Chapter 11
Review
Problems similar to what
you will see on the
chapter 11 exam.
1 Exam
4/22/13 and retest 4/24/13
11-1
2
1. Use Euler’s Formula to find the missing
number.
A.
Faces: 23
Vertices: 12
Edges: ?
B.
C.
D.
36
34
33
32
69%
15%
15%
0%
A.
B.
C.
D.
11-1
3
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
11-1
4
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.
A.
B.
C.
D.
hexagon
pentagon
pyramid
rectangle
86%
14%
0%
A.
B.
C.
0%
D.
5
11-1
Cross sections are always 2 dimensional.
This one has 5 sides making it a pentagon.
11-2
6
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.
A.
5 cm
1.34 cm
B.
13 cm
C.
Not drawn to scale
D.
24 cm2; 243 cm2
24 cm2; 178 cm2
48 cm2; 178 cm2
48 cm2; 243 cm2
43%
21%
21%
14%
A.
B.
C.
D.
7
11-2
5 cm
1.34 cm
13 cm
Not drawn to scale
LA = ph
LA = (13+5+13+5)(1.34)
LA = 48.24  48 cm2
SA = ph + 2B
SA = 48.24 + 2(13*5)
SA = 48.24 + 130
SA = 178.24  178 cm2
11-2
8
4. Find the surface area of the cylinder in
terms of .
2
A.
1188
cm
18 cm
B. 702 cm2
15 cm
C. 918 cm2
D. 432 cm2
50%
Not drawn to scale
29%
14%
7%
A.
B.
C.
D.
9
11-2
18 cm
15 cm
Not drawn to scale
SA = 2rh + 2r2
SA = 2 (9)(15) + 2 (9)2
SA = 270  + 162 
SA = 432  cm2
11-3
10
5. The lateral area of a cone is 646 cm2.
The radius is 38 cm. Find the slant height
to the nearest tenth.
A. 17 cm
B.
C.
D.
64%
21%
7%
A.
B.
C.
7%
D.
15.6 cm
10.4 cm
9.7 cm
11
LA = 𝑟𝑙
646 = (38)𝑙
646 = 38𝑙
17 = 𝑙
11-3
11-3
12
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.
A.
B.
C.
D.
1124 m2
1325 m2
2449 m2
3052 m2
64%
14%
14%
7%
A.
B.
C.
D.
13
𝑎2 + 𝑏2 = 𝑙2
442 + 82 = 𝑙2
2000 = 𝑙2
𝑙 = 2000
SA = π𝑟𝑙 + 𝜋𝑟 2
SA = π(8)( 2000) + 𝜋(8)2
SA = 1325.032287
SA  1325 m2
11-3
11-3
14
7. Find the surface area of the regular
pyramid shown to the nearest whole
number.
A. 540 m2
B. 889 m2
C. 574 m2
D. 914 m2
50%
36%
7%
A.
B.
7%
C.
D.
15
11-3
1
1
𝑆𝐴 = 𝑝𝑙 + 𝑎𝑝
2
2
1
1
𝑆𝐴 = (48)(17) + (4 3 )(48)
2
2
𝑆𝐴 = 574.2768775 ≈ 574𝑚2
11-4
16
8. Find the volume of the given prism.
Round to the nearest tenth if necessary.
A. 333.5 cm3
B. 341.1 cm3
C. 311.7 cm3
D. 345.5 cm3
79%
14%
7%
0%
A.
B.
C.
D.
17
11-4
𝑉 = 𝐵ℎ
𝑉 = 10.2 4.4 7.6
3
𝑉 = 341.088 ≈ 341.1𝑐𝑚
11-4
18
9. Find the volume of the given prism.
Round to the nearest tenth if necessary.
A. 17 m3
B. 34 m3
C. 8.5 m3
D. 1 m3
50%
21%
21%
7%
A.
B.
C.
D.
19
11-4
𝑉 = 𝐵ℎ
1
𝑉 = ( 𝑏ℎ )ℎ
2
1
𝑉=
1)(2 17
2
3
𝑉 = 17 𝑚
11-4
20
10. Find the volume of the cylinder in
terms of  . 𝒉 = 𝟔 𝒊𝒏. 𝒂𝒏𝒅 𝒓 = 𝟑 𝒊𝒏.
A.
B.
C.
D.
27 in.3
108 in.3
54 in.3
324 in.3
79%
7%
A.
7%
B.
7%
C.
D.
11-4
21
𝒉 = 𝟔 𝒊𝒏. 𝒂𝒏𝒅 𝒓 = 𝟑 𝒊𝒏.
2
𝑉 = 𝜋𝑟 ℎ
2
𝑉=𝜋 3 6
3
𝑉 = 54𝜋 𝑖𝑛
11-5
22
11. Find the volume of the square pyramid shown.
Round to the nearest tenth if necessary.
5 cm
A.
B.
C.
D.
258.7 cm3
1152.3 cm3
48.1 cm3
201.7 cm3
64%
14%
14%
7%
A.
B.
C.
D.
23
5 cm
11-3
1
𝑉 = 𝐵ℎ
3
1
𝑉 = (𝑏ℎ)ℎ
3
1
𝑉 = 11 11 5
3
𝑉 = 201.6666667
3
≈ 201.7 𝑐𝑚
11-5
24
12. Find the volume of a square pyramid with base
edges of 48 cm and a slant height of 26 cm.
A.
B.
C.
D.
768 cm3
23,040 cm3
11,520 cm3
7,680 cm3
43%
21%
21%
14%
A.
B.
C.
D.
25
2
2
2
𝑎 + 𝑏 =𝑐
2
2
2
ℎ + (24) = (26)
2
2
2
ℎ = (26) − 24 = 100
ℎ = 10
1
𝑉 = 𝐵ℎ
3
1
𝑉 = 48 48 10
3
𝑉 = 7680
2
𝑉 = 7680 𝑐𝑚
11-5
11-5
26
13. Find the volume of the oblique
cone shown in terms of  .
A.
B.
C.
D.
672 𝑖𝑛.3
1344 𝑖𝑛.3