1. Final REVIEW Geometry B
Questions similar to what will be on the final exam

2. 7-1 𝟑𝟔 𝒐 and 𝟏𝟒𝟒 𝒐 𝟏𝟓 𝒐 and 𝟕𝟓 𝒐 𝟑𝟎 𝒐 and 𝟏𝟓𝟎 𝒐 𝟏𝟖 𝒐 and 𝟕𝟐 𝒐
1. The measure of two complementary angles are in the ratio 1 : 5. What are the degree measures of the two angles?
𝟑𝟔 𝒐 and 𝟏𝟒𝟒 𝒐
𝟏𝟓 𝒐 and 𝟕𝟓 𝒐
𝟑𝟎 𝒐 and 𝟏𝟓𝟎 𝒐
𝟏𝟖 𝒐 and 𝟕𝟐 𝒐
7-1 89

3. 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

4. 2. The polygons are similar, but not necessarily drawn to scale
2. The polygons are similar, but not necessarily drawn to scale. Find the value of X.
220
27.5
110
15.8
7-1 88

5. 8 55 = 4 𝑥
Cross multiply 8𝑥=220
Divide by 8 𝑥=27.5
Answer is 27.5

6. 3. Are the two triangles similar? How do you know?
Yes, by SSS
Yes, by AA
Yes, by SAS
7-3 88

7. Angles at M are vertical angles so ∠𝐻𝑀𝐺≅∠𝐽𝑀𝐾
This makes the two triangles similar by AA

8. 4. Find the geometric mean of the pair of numbers. 242 and 844 49
1872 54
7-4 88

9. 242 𝑥 = 𝑥 8
Cross multiply 𝑥 2 =1936
Take the square root of both sides
𝑥=44
Answer is 44

10. 5. What is the value of x, given that 𝑃𝑄 ∥ 𝐵𝐶 ?8 10 11 16
7-5 88

11. 7 35 = 𝑥 40
Cross multiply 280=35𝑥
Divide by 35 8=𝑥
Answer is 8

12. 6. Find the length of the missing side
6. Find the length of the missing side. The triangle is not drawn to scale. 𝟖 64 4 𝟐𝟑
8-1 88

13. 𝑎 2 + 𝑏 2 = 𝑐 2
Substitute 𝑎 2 + (15) 2 = (17) 2
Simplify 𝑎 =289
Subtract 225 from both sides 𝑎 2 =64
Take the square root of both sides 𝑎=8
Side length is 8.

14. 7. Find the length of the leg
7. Find the length of the leg. If your answer is not an integer, leave it in simplest radical form.
𝟐 𝟑
𝟐𝟖𝟖 𝟐𝟒
𝟏𝟐 𝟐
8-2 88

15. The legs of a 45-45-90 triangle are the same.
The hypotenuse is leg * 2
Divide 24 by 2
Simplify = 12 2
The answer is 12 2

16. 8. Find the value of the variable(s)
8. Find the value of the variable(s). If your answer is not an integer, leave it in simplest radical form.
𝟓 𝟑
𝟏 𝟐
𝟏𝟎 𝟑 2
8-2 87

17. The long leg of a 30-60-90 triangle is the short leg * 3 .
Multiply 5 by 3
The answer is 5 3.

18. 9. Find the missing value to the nearest hundredth.
8-3 86

19. To find degrees you use the inverse function:
Cos-1( 𝟐 𝟓 ) =
The answer is 66.42o

20. 10. Find the value of x. Round the length to the nearest tenth.
42.7 m
6.5 m
14.8 m
6 m
8-4 89

21. 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 ∗ sin 22=𝑥
Simplify 𝑥=
The answer is 6 m.

