1. THE MEASURE OF TWO COMPLEMENTARY ANGLES
ARE IN THE RATIO 1 : 5. WHAT ARE THE DEGREE
MEASURES OF THE TWO ANGLES?
A.
B.
C.
D.
𝟑𝟔𝒐 and 𝟏𝟒𝟒𝒐
𝟏𝟓𝒐 and 𝟕𝟓𝒐
𝟑𝟎𝒐 and 𝟏𝟓𝟎𝒐
𝟏𝟖𝒐 and 𝟕𝟐𝒐
57%
43%
7-1
0%
A.
0%
B.
C.
D.
2
1
5
=
𝑥
90−𝑥
Cross multiply
90 − 𝑥 = 5𝑥
Add x
90 = 6𝑥
Divide by 6
15 = 𝑥
Answers are 15 and 90  15 = 75
3
2. THE POLYGONS ARE SIMILAR, BUT NOT
NECESSARILY DRAWN TO SCALE. FIND THE
K
VALUE OF X.
A.
B.
C.
D.
B
J
A
L
C
8
4
55
x
D
220
27.5
110
15.8
88%
M
7-1
13%
0%
A.
0%
B.
C.
D.
4
8
55
=
4
𝑥
Cross multiply
8𝑥 = 220
Divide by 8
𝑥 = 27.5
Answer is 27.5
5
3. ARE THE TWO TRIANGLES SIMILAR?
HOW DO YOU KNOW?
A.
B.
C.
D.
J
H
39 °
M
39 º
no
Yes, by SSS
Yes, by AA
Yes, by SAS
K
50%
G
38%
7-3
13%
0%
A.
6
B.
C.
D.
Angles at M are vertical angles so ∠𝐻𝑀𝐺 ≅ ∠𝐽𝑀𝐾
This makes the two triangles similar by AA
7
4. FIND THE GEOMETRIC MEAN OF THE PAIR
OF NUMBERS.
242 AND 8
A.
B.
C.
D.
44
49
1872
54
63%
7-4
25%
13%
0%
A.
B.
C.
D.
8
242
𝑥
Cross multiply
=
𝑥
8
𝑥 2 = 1936
Take the square root of both sides
𝑥 = 44
Answer is 44
9
5. WHAT IS THE VALUE OF X, GIVEN THAT 𝑃𝑄 ∥ 𝐵𝐶?
A
7
P
35
B
A.
B.
C.
D.
x
Q
40
8
10
11
16
C
63%
7-5
25%
13%
0%
A.
B.
C.
D.
10
7
35
=
𝑥
40
Cross multiply
280 = 35𝑥
Divide by 35
8=𝑥
Answer is 8
11
6. FIND THE LENGTH OF THE MISSING SIDE.
THE TRIANGLE IS NOT DRAWN TO SCALE.
A.
B.
C.
D.
17
15
𝟖
64
4
𝟐𝟑
63%
8-1
25%
13%
0%
A.
B.
12
C.
D.
𝑎2 + 𝑏 2 = 𝑐 2
Substitute
𝑎2 + (15)2 = (17)2
Simplify
𝑎2 + 225 = 289
Subtract 225 from both sides
𝑎2 = 64
Take the square root of both sides
𝑎=8
Side length is 8.
13
7. FIND THE LENGTH OF THE LEG. IF YOUR
ANSWER IS NOT AN INTEGER, LEAVE IT IN
SIMPLEST RADICAL FORM.
A. 𝟐 𝟑
B. 𝟐𝟖𝟖
C. 𝟐𝟒
24
D. 𝟏𝟐 𝟐
45°
63%
Not drawn to scale
8-2
25%
13%
0%
A.
B.
14
C.
D.
The legs of a 45-45-90 triangle are the same.
The hypotenuse is leg * 2
Divide 24 by 2
Simplify
24
2
= 12 2
The answer is 12 2
15
8. FIND THE VALUE OF THE VARIABLE(S). IF YOUR
ANSWER IS NOT AN INTEGER, LEAVE IT IN SIMPLEST
RADICAL FORM.
A. 𝟓 𝟑
5
x
60°
10
B.
𝟏
𝟐
C. 𝟏𝟎 𝟑
D. 2
Not drawn to scale
88%
8-2
13%
0%
0%
16
The long leg of a 30-60-90 triangle is the short leg * 3.
Multiply 5 by 3
The answer is 5 3.
