Geometry – Chapter 7 – Notes and Examples
Sections 1 & 2 Ratios and Proportions and Ratios in Similar Polygons
ratio
a to b
a:b
division
A _________________ compares two numbers by ___________________. A ratio can be written as
𝒂
_____________, ___________ or ____________
where b ≠ 0.
𝒃
The slope of a line can be expressed as a ratio.
A ratio can involve more than two numbers such as the side lengths of a
3:7:3:7
rectangle can be shown as ____________________.
Problem 1
Given that two points on line l are A(–1, 3) and
B(2, –2), write a ratio expressing the slope of line l.
Slope = m =
𝒓𝒊𝒔𝒆
𝒓𝒖𝒏
=
=
Problem 2
The ratio of the side lengths of a triangle is 4:7:5,
and its perimeter is 96 cm. What is the length of the
shortest side?
(𝒚𝟐 −𝒚𝟏 )
Let the side lengths be 4x, 7x, and 5x.
(𝒙𝒙 −𝒙𝟏 )
−𝟐−𝟑
𝟐−(−𝟏)
=
−𝟓
𝟑
=−
𝟓
𝟑
4x + 7x + 5x = 96
16x = 96
x =6
The shortest side is 4x = 4(6) = 24 cm
Problem 3
The ratio of the angle measures in a triangle is 1:6:13. What is the measure of each angle?
Let the measures of the angles = x, y, and z.
x + y + z = 180°
y = 6x = 6(9°) = 54°
x + 6x + 13x = 180°
20x = 180°
z = 13x = 13(9°) = 117°
x = 9°
The measures of the angles are 9°, 54°, and 117°.
proportion
A ______________________is an equation stating that two ratios are equal. In the proportion
a
d
c
𝑎
𝑏
=
𝑐
𝑑
, the
d
values _____ and _____ are the extremes. The values _____ and _____ are the means. When the proportion
a:b
c:d
is written as ________ = ________, the extremes are in the first and last positions. The means are in the
two middle positions.
In Algebra 1 you learned the Cross Products Property. The product of the extremes ad and the product of
the means bc are called the cross products.
Problem 4
Solve the proportion.
Problem 5
7
𝑥
=
56
72
Solve the proportion.
7(72) = 56x
9 =x
3(64) = 8x
24 = x
Problem 6
Solve the proportion.
Problem 7
Solve the proportion.
2𝑦
9
=
8
𝑧−4
4𝑦
5
2y(4y) = 9(8)
8y2 = 72
y2 = 9
y = 3
y = 3 or y = –3
=
3
8
=
20
𝑧−4
(z – 4)2 = 5(20)
(z – 4)2 = 100
(z – 4) = 10
(z – 4) = 10 or (z – 4) = –10
z = 14 or z = –6
𝑥
64
Problem 8
Given that 18c = 24d, find the ratio of d to c in
simplest form. Hint: use Properties of Proportions.
Problem 9
Given that 16s = 20t, find the ratio t:s in simplest
form. Hint: use Properties of Proportions.
18c = 24d
16s = 20t
𝟏𝟖
𝟐𝟒
=
𝒅
thus
𝒄
𝟑
𝟒
=
𝒅
𝟏𝟔
𝒄
𝟐𝟎
=
𝒕
𝟒
thus
𝒔
𝟓
=
𝒕
𝒔
similar
Figures that are ____________________ have the same shape but not necessarily the same size. The
~
symbol for similar is ________.
Problem 10
Identify the pairs of congruent angles and
corresponding sides.
Problem 11
Identify the pairs of congruent angles and
corresponding sides.
Q  N and R  P. By the Third Angles
Theorem, T  M.
B  G and C  H. By the Third Angles
Theorem, A  J.
