides of sinθ = cosθ tanθ by tanθ, you get the identity cosθ = ____
.
tanθ
You can use the first of these identities to find the sine and cosine of an angle when you know the tangent.
Example 2

Finding the Value of the Other
Trigonometric Functions Given the
Value of tan θ
Find the approximate value of each trigonometric function.
π
Given that tanθ ≈ -2.327 where __
< θ < π, find the values of sinθ and cosθ.
2
First, write sinθ in terms of cosθ.
Use the identity sinθ = cosθ tanθ.
QUESTIONING STRATEGIES
sinθ = cosθ tanθ
≈ -2.327cosθ
Substitute the value of tanθ.
Is there another way to begin the problem
besides writing sinθ in terms of cosθ? Yes,
you can write cosθ in terms of sinθ.
Now use the Pythagorean Identity to find cosθ. Then find sinθ.
sin2θ = cos θ = 1
2
Use the Pythagorean Identity.
Substitute for sinθ.
(-2327 cosθ)2 + cos2θ ≈ 1
How can you rewrite the Pythagorean identity
with cosθ written in terms of sinθ and tanθ?
5.415 cos2θ + cos2θ ≈ 1
Square.
(_) = 1
6.415 cos2θ ≈ 1
Combine like terms.
Solve for cos2θ.
cos2 ≈ 0.156
Solve for cosθ.
cosθ ≈ ±0.395
sin 2θ + sinθ
tanθ
INTEGRATE TECHNOLOGY
The terminal side of θ lies in Quadrant II, where cosθ < 0. Therefore,
cosθ ≈ −0.395 and sinθ ≈ −2.327cosθ ≈ 0.919.

Students can use a graphing calculator to
check that the sum of the squares of the values
that they find for sinθ and cosθ is 1. Note that,
because the values are approximations, the calculator
may return a sum that is not exactly 1.
3π
Given that tanθ ≈ -4.366 where ___
< θ < 2π, find the values of sinθ and cosθ.
2
First, write sinθ in terms of cosθ.
Use the identity sinθ = cosθ tanθ.
sinθ = cosθ tanθ
© Houghton Mifflin Harcourt Publishing Company
≈ -4.366 cosθ
Substitute the value of tanθ.
Now use the Pythagorean Identity to find cosθ . Then find sinθ .
Use the Pythagorean Identity.
sin2θ + cos2θ = 1
( -4.366 cosθ) + cos θ ≈ 1
2
Substitute for sinθ.
2
19.062 cos2θ + cos2θ ≈ 1
Square.
Combine like terms.
2
20.062 cos2θ ≈ 1
Solve for cos2θ.
cos2θ ≈
0.050
Solve for cosθ.
cosθ ≈
0.223
The terminal side of θ lies in Quadrant IV , where cosθ > 0. Therefore, cosθ ≈
0.223
and sinθ ≈ -4.366 cosθ ≈ -0.974 .
Module 18
984
Lesson 5
DIFFERENTIATE INSTRUCTION
IN2_MNLESE389847_U7M18L5.indd 984
19/04/14 12:26 AM
Multiple Representations
Students may benefit from relating the concepts in this lesson to right triangle
trigonometry. For example, if students are given that sinθ = 0.766 and 0 < θ < __π2 , they
can draw a right triangle in the first quadrant, with angle θ at the origin and the
side opposite
to label
adjacent leg on the x-axis. They can then use the definition sinθ = _________
hypotenuse
the vertical leg 766 units and the hypotenuse 1000 units, and use the Pythagorean
Theorem to find the length of the missing leg. Finally, the definitions cosθ = _________
hypotenuse
side oppositte
can be used to find these values. Students can check their
and tanθ = _________
results using the methods from the lesson. In doing so, students will likely see the
connection between the two methods of solution.
Using a Pythagorean Identity
984
Reflect
ELABORATE
9.
QUESTIONING STRATEGIES
In part A of this Example, when you multiplied the given value of tanθ by the
calculated value of cosθ in order to find the value of sinθ, was the product positive
or negative? Explain why this is the result you would expect.
The product was positive. This is expected, since sinθ should be positive in Quadrant II.
Suppose you know the sine and tangent of an
angle. Would you also need to be given
terminates in order to find the cosine of the angle?
Explain. No; the signs of the sine and tangent of the
angle would be enough information to determine
the quadrant in which the angle terminates and,
thus, the sign of the cosine.
π , show that you can solve for sinθ and cosθ exactly
10. If tanθ = 1 where 0 < θ < _
2
using the Pythagorean identity. Why is this so?
If tanθ = 1, then sinθ = cosθ , so sin2θ + cos2 θ = 1 becomes 2 sin2 θ = 1, which gives the
_
√2
π
result sinθ = cosθ = __
. This occurs because θ is the special angle __
.
4
2
3π , find the values of sinθ and cosθ.
11. Given that tanθ ≈ 3.454 where π < θ < _
2
Write sinθ in terms of cosθ.
sinθ = cosθ tanθ ≈ 3.454 cosθ
Solve for cosθ and sinθ.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Challenge students to use the identities in the
sin2θ + cos2θ = 1
(3.454 cosθ )2+ cos2θ ≈ 1
11.930 cos2θ + cos2θ ≈ 1
12.930 cos2θ ≈ 1
lesson to write a quotient identity that defines in
cos2θ ≈ 0.0773
1 - cos θ
1
or tan 2θ = ____
-1
terms of cosθ. tan 2θ = _______
cos 2θ
cos 2θ
2
cosθ ≈ ±0.278
Since cosine is negative and sine is negative in Quadrant III, cosθ = 0.278 and
If you know the sine of an angle, how can you
find the cosine and tangent of the angle? To
find the cosine, you can use the identity
sin θ + cos θ = 1, substitute the given value for
2
2
sin θ, and solve for cos θ. You can then substitute
the values of sin θ and cos θ in the identity
sinθ
tan θ = ____
to find the tangent.
cosθ
sinθ ≈ -3.454 cosθ ≈ -0.960.
