wccusd geometry benchmark 2 study guide · wccusd geometry benchmark 2 study guide page 5 of 12...

WCCUSD Geometry Benchmark 2 Study Guide

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


origin.



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:



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


point C and a scale factor of 3.

D. A rotation of 180° clockwise point C


point C and a scale factor of 1

3.



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.



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



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



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.



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:



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



You Try Solutions:

1´ A. Dilate the following figure using a scale


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÷ =

∴ ∆��~∆��.



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.



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



8´ You try:



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:



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

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