𝑻𝑹
𝑴𝑷
𝑸𝑹
𝑵𝑷
=
=
𝟏.𝟏
𝟐.𝟐
𝟏
𝟎.𝟓
=
=
𝟏
𝑻𝑸
𝟐
𝑴𝑵
𝟏
𝟐
=
𝟏
𝑨𝑩
𝟐
𝑱𝑮
𝑨𝑪
=
=
𝑱𝑯
𝟏𝟎
𝟓
=2
𝟏𝟏.𝟔
𝟓.𝟖
=2
𝑩𝑪
=
𝑮𝑯
𝟔
𝟑
=2
similarity
ratio
A __________________ ____________is the ratio of the lengths of the corresponding sides of two similar
polygons.
3
1
6
2
The similarity ratio of ∆ABC to ∆DEF is , or
6
. The similarity ratio of ∆DEF to ∆ABC is , or 2.
3
Writing a similarity statement is like writing a congruence statement — be sure to list corresponding
vertices in the same order.
Problem 12
Determine whether the polygons are similar. If
so, write the similarity ratio and a similarity
statement.
Problem 13
Determine if ∆JLM ~ ∆NPS. If so, write the
similarity ratio and a similarity statement.
Step 1: Identify
pairs of congruent
angles.
N  M, L 
P, S  J
Step 1: Identify pairs of
congruent angles.
P  R and S  W
Step 2: Compare corresponding angles.
mP = mR =
𝟏
𝟐
(180° – 36°) = 72°
mW = mS = 62°
mT = 180° – 2(62°) = 56°
Since no pairs of angles are congruent, the
triangles are not similar.
Step 2 Compare corresponding sides
𝑳𝑱
𝑷𝑺
𝑱𝑴
𝑺𝑵
=
=
𝟕𝟓
𝟑𝟎
𝟒𝟓
𝟏𝟖
=
=
𝟓
𝑳𝑴
𝟐
𝑷𝑵
=
𝟔𝟎
𝟐𝟒
=
𝟓
𝟐
𝟓
𝟐
Thus the similarity ratio is
𝟓
𝟐
, and ∆LMJ ~ ∆PNS.
Geometry – Chapter 7 – Notes and Examples
Section 3 – Triangle Similarity: AA, SSS, and SAS
There are several ways to prove certain triangles are similar. These ways are:
Hypothesis
Conclusion
Angle-Angle Similarity AA ~
If two angles of one triangle are
congruent o two angles of
another triangle, then the
triangles are similar.
Side-Side-Side Similarity SSS~
If the three sides of one triangle
are proportional to the three
corresponding sides of another
triangle, then the triangles are
similar.
Side-Angle-Side Similarity
SAS ~
If two sides of one triangle are
proportional to two sides of
another triangle and their
included angles are congruent,
then the triangles are similar
Problem 1
Problem 2
Explain why the triangles are similar and write a Verify that the
similarity statement.
triangles are
similar.
𝑷𝑸
By the Triangle Sum Theorem, mC = 47°, so
C  F. B  E by the Right Angle
Congruence Theorem.
Therefore, ∆ABC ~ ∆DEF by AA ~.
𝑺𝑻
𝑷𝑹
𝑺𝑼
=
=
𝟑
𝟒.𝟓
=
𝟐
𝑸𝑹
𝟑
𝑻𝑼
=
𝟑
𝟒.𝟓
𝟐
𝟑
Therefore ∆PQR ~ ∆STU by SSS ~.
=
𝟐
𝟑
Problem 3
Verify that the triangles are similar.
Problem 4
Explain why ∆RSV ~ ∆RTU and then find RT.
D  H by the Definition of Congruent Angles.
𝑫𝑬
𝑯𝑱
=
𝟐
𝟏
=2
𝑫𝑭
𝑯𝑲
=
𝟐
𝟏
Therefore ∆DEF ~ ∆HJK by SAS ~.
=2
It is given that S  T.
R  R by Reflexive Property of .
Therefore ∆RSV ~ ∆RTU by AA ~.