© Houghton Mifflin Harcourt Publishing Company
SUMMARIZE THE LESSON
Elaborate
12. What conclusions can you draw if you are given only the information that tanθ = −1?
Possible Answer: Since the tangent is negative, the terminal angle must be either in
Quadrant II, where sine is positive and cosine is negative, or in Quadrant IV, where sine
is negative and cosine is positive. More specifically, sinθ = cosθ , so sin 2θ + cos 2θ = 1
―
―
―
―
becomes 2 cos 2 = 1, so cosθ = ± ____
. Then sinθ = ____
= and cosθ = -____
or sinθ = -____
2
2
2
2
―
√
2
and cosθ = ____
.
2
√2
Module 18
√2
985
√2
√2
Lesson 5
LANGUAGE SUPPORT
IN2_MNLESE389847_U7M18L5.indd 985
Communicate Math
Have students work in pairs. The first student explains what the Pythagorean
Theorem is while the second student takes notes. They switch roles and repeat the
procedure with the Pythagorean identity. Have them collaborate to describe the
relationship between the two, as well as similarities and differences. Encourage
students to use pictures and symbols as well as verbal descriptions when
explaining to their partners.
985
Lesson 18.5
19/04/14 12:26 AM
13. Discussion Explain in what way the process of finding the sine and cosine of an
angle from the tangent ratio is similar to the process of solving a linear equation in
two variables by substitution.
When solving a linear equation in two variables, you use one equation to find an
EVALUATE
expression for one variable in terms of the other, then substitute that expression in
the second equation to obtain an equation in one variable that you can solve for that
variable and use to find the value of the other variable. When finding the sine and cosine
of an angle from the tangent ratio, you use the known tangent value and the identity
sinθ = cosθ tanθ to write an expression for sinθ in terms of cosθ, then substitute that
ASSIGNMENT GUIDE
expression in the second identity sin 2θ + cos 2θ = 1 to obtain an equation that you can
solve for cosθ, and then use the value of cosθ to find the value of sinθ.
14. Essential Question Check-In If you know only the sine or cosine of an angle
and the quadrant in which the angle terminates, how can you find the other
trigonometric ratios?
If you know only sinθ or cosθ, you can find the other by substituting the known value
in the Pythagorean identity sin θ + cos θ = 1 and solving for the unknown value. Then
2
2
sinθ
you can find the tangent of the angle using the identity tanθ = ____
. In all cases, use the
cosθ
quadrant in which the angle terminates to choose the correct sign for the ratio.
Evaluate: Homework and Practice
• Online Homework
• Hints and Help
• Extra Practice
Find the approximate value of each trigonometric function.
1.
π
Given that sinθ = 0.515 where 0 < θ < __
, find cosθ.
―――
2
―――――
cosθ = ± √1 - sin 2θ = ± √1 - (0.515) ≈ ±0.857
Since θ lies in Quadrant I, where cosθ > 0, cosθ ≈ 0.857.
2.
3π
Given that cosθ = 0.198 where ___
< θ < 2π, find sinθ.
2
――――
―――――
sinθ = ± √1 - cos 2θ = ± √1 - (0.198) ≈ ±0.980
2
Since θ lies in Quadrant IV, where sinθ < 0, sinθ ≈ −0.980.
3.
3π
Given that sinθ = −0.447 where ___
< θ < 2π, find cosθ.
2
―――――
―――
cosθ = ± √1 - sin θ = ± √1 - (-0.447) ≈ ±0.895
2
2
―――――
2
Since θ lies in Quadrant II, where sinθ > 0, sinθ ≈ 0.839.
Exercise
IN2_MNLESE389847_U7M18L5.indd 986
Lesson 5
986
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–16
1 Recall of Information
MP.7 Using Structure
17
1 Recall of Information
MP.6 Precision
18–19
1 Recall of Information
MP.7 Using Structure
20
2 Skills/Concepts
MP.4 Modeling
21
3 Strategic Thinking
MP.4 Modeling
22–23
2 Skills/Concepts
MP.3 Logic
24–25
3 Strategic Thinking
MP.3 Logic
Example 1
Finding the Value of the Other
Trigonometric Functions Given the
Value of sin θ or cos θ
Exercises 1–8, 20
Example 2
Finding the Value of the Other
Trigonometric Functions Given the
Value of tan θ
Exercises 9–17,
21
Suggest that students draw a sketch of the angle
described in the problem, placing the terminal side of
the angle in the correct quadrant. Students can then
write the signs of each of the functions for that
quadrant so that they will remember to affix the
proper signs to the calculated values.
sinθ = ± √1 - cos 2θ = ± √1 - (-0.544) ≈ ±0.839
Module 18
Exercises 18–19
VISUAL CUES
π
Given that cosθ = −0.544 where __
< θ < 2π, find sinθ.
2
―――
Explore
Proving a Pythagorean Identity
If you are given the cosine of an angle, why is
it necessary to know in which quadrant the
terminal side of the angle lies in order to determine
the sine of the angle? Knowing the quadrant
enables you to determine the sign of the sine.
Since θ lies in Quadrant IV, where cosθ > 0, cosθ ≈ 0.895.
4.
Practice
QUESTIONING STRATEGIES
© Houghton Mifflin Harcourt Publishing Company
2
Concepts and Skills
19/04/14 12:26 AM
Using a Pythagorean Identity
986
5.
AVOID COMMON ERRORS
3π
Given that sinθ = −0.908 where π < θ < ___
, find cosθ.
2
―――
―――――
cosθ = ± √1 - sin 2θ = ± √1 - (-0.908) ≈ ±0.419
2
Since θ lies in Quadrant III, where cosθ > 0, cosθ ≈ −0.419.
When making a substitution from the transformation
sinθ
into the Pythagorean
of the identity tanθ = ___
cosθ
6.
identity, some students may forget to square the
coefficient of the substituted expression. Review that
π
< θ < π, find cosθ.