17
9. FIND THE MISSING VALUE TO THE NEAREST
HUNDREDTH.
cos
2
= 5
A.
B.
C.
D.
23.58o
66.42o
21.8o
63.21o
50%
50%
8-3
0%
A.
0%
B.
C.
D.
18
To find degrees you use the inverse function:
𝟐
-1
Cos ( )
𝟓
= 66.42182152
The answer is 66.42o
19
10. FIND THE VALUE OF X. ROUND THE LENGTH
TO THE NEAREST TENTH.
A.
B.
C.
D.
16 m
x

42.7 m
6.5 m
14.8 m
6m
Not drawn to scale
38%
25%
8-4
25%
13%
20
A.
B.
C.
D.
Use trigonometry to find the missing side.
Set of trig function
sin 22 =
𝑥
16
Variable is in the top so we multiply by the trig function
16 ∗ sin 22 = 𝑥
Simplify
𝑥 = 5.993705495
The answer is 6 m.
21
11. IN THE DIAGRAM, FIGURE RQTS IS THE IMAGE
OF FIGURE DEFC AFTER A RIGID MOTION.
NAME THE IMAGE OF ∠F.
A.
B.
C.
D.
D
R
E
S
C
F
∠T
∠R
∠Q
∠S
Q
T
63%
9-1
13%
13%
13%
22
A.
B.
C.
D.
D
R
E
S
C
F
T
Q
The image of ∠F is ∠S.
23
12. FIND THE IMAGE OF C UNDER THE TRANSLATION
DESCRIBED BY THE TRANSLATION RULE 𝑇 4,5 𝐶 .
y
E8
A
4
C
–8
A.
B.
C.
D.
–4
4
–4
8
x
D
B
E
A
D
50%
B
–8
38%
9-1
13%
0%
A.
24
B.
C.
D.
1.
𝑇 4,5 𝐶
y
E8
4
C
–8
Move C four units to the right.
A
–4
Then move five units up.
4
–4
8
x
New point is E(-3, -7)
D
B
–8
25
13. FIND THE AREA. THE FIGURE IS NOT DRAWN
TO SCALE.
36 in.
40 in.
A.
B.
C.
D.
33 in.
1188 in2
69 in2
138 in2
1440 in2
50%
38%
10-1
13%
0%
A.
B.
26
C.
D.
To find area of a parallelogram:
36 in.
40 in.
33 in.
Base x Height
(36)(33) = 1188
The answer is 1188 in2
27
14. FIND THE AREA. THE FIGURE IS NOT DRAWN
TO SCALE.
A.
B.
C.
D.
2 cm
4.4 cm
10-1
4.4 cm2
6.4 cm2
8.8 cm2
17.6 cm2
75%
25%
0%
A.
B.
0%
C.
D.
28
To find area of a triangle:
2 cm
𝟏
(Base
𝟐
x Height)
𝟏
(2)(4.4)
𝟐
4.4 cm
= 4.4
The answer is 4.4 cm2
29
15. FIND THE AREA OF THE TRAPEZOID. LEAVE
YOUR ANSWER IN SIMPLEST RADICAL FORM.
A.
B.
C.
D.
5 cm
9 cm
45°
31.5 cm2
7 cm2
81 cm2
94.5 cm2
2 cm
50%
Not drawn to scale
10-2
25%
25%
0%
A.
B.
30
C.
D.
5 cm
To find area of a trapezoid:
9 cm
𝟏
(Sum
𝟐
45°
2 cm
Not drawn to scale
𝟏
(5
𝟐
of Bases)(Height)
+ 2 + 5 + 9)(9) = 94.5
The answer is 94.5 cm2
31
16. WHAT IS THE AREA OF THE KITE?
6 ft
2 ft
2 ft
12 ft
A.
B.
C.
D.
11 ft2
72 ft2
36 ft2
44 ft2
50%
10-2
25%
13%
13%
32
A.
B.
C.
D.
To find area of a kite:
6 ft
2 ft
2 ft
𝟏
(diagonal
𝟐
1)(diagonal 2)
12 ft
𝟏
(4)(18)
𝟐
= 36
The answer is 36 ft2
33
17. FIND THE AREA OF A REGULAR HEXAGON WITH AN
APOTHEM 17.3 MILES LONG AND A SIDE 20 MILES
LONG. ROUND YOUR ANSWER TO THE NEAREST TENTH.
