1. Problems similar to what you will see on the chapter 11 exam.
Chapter 11 Review
Problems similar to what you will see on the chapter 11 exam.
Exam 4/22/13 and retest 4/24/13

2. 11-1
1. Use Euler’s Formula to find the missing number. Faces: 23 Vertices: 12 Edges: ? 36 34 33 32

3. 𝐹+𝑉=𝐸+2 23+12=𝐸+2 35=𝐸+2 33=𝐸 The correct answer is C. 33 11-1
1. Use Euler’s Formula to find the missing number. Faces: 23 Vertices: 12 Edges: ?
𝐹+𝑉=𝐸+2
23+12=𝐸+2
35=𝐸+2
33=𝐸
The correct answer is C. 33

4. hexagon pentagon pyramid rectangle 11-1
2. Pierre built the model shown in the diagram below for a social studies project. He wants to be able to show the inside of his model, so he sliced the figure as shown. Describe the cross section he created.
hexagon
pentagon
pyramid
rectangle

5. 11-1
Cross sections are always 2 dimensional. This one has 5 sides making it a pentagon.

6. 24 cm2; 243 cm2 24 cm2; 178 cm2 48 cm2; 178 cm2 48 cm2; 243 cm2 11-2
3. A jewelry store buys small boxes in which to wrap items that they sell. The diagram below shows one of the boxes. Find the lateral area and the surface area of the box to the nearest whole number.
24 cm2; 243 cm2
24 cm2; 178 cm2
48 cm2; 178 cm2
48 cm2; 243 cm2

7. 11-2 LA = ph LA = (13+5+13+5)(1.34) LA = 48.24  48 cm2 SA = ph + 2B
SA =  178 cm2

8. 4. Find the surface area of the cylinder in terms of .
11-2
4. Find the surface area of the cylinder in terms of .
1188 cm2
702 cm2
918 cm2
432 cm2

9. SA = 2rh + 2r2 SA = 2 (9)(15) + 2 (9)2 SA = 270  + 162 
11-2
SA = 2rh + 2r2
SA = 2 (9)(15) + 2 (9)2
SA = 270  
SA = 432  cm2

10. 11-3
5. The lateral area of a cone is 646 cm2. The radius is 38 cm. Find the slant height to the nearest tenth.
17 cm
15.6 cm
10.4 cm
9.7 cm

11. 11-3
LA = 𝑟𝑙
646 = (38)𝑙
646 = 38𝑙
17 = 𝑙

12. 11-3
6. Find the surface area of a conical grain storage tank that has a height of 44 meters and a diameter of 16 meters. Round the answer to the nearest square meter.
1124 m2
1325 m2
2449 m2
3052 m2

13. 𝑎 2 + 𝑏 2 = 𝑙 2 44 2 + 8 2 = 𝑙 2 2000= 𝑙 2 𝑙= 2000 SA = π𝑟𝑙+ 𝜋 𝑟 2
11-3
𝑎 2 + 𝑏 2 = 𝑙 2
= 𝑙 2
2000= 𝑙 2
𝑙= 2000
SA = π𝑟𝑙+ 𝜋 𝑟 2
SA = π(8)( )+ 𝜋 (8) 2
SA =
SA  1325 m2

14. 11-3
7. Find the surface area of the regular pyramid shown to the nearest whole number.
540 m2
889 m2
574 m2
914 m2

15. 𝑆𝐴= 1 2 𝑝𝑙+ 1 2 𝑎𝑝 𝑆𝐴= 1 2 (48)(17)+ 1 2 (4 3 )(48)
11-3
𝑆𝐴= 1 2 𝑝𝑙 𝑎𝑝
𝑆𝐴= 1 2 (48)(17) (4 3 )(48)
𝑆𝐴= ≈574 𝑚 2

16. 11-4
8. Find the volume of the given prism. Round to the nearest tenth if necessary.
333.5 cm3
341.1 cm3
311.7 cm3
345.5 cm3

17. 11-4
𝑉=𝐵ℎ 𝑉=
𝑉= ≈341.1 𝑐𝑚 3

18. 11-4
9. Find the volume of the given prism. Round to the nearest tenth if necessary.
17 m3
34 m3
8.5 m3
1 m3

19. 11-4
𝑉=𝐵ℎ
𝑉=( 1 2 𝑏ℎ )ℎ
𝑉= )(2 17
𝑉=17 𝑚 3

20. 11-4
10. Find the volume of the cylinder in terms of  . 𝒉=𝟔 𝒊𝒏. 𝒂𝒏𝒅 𝒓=𝟑 𝒊𝒏.
27 in.3
108 in.3
54 in.3
324 in.3

21. 11-4
𝒉=𝟔 𝒊𝒏. 𝒂𝒏𝒅 𝒓=𝟑 𝒊𝒏.
𝑉=𝜋 𝑟 2 ℎ
𝑉=𝜋
𝑉=54𝜋 𝑖𝑛 3

22. 11-5
11. Find the volume of the square pyramid shown. Round to the nearest tenth if necessary.
258.7 cm3
cm3
48.1 cm3
201.7 cm3
5 cm

23. 11-3
5 cm
𝑉= 1 3 𝐵ℎ
𝑉= 1 3 (𝑏ℎ)ℎ 𝑉=
𝑉= ≈ 𝑐𝑚 3

24. 11-5
12. Find the volume of a square pyramid with base edges of 48 cm and a slant height of 26 cm.
768 cm3
23,040 cm3
11,520 cm3
7,680 cm3

