Download WCCUSD Geometry Benchmark 2 Study Guide

WCCUSD Geometry
1
Benchmark 2 Study Guide
G.SRT.1-Some Things To Know
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 −1 > x > 1 enlarges it. A
dilation of −1 < x < 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).
B. Dilate the following figure using a scale
1
factor of
with center at (4,-2).
2
One Solution: Plot (5, −6) . Draw lines from
center of dilation through vertices of the preimage. 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.
C. Dilate the following figure using a scale
factor of -2 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
G.SRT.1
Page 1 of 11
MCC@WCCUSD 12/11/14
WCCUSD Geometry
2
Benchmark 2 Study Guide
G.SRT.2-Some Things To Know
2´
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.
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.
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.
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
. Then
translate Figure 1 four units right and four units down.
Ex #2: If 𝑨𝑩𝑪𝑫~𝑬𝑭𝑮𝑯, find
.
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
∴ 𝐶𝐷 = 80 𝑐𝑚
G.SRT.2
G.SRT.2
Page 2 of 11
MCC@WCCUSD 12/11/14
WCCUSD Geometry
3
Benchmark 2 Study Guide
G.SRT.3-Some Things To Know
3´
This standard asks the student to establish that
AA is a similarity criterion for two triangles
using a series of transformations.
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.
Ex: 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
Page 3 of 11
Which set of transformations maps ΔABC to
ΔDEC 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
MCC@WCCUSD 12/11/14
WCCUSD Geometry
Benchmark 2 Study Guide
G.SRT.5-Some Things To Know
4
When solving problems regarding similarity, remember
that corresponding sides of similar figures are
proportional.
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?
A. A flagpole 4 meters tall casts a 6-meter shadow.
At the same time of day, a nearby building casts a 24meter shadow. How tall is the building?
Solution: Draw a picture:
h
4m
24 m
6m
Write a proportion and solve:
h 24
=
4 6
h 24
(6)(4) = (6)(4)
4 6
B. Find the value of x in the figure below.
6h = 24(4)
6h 24(4)
=
6
6
h = (4)(4)
h = 16
The building is 16 meters tall.
B. Find mBE .
Let BE = x.
Or using SideSplitting Theorem:
G.SRT.5
Page 4 of 11
G.SRT.5
MCC@WCCUSD 12/11/14
WCCUSD Geometry
5
Benchmark 2 Study Guide
G.SRT.6-Some Things To Know
Similar right triangles have side ratios that are
equal to each other. For example, every 3060-90 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 :
Adjacent
Hypotenuse
tan θ =
Opposite
Adjacent
You try:
Given ΔMAT, match each trigonometric ratio
to its equivalent value in the box.
17
M
T
8
15
A
Opposite
sin θ =
Hypotenuse
cos θ =
5´
Ex: Write each trigonometric ratio using the
side lengths of ΔABC below.
A.
B.
C.
D.
E.
Solutions:
A.
B.
C.
D.
E.
4
5
3
5
3
4
4
3
4
5
G.SRT.6
Page 5 of 11
G.SRT.6
MCC@WCCUSD 12/11/14
WCCUSD Geometry
6
Benchmark 2 Study Guide
G.SRT.7-Some Things To Know
6´
The sine of an angle is equal to the cosine of
its complement: sin θ = cos(90 − θ ) .
You try:
The table below shows the approximate
values of sine and cosine for selected angles.
A. Fill in the rest of the table.
Q
12
5
P
Angle
Value of Sine
Value of Cosine
0°
0
0.2588
0.5000
0.7071
1
0.9659
0.8660
15°
13
R
30°
45°
According to the figure above:
5
5
sin P = ≈ 0.3846 and cos R = ≈ 0.3846
13
13
60°
75°
90°
So, sin P = cos R .
B. Explain how you determined the values
you used.
If sin 32° ≈ 0.5514 then the cosine of its
complement is equivalent:
sin 32° = cos(90° − 32°) = cos58° ≈ 0.5514
Ex: Determine whether the following
statements are true or false:
A. sin 43° = cos 47°
B. sin 43° = cos 43°
C. sin 45° = cos 45°
D. sin17° = cos(90 −17)°
E. cosθ = sin(90 − θ )
Solutions: A—True, B—False, C—True,
D—True, E—True
G.SRT.7
G.SRT.7
Page 6 of 11
MCC@WCCUSD 12/11/14
WCCUSD Geometry
Benchmark 2 Study Guide
G.SRT.8-Some Things To Know
7
You try:
7´
A. 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.
Drawing and labeling pictures are a great way to solve
problems using the trigonometric ratios. Don’t forget
the Pythagorean Theorem (𝑎! + 𝑏! = 𝑐 ! )!