22. 11. In the diagram, figure RQTS is the image of figure DEFC after a rigid motion. Name the image of ∠F. ∠T ∠R ∠Q ∠S
9-1 86

23. The image of ∠F is ∠S.

24. 12. Find the image of C under the translation described by the translation rule 𝑇 4, 5 𝐶 .
9-1 89

25. B 𝑇 4, 5 𝐶 1. Move C four units to the right. Then move five units up.
𝑇 4, 5 𝐶 1. B
Move C four units to the right.
Then move five units up.
New point is E(-3, -7)

26. 13. Find the area. The figure is not drawn to scale.
10-1 88

27. To find area of a parallelogram:
Base x Height
(36)(33) = 1188
The answer is 1188 in2

28. 14. Find the area. The figure is not drawn to scale.
4.4 cm2
6.4 cm2
8.8 cm2
17.6 cm2
10-1 89

29. To find area of a triangle:
𝟏 𝟐 (Base x Height)
𝟏 𝟐 (2)(4.4) = 4.4
The answer is 4.4 cm2

30. 15. Find the area of the trapezoid
15. Find the area of the trapezoid. Leave your answer in simplest radical form.
31.5 cm2
7 cm2
81 cm2
94.5 cm2
10-2 88

31. To find area of a trapezoid:
𝟏 𝟐 (Sum of Bases)(Height)
𝟏 𝟐 ( )(9) = 94.5
The answer is 94.5 cm2

32. 16. What is the area of the kite?
10-2 87

33. To find area of a kite:
𝟏 𝟐 (diagonal 1)(diagonal 2)
𝟏 𝟐 (4)(18) = 36
The answer is 36 ft2

34. 17. Find the area of a regular hexagon with an apothem 17
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.  
173.2 mi2
mi2
692.8 mi2
1038 mi2
10-3 88

35. To find area of a polygon: 𝟏 𝟐 (apothem)(perimeter)
𝟏 𝟐 (17.3)(120) = 1038
The answer is 1038 mi2
17.3 miles
20 miles

36. 18. The widths of two similar rectangles are 21 ft and 18 ft
18. The widths of two similar rectangles are 21 ft and 18 ft. What is the ratio of the perimeters? Of the areas?
8 : 7 and 64 : 49
8 : 7 and 49 : 36
7 : 6 and 64 : 49
7 : 6 and 49 : 36
10-4 87

37. Ratio of perimeters is a : b: 𝟐𝟏 :𝟏𝟖 Which reduces to: 𝟕 :𝟔
Ratio of areas is a2 : b2:
72 : 62
Which simplifies to: 𝟒𝟗 :𝟑𝟔
18 feet
21 feet

38. 19. Find the similarity ratio and the ratio of perimeters for two regular pentagons with areas of 𝟏𝟔 𝒄𝒎 𝟐 and 𝟒𝟗 𝒄𝒎 𝟐 .
4 : 7 ; 4 : 7
16 : 49 ; 4 : 7
16 : 49; 16 : 49
4 : 7; 16 : 49
10-4 84

39. Taking the square root gives you a : b:
Ratio of areas is a2 : b2:
49 : 𝟏𝟔
Taking the square root gives you a : b:
𝟕 :𝟒
Ratio of perimeters is also a : b:
𝟏𝟔 𝒄𝒎 𝟐
𝟒𝟗 𝒄𝒎 𝟐

40. 20. Find the measure of 𝐶𝐷𝐸 . The figure is not drawn to scale.
188
182
162
172
10-6 81

41. 𝒎 𝑫𝑪 =𝟑𝟔𝟎 −𝒎 𝑬𝑫 −𝒎 𝑬𝑨 −𝒎 𝑨𝑩 −𝒎 𝑩𝑪
𝒎 𝑫𝑪 =𝟑𝟔𝟎 −𝟐𝟕−𝟏𝟎𝟑−𝟓𝟎−𝟑𝟓
𝒎 𝑫𝑪 =𝟑𝟔𝟎 −𝟐𝟏𝟓
𝒎 𝑫𝑪 =𝟏𝟒𝟓
𝐶𝐷𝐸 =𝒎 𝑬𝑫 +𝒎 𝑫𝑪
𝐶𝐷𝐸 =𝟐𝟕+𝟏𝟒𝟓=𝟏𝟕𝟐

42. 21. Find the circumference. Leave your answer in terms of .
11.4 cm
8.55 cm
2.85 cm
5.7 cm
10-6 89