25. 𝑎 2 + 𝑏 2 = 𝑐 2 ℎ 2 + (24) 2 = (26) 2 ℎ 2 = (26) 2 − 24 2 =100 ℎ=10
11-5
𝑎 2 + 𝑏 2 = 𝑐 2
ℎ 2 + (24) 2 = (26) 2
ℎ 2 = (26) 2 − =100
ℎ=10
𝑉= 1 3 𝐵ℎ 𝑉=
𝑉=7680
𝑉=7680 𝑐𝑚 2

26. 13. Find the volume of the oblique cone shown in terms of  .
11-5
13. Find the volume of the oblique cone shown in terms of  .
672 𝑖𝑛. 3
1344 𝑖𝑛. 3
56 𝑖𝑛. 3
441 𝑖𝑛. 3

27. 11-5
𝑉= 1 3  𝑟 2 ℎ
𝑉= 1 3  (27)
𝑉= 441 𝑖𝑛. 3

28. 11-6
14. Find the surface area of the sphere with the given dimension. Leave your answer in terms of . radius of 70 m
2,450 m2
19,600 m2
9,800 m2
4,900 m2

29. 11-6
radius of 70 m
𝑆𝐴=4𝜋 𝑟 2
𝑆𝐴=4𝜋 (70) 2
𝑆𝐴=19600𝜋 𝑚 2

30. 11-6
15. Find the surface area of the sphere with a circumference of 67 cm. Round to the nearest tenth.
cm2
714.4 cm2
10.7 cm2
113.7 cm2

31. 11-6
surface area of the sphere with a circumference of 67 cm
𝐶=2𝜋𝑟
67=2𝜋𝑟
𝑟= 67 2𝜋
𝑆𝐴=4𝜋 𝑟 2
𝑆𝐴=4𝜋 𝜋 2
𝑆𝐴=
𝑆𝐴= 𝑐𝑚 2

32. 11-6
16. Find the volume of the sphere shown. Give each answer rounded to the nearest cubic unit.
201 mm3
268 mm3
67 mm3
134 mm3

33. 11-6
𝑉= 4 3 𝜋 𝑟 3
𝑉= 4 3 𝜋 (4) 3 𝑉=
𝑉 ≈268 𝑚𝑚 3

34. 11-6
17. The volume of a sphere is 2410 m3. What is the surface area of the sphere to the nearest tenth?
30,285 m2
153.1 m2 m2
932.3 m2

35. 11-6
𝑉= 4 3 𝜋 𝑟 3
2410𝜋= 4 3 𝜋 𝑟 3
= 4 3 𝑟
1807.5= 𝑟 3 𝑟=
𝑆𝐴=4𝜋 𝑟 2
𝑆𝐴=4𝜋
𝑆𝐴=
𝑆𝐴≈ 𝑚 2

36. 11-7
18. Are the two figures similar? If so, give the similarity ratio of the smaller figure to the larger figure.
Yes,
Yes,
Yes, no

37. 11-7
Yes; = = 5 9

38. 11-7
19. A glass vase weighs 0.42 lb. How much does a similarly shaped vase of the same glass weigh if each dimension is 3 times as large?
3.78 lb
34.02 Ib
1.26 lb
11.34 lb

39. 11-7
𝑎 𝑏 = 1 3
𝑎 3 𝑏 3 = 1 27 = .42 𝑥
𝑥=11.34 𝑙𝑏

40. 11-7
20. The surface areas of two similar solids are 385 yd2 and 1149 yd2. The volume of the larger solid is 1743 yd3. What is the volume of the smaller solid?
338 yd3
1,743 yd3
1,009 yd3
876 yd3

41. 11-7
𝑎 2 𝑏 2 =
𝑎 𝑏 =
= 𝑥 1743
𝑥= 1743∗
x=  338 yd3

42. How to study: Chapter 11 Review
Access this powerpoint on my webpage and make sure you can answer these problems correctly
Go back and work problems from the sections you are having trouble with (powerpoints have answers worked out).
Review your vocabulary.
Finish your missing lessons!
Mrs. Marrara

43. 𝑭+𝑽=𝑬+𝟐
Prism:
𝑳𝑨=𝒑𝒉
𝑺𝑨=𝒑𝒉+𝟐𝑩
𝑽=𝑩𝒉
Cylinder:
𝑳𝑨=𝟐𝝅𝒓𝒉
𝑺𝑨=𝟐𝝅𝒓𝒉+𝟐𝝅 𝒓 𝟐
𝑽= 𝝅 𝒓 𝟐 𝒉
Pyramid:
𝑳𝑨= 𝟏 𝟐 𝒑𝒍
𝑺𝑨= 𝟏 𝟐 𝒑𝒍+𝑩
𝑽= 𝟏 𝟑 𝑩𝒉
Cone:
𝑳𝑨= 𝝅𝒓𝒍
𝑺𝑨= 𝝅𝒓𝒍+ 𝝅 𝒓 𝟐
𝑽= 𝟏 𝟑 𝝅 𝒓 𝟐 𝒉
Sphere:
𝑪=𝟐𝝅𝒓
𝑺𝑨=𝟒𝝅 𝒓 𝟐
𝑽= 𝟒 𝟑 𝝅 𝒓 𝟑
Similarity ratio:
𝒂 𝒃
Area: 𝒂 𝟐 𝒃 𝟐
Volume: 𝒂 𝟑 𝒃 𝟑