A. The angle of elevation from a landscaped rock
to the top of a 30-foot tall flagpole is
.
Select all equations that can be used to solve
for the angle of depression (θ) from the plane
to the runway.
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. cos57° =
C. tan 57° =
D. tan 33° =
30
x
x
30
30
x
x
30
Solution:
A. sin θ =
900
1500
B. cosθ =
1200
1500
C. tan θ =
1200
900
D. sin θ =
1200
1500
E. cosθ =
900
1500
F. tan θ =
900
1200
B. Determine the missing side lengths of the
following triangles:
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
opposite
ratios of those angles do show
.
hypotenuse
y
x
60°
18 ft
30°
22
60°
18 ft
Also, here are some special right triangle
ratios to memorize:
45-45-90
18 ft
h
30-60-90
1
2
1
1
G.SRT.8
G.SRT.8
End of Study Guide
Page 7 of 11
MCC@WCCUSD 12/11/14
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 dilated ΔABC by a
scale factor of 3 with center of dilation at
(2, 1) and then translated 4 units right and 2
units up, then ΔABC maps onto ΔA' B'C ' .
B. Dilate the following figure using a scale
1
factor of
with center at (4,-2).
2
Another solution could be dilating ΔABC 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 -2 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
∴ ∆𝑇𝑈𝑉~∆𝑊𝑋𝑌.
G.SRT.1
Page 8 of 11
G.SRT.2
MCC@WCCUSD 12/11/14
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 ΔABC
to ΔDEC and supports Omar’s thinking?
h 15
(4)(6) = (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.
G.SRT.3
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
3x + 30 = 34
3x = 4
4
x=
3
G.SRT.5
Page 9 of 11
MCC@WCCUSD 12/11/14
WCCUSD Geometry
5´
Benchmark 2 Study Guide
You try:
7´
Given ΔMAT, match each trigonometric
ratio to its equivalent value in the box.
17
M
T
A. 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.
8
15
You try:
A
A.
B.
C.
D.
E.
F.
A
B
D
C
A
B
900 m
1200 m
G.SRT.6
6´
Solution: Use the Pythagorean Theorem to
find the length of the hypotenuse:
You try:
The table below shows the approximate
values of sine and cosine for selected
angles.
A. Fill in the rest of the table.
Value of Sine
Value of Cosine
0°
0
0.2588
0.5000
1
0.9659
0.8660
0.7071
0.8660
0.9659
1
0.7071
0.5000
0.2588
0
30°
45°
60°
75°
90°
1200
!
+ 900
!
= 𝑐!
1440000 + 810000 = 𝑐 !
Angle
15°
𝑎! + 𝑏! = 𝑐 !
2250000 = 𝑐 !
1500 = c
B. 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° = cos60° , sin 45° = cos 45° and
sin 0° = cos90° .
C. tan θ =
1200
900
D. sin θ =
1200
1500
E. cosθ =
900
1500
G.SRT.8
G.SRT.7
Page 10 of 11
MCC@WCCUSD 12/11/14
WCCUSD Geometry
7´
Benchmark 2 Study Guide
Continued…
B. Determine the missing side lengths of the
following triangles:
y
x
60°
30°
22
The shorter side’s length is half the
hypotenuse. So x = 11 . The length of the
longer leg is 3 longer than the shorter leg,
so y = 11 3 .
18 ft
h
18 ft
60°
18 ft
The altitude of the triangle creates a 30-60-90
triangle with the shorter leg being half of
18 ft, which is 9. Therefore the longer leg is
3 longer than the shorter leg, so h = 9 3 .
Since two sides are equal, this is an isosceles
triangle, which means that the base angles are
equal. Each base angle is 45° , so this is a
45-45-90 triangle. The hypotenuse is 2
longer than the sides, which means the side
lengths are 7 units.
G.SRT.8
Page 11 of 11
MCC@WCCUSD 12/11/14