Given that sinθ = 0.313 where __
2
―――
―――――
cosθ = ± √1 - sin 2θ = ± √1 - (-0.313) ≈ ±0.950
2
Since θ lies in Quadrant II, where cosθ < 0, cosθ ≈ −0.950.
sin 2θ means (sinθ) , and the same for cosine, and
encourage students to use parentheses when making
the substitution.
2
7.
π
, find sinθ.
Given that cosθ = 0.678 where 0 < θ < __
2
――――
―――――
sinθ = ± √1 - cos 2θ = ± √1 - (-0.678) ≈ ±0.735
2
Since θ lies in Quadrant I, where sinθ > 0, sinθ ≈ 0.735.
INTEGRATE TECHNOLOGY
8.
3π
, find sinθ.
Given that cosθ = -0.489 where π < θ < ___
2
――――
―――――
sinθ = ± √1 - cos 2θ = ± √1 - (-0.489) ≈ ±0.872
Encourage students to use their calculators to
check their results, checking that for the given
2
Since θ lies in Quadrant III, where sinθ < 0, sinθ ≈ -0.872.
sinθ
.
angle θ, tanθ = ___
Find the approximate value of each trigonometric function.
cosθ
9.
__ < θ < π, find the values of sinθ and cosθ.
Given that tanθ ≈ −3.966 where π
2
sinθ = cosθ tanθ ≈ -3.966 cosθ
sin 2θ + cos 2θ = 1
(-3.966 cosθ)2 + cos 2θ ≈ 1
© Houghton Mifflin Harcourt Publishing Company
cos 2θ ≈ 0.060 → cosθ ≈ ±0.245
In Quad. II, cosine is neg. and sine is pos., so cosine is negative and sine is
positive. cosθ ≈ −0.245 and sinθ ≈ −3.966 cosθ ≈ 0.972.
3π
< θ < 2π, find the values of sinθ and cosθ.
10. Given that tanθ ≈ −4.580 where ___
2
sinθ = cosθ tanθ ≈ -4.580 cosθ
sin 2θ + cos 2θ = 1
(-4.580 cosθ)2 + cos 2θ ≈ 1
cos 2θ ≈ 0.046 → cosθ ≈ ±0.214
Since cosine is positive in Quadrant IV, cosθ ≈ 0.214 and
sinθ ≈ −4.580 cosθ ≈ 0.980.
Module 18
IN2_MNLESE389847_U7M18L5.indd 987
987
Lesson 18.5
987
Lesson 5
19/04/14 12:26 AM
π
11. Given that tanθ ≈ 7.549 where 0 < θ < __
, find the values of sinθ and cosθ.
2
Write sinθ in terms of cosθ.
sinθ = cosθ tanθ ≈ 7.549 cosθ
Solve for sinθ and cosθ.
sin 2θ + cos 2θ = 1
(7.549 cosθ)2 + cos 2θ ≈ 1
56.987 cos 2θ + cos 2θ ≈ 1
57.987 cos 2θ ≈ 1
cos 2θ ≈ 0.017
cosθ ≈ ±0.130
Since cosine is positive in Quadrant I, cosθ ≈ −0.130 and sinθ ≈ 7.549 cosθ ≈ 0.981.
3π
, find the values of sinθ and cosθ.
12. Given that tanθ ≈ 4.575 where π < θ < ___
2
Write sinθ in terms of cosθ.
sinθ = cosθ tanθ ≈ 4.575 cosθ
Solve for sinθ and cosθ.
sin 2θ + cos 2θ = 1
(4.575 cosθ)2 + cos 2θ ≈ 1
20.931 cos 2θ + cos 2θ ≈ 1
21.931 cos 2θ ≈ 1
cos 2θ ≈ 0.046
cosθ ≈ ±0.214
Since cosine is negative in Quadrant III, cosθ ≈ −0.214 and sinθ ≈ 4.575 cosθ ≈ -0.979.
© Houghton Mifflin Harcourt Publishing Company
3π
, < θ < 2π find the values of sinθ and cosθ.
13. Given that tanθ ≈ -1.237 where ___
2
Write sinθ in terms of cosθ.
sinθ = cosθ tanθ ≈ −1.237 cosθ
Solve for sinθ and cosθ.
sin 2θ + cos 2θ = 1
(-1.237 cosθ)2 + cos 2θ ≈ 1
1.530 cos 2θ + cos 2θ ≈ 1
2.530 cos 2θ ≈ 1
cos 2θ ≈ 0.395
cosθ ≈ ±0.628
Since cosine is negative in Quadrant IV, cosθ ≈ 0.628 and sinθ ≈ -1.237 cosθ ≈ -0.777.
Module 18
IN2_MNLESE389847_U7M18L5.indd 988
988
Lesson 5
19/04/14 12:26 AM
Using a Pythagorean Identity
988
3π
14. Given that tanθ ≈ 5.632 where π < θ < ___
, find the values of sinθ and cosθ.
2
Write sinθ in terms of cosθ.
sinθ = cosθ tanθ ≈ 5.632 cosθ
Solve for sinθ and cosθ.
sin 2θ + cos 2θ = 1
(5.632 cosθ)2 + cos 2θ ≈ 1
31.719 cos 2θ + cos 2θ ≈ 1
32.719 cos 2θ ≈ 1
cos 2θ ≈ 0.031
cosθ ≈ ±0.176
Since cosine is negative in Quadrant III, cosθ ≈ -0.176 and sinθ ≈ 5.632 cosθ ≈ -0.991.
π
, find the values of sinθ and cosθ.
15. Given that tanθ ≈ 6.653 where 0 < θ < __
2
Write sinθ in terms of cosθ.
sinθ = cosθ tanθ ≈ 6.653 cosθ
Solve for sinθ and cosθ.
sin 2θ + cos 2θ = 1
(6.653 cosθ)2 + cos 2θ ≈ 1
44.262 cos 2θ + cos 2θ ≈ 1
45.262 cos 2θ ≈ 1
cos 2θ ≈ 0.022
cosθ ≈ ±0.148
© Houghton Mifflin Harcourt Publishing Company
Since cosine is positive in Quadrant I, cosθ ≈ -0.148 and sinθ ≈ 6.653 cosθ ≈ 0.985.
