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WCCUSD Geometry Benchmark 2 Study Guide
Page 1 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
1 Dilations affect the size of the pre-image.
The pre-image will enlarge or reduce by the
ratio given by the scale factor. A dilation with
a scale factor of 1k > enlarges it. A dilation
of 0 1k< < reduces it.
Ex: Dilate the following figure using a scale
factor of 3 with center of dilation at (5,-6).
One Solution: Plot (5, 6)− . Draw lines from
the center of dilation through vertices of the
pre-image. Since the scale factor is 3, each
distance from the center of dilation to the
image will triple. Plot image’s vertices and
connect them to complete the image.
For all dilations centered at (a, b) with a scale
factor of k, the image’s coordinates can be
found using ( ( ), ( ))a k x a b k y b+ − + − . If the
center of dilation is at the origin, then a and b
are zero, resulting in the new image location
coordinates as ( , )k a k b⋅ ⋅ .
G.SRT.1
1´ You Try: A. Dilate the following figure using a scale
factor of 2 with center of dilation at the
origin.
B. Dilate the following figure using a scale
factor of 1
2 with center at (4,-2).
C. Dilate the following figure using a scale
factor of 3 with center of dilation at the
origin.
WCCUSD Geometry Benchmark 2 Study Guide
Page 2 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
2
BD
FG=
CD
HG
30
6=
CD
16
G.SRT.2
2´ You try:
A. Prove the following figures are similar by
describing a series of transformations that
will map the smaller triangle to the larger
triangle.
B. Are these triangles similar? Justify your
reasoning.
∴ �� = 80 �
When a figure is dilated to make an image,
corresponding angles are equal and
corresponding sides are proportional relative
to the scale factor used to dilate.
Two different-sized figures can be shown to be
similar by using transformations if one of the figures
can be mapped onto the other using a series of
transformations, one of which is a dilation and the
other(s) a reflection, rotation and/or translation.
Ex #1: Prove the following figures are similar by
describing a series of transformations that will
map Figure 1 onto Figure 2.
One possible solution: Dilate Figure 1 by a scale factor of
2 with center of dilation at ( 3, 2)− . Then translate the
resulting image four units right and four units down.
Ex #2: If �� ~����, find CD .
Solution: Since the figures are similar, then a dilation
has occurred using a scale factor. This creates
corresponding sides that are proportional:
WCCUSD Geometry Benchmark 2 Study Guide
Page 3 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
3 Phillip draws two triangles. Two pairs of
corresponding angles are congruent. Select
each statement that is true for all such pairs
of triangles.
A. A sequence of rigid motions carries one
triangle onto the other.
B. A sequence of rigid motions and dilations
carries one triangle onto the other.
C. The two triangles are similar because the
triangles satisfy Angle-Angle criterion.
D. The two triangles are congruent because
the triangles satisfy Angle-Angle criterion.
E. All pairs of corresponding angles are
congruent because triangles must have an
angle sum of 180° .
F. All pairs of corresponding sides are
congruent because of the proportionality
of corresponding side lengths.
Solutions: True—B, C, and E.
A is only true for congruent triangles and not
for similar triangles.
D is not true because Angle-Angle does not
prove triangle congruency.
F is not true because corresponding sides are
not congruent for similar triangles.
G.SRT.3
3´ You try:
Omar thinks that if two angles of one triangle
are congruent to two angles of another
triangle, then the triangles are similar. To
show this, he drew the figure below.
Which set of transformations maps ABC∆ to
DEC∆ and supports Omar’s thinking?
A. A rotation of 180° clockwise about point
C followed by a dilation with a center of
point C and a scale factor of 2.
B. A rotation of 180° clockwise point C
followed by a dilation with a center of
point C and a scale factor of 1
2.
C. A rotation of 180° clockwise point C
followed by a dilation with a center of
point C and a scale factor of 3.
D. A rotation of 180° clockwise point C
followed by a dilation with a center of
point C and a scale factor of 1
3.
WCCUSD Geometry Benchmark 2 Study Guide
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4
Write a proportion and solve:
( ) ( )
24
4 6
44
4 4 44
16
h
h
h
h
=
=
=
=
Ex #2: Find BE .
G.SRT.5
4´ You try:
A. Mark stands next to a tree that casts a 15-
foot shadow. If Mark is 6 feet tall and
casts a 4-foot shadow, how tall is the tree?
B. Find the value of x in the figure below.
4 m
6 m 24 m
h
The building is 16 meters tall.
Ex #1: A flagpole 4 meters tall casts a 6-meter
shadow. At the same time of day, a nearby
building casts a 24-meter shadow. How tall is
the building?
Solution: Draw a picture:
Or using the
Parallel/Proportionality
Conjecture:
( ) ( )
4
8 9
4 9 8
36 8
4.5
x
x
x
x
=
=
=
=
Let BE = x.
WCCUSD Geometry Benchmark 2 Study Guide
Page 5 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
5 Similar right triangles have side ratios that are
equal to each other. For example, every 30-60-
90 triangle, no matter what size, has a small
side to hypotenuse ratio of 1:2 or 1
2 (or 0.5) .