𝑹𝑻
𝑹𝑺
𝑹𝑻
𝟏𝟎
=
𝑻𝑼
=
𝑺𝑽
𝟏𝟐
𝟖
RT(8) = 10(12)
8RT =120
RT = 15
Geometry – Chapter 7 – Notes and Examples
Section 4 Applying Properties of Similar Triangles
Problem 1
Find US.
𝑼𝑺
𝟏𝟒
=
𝑼𝑺
𝑹𝑼
=
𝑽𝑻
Problem 2
Find PN.
𝑹𝑽
𝑳𝑷
𝑷𝑵
𝟒
𝟑
𝟏𝟎
𝑷𝑵
US(10) = 56
US =
=
=
𝑴𝑸
𝑸𝑵
𝟐
𝟓
PN(2) = 15
𝟓𝟔
Or 5.6
𝟏𝟎
PN = 7.5
Problem 3
Problem 4
Verify that DE ∥BC.
Verify that DE ∥AB.
𝑨𝑫
𝟐
𝑫𝑪
𝟑
𝑨𝑫
𝟐
𝑬𝑪
𝑫𝑩
𝑨𝑬
𝑬𝑪
=
=
Since
𝟏𝟎
𝟏𝟓
𝟖
𝟏𝟐
𝑨𝑫
𝑫𝑩
=
=
=
𝟑
𝑩𝑬
=
=
𝑨𝑬
, then DE ∥BC by the Converse
𝑬𝑪
of the Triangle Proportionality Theorem.
Since
𝟐𝟎
𝟏𝟔
𝟏𝟓
𝟏𝟐
𝑫𝑪
𝑨𝑫
=
=
=
𝟓
𝟒
𝟓
𝟒
𝑬𝑪
𝑩𝑬
, then DE ∥AB by the Converse
of the Triangle Proportionality Theorem.
Problem 5
Use the diagram to find LM and MN to the nearest tenth.
AK ∥BL ∥ CM ∥DN
𝑲𝑳
=
𝑳𝑴
𝑨𝑩
𝟐.𝟔
𝑩𝑪
𝑳𝑴
=
𝟐.𝟒
𝟏.𝟒
2.4 (LM) = 1.4(2.6)
LM ≈ 1.5 cm
𝑲𝑳
𝑴𝑵
=
𝑨𝑩
𝟐.𝟔
𝑪𝑫
𝑴𝑵
=
𝟐.𝟒
𝟐.𝟐
2.4 (MN) = 2.2(2.6)
MN ≈ 2.4 cm
Problem 6
Find the length of XZ.
Problem 7
Find PS and SR.
𝒀𝒁
𝑾𝒀
𝑷𝑺
𝑾𝑿
𝑺𝑹
𝟗
𝒙− 𝟐
𝒁𝑿
𝟔
𝒁𝑿
=
=
𝟏𝟐
ZX (9) = 72
ZX = 8
=
𝒙+𝟓
𝑸𝑷
𝑸𝑹
=
𝟑𝟐
𝟒𝟎
40(x – 2) = 32(x + 5)
40x – 80 = 32x + 160
8x = 240
x = 30
PS = x – 2 = 30 – 2 = 28
SR = x + 5 = 30 + 5 = 35
Geometry – Chapter 7 – Notes and Examples
Section 5 – Using Proportional Relationships
Indirect
measurement
__________________ ____________________________ is any method that uses formulas, similar figures, and/or
proportions to measure an object.
scale drawing
The drawings _____________
scale is the ratio of the length in the drawing to the corresponding actual length.
A ________ __________________ represents an object as smaller than or larger than its actual size.
For example, on a map with a scale of 1 cm : 1500 m, one centimeter on the map represents 1500 m in actual
distance.
Problem 1
A fire hydrant 2.5 feet high casts
a 5-foot shadow. How tall is a street light that
casts a 26-foot shadow at the same time?
x
2.5 ft
26 ft
𝒙
𝟐.𝟓
=
5 ft
𝟐𝟔
𝟓
𝑩𝑪
5x = (2.5)(26)
x = 13
The height x of the street light is 13 ft.