A.
B.
C.
D.
173.2 mi2
2078.5 mi2
692.8 mi2
1038 mi2
43%
10-3
43%
14%
0%
A.
34
B.
C.
D.
To find area of a polygon:
𝟏
(apothem)(perimeter)
𝟐
𝟏
(17.3)(120)
𝟐
= 1038
17.3 miles
The answer is 1038 mi2
20 miles
35
18. THE WIDTHS OF TWO SIMILAR RECTANGLES
ARE 21 FT AND 18 FT. WHAT IS THE RATIO OF THE
PERIMETERS? OF THE AREAS?
57%
A.
B.
C.
D.
8 : 7 and 64 : 49
8 : 7 and 49 : 36
7 : 6 and 64 : 49
7 : 6 and 49 : 36
29%
14%
0%
A.
B.
C.
D.
10-4
36
Ratio of perimeters is a : b:
𝟐𝟏 ∶ 𝟏𝟖
Which reduces to:
21 feet
18 feet
𝟕∶𝟔
Ratio of areas is a2 : b2:
72 : 62
Which simplifies to:
𝟒𝟗 ∶ 𝟑𝟔
37
19. FIND THE SIMILARITY RATIO AND THE RATIO OF
PERIMETERS FOR TWO REGULAR PENTAGONS
WITH AREAS OF 𝟏𝟔 𝒄𝒎𝟐 AND 𝟒𝟗 𝒄𝒎𝟐 .
A.
B.
C.
D.
4:7;4:7
16 : 49 ; 4 : 7
16 : 49; 16 : 49
4 : 7; 16 : 49
43%
43%
10-4
14%
0%
A.
B.
C.
D.
38
Ratio of areas is a2 : b2:
𝟒𝟗
𝒄𝒎𝟐
𝟏𝟔 𝒄𝒎𝟐
49 : 𝟏𝟔
Taking the square root gives you a : b:
𝟕∶𝟒
Ratio of perimeters is also a : b:
𝟕∶𝟒
39
20. FIND THE MEASURE OF 𝐶𝐷𝐸.
THE FIGURE IS NOT DRAWN TO SCALE.
A
A. 188
E
103°
D
27°
O
50°
B. 182
C. 162
D. 172
B
35°
C
57%
10-6
29%
14%
0%
A.
40
B.
C.
D.
𝒎𝑫𝑪 = 𝟑𝟔𝟎 − 𝒎𝑬𝑫 − 𝒎𝑬𝑨 − 𝒎𝑨𝑩 − 𝒎𝑩𝑪
𝒎𝑫𝑪 = 𝟑𝟔𝟎 − 𝟐𝟕 − 𝟏𝟎𝟑 − 𝟓𝟎 − 𝟑𝟓
A
𝒎𝑫𝑪 = 𝟑𝟔𝟎 − 𝟐𝟏𝟓
E
103°
D
27°
O
𝒎𝑫𝑪 = 𝟏𝟒𝟓
50°
B
35°
C
𝐶𝐷𝐸 = 𝒎𝑬𝑫 + 𝒎𝑫𝑪
𝐶𝐷𝐸 = 𝟐𝟕 + 𝟏𝟒𝟓 = 𝟏𝟕𝟐
41
21. FIND THE CIRCUMFERENCE. LEAVE YOUR
ANSWER IN TERMS OF .
5.7 cm
A.
B.
C.
D.
11.4 cm
8.55 cm
2.85 cm
5.7 cm
43%
29%
10-6
14%
14%
42
A.
B.
C.
D.
To find circumference of a circle:
𝟐()(radius) or ()(diameter)
5.7 cm
()(5.7) = 5.7
The answer is 5.7 cm2
43
22. FIND THE LENGTH OF 𝑌𝑃𝑋. LEAVE YOUR
ANSWER IN TERMS OF .
X
10 m
Y
A.
B.
C.
D.
30 m
15 m
5 m
900 m
P
86%
10-6
14%
0%
A.
B.
C.
0%
D.
44
X
To find the length of an arc:
𝒎𝒂𝒓𝒄
𝟐()(𝑟𝑎𝑑𝑖𝑢𝑠)
𝟑𝟔𝟎
10 m
P
Y
𝟐𝟕𝟎
𝟐( )(𝟏𝟎)
𝟑𝟔𝟎
The answer is 15 m.