These are the side length ratio definitions of the
acute angle, θ :
sinOpposite
Hypotenuseθ =
cosAdjacent
Hypotenuseθ =
tanOpposite
Adjacentθ =
Ex: Write each trigonometric ratio using the
side lengths of below.
Solutions:
Answers:
G.SRT.6
5´ You try:
Given ∆MAT, match each trigonometric ratio
to its equivalent value in the box.
___ 1) cosT =
___ 2) cos M =
___ 3) tan M =
___ 4) tanT =
___ 5) sin M =
___ 6) sinT =
∆ABC
T M
A
15 8
17
A. 8
17
B. 15
17
C. 15
8
D. 8
15
A. sin C =
B. cosC =
C. tan A =
D. tan C =
E. cos A =
A. 4
5 B.
3
5 C.
3
4 D.
4
3 E.
4
5
WCCUSD Geometry Benchmark 2 Study Guide
Page 6 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
6 The sine of an angle is equal to the cosine of
its complement: sinθ = cos(90 −θ ).
According to the figure above:
sin P =5
13≈ 0.3846 and cosR =
5
13≈ 0.3846
So, sin P = cosR .
If sin32° ≈ 0.5514 , then the cosine of its
complement is equivalent:
sin32° = cos(90° − 32°) = cos58° ≈ 0.5514
Ex: Determine whether the following
statements are true or false:
1. sin 43° = cos47°
2. sin 43° = cos43°
3. sin 45° = cos45°
4. sin17° = cos(90 −17)°
5. cosθ = sin(90 −θ )
Solutions: 1—True, 2—False, 3—True, 4—
True, 5—True
G.SRT.7
6´ You try:
The table below shows the approximate
values of sine and cosine for selected angles.
A. Fill in the rest of the table without a
calculator.
Angle Value of Sine Value of Cosine
0° 0 1
15° 0.2588 0.9659
30° 0.5000 0.8660
45° 0.7071
60°
75°
90°
B. Explain how you determined the values
you used.
P
R
Q
13
12
5
WCCUSD Geometry Benchmark 2 Study Guide
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7
A. 30
sin 57x
° =
B. cos5730
x° =
C. 30
tan 57x
° =
D. tan 3330
x° =
Solution:
All the angles in a triangle add up to 180° , so
the acute angle near the flag at the top of the
triangle must measure 33° . A and B are not
correct because the ratios do not correspond
to the definitions of the trig ratios. C and D
are correct since the tangent ratios of those
angles do show opposite
hypotenuse.
G.SRT.8
7´ You try:
A plane is flying at an elevation of 900
meters. From a point directly underneath the
plane, the plane is 1200 meters away from a
runway.
Select all equations that can be used to solve
for the angle of depression (θ) from the plane
to the runway.
A. 900
sin1500
θ =
B. 1200
cos1500
θ =
C. 1200
sin1500
θ =
D. 900
cos1500
θ =
E. 900
tan1200
θ =
Drawing and labeling pictures are a great way
to solve problems using the trigonometric ratios.
Don’t forget the Pythagorean Theorem
(�� + �� = ��)!
The angle of elevation from a landscaped rock to
the top of a 30-foot tall flagpole is 57° .
Which of the following equations could be used to
find the distance between the rock and the base of
the flagpole? Select all that apply.
WCCUSD Geometry Benchmark 2 Study Guide
Page 8 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
8 Solve for the variables.
This is a 45 45 90° − ° − ° triangle.
Using 45°-45°-90°
theorem:
9
leg leg
x
=
=
2
9 2
9 2
hypotenuse leg
y
y
=
=
=
i
i
Using 45°-45°-90°
leg:leg:hyp. ratios:
1:1: 2
leg
leg
1
1 9
9
x
x
=
=
.
leg
hyp
1 9
2
9 2
y
y
=
=
G.SRT.8.1
9 Solve for the variables.
This is a 30 60 90° − ° − ° triangle.
Using 30°-60°-90°
theorem:
( )2
24 2
12
hypotenuse short leg
m
m
=
=
=
i
( ) ( ) 3
12 3
12 3
long leg short leg
n
n
=
=
=
i
i
Using 30°-60°-90°
short leg:long
leg:hypotenuse ratios:
1: 3 : 2
short leg
hypotenuse
1
2 24
2 24
12
m
m
m
=
=
=
long leg
hypotenuse
3
2 24
2 24 3
12 3
n
n
n
=
=
=
G.SRT.8.1
8´ You try:
Determine whether each of the following
statements is true or false.
A. a b=
B. JKL∆ is a right scalene triangle
C. Area of 25 2 sq. un.JKL∆ =
D. Perimeter of ( )10 10 2 un.JKL∆ = +
9´ You try:
Determine whether each of the following
statements is true or false.