π
, < θ < π find the values of sinθ and cosθ.
16. Given that tanθ ≈ -9.366 where __
2
Write sinθ in terms of cosθ.
sinθ = cosθ tanθ ≈ -9.366 cosθ
Solve for sinθ and cosθ.
sin 2θ + cos 2θ = 1
(-9.366 cosθ)2 + cos 2θ ≈ 1
87.722 cos 2θ + cos 2θ ≈ 1
88.722 cos 2θ ≈ 1
cos 2θ ≈ 0.011
cosθ ≈ ±0.105
Since cosine is negative in Quadrant II, cosθ ≈ -0.105 and sinθ ≈ -9.366 cosθ ≈ 0.983.
Module 18
IN2_MNLESE389847_U7M18L5.indd 989
989
Lesson 18.5
989
Lesson 5
19/04/14 12:26 AM
17. Given the trigonometric function and the location of the terminal angle, state whether
the function value will be positive or negative.
PEERTOPEER DISCUSSION
A. Positive
B. Negative
C. Negative
D. Negative
E. Positive
Ask students to discuss with a partner how they
could use what they know about the trigonometry of
sinθ
.
a right triangle to prove the identity tanθ = ___
cosθ
Have them discuss their proofs with the class.
Students’ work should show that since
side opposite
sinθ = _____________ and
hypotenuse
cosθ = ____________ , then
hypotenuse
7π
18. Confirm the Pythagorean identity sin 2θ + cos 2θ = 1 for θ = ___
.
4
√―
√―
2
2
7π
7π
= -____
and cos __
= ____
. Therefore,
sin __
4
4
2
2
( )
( )
7π
7π
sin θ + cos θ = sin (__
) + cos2(__
)=
2
2
2
4
4
(-____― ) + (____― ) = __ + __ = 1.
√2
2
2
√2
2
2
1
2
1
2
side opposite
_________
hypotenuse
sinθ = __________
_____
cosθ
_________
hypotenuse
hypotenuse
side opposite ____________
____________
=
·
19. Recall that the equation of a circle with radius r centered at the origin
is x 2 + y 2 = r 2. Use this fact and the fact that the coordinates of a point on this circle
are (x, y) = (rcosθ, r sinθ) for a central angle θ to show that the Pythagorean identity
derived above is true.
2
2
x 2 + y 2 = r 2 ⇒ (r cosθ) + (rsinθ) = r 2
r 2 cos 2θ + r 2 sin 2θ = r 2 ⇒ cos 2θ + sin 2θ = 1
hypotenuse
side opposite
=
____________
20. Sports A ski supply company is testing the friction of a new ski wax
by placing a waxed block on an inclined plane covered with wet snow.
The incline plane is slowly raised until the block begins to slide. At the
instant the block starts to slide, the component of the weight of the
block parallel to the incline, mgsinθ, and the resistive force of friction,
μmgcosθ, are equal, where μ is the coefficient of friction. Find the value
of μ to the nearest hundredth if sinθ = 0.139 at the instant the block
begins to slide.
Find cosθ. Because cosθ will be between 0 and 90 degrees,
cosθ will be positive.
= tanθ
Photos/Alamy
―――
cosθ = √1 - sin 2θ =
―――――
√1 - (0.139) 2 ≈ 0.928
mg sinθ = μ mg cosθ ⇒ sinθ = μ cosθ
0.139 = μ cosθ
0.139
0.139
μ=
≈
≈ 0.15
0.928
cosθ
21. Driving Tires and roads are designed so that the coefficient of friction between
the rubber of the tires and the asphalt of the roads is very high, which gives plenty
of traction for starting, stopping, and turning. For a particular road surface and tire,
the steepest angle for which a car could rest on the slope without starting to slide has
a sine of 0.643. This value satisfies the equation mgsinθ = μmgcosθ where μ is the
coefficient of friction. Find the value of μ to the nearest hundredth.
Because cosθ will be between 0 and 90 degrees, cosθ will be positive.
_ _
―――
―――――
cosθ = √1 - sin 2θ = √1 - (0.643) ≈ 0.766
mg sinθ = μ mg cosθ ⇒ sinθ = μ cosθ
0.643 = μ cosθ ⇒
0.643
0.643
≈
≈ 0.84
μ=
0.766
cosθ
2
_ _
Module 18
IN2_MNLESE389847_U7M18L5.indd 990
990
Lesson 5
19/04/14 12:26 AM
Using a Pythagorean Identity
990
H.O.T. Focus on Higher Order Thinking
JOURNAL
3π
22. Explain the Error Julian was given that sinθ = -0.555 where ___
, < θ < 2π and
2
told to find cosθ . He produced the following work:
Have students explain how knowing that the sine of
an angle is approximately 0.2588 enables them to find
two possible values for the cosine of the angle. Have
them find these values, and tell what information
would be necessary to decide which of the two
possible values is the correct cosine of the angle.
__
cosθ = ±√1 - sin 2θ
__
= ±√1 - (-0.555 2)
≈ ±1.144
3π < θ < 2π, cosθ ≈ 1.144.
Since cosθ > 0 when _
2
Explain his error and state the correct answer.
Julian added the two quantities inside of the square root together instead
of subtracting. He should have noticed, since the cosine of a number is
never more than 1. He should have done the following:
――――
―――――
= ± √1 - (-0.555)
cosθ = ± √1 - sin 2θ
2
≈ ±0.832
Since cosθ > 0 when
3π
_
< θ < 2π, cosθ ≈ 0.832.