45
23. FIND THE AREA OF THE CIRCLE. LEAVE YOUR
ANSWER IN TERMS OF .
4.1 m
A.
B.
C.
D.
4.2025 cm2
8.405 cm2
16.81 cm2
11.2 cm2
10-7
0%
A.
0%
B.
0%
C.
0%
D.
46
90
To find area of a circle:
()(radius)2
4.1 m
()(
𝟒.𝟏 2
)
𝟐
= 4.2025
The answer is 4.2025 cm2
47
24. THE AREA OF SECTOR AOB IS 𝟐𝟎. 𝟐𝟓𝝅𝒇𝒕𝟐 .
FIND THE EXACT AREA OF THE SHADED REGION.
A
A. (𝟐𝟎. 𝟐𝟓𝝅 − 𝟒𝟎. 𝟓)𝒇𝒕𝟐
B. (𝟐𝟎. 𝟐𝟓𝝅 − 𝟖𝟏)𝒇𝒕𝟐
O
9 ft
B
C. (𝟐𝟎. 𝟐𝟓𝝅 − 𝟒𝟎. 𝟓 𝟐)𝒇𝒕𝟐
D. None of these
10-7
0%
A.
0%
B.
0%
C.
0%
D.
48
90
To find area of a segment of a circle:
A
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒔𝒆𝒄𝒕𝒐𝒓 − 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆
𝒎𝒂𝒓𝒄 𝟐 𝟏
𝝅𝒓 − 𝒃𝒉
𝟑𝟔𝟎
𝟐
O
9 ft
B
𝟗𝟎
𝟏
𝟐
𝝅(𝟗) − 𝟗 𝟗
𝟑𝟔𝟎
𝟐
𝟐𝟎. 𝟐𝟓𝝅 − 𝟒𝟎. 𝟓
The answer is (𝟐𝟎. 𝟐𝟓𝝅 − 𝟒𝟎. 𝟓) ft2
49
25. FIND THE PROBABILITY THAT A POINT CHOSEN AT
𝟏
RANDOM FROM 𝐽𝑃 IS ON THE SEGMENT 𝐾𝑂 .
A.
J K L M N O P Q R S T
0
1
2
3
4
5
6
7
8
9
B.
10
C.
D.
𝟐
𝟒
𝟓
𝟓
𝟔
𝟐
𝟑
10-8
0%
A.
0%
B.
0%
C.
0%
D.
50
90
J K L M N O P Q R S T
0
1
2
3
4
5
To find probability:
6
7
8
9
10
𝒘𝒉𝒂𝒕 𝒚𝒐𝒖 𝒘𝒂𝒏𝒕
𝒘𝒉𝒂𝒕 𝒊𝒔 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆
𝑲𝑶 𝟒
=
𝑱𝑷 𝟔
The answer is
𝟐
𝟑
51
26. USE EULER’S FORMULA TO FIND THE MISSING
NUMBER.
A. 43
FACES: 25
VERTICES: 17
B. 40
EDGES: ?
C. 39
D. 41
11-1
0%
A.
0%
B.
0%
C.
0%
D.
52
90
Faces + Vertices = Edges + 2
F+V=E+2
25 + 17 = E + 2
Plug in values
42= E + 2
Simplify
40= E
Subtract 2 from both sides
The answer is 40 𝑒𝑑𝑔𝑒𝑠
53
27. USE FORMULAS TO FIND THE LATERAL AREA AND
SURFACE AREA OF THE GIVEN PRISM. ROUND YOUR
ANSWER TO THE NEAREST WHOLE NUMBER.
4m
4m
13 m
11-2
A.
B.
C.
D.
𝟐
𝟐
𝟐𝟎𝟖
𝒎
;
𝟏𝟖𝟖
𝒎
Not drawn to scale
𝟏𝟑𝟔 𝒎𝟐 ; 𝟏𝟖𝟖 𝒎𝟐
𝟏𝟑𝟔 𝒎𝟐 ; 𝟐𝟒𝟎 𝒎𝟐
𝟐𝟎𝟖 𝒎𝟐 ; 𝟐𝟒𝟎 𝒎𝟐
0%
A.
0%
B.
0%
C.
0%
D.