43. To find circumference of a circle:
𝟐()(radius) or ()(diameter)
()(5.7) = 5.7
The answer is 5.7 cm2

44. 22. Find the length of 𝑌𝑃𝑋 . Leave your answer in terms of .
10-6 39

45. To find the length of an arc:
𝒎 𝒂𝒓𝒄 𝟑𝟔𝟎 𝟐()(𝑟𝑎𝑑𝑖𝑢𝑠)
𝟐𝟕𝟎 𝟑𝟔𝟎 𝟐()(𝟏𝟎)
The answer is 15 m.

46. 23. Find the area of the circle. Leave your answer in terms of .
4.2025 cm2
8.405 cm2
16.81 cm2
11.2 cm2
10-7 90

47. To find area of a circle:
()(radius)2
()( 𝟒.𝟏 𝟐 ) 2 = 
The answer is  cm2

48. 24. The area of sector AOB is 𝟐𝟎. 𝟐𝟓𝝅 𝒇𝒕 𝟐
24. The area of sector AOB is 𝟐𝟎.𝟐𝟓𝝅 𝒇𝒕 𝟐 . Find the exact area of the shaded region.
(𝟐𝟎.𝟐𝟓𝝅−𝟒𝟎.𝟓) 𝒇𝒕 𝟐
(𝟐𝟎.𝟐𝟓𝝅−𝟖𝟏) 𝒇𝒕 𝟐
(𝟐𝟎.𝟐𝟓𝝅−𝟒𝟎.𝟓 𝟐 ) 𝒇𝒕 𝟐
None of these
10-7 90

49. To find area of a segment of a circle:
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒔𝒆𝒄𝒕𝒐𝒓 −𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆
𝒎 𝒂𝒓𝒄 𝟑𝟔𝟎 𝝅 𝒓 𝟐 − 𝟏 𝟐 𝒃𝒉
𝟗𝟎 𝟑𝟔𝟎 𝝅 (𝟗) 𝟐 − 𝟏 𝟐 𝟗 𝟗
𝟐𝟎.𝟐𝟓𝝅−𝟒𝟎.𝟓
The answer is (𝟐𝟎.𝟐𝟓𝝅−𝟒𝟎.𝟓) ft2

50. 25. Find the probability that a point chosen at random from 𝐽𝑃 is on the segment 𝐾𝑂 .
𝟏 𝟐
𝟒 𝟓
𝟓 𝟔
𝟐 𝟑
10-8 90

51. To find probability: 𝒘𝒉𝒂𝒕 𝒚𝒐𝒖 𝒘𝒂𝒏𝒕 𝒘𝒉𝒂𝒕 𝒊𝒔 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆
𝑲𝑶 𝑱𝑷 = 𝟒 𝟔
The answer is 𝟐 𝟑

52. 26. Use Euler’s Formula to find the missing number. Faces: 25
26. Use Euler’s Formula to find the missing number. Faces: 25 Vertices: 17 Edges: ? 43 40 39 41
11-1 90

53. Faces + Vertices = Edges + 2
F + V = E + 2
= E + 2 Plug in values
42= E + 2 Simplify
40= E Subtract 2 from both sides
The answer is 40 𝑒𝑑𝑔𝑒𝑠

54. 27. Use formulas to find the lateral area and surface area of the given prism. Round your answer to the nearest whole number.
𝟐𝟎𝟖 𝒎 𝟐 ; 𝟏𝟖𝟖 𝒎 𝟐
𝟏𝟑𝟔 𝒎 𝟐 ; 𝟏𝟖𝟖 𝒎 𝟐
𝟏𝟑𝟔 𝒎 𝟐 ; 𝟐𝟒𝟎 𝒎 𝟐
𝟐𝟎𝟖 𝒎 𝟐 ; 𝟐𝟒𝟎 𝒎 𝟐
11-2 90

55. LA = (perimeter of base)(height)
SA = LA + 2*B
SA = (13*4)
SA = = 240

56. 28. Find the surface area of the cylinder in terms of .
504 cm2
333 cm2
382.5 cm2
211.5 cm2
11-2 90