2
Critical Thinking Rewrite each trigonometric expression in terms of cosθ and
simplify.
sin 2θ
23. _
1 - cosθ
sin 2θ
1 - cosθ
1 - cos 2θ
=
1 - cosθ
(1- cosθ)(1+ cosθ)
=
1- cosθ
_
_
__
© Houghton Mifflin Harcourt Publishing Company
= 1 + cosθ
24. cosθ + sinθ cosθ - tanθ + tanθ sin 2θ (Hint: Begin by factoring tanθ from the last
two terms.)
cosθ + sinθ cosθ - tanθ + tanθ sin 2θ
= cosθ + sinθ cosθ - tanθ(1 - sin 2θ)
= cosθ + sinθ cosθ - tanθ cos 2θ
= cosθ + sinθ cosθ -
sinθ
_
cos θ
2
cosθ
= cosθ + sinθ cosθ - sinθ cosθ
__
= cosθ
√ 1 - cos 2θ
__
25. Critical Thinking To what trigonometric function does the expression _________
√ 1 - sin 2θ
――――
――――
The expression __________
simplifies to tanθ. If you rewrite __________
―――
―――
2
2
√1 - cos 2θ
√1 - sin θ
――
√1 - cos 2θ
√1 - sin θ
using the Pythagorean Identity, you get the result _______
. Taking the
――
2
√sin 2θ
√cos θ
sinθ
square root results in the expression ____
, which is equivalent to tanθ.
cosθ
Module 18
IN2_MNLESE389847_U7M18L5.indd 991
991
Lesson 18.5
991
Lesson 5
19/04/14 12:25 AM
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Students likely have analyzed the relationships
A tower casts a shadow that is 160 feet long at a particular time one
morning. With the base of the tower as the origin, east as the positive
x-axis, and north as the positive y-axis, the shadow at this time is
in the northwest quadrant formed by the axes. Also at this time, the
tangent of the angle of rotation measured so that the shadow lies on
the terminal ray is tanθ = −2.545. What are the coordinates of the
tip of the shadow to the nearest foot, and what do they indicate?
between shadow length and other measures, such as
object height, in two dimensions. Remind them that
they have used indirect measurement and proportion
as well as the Pythagorean Theorem to approach
these problems. Note that the Lesson Performance
Task expands these analyses to allow for positions in
all four quadrants using positive and negative
coordinate positioning.
The coordinates of a point on a circle centered at the
origin with radius r are r cosθ and r sinθ. Use identities
relating the sine, cosine, and tangent of an angle to find
cosθ and sinθ, then use those values to find the coordinates.
First use the value of the tangent and the identity sinθ = cosθ tanθ to find an expression for sinθ
in terms of cosθ.
CURRICULUM INTEGRATION
sinθ = cosθ tanθ = -2.545 cosθ
Note that a counterclockwise movement of the sun’s
shadows occurs only in the Southern Hemisphere.
Most sundials, which were first introduced in the
Northern Hemisphere, assume that shadows move
clockwise, and this is also the reason for the
development of clockwise-turning clock hands! Point
out to students that the sun appears to move in the
opposite direction from its shadows—the sun moves
counterclockwise in the Northern Hemisphere, so its
shadows move in the opposite direction.
Substitute this expression for sinθ into the Pythagorean identity
which means that the value of cosθ will be negative.
sin 2θ + cos 2θ = 1
(-2.545 cosθ)2 + cos 2θ ≈ 1
6.477 cos 2θ + cos 2θ ≈ 1
7.477 cos 2θ ≈ 1
cos 2θ ≈ 0.1337
cosθ ≈ -0.366
Now substitute to find sinθ.
sinθ = -2.545 cosθ ≈ -2.545(-0.366) ≈ 0.931
Use the fact that r = 160 to find the coordinates of the tip of the shadow.
r cosθ = 160(-0.366) = -58.56
r sinθ = 160(0.931) = 148.96
Shutterstock
sin 2θ + cos 2θ = 1 to solve for cosθ. Since the shadow is in the northwest, it is in Quadrant II,
The coordinates are (-59,149). This indicates that the tip of the shadow is about 58 feet west
and 149 feet north of the base of the tower.
Module 18
992
Lesson 5
EXTENSION ACTIVITY
IN2_MNLESE389847_U7M18L5.indd 992
Your location with respect to two cell towers can be determined by a process
called triangulation. Have students research how triangulation, a use of angles to
find distances, is also employed in other activities, such as mapmaking. Students
will encounter and may present the Law of Cosines, which states that
c 2 = a 2 + b 2 - 2ab cos C.
19/04/14 12:25 AM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Using a Pythagorean Identity
992
MODULE
18
MODULE
STUDY GUIDE REVIEW
Study Guide Review
Essential Question: How can you use trigonometry with right
triangles to solve real-world problems?
ASSESSMENT AND INTERVENTION
KEY EXAMPLE
(Lesson 18.1)
C
60
Assign or customize module reviews.
11
MODULE
A
61
length of leg opposite ∠A
tanA = ___
length of leg adjacent to ∠A
60 ≈ 5.45
tanA = _
11
KEY EXAMPLE
Mathematical Practices: MP.1, MP.2, MP.3, MP.4, MP.7
G-SRT.C.8, G-MG.A.1
• How to find the total height of layers of
staggered pipes: Students need to realize that the
centers of three staggered pipes form an
equilateral triangle. The height of the triangle is
the height of three rows of staggered pipes. The
height can be found in multiple ways, including
the Pythagorean Theorem or trigonometry, or
you can provide an algorithm to find the height.
• How to find the height of n layers of staggered
pipes: Students need to generalize the
relationship from the height of 10 layers of
staggered pipes to n layers.
• If the thickness of the pipes matters: Students
can brainstorm possibilities for both a stack and
a staggered stack of pipes. The thickness adds to
the height of the stack, but will add less to the
height of the staggered stack.