54
90
4m
4m
13 m
Not drawn to scale
LA = (perimeter of base)(height)
SA = LA + 2*B
LA = (34)(4)
SA = 136 + 2(13*4)
LA = 136
SA = 136 + 104 = 240
55
28. FIND THE SURFACE AREA OF THE CYLINDER IN
TERMS OF .
A.
B.
C.
D.
9 cm
19 cm
504 cm2
333 cm2
382.5 cm2
211.5 cm2
Not drawn to scale
11-2
0%
A.
0%
B.
0%
C.
0%
D.
56
90
SA = 2𝜋𝑟ℎ + 2𝜋𝑟 2
9 cm
19 cm
9
2
9
2
SA = 2𝜋( )(19) + 2𝜋( )2
SA = 171𝜋 + 40.5𝜋
SA = 211.5𝜋 𝑐𝑚2
Not drawn to scale
57
29. FIND THE SURFACE AREA OF THE REGULAR
PYRAMID SHOWN TO THE NEAREST WHOLE NUMBER.
𝟏𝟏𝟒𝟗 𝒎𝟐
𝟓𝟓𝟎 𝒎𝟐
𝟓𝟕𝟒 𝒎𝟐
𝟒𝟎𝟖 𝒎𝟐
A.
B.
C.
D.
11-3
0%
A.
0%
B.
0%
C.
0%
D.
58
90
SA =
1
2
𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 + 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒
SA
1
2
SA = 8 ∗ 6 17 +
1
2
1
= 𝑝𝑙
2
1
+ 𝑎𝑝
2
4 3 8∗6
SA =408 + 166.2768775
SA = 574.2768775 𝑚2
59
30. FIND THE SURFACE AREA OF THE CONE TO THE
NEAREST TENTH.
A.
B.
C.
D.
𝟐𝟎𝟑 𝒄𝒎𝟐
𝟏𝟑𝟒𝟏. 𝟓 𝒄𝒎𝟐
𝟕𝟒𝟕. 𝟕 𝒄𝒎𝟐
𝟐𝟏𝟑. 𝟓 𝒄𝒎𝟐
11-3
0%
A.
0%
B.
0%
C.
0%
D.
60
90
SA = 𝜋)(𝑟𝑎𝑑𝑖𝑢𝑠 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 + 𝜋𝑟 2
SA = 𝜋)(7 27 + 𝜋(7)2
SA = 189𝜋 + 49𝜋
SA =238𝜋
SA = 747.6990516 𝑚2
61
31. FIND THE VOLUME OF THE GIVEN PRISM. ROUND
TO THE NEAREST TENTH IF NECESSARY.
𝟑𝟎𝟖. 𝟗 𝒄𝒎𝟑
𝟑𝟎𝟖. 𝟐 𝒄𝒎𝟑
𝟑𝟏𝟐. 𝟖 𝒄𝒎𝟑
𝟑𝟎𝟐. 𝟗 𝒄𝒎𝟑
A.
B.
C.
D.
11-4
0%
A.
0%
B.
0%
C.
0%
D.
62
90
Volume= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡
V= 𝐵ℎ = 𝑏 ∗ ℎ ℎ
V= 13.2 ∗ 3.9 (6)
V = 308.88 𝑐𝑚3
63
32. FIND THE VOLUME OF THE CYLINDER IN
TERMS OF  .
A.
B.
C.
D.
𝟔𝟎. 𝟖𝝅 𝒎𝟑
𝟏𝟏𝟓. 𝟓𝟐𝝅 𝒎𝟑
𝟒𝟑𝟖. 𝟗𝟖𝝅 𝒎𝟑
𝟓𝟕. 𝟕𝟔𝝅 𝒎𝟑
11-4
0%
A.
0%
B.
0%
C.
0%
D.
64
90
Volume= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡
V= 𝜋𝑟 2 ℎ
V= 𝜋 3.82 (8)
V = 115.52𝜋 𝑐𝑚3
65
33. FIND THE VOLUME OF THE SQUARE PYRAMID
SHOWN. ROUND TO THE NEAREST TENTH IF
NECESSARY.
A. 𝟏𝟐𝟔 𝒄𝒎𝟑
B. 𝟗𝟎𝟕. 𝟓 𝒄𝒎𝟑
C. 𝟔𝟎𝟓 𝒄𝒎𝟑
D. 𝟓𝟓 𝒄𝒎𝟑
11-5
0%
A.
0%
B.
0%
C.
0%
D.