57. SA = 2𝜋𝑟ℎ+2𝜋 𝑟 2
SA = 2𝜋( 9 2 )(19)+2𝜋 ( 9 2 ) 2
SA = 171𝜋+40.5𝜋
SA = 211.5𝜋 𝑐𝑚 2

58. 29. Find the surface area of the regular pyramid shown to the nearest whole number.
𝟏𝟏𝟒𝟗 𝒎 𝟐
𝟓𝟓𝟎 𝒎 𝟐
𝟓𝟕𝟒 𝒎 𝟐
𝟒𝟎𝟖 𝒎 𝟐
11-3 90

59. SA = 1 2 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 +𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒

60. 30. Find the surface area of the cone to the nearest tenth.
𝟐𝟎𝟑 𝒄𝒎 𝟐
𝟏𝟑𝟒𝟏.𝟓 𝒄𝒎 𝟐
𝟕𝟒𝟕.𝟕 𝒄𝒎 𝟐
𝟐𝟏𝟑.𝟓 𝒄𝒎 𝟐
11-3 90

61. SA = 𝜋)(𝑟𝑎𝑑𝑖𝑢𝑠 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 +𝜋 𝑟 2

62. 31. Find the volume of the given prism
31. Find the volume of the given prism. Round to the nearest tenth if necessary.  
𝟑𝟎𝟖.𝟗 𝒄𝒎 𝟑
𝟑𝟎𝟖.𝟐 𝒄𝒎 𝟑
𝟑𝟏𝟐.𝟖 𝒄𝒎 𝟑
𝟑𝟎𝟐.𝟗 𝒄𝒎 𝟑
11-4 90

63. Volume= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡

64. 32. Find the volume of the cylinder in terms of  .
𝟔𝟎.𝟖𝝅 𝒎 𝟑
𝟏𝟏𝟓.𝟓𝟐𝝅 𝒎 𝟑
𝟒𝟑𝟖.𝟗𝟖𝝅 𝒎 𝟑
𝟓𝟕.𝟕𝟔𝝅 𝒎 𝟑
11-4 90

65. Volume= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡

66. 33. Find the volume of the square pyramid shown
33. Find the volume of the square pyramid shown. Round to the nearest tenth if necessary.
𝟏𝟐𝟔 𝒄𝒎 𝟑
𝟗𝟎𝟕.𝟓 𝒄𝒎 𝟑
𝟔𝟎𝟓 𝒄𝒎 𝟑
𝟓𝟓 𝒄𝒎 𝟑
11-5 90

67. Volume= 1 3 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡

68. 34. Find the volume of the square pyramid shown
34. Find the volume of the square pyramid shown. Round to the nearest tenth if necessary.
𝟔𝟎𝟎 𝒇𝒕 𝟑
𝟑𝟑.𝟑 𝒇𝒕 𝟑
𝟏𝟐𝟎𝟎 𝒇𝒕 𝟑
𝟒𝟎𝟎 𝒇𝒕 𝟑
11-5 90

69. Volume= 1 3 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 =V= 1 3 𝑏ℎ ℎ
𝐹𝑖𝑟𝑠𝑡 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡: 𝑎 2 + 𝑏 2 = 𝑐 2
25+ 𝑏 2 =169
𝑏 2 = 𝑏=12
V= ∗10 12
V =400 𝑓𝑡 3

70. 35. Find the volume of the oblique cone shown in terms of  .
𝟔𝟔𝟏.𝟓𝝅 𝒊𝒏 𝟑
𝟏𝟑𝟐𝟑𝝅 𝒊𝒏 𝟑
𝟔𝟑𝝅 𝒊𝒏 𝟑
𝟒𝟒𝟏𝝅 𝒊𝒏 𝟑
11-5 90

71. Volume= 1 3 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 =V= 1 3 𝜋 𝑟 2 ℎ

72. 36. Find the surface area of the sphere with the given dimension
36. Find the surface area of the sphere with the given dimension. Leave your answer in terms of 𝝅 . diameter of 14 cm
𝟕𝟖𝟒𝝅 𝒄𝒎 𝟐
𝟏𝟗𝟔𝝅 𝒄𝒎 𝟐
𝟐𝟖𝝅 𝒄𝒎 𝟐
𝟗𝟖𝝅 𝒄𝒎 𝟐
11-6 90