993
Module 18
© Houghton Mifflin Harcourt Publishing Company
SUPPORTING STUDENT REASONING
Definition of tangent
Substitute and simplify
(Lesson 18.2)
Find the sine and cosine of angle A.
length of leg opposite ∠A
sinA = ___
length of hypotenuse
60
_
sinA =
≈ .98
61
length of leg adjacent to ∠A
cosA = ___
length of hypotenuse
11
_
cosA =
≈ .18
61
KEY EXAMPLE
opposite (opuesto)
tangent (tangente)
cosine (coseno)
inverse trigonometric ratios
(proporción inversa
trigonométrica)
Pythagorean triple
(terna pitagórica)
Find the tangent of angle A.
COMMON
CORE
Key Vocabulary
tan -1 A (tan -1 A )
inverse tangent of ∠A
(tangente inversa de ∠A)
trigonometric ratio
(proporción trigonométrica)
sine (seno)
B
Students should begin this problem by focusing on
what information they will need. Here are some
issues they might bring up.
18
Trigonometry with
Right Triangles
Definition of sine
11
C
60
A
61
Simplify.
Definition of cosine
B
Simplify.
(Lesson 18.3)
_
Given an isosceles right triangle ∠DEF with ∠F = 90° and DE = 7, find the length
of the other two sides.
_
DE = DF√ 2
Apply the relationship of 45°-45°-90° triangles.
_
7 = DF√ 2
7_ = DF
_
√2
Substitute.
Simplify.
_
7√ 2
_
7_ =
DE = EF = _
2
√2
Module 18
Apply properties of isosceles triangles.
993
Study Guide Review
SCAFFOLDING SUPPORT
IN2_MNLESE389847_U7M18MC 993
• Suggest that students use the strategy of solving a simpler problem to help
them find the height of stacked pipes. Have them first find the height of two
layers, then find the height as another layer is added until they recognize
a pattern.
• One way to determine the height for two layers is to look at three pipes that
form an equilateral triangle, and determine the total height using the students’
knowledge of triangles. They can continue this process, adding more layers to
form larger triangles.
19/04/14 12:34 AM
EXERCISES
Given a right triangle △XYZ where ∠Z is a right angle, XY = 53, YZ = 28, and
XZ = 45, find the following rounded to the nearest hundredth. (Lessons 18.1, 18.2)
1.
sinX
0.53
2.
cosX
0.85
tanX
3.
SAMPLE SOLUTION
Methodology:
0.62
The centers of three staggered pipes form an
equilateral triangle. Find an expression for the
height of this triangle, which is two layers. Next, add
another layer to form an equilateral triangle using 6
pipes. Find an expression for the height of this
triangle, which is three layers. Continue the process
until a clear pattern emerges, then generalize to n
layers.
Find the area of the following triangle, rounding to the nearest tenth.
4.
triangle △ABC, where C = 127°, AC = 5, and BC = 9
18.0
Find the other two sides of the following triangle. Find exact answers in order of least to
greatest. (Lesson 18.3)
_
5.
7, 7√ 3
30°-60°-90° triangle with a hypotenuse of 14
The height of an equilateral triangle formed by the
centers of three pipes can be found using the
Pythagorean Theorem, a 2 + b 2 = c 2, where a = 1
in., c = 2 in.
How Much Shorter Are Staggered Pipe Stacks?
How much space can be saved by stacking pipe in a staggered pattern?
The illustration shows you the difference between layers of pipe stacked
directly on top of each other (left) and in a staggered pattern (right).
Suppose you have pipes that are 2 inches in diameter. How much
shorter will a staggered stack of 10 layers be than a non-staggered stack
with the same number of layers? In general, how much shorter are n
layers of staggered pipe?
2 in.
b
Start by listing in the space below how you plan to tackle the problem. Then use your own
paper to complete the task. Be sure to write down all your data and assumptions. Then use
numbers, graphs, diagrams, or algebraic equations to explain how you reached your conclusion.
© Houghton Mifflin Harcourt Publishing Company
1 in.
The total height of the three pipes is the height of
the triangle plus two times the radius of the pipes.
―
h 2 layers = 2 + √3
Continuing the calculations, we get:
―
= 2 + 3 √―
3
h 3 layers = 2 + 2 √3
h 4 layers
Or, generalizing,
Module 13
994
Study Guide Review
DISCUSSION OPPORTUNITIES
IN2_MNLESE389847_U7M18MC 994
• What assumptions are made when finding the height of a stack of pipes?
• What could be sources of error in finding the height of a staggered stack
of pipes?
―
h n layers = 2 + (n - 1) √3
19/04/14 12:34 AM
So the difference in the heights of 10 layers
staggered versus non-staggered is
―
10(2) - ⎡⎣2 + (10 - 1) √3 ⎤⎦≈ 2.41 in.
Assessment Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain.
0 points: Student does not demonstrate understanding of the problem.
Study Guide Review 994
18.1–18.5 Trigonometry with
Right Triangles
ASSESS MASTERY
students have mastered the concepts and standards
covered in this module.
• Online Homework
• Hints and Help
• Extra Practice
Solve the problem. (Lesson 18.1)
1.
ASSESSMENT AND INTERVENTION
A painter is placing a ladder to reach the third story window, which is 20 feet above the ground and
makes an angle with the ground of 70°. How far out from the building does the base of the ladder
need to be positioned?
20
20
20
_________
_
tan70° = __
x , so x = tan70° ≈ 2.747477419 ≈ 7.3 feet
2.
Access Ready to Go On? assessment online, and
receive instant scoring, feedback, and customized
intervention or enrichment.
Given the value of cos 30° = 1, write the sine of a complementary angle. Use an expression relating
trigonometric ratios of complementary angles. (Lesson 18.2)
cos θ° = sin(90 - θ)°
Substitute 30 into both sides.
cos30° = sin(90 - 30)°
Simplify the left side.
cos30° = sin60°
Substitute for cos30°.
1 = sin60°
Find the area of the regular polygon. (Lesson 18.3)
3.
What is the area of a regular hexagon with a distance from its center to a vertex
of 1 cm? (Hint: A regular hexagon can be divided into six equilateral triangles.)