66
90
Volume=
1
3
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡
V=
V=
1
3
1
3
𝑏ℎ ℎ
11 ∗ 11 15
V = 605 𝑐𝑚3
67
34. FIND THE VOLUME OF THE SQUARE PYRAMID
SHOWN. ROUND TO THE NEAREST TENTH IF NECESSARY.
A.
B.
C.
D.
𝟔𝟎𝟎 𝒇𝒕𝟑
𝟑𝟑. 𝟑 𝒇𝒕𝟑
𝟏𝟐𝟎𝟎 𝒇𝒕𝟑
𝟒𝟎𝟎 𝒇𝒕𝟑
11-5
0%
A.
0%
B.
0%
C.
0%
D.
68
90
Volume=
1
3
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 = V =
1
3
𝑏ℎ ℎ
𝐹𝑖𝑟𝑠𝑡 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡: 𝑎2 + 𝑏 2 = 𝑐 2
25 + 𝑏 2 = 169
𝑏 2 = 144
V=
1
3
𝑏 = 12
10 ∗ 10 12
V = 400 𝑓𝑡 3
69
35. FIND THE VOLUME OF THE OBLIQUE CONE
SHOWN IN TERMS OF  .
𝟔𝟔𝟏. 𝟓𝝅 𝒊𝒏𝟑
𝟏𝟑𝟐𝟑𝝅 𝒊𝒏𝟑
𝟔𝟑𝝅 𝒊𝒏𝟑
𝟒𝟒𝟏𝝅 𝒊𝒏𝟑
A.
B.
C.
D.
11-5
0%
A.
0%
B.
0%
C.
0%
D.
70
90
Volume=
1
3
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 = V =
V=
1
3
1
3
𝜋𝑟 2 ℎ
𝜋72 27
V = 441𝜋 𝑖𝑛3
71
36. FIND THE SURFACE AREA OF THE SPHERE WITH
THE GIVEN DIMENSION. LEAVE YOUR ANSWER IN
TERMS OF 𝝅 .
DIAMETER OF 14 CM
A.
B.
C.
D.
𝟕𝟖𝟒𝝅 𝒄𝒎𝟐
𝟏𝟗𝟔𝝅 𝒄𝒎𝟐
𝟐𝟖𝝅 𝒄𝒎𝟐
𝟗𝟖𝝅 𝒄𝒎𝟐
11-6
0%
A.
0%
B.
0%
C.
0%
D.
72
90
SA of a sphere = 4 𝜋 𝑟 2
SA = 4 𝜋 72
14 cm
SA = 196𝜋 𝑐𝑚2
73
37. FIND THE VOLUME OF THE SPHERE SHOWN. GIVE
EACH ANSWER ROUNDED TO THE NEAREST CUBIC UNIT.
9 mm
𝟏𝟓𝟐𝟕 𝒎𝒎𝟑
𝟑𝟑𝟗 𝒎𝒎𝟑
𝟑𝟎𝟓𝟒 𝒎𝒎𝟑
𝟏𝟎𝟏𝟖 𝒎𝒎𝟑
A.
B.
C.
D.
11-6
0%
A.
0%
B.
0%
C.
0%
D.
74
90
Volume of a sphere =
4
3
𝜋 𝑟3
4
𝑉 = 𝜋 93
3
9 mm
𝑉 = 972𝜋 𝑚𝑚3
𝑉 = 3053.628059
𝑉 = 3054 𝑚𝑚3
75
38. ASSUME THAT LINES THAT APPEAR TO BE TANGENT
ARE TANGENT. O IS THE CENTER OF THE CIRCLE. FIND
THE VALUE OF X. (FIGURES ARE NOT DRAWN TO SCALE.)
𝒎∠𝑶 = 𝟏𝟑𝟓
A. 45
x°
B. 67.5
C. 315
D. 270
O
12-1
0%
A.
0%
B.
0%
C.
0%
D.
76
90
Tangent lines are perpendicular to the radius of the circle so the
red angles are 90 degrees.
135 + 90 + 90 = 315
360  315 = 45
x = 45o
𝒎∠𝑶 = 𝟏𝟑𝟓
x°
O
77
39. FIND THE VALUE OF X. IF NECESSARY, ROUND
YOUR ANSWER TO THE NEAREST TENTH. O IS THE
CENTER OF THE CIRCLE. THE FIGURE IS NOT DRAWN
TO SCALE.
12
9
O
x
A.