Problem 3
On a Wisconsin road map, Kristin measured a
distance of 11 in. from Madison to Wausau. The
scale of this map is 1 inch :13 miles. What is the
actual distance between Madison and Wausau to
the nearest mile?
𝟏 𝒊𝒏
𝟏𝟑 𝒎𝒊
=
Problem 2
Tyler wants to
find the height
of a telephone
pole. He
measured the
pole’s shadow
and his own
shadow and then made a diagram. What is the
height h of the pole?
AB = 7 ft 8 in. = (7  12) in. + 8 in. = 92 in.
BC = 5 ft 9 in. = (5  12) in. + 9 in. = 69 in.
FG = 38 ft 4 in. = (38  12) in. + 4 in. = 460 in.
𝑮𝑯
=
𝑨𝑩
𝟔𝟗
𝑭𝑮
𝒉
=
𝟗𝟐
𝟒𝟔𝟎
92h = (69)460
h = 345
The height h of the pole is 345 inches, or 28 feet 9
inches.
Problem 4
The rectangular central chamber of the Lincoln
Memorial is 74 ft long and 60 ft wide. What would
be the measurements of a scale drawing of the
floor of the chamber using a scale of 1 in : 20 ft?
𝟏𝟏 𝒊𝒏
𝟏 𝒊𝒏
𝒙
𝟐𝟎 𝒇𝒕
=
𝒍
𝟏 𝒊𝒏
𝟕𝟒 𝒇𝒕
𝟐𝟎 𝒇𝒕
=
𝒘
𝟔𝟎 𝒇𝒕
x = 143
l = 3.7 in
w = 3 in
Problem 5
Given that
∆LMN:∆QRS, find
the perimeter P
and area A of
∆QRS.
Problem 6
∆ABC ~ ∆DEF. Find
the perimeter and
area of ∆ABC.
The similarity ratio of ∆ABC to ∆DEF is
The similarity ratio of ∆LMN to ∆QRS is
Perimeter
Area
𝟑𝟔
𝑷
=
𝟏𝟑
𝟔𝟎
𝟗.𝟏
𝑨
13P = (36)(9.1)
P = 25.2 cm
=
𝟏𝟑
𝟗.𝟏
𝟏𝟑𝟐
Perimeter
Area
𝑷
𝟏
𝑨
𝟐
𝟏𝟐𝟔
𝟓𝟒
𝟗.𝟏𝟐
=
2P = (54)(1)
P = 27 in
132 A = (9.1)2 (60)
A = 29.4 cm2
Problem 7
Parallelogram PQRS ~ Parallelogram TUVW.
Find the perimeter and area of parallelogram
TUVW.
=
𝟕𝟐
𝑷
=
𝟑
𝟐
𝟑𝟏𝟓
𝑨
=
𝟐𝟏
𝟏𝟒
=
𝟐𝟐
22 A = (1)2 (126)
A = 31.5 in2
Problem 8
 EFG ~  HJK. Find the perimeter
and area of  HJK.
The similarity ratio
of ∆EFG to ∆HJK is
𝟖
𝟑𝟐
Perimeter
Area
𝟐𝟐
𝟐𝟕
𝟏
𝟑𝟐
𝟑
𝑨
32 A = (2)2 (315)
A = 140 cm2
=
𝟏
𝟑
𝟑
𝟐
𝑷
3P = (72)(2)
P = 48 cm
𝟏𝟑
𝟏
=𝟐
𝟏𝟐
𝟐𝟒
The similarity ratio of PQRS to TUVW is
Perimeter
Area
𝟔.𝟓
=
1P = (27)(3)
P = 81 m
=
𝟏𝟐
𝟑𝟐
12 A = (3)2 (32)
A = 288 m2