Each of the six equilateral triangles has sides of length 1.
1
You can split each of these triangles into two congruent
Response to Intervention Resources
30-60-90 triangles. These make twelve 30-60-90 triangles.
Differentiated Instruction Resources
• Success for English Learners
• Challenge Worksheets
Assessment Resources
© Houghton Mifflin Harcourt Publishing Company
• Reteach Worksheets
―
3
1
Each of these triangles has a base of _
and a height of ___
.
2
2
√
The area of each of these equilateral triangles is
√―
√―
3
3
_1bh = _1 · _1 · ___
= ___
. The area of the hexagon is twelve
2
2 2
2
8
―
12 √3
3 √―
3
_____
times that, or
= ____
.
8
2
ESSENTIAL QUESTION
4.
How would you go about finding the area of a regular pentagon given the distance from its center to
the vertices?
Answers may vary. Sample: You can determine the area of a regular pentagon by
splitting it into 5 congruent triangles (each including a 360/5 = 72 degree angle).
Find the angles of the triangles and use trigonometric ratios to find their height and
base. With this information we can find the area of one triangle and multiply by 5 to
get the area of the polygon.
• Leveled Module Quizzes
Module 18
COMMON
CORE
IN2_MNLESE389847_U7M18MC 995
995
Module 18
Study Guide Review
995
Common Core Standards
19/04/14 12:34 AM
Content Standards Mathematical Practices
Lesson
Items
18.1
1
G-SRT.C.6, G-SRT.C.8
MP.2
18.2
2
G-SRT.C.7
MP.6
18.4
3
G-SRT.C.9
MP.6
MODULE
MODULE 18
MIXED REVIEW
MIXED REVIEW
1. Julia is standing 2 feet away from a lamppost. She casts a shadow of 5 feet and the light
makes a 20° angle relative to the ground from the top of her shadow. Consider each
expression. Does the expression give you the height of the lamppost?
Select Yes or No for A–C.
A. 7 sin20°
Yes
No
B. 7 tan20°
Yes
No
C. 7 tan70°
Yes
No
ASSESSMENT AND INTERVENTION
2. A right triangle has two sides with lengths 10 and 10. Choose True or False for each
statement.
A. The triangle has two angles that measure
True
False
45° each.
B. The triangle is equilateral.
True
False
C.
_
The length of the third side is 10√2 .
True
prepare students for high-stakes tests.
False
3. The measure of angle 1 is 125° and the measure of angle
2 is 55°. State two different relationships that can be
used to prove m∠3 = 125°.
4
Possible Answer: ∠2 and ∠3 are supplementary
1
3
2
Assessment Resources
angles, ∠1 and ∠3 are vertical angles.
• Leveled Module Quizzes: Modified, B
4. For the rhombus, specify how to find its area using the four
congruent right triangles with variable angle θ.
θ
AVOID COMMON ERRORS
Answers may vary. Sample: Find the base and height of each
right triangle using trigonometric ratios of θ. Multiply by 4 to
2
1
A r = 4 · __
bh = 2h cos θ sin θ
2
COMMON
CORE
Item 1 Some students will have a hard time
visualizing the problem. Encourage students to draw
and label a quick sketch to help them see which
trigonometric function best fits the situation.
© Houghton Mifflin Harcourt Publishing Company
get the area of the rhombus, as all the triangles are congruent.
Module 18
18
Study Guide Review
996
Common Core Standards
IN2_MNLESE389847_U7M18MC 996
19/04/14 12:34 AM
Content Standards Mathematical Practices
Lesson
Items
18.1
1
G-SRT.C.8
MP.4
18.3, 15.2
2*
G-SRT.C.8
MP.2
14.1
3*
G-CO.C.9
MP.6
18.2
4
G-SRT.C.8
MP.5
* Item integrates mixed review concepts from previous modules or a previous course.
Study Guide Review 996
UNIT
7
UNIT 7 MIXED REVIEW
MIXED REVIEW
1. Determine whether the statements are true.
Select True or False for each statement.
A. Dilations preserve angle measure.
ASSESSMENT AND INTERVENTION
B. Dilations preserve distance.
C. Dilations preserve collinearity.
D. Dilations preserve orientation.
B. Reflection across the x-axis
C. Dilation
D. Translation
• Leveled Unit Tests: Modified, A, B, C
• Performance Assessment
© Houghton Mifflin Harcourt Publishing Company
AVOID COMMON ERRORS
Yes
Yes
Yes
No
No
No
8
S
-4
T
Yes
No
B. Quadrilateral JKLM is a rhombus.
Yes
No
C. Quadrilateral JKLM is a rectangle.
Yes
No
Yes
Yes
No
No
Yes
No
4. Will the transformation produce similar figures?
Select Yes or No for each statement.
A. (x, y) → (x - 5, y + 5) → (-x, -y) → (3x, 3y)
B. (x, y) → (3x, y + 5) → (x, 3y) → (x - 1, y - 1)
C. (x, y) → (x, y + 5) → (2x, y) → (x + 5, y)
4
D
y
A
0
x
B
4
8
R C
Q
A. Quadrilateral JKLM is a parallelogram.
-8
5. Is ABC similar to DEF? Select Yes or No for each statement.
A. A (-1, -3), B (1, 3), C (3, -5)
D (2, -6), E (3, 0), F (6, -8)
B. A (-1, -3), B (1, 3), C (3, -5)
D (-5, -1), E (-4, 2), F (-3, -2)
C. A (-1, -3), B (1, 3), C (3, -5)
D (-2, -2), E (2, 4), F (2, -4)
Unit 7
COMMON
CORE
IN2_MNLESE389847_U7UC 997
Yes
No
Yes
No
Yes
No
997
Common Core Standards
Items
Unit 7
False
False
False
3. The vertices of quadrilateral JKLM are J(-2, 0), K(-1, 2), L(1, 3), and
M(0, 1). Can you use slopes and/or the distance formula to prove
each statement?
Select Yes or No for A–C.