B.
C.
D.
12
9
15
5
12-2
0%
0%
0%
0%
78
90
Rotate the radius around until it makes a right triangle.
𝑎2 + 𝑏 2 = 𝑐 2
122 + 92 = 𝑐 2
225 =
𝑐2
12
9
O
15 = 𝑐
x
79
40. FIND THE MEASURE OF ∠BAC IN CIRCLE O.
(THE FIGURE IS NOT DRAWN TO SCALE.)
A
A.
B.
C.
D.
O
40º
50
80
20
40
C
B
12-3
0%
A.
0%
B.
0%
C.
0%
D.
80
90
Inscribed angles are half their intercepted arcs measure.
∠𝐵𝐴𝐶 = 1/2𝐵𝐶
A
∠𝐵𝐴𝐶 = 1/2(40)
O
∠𝐵𝐴𝐶 = 20
40º
C
B
81
41. 𝐴𝐶 IS TANGENT TO CIRCLE O AT A. IF 𝑚𝐵𝑌 = 24,
WHAT IS 𝑚∠YAC? (THE FIGURE IS NOT DRAWN TO
B
SCALE.)
Y
A.
B.
C.
D.
O
A
132
48
78
156
C
12-3
0%
A.
0%
B.
0%
C.
0%
D.
82
90
Measure of ∠𝑌𝐴𝐶 is half of its intercepted arc 𝑌𝐴
𝐵𝑌𝐴 𝑖𝑠 𝑎 𝑠𝑒𝑚𝑖𝑐𝑖𝑟𝑐𝑙𝑒 = 180
B
24
Y
𝐵𝑌𝐴 = 𝐵𝑌 + 𝑌𝐴
180 = 24 + 𝑌𝐴
O
156 = 𝑌𝐴
∠𝑌𝐴𝐶
A
C
1
= 𝑌𝐴
2
=
1
(156)
2
∠𝑌𝐴𝐶 = 78
83
42. 𝑚𝐷𝐸 = 120 AND 𝑚𝐵𝐶 = 53. FIND 𝑚∠𝐴.
(THE FIGURE IS NOT DRAWN TO SCALE.)
A.
B.
C.
D.
D
B
A
C
33.5
93.5
86.5
67
E
12-4
0%
A.
0%
B.
0%
C.
0%
D.
84
90
Measure of ∠𝐴 is half of the difference of its
intercepted arcs 𝐷𝐸 and 𝐵𝐶
1
(𝐷𝐸 − 𝐵𝐶) = 𝑚∠𝐴
2
D
B
120
53
A
1
(120 − 53) = 𝑚∠𝐴
2
C
E
33.5 = 𝑚∠𝐴
85
43. FIND THE VALUE OF X. IF NECESSARY, ROUND
YOUR ANSWER TO THE NEAREST TENTH. THE
FIGURES ARE NOT DRAWN TO SCALE. THE FIGURE
CONSISTS OF A TANGENT AND A SECANT TO THE
x
CIRCLE
5
12
A.
B.
C.
D.
9.2
7.7
14.3
85
0%
A.
0%
B.
0%
C.
0%
D.
12-4
86
90
x
5
The formula is: 𝑥 2 = 12 + 5 5
𝑥 2 = 17 5
𝑥 2 = 85
12
𝑥 = 9.219544457
𝑥 = 9.2
87
44. WRITE THE STANDARD EQUATION FOR THE CIRCLE.
CENTER (10, –6), R = 6
A.
B.
C.
D.
(𝒙 + 𝟏𝟎)𝟐 +(𝒚 − 𝟔)𝟐 = 𝟔
(𝒙 + 𝟔)𝟐 +(𝒚 − 𝟏𝟎)𝟐 = 𝟑𝟔
(𝒙 − 𝟏𝟎)𝟐 +(𝒚 + 𝟔)𝟐 = 𝟑𝟔
(𝒙 − 𝟏𝟎)𝟐 +(𝒚 + 𝟔)𝟐 = 𝟔
12-5
0%
A.
0%
B.
0%
C.
0%
D.
88
90
The equation for a circle is :
𝑥 − 𝑥1
2
+ 𝑦 − 𝑦1
2
= 𝑟2
center (10, –6), r = 6
(𝒙 − 𝟏𝟎)𝟐 +(𝒚 + 𝟔)𝟐 = 𝟑𝟔
89