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WCCUSD Geometry
1
Benchmark 2 Study Guide
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 k > 1 enlarges it. A dilation
of 0 < k < 1 reduces it.
1´
You Try:
A. Dilate the following figure using a scale
factor of 2 with center of dilation at the
origin.
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.
B. Dilate the following figure using a scale
1
factor of
with center at (4,-2).
2
C. Dilate the following figure using a scale
factor of 3 with center of dilation at the
origin.
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
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WCCUSD Geometry
2
Benchmark 2 Study Guide
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.
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.
Ex #1: Prove the following figures are similar by
describing a series of transformations that will
map Figure 1 onto Figure 2.
B. Are these triangles similar? Justify your
reasoning.
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 ࡭࡮࡯ࡰ~ࡱࡲࡳࡴ, find CD .
Solution: Since the figures are similar, then a dilation
has occurred using a scale factor. This creates
corresponding sides that are proportional:
BD CD
=
FG HG
30 CD
=
6 16
∴ ‫ = ܦܥ‬80 ܿ݉
G.SRT.2
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WCCUSD Geometry
3
Benchmark 2 Study Guide
Phillip draws two triangles. Two pairs of
corresponding angles are congruent. Select
each statement that is true for all such pairs
of triangles.
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.
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.
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
1
point C and a scale factor of .
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
1
point C and a scale factor of .
3
G.SRT.3
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WCCUSD Geometry
4
Benchmark 2 Study Guide
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?
4´
You try:
A. Mark stands next to a tree that casts a 15foot shadow. If Mark is 6 feet tall and
casts a 4-foot shadow, how tall is the tree?
Solution: Draw a picture:
h
4m
24 m
6m
Write a proportion and solve:
h 24
=
4 6
h
=4
4
B. Find the value of x in the figure below.
h
( 4) = 4 ( 4)
4
h = 16
The building is 16 meters tall.
Ex #2: Find BE .
Let BE = x.
Or using the
Parallel/Proportionality
Conjecture:
4 x
=
8 9
4 (9) = 8 ( x )
36 = 8 x
4.5 = x
G.SRT.5
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WCCUSD Geometry
5
Benchmark 2 Study Guide
Similar right triangles have side ratios that are
equal to each other. For example, every 30-6090 triangle, no matter what size, has a small
1
side to hypotenuse ratio of 1:2 or (or 0.5) .
2
These are the side length ratio definitions of the
acute angle, θ :
5´
You try:
Given ∆MAT, match each trigonometric ratio
to its equivalent value in the box.
17
M
T
8
15
sin θ =
Opposite
Hypotenuse
A
___ 1) cosT =
Adjacent
cos θ =
Hypotenuse
___ 2) cos M =
___ 3) tan M =
tan θ =
Opposite
Adjacent
___ 4) tan T =
___ 5) sin M =
Ex: Write each trigonometric ratio using the
side lengths of ∆ABC below.
___ 6) sin T =
8
17
15
B.
17
15
C.
8
8
D.
15
A.
A. sin C =
B. cos C =
C. tan A =
D. tan C =
E. cos A =
Solutions:
Answers:
A.
4
5
B.
3
5
C.
3
4
D.
4
3
E.
4
5
G.SRT.6
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WCCUSD Geometry
6
Benchmark 2 Study Guide
The sine of an angle is equal to the cosine of
its complement: sin θ = cos(90 − θ ) .
Q
12
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.
5
P
13
6´
R
Angle
Value of Sine
Value of Cosine
0°
0
1
15°
0.2588
0.9659
According to the figure above:
5
5
sin P = ≈ 0.3846 and cos R = ≈ 0.3846
13
13
30°
0.5000
0.8660
45°
0.7071
So, sin P = cos R .
75°
60°
90°
If sin32° ≈ 0.5514 , then the cosine of its
complement is equivalent:
B. Explain how you determined the values
you used.
sin 32° = 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
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WCCUSD Geometry
7
Benchmark 2 Study Guide
Drawing and labeling pictures are a great way
to solve problems using the trigonometric ratios.
Don’t forget the Pythagorean Theorem
(ܽଶ + ܾ ଶ = ܿ ଶ )!
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.
A. sin 57° =
B. cos 57° =
C. tan 57° =
D. tan 33° =
30
x
x
30
30
x
x
30
Solution:
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. sin θ =
900
1500
B. cos θ =
1200
1500
C. sin θ =
1200
1500
D. cos θ =
900
1500
E. tan θ =
900
1200
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
opposite
angles do show
.
hypotenuse
G.SRT.8
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WCCUSD Geometry
8
Benchmark 2 Study Guide
Solve for the variables.
8´
You try:
a
J
b
K
10
45°
This is a 45° − 45° − 90° triangle.