Assessment Resources
997
False
True
True
True
2. Was the given transformation used to map ABCD to QRST?
Select Yes or No for each statement.
Yes
No
A. Reflection across the y-axis
prepare students for high-stakes tests.
Item 6 Some students will have trouble deciding
which sides correspond on the triangles, especially
when the triangles have different orientations.
Remind students that the smallest side on one shape
will always correspond to the smallest side on the
other shape. The same goes for the largest side.
True
• Online Homework
• Hints and Help
• Extra Practice
Content Standards Mathematical Practices
1
G-SRT.A.1
MP.6
2
G-SRT.A.1
MP.7
3*
G-CO.C.11
MP.2
4
G-SRT.A.2
MP.1
5*
G-SRT.A.1, S-ID.B.6
MP.5
19/04/14 12:37 AM
6. Are the triangles similar? Select Yes or No for each statement.
A.
Yes
No
N
There are three different levels of performance tasks:
R
14
12
*Novice: These are short word problems that
require students to apply the math they have learned
in straightforward, real-world situations.
10.5
O
P
M
Q
9
B.
Yes
T
12.8
S
**Apprentice: These are more involved problems
that guide students step-by-step through more
complex tasks. These exercises include more
complicated reasoning, writing, and open ended
elements.
No
16
U
V
20
C.
Yes
***Expert: These are open-ended, nonroutine
problems that, instead of stepping the students
through, ask them to choose their own methods for
solving and justify their answers and reasoning.
No
X
6
B
3
W
1
6
3
6
1
Y
7. Find the missing heights.
(ST) + (TU) = (SU)
2
2
P
T
―
45
2
Q
(45) 2 + (30) 2 = (SU) 2
――
50
R
S
30
U
5 √117 = (SU)
Unit 7
© Houghton Mifflin Harcourt Publishing Company
Using AA Similarity, the triangles are similar.
45
x = ___
___
(PQ) 2 + (QR) 2 = (PR) 2
50 30
45 . 50 (75) 2 + (50) 2 = (PR) 2
x = ___
30
x = 75
25 √13 = PR
998
COMMON
CORE
IN2_MNLESE389847_U7UC 998
Common Core Standards
Items
19/04/14 12:37 AM
Content Standards Mathematical Practices
6
G-CO.C.10, G-SRT.A.2
MP.5
7
G-CO.C.10
MP.2
* Item integrates mixed review concepts from previous modules or a previous course.
Unit 7
998
2.4 3.2
x = 2.1
Item 9 (6 points)
i
m
B Street
Distance: 2.1 + 2.4 = 4.5; 4.5 miles
1 point for finding each coordinate point, 3 in all
5th Ave
9. A city has a walkway between the middle school and the library
that can be represented in the image given. The city decides it
wants to place three trash cans, equally spaced along the walkway,
to help reduce any littering. Find the coordinates of the points at
which the trash cans should be placed, and then plot them on the
graph.
Subdivide the line segment SL into four equal parts. The
horizontal distance from point S to point L is 16, and one-fourth
of this is 4. So each trash can should be placed at horizontal
intervals of 4 units. The vertical distance from point S to point L
is 8, and one-fourth of this is 2, so each trash can should be
placed at vertical intervals of 2 units. The coordinates are (-4, 4),
(0, 2), and (4, 0).
Item 10 (6 points)
3 points for correct cost
3 points for explanation
© Houghton Mifflin Harcourt Publishing Company
The distance from K to L is 35 - (15 + 10) = 10 yd, so
8
School
S
y
4
x
-8
-4
0
-4
L
4
Library
-8
(-4, 4), (0, 2), and (4, 0)
10. Sam is planning to fence his backyard. Every side of the yard
except for the side along the house is to be fenced, and fencing
costs \$3/yd and can only be bought in whole yards. (Note that
m∠NPM = 28°, and the side of his yard opposite the house
measures 35 yd.) How much will the fencing cost? Explain how
House
J
P
Yard
K
triangles JKL and MNP are congruent. The distance JK (and PN)
L
15 yd
M
10 yd
N
10
is _____
= 21.3 yd. So the total length of fencing needed is
sin28°
(2)(21.3) + 35 = 77.6 yd. Because the fencing can only be bought in whole
yards, Sam will need to buy 78 yd. The cost is (78 yd)(\$3/yd) = \$234.
Unit 7
IN2_MNLESE389847_U7UC 999
Unit 7
6th Ave
The Triangle Proportionality Theorem
1 point for correctly graphing each coordinate point,
3 in all
999
3.2
A Street
mi
2.8
x = ____
____
2.4 mi
8. The map shows that A Street and B Street are parallel. Find the
distance on 6th Ave between A Street and the library. Explain any
theorems that come into play here.
Item 8 (2 points) Award the student 1 point for the
correct distance of 4.5 miles, and 1 point for correctly
naming the Triangle Proportionality Theorem.
2.8
SCORING GUIDES
999
19/04/14 12:37 AM
math in careers
MATH IN CAREERS
special effects engineers A special effects engineer is helping create a movie and
needs to add a shadow to a tall totem pole that is next to a 6-foot-tall man. The totem pole
is 48 feet tall and is next to the man, who has a shadow that is 2.5 feet long. Create an image
with the given information and then use the image to find the length of the shadow that the
engineer needs to create for the totem pole.
Special Effects Engineer In this Unit Performance
Task, students can see how a special effects engineer
uses mathematics on the job.
as well as various mathematics appreciation topics,
visit the American Mathematical Society
http://www.ams.org
Totem Pole
Man
48 ft
6 ft
SCORING GUIDES
X
2.5 ft
2.5
x = ____
___
6
48
x = 20
3 points for creating a diagram that represents the
situation, with distances labeled
3 points for correctly setting up the proportion and
solving
© Houghton Mifflin Harcourt Publishing Company
Unit 7
IN2_MNLESE389847_U7UC 1000
1000
19/04/14 12:37 AM
Unit 7
1000
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