73. SA of a sphere=4 𝜋 𝑟 2
SA=4 𝜋 7 2
SA =196𝜋 𝑐𝑚 2
14 cm

74. 37. Find the volume of the sphere shown
37. Find the volume of the sphere shown. Give each answer rounded to the nearest cubic unit.
𝟏𝟓𝟐𝟕 𝒎𝒎 𝟑
𝟑𝟑𝟗 𝒎𝒎 𝟑
𝟑𝟎𝟓𝟒 𝒎𝒎 𝟑
𝟏𝟎𝟏𝟖 𝒎𝒎 𝟑
11-6 90

75. Volume of a sphere= 4 3 𝜋 𝑟 3
𝑉= 4 3 𝜋 9 3
𝑉=972𝜋 𝑚𝑚 3 𝑉=
𝑉=3054 𝑚𝑚 3

76. 38. Assume that lines that appear to be tangent are tangent
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 45
67.5
315
270
12-1 90

77. Tangent lines are perpendicular to the radius of the circle so the red angles are 90 degrees.
= 315
360  315 = 45
x = 45o
𝒎∠𝑶=𝟏𝟑𝟓

78. 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 15 5
12-2 90

79. Rotate the radius around until it makes a right triangle.
𝑎 2 + 𝑏 2 = 𝑐 2
= 𝑐 2
225= 𝑐 2
15=𝑐

80. 40. Find the measure of ∠BAC in circle O
40. Find the measure of ∠BAC in circle O. (The figure is not drawn to scale.) 50 80 20 40
12-3 90

81. Inscribed angles are half their intercepted arcs measure.
∠𝐵𝐴𝐶=1/2 𝐵𝐶
∠𝐵𝐴𝐶=1/2(40)
∠𝐵𝐴𝐶=20

82. 41. 𝐴𝐶 is tangent to circle O at A. If 𝑚 𝐵𝑌 =24, what is 𝑚∠YAC
41. 𝐴𝐶 is tangent to circle O at A. If 𝑚 𝐵𝑌 =24, what is 𝑚∠YAC? (The figure is not drawn to scale.)
132 48 78
156
12-3 90

83. Measure of ∠𝑌𝐴𝐶 is half of its intercepted arc 𝑌𝐴
𝐵𝑌𝐴 𝑖𝑠 𝑎 𝑠𝑒𝑚𝑖𝑐𝑖𝑟𝑐𝑙𝑒=180
𝐵𝑌𝐴 = 𝐵𝑌 + 𝑌𝐴
180=24+ 𝑌𝐴
156= 𝑌𝐴
∠𝑌𝐴𝐶 = 𝑌𝐴 = 1 2 (156)
∠𝑌𝐴𝐶 = 78 24

84. 42. 𝑚 𝐷𝐸 =120 and 𝑚 𝐵𝐶 =53. Find 𝑚∠𝐴. (The figure is not drawn to scale.)
33.5
93.5
86.5 67
12-4 90

85. Measure of ∠𝐴 is half of the difference of its intercepted arcs 𝐷𝐸 and 𝐵𝐶
1 2 ( 𝐷𝐸 − 𝐵𝐶 )=𝑚∠𝐴
1 2 (120−53)=𝑚∠𝐴
33.5=𝑚∠𝐴
120 53

86. 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 circle
9.2
7.7
14.3 85
12-4 90

87. The formula is: 𝑥 2 =
𝑥 2 = 17 5
𝑥 2 =85 𝑥=
𝑥=9.2

88. 44. Write the standard equation for the circle. center (10, –6), r = 6
(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 90

89. The equation for a circle is :
𝑥− 𝑥 𝑦− 𝑦 = 𝑟 2
center (10, –6), r = 6
(𝒙−𝟏𝟎) 𝟐 + (𝒚+𝟔) 𝟐 =𝟑𝟔