Using 45°-45°-90°
theorem:
Using 45°-45°-90°
leg:leg:hyp. ratios:
leg = leg
1:1: 2
x=9
hypotenuse = leg i 2
y = 9i 2
y=9 2
leg
leg
leg
hyp.
1 x
=
1 9
1
9
=
2 y
x=9
L
Determine whether each of the following
statements is true or false.
A. a = b
T
F
B. ∆JKL is a right scalene triangle
T
F
C. Area of ∆JKL = 25 2 sq. un.
T
F
D. Perimeter of ∆JKL = 10 + 10 2 un.
T
F
y=9 2
(
)
G.SRT.8.1
9
Solve for the variables.
9´
You try:
m
Y
60°
n
24
p
n
30°
X
12
Z
This is a 30° − 60° − 90° triangle.
Using 30°-60°-90°
theorem:
Using 30°-60°-90°
short leg:long
leg:hypotenuse ratios:
hypotenuse = 2i( short leg )
1: 3 : 2
24 = 2m
Determine whether each of the following
statements is true or false.
A. m∠X = 30°
T
F
B. n = 6
T
F
T
F
T
F
T
F
12 = m
long leg
hypotenuse
C. n = 6 3
3 n
=
2
24
D. p = 8 3
2m = 24
2n = 24 3
E. Area of ∆XYZ = 24 3 sq. un.
m = 12
n = 12 3
short leg
hypotenuse
( long leg ) = ( short leg )i
n = 12i 3
n = 12 3
3
1 m
=
2 24
G.SRT.8.1
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End of Study Guide
MCC@WCCUSD (WCCUSD) 12/17/15
WCCUSD Geometry
Benchmark 2 Study Guide
You Try Solutions:
1´
A. Dilate the following figure using a scale
factor of 2 with center of dilation at the
origin.
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.
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)
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 ' .
B. Dilate the following figure using a scale
1
factor of
with center at (4,-2).
2
Another solution could be dilating ∆ABC by a
scale factor of 3 with center of dilation at the
origin.
2´
B. Are these triangles similar? Justify
your reasoning.
C. Dilate the following figure using a scale
factor of 3 with center at the origin.
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
∴ ∆ܷܸܶ~∆ܹܻܺ.
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WCCUSD Geometry
3´
Benchmark 2 Study Guide
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.
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?
h
6 ft
15 ft
4 ft
h 15
=
6 4
Which set of transformations maps ∆ABC
to ∆DEC and supports Omar’s thinking?
(4)(6)
h 15
= (4)(6)
6 4
4h = 90
The scale factor is 2 since the corresponding
sides have a ratio of 6:3, or 2:1. Therefore, A
is the correct answer.
h = 22.5
The tree is 22.5 ft high.
B. Find the value of x in the figure below.
( x + 2) + 8 12 + 5
=
8
12
x + 10 17
=
8
12
x + 10 17
(24)
= (24)
8
12
3( x + 10) = 34
3 x + 30 = 34
3x = 4
4
x=
3
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WCCUSD Geometry
5´
Benchmark 2 Study Guide
You try:
7´
Given ∆MAT, match each trigonometric
ratio to its equivalent value in the box.
17
M
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.
T
Select all equations that can be used to
solve for the angle of depression (θ) from
the plane to the runway.
angle of
depression
8
15
A
1.
2.
3.
4.
5.
6.
You try:
A
B
D
C
A
B
θ
Alternate
Interior Angles
900 m
θ
1200 m
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.
Solution: Use the Pythagorean Theorem to
find the length of the hypotenuse:
ܽଶ + ܾ ଶ = ܿ ଶ
(1200)ଶ + (900)ଶ = ܿ ଶ
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°
75°
0.8660
0.9659
0.5000
0.2588
90°
1
0
1440000 + 810000 = ܿ ଶ
2250000 = ܿ ଶ
1500 = c
Based on this information, the following are
equations that can be used to solve for the
angle of depression:
B. Explain how you determined the
values you used.
The sine of an angle is equal to the cosine of
its complement. So, sin15° = cos 75° ,
sin 30° = cos 60° , sin 45° = cos 45° and
sin 0° = cos 90° .
Page 11 of 12
A. sin θ =
900
1500
B. cos θ =
1200
1500
E. tan θ =
900
1200
MCC@WCCUSD (WCCUSD) 12/17/15
WCCUSD Geometry
8´
Benchmark 2 Study Guide
You try:
Determine whether each of the following
statements is true or false.
A. a = b
T
F
B. ∆JKL is a right scalene triangle
T
F
C. Area of ∆JKL = 25 2 sq. un.
T
F
T
F
(
)
D. Perimeter of ∆JKL = 10 + 10 2 un.
9´
You try:
Determine whether each of the following
statements is true or false.
A. m∠X = 30°
T
F
B. n = 6
T
F
C. n = 6 3
T
F
D. p = 8 3
T
F
E. Area of ∆XYZ = 24 3 sq. un.
T
F
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