A. 30m X∠ = °
B. 6n =
C. 6 3 n =
D. 8 3 p =
E. Area of 24 3 sq. un.XYZ∆ =
End of Study Guide
30°
m
n 24
K J a
L
b
45°
10
F T
F T
F T
F T
Z X
n
Y
p 60°
12
F T
F T
F T
F T
F T
WCCUSD Geometry Benchmark 2 Study Guide
Page 9 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
You Try Solutions:
1´ A. Dilate the following figure using a scale
factor of 2 with center of dilation at the
origin.
OR multiply each vertex’s coordinate by the
scale factor of 2 to find the image’s
coordinates:
( 1, 1) (2 1,2 1) ( 2, 2)− − → ⋅− ⋅− → − −
( 2, 3) (2 2,2 3) ( 4, 6)− − → ⋅ − ⋅ − → − −
(0, 3) (2 0, 2 3) (0, 6)− → ⋅ ⋅− → −
B. Dilate the following figure using a scale
factor of 1
2 with center at (4,-2).
C. Dilate the following figure using a scale
factor of 3 with center at the origin.
2´ You try:
A. Prove the following figures are similar
by describing a series of
transformations that will map the
smaller triangle to the larger triangle.
One solution could be dilating ABC∆ by a
scale factor of 3 with center of dilation at
(2, 1) and then translated 4 units right and 2
units up, then ABC∆ maps onto ' ' 'A B C∆ .
Another solution could be dilating ABC∆ by a
scale factor of 3 with center of dilation at the
origin.
B. Are these triangles similar? Justify
your reasoning.
If the triangles are similar, then all
corresponding side ratios must be equal since
a dilation has occurred.
12 4 3
8 4 2÷ =
18 6 3
12 6 2÷ =
24 8 3
16 8 2÷ =
∴ ∆���~∆���.
WCCUSD Geometry Benchmark 2 Study Guide
Page 10 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
3´ You try:
Omar thinks that if two angles of one
triangle are congruent to two angles of
another triangle, then the triangles are
similar. To show this, he drew the figure
below.
Which set of transformations maps ABC∆
to DEC∆ and supports Omar’s thinking?
The scale factor is 2 since the corresponding
sides have a ratio of 6:3, or 2:1. Therefore, A
is the correct answer.
4´ You try:
A. Mark stands next to a tree that casts a
15-foot shadow. If Mark is 6 feet tall and
casts a 4-foot shadow, how tall is the tree?
15
6 4
h=
15(4)(6) (4)(6)
6 4
h=
4 90h =
22.5h =
B. Find the value of x in the figure below.
( 2) 8 12 5
8 12
x + + +=
10 17
8 12
x +=
10 17(24) (24)
8 12
x +=
3( 10) 34x + =
3 30 34x + =
3 4x =
4
3x =
15 ft 4 ft
6 ft
h
The tree is 22.5 ft high.
WCCUSD Geometry Benchmark 2 Study Guide
Page 11 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
5´ You try:
Given ∆MAT, match each trigonometric
ratio to its equivalent value in the box.
1. A
2. B
3. D
4. C
5. A
6. B
G.SRT.6
6´ You try:
The table below shows the approximate
values of sine and cosine for selected
angles.
A. Fill in the rest of the table.
Angle Value of Sine Value of Cosine
0° 0 1
15° 0.2588 0.9659
30° 0.5000 0.8660
45° 0.7071 0.7071
60° 0.8660 0.5000
75° 0.9659 0.2588
90° 1 0
B. Explain how you determined the
values you used.
The sine of an angle is equal to the cosine of
its complement. So, sin15 cos75° = ° ,
sin 30 cos 60° = ° , sin 45 cos 45° = ° and
sin 0 cos90° = ° .
7´ You try:
A plane is flying at an elevation of 900
meters. From a point directly underneath
the plane, the plane is 1200 meters away
from a runway.
Select all equations that can be used to
solve for the angle of depression (θ) from
the plane to the runway.
Solution: Use the Pythagorean Theorem to
find the length of the hypotenuse:
�� + �� = ��
(1200)� + (900)� = ��
1440000 + 810000 = ��
2250000 = ��
1500 = c
Based on this information, the following are
equations that can be used to solve for the
angle of depression:
A. 900
sin1500
θ =
B. 1200
cos1500
θ =
E. 900
tan1200
θ =
T M
A
15 8
17
900 m
1200 m
θ
θ
Alternate
Interior Angles
angle of
depression
WCCUSD Geometry Benchmark 2 Study Guide
Page 12 of 12 MCC@WCCUSD (WCCUSD) 12/17/15
8´ You try:
Determine whether each of the following
statements is true or false.
A. a b=
B. JKL∆ is a right scalene triangle
C. Area of 25 2 sq. un.JKL∆ =
D. Perimeter of ( )10 10 2 un.JKL∆ = +
9´ You try:
Determine whether each of the following
statements is true or false.
A. 30m X∠ = °
B. 6n =
C. 6 3 n =
D. 8 3 p =
E. Area of 24 3 sq. un.XYZ∆ =
F T
F T
F T
F T
F T
F T
F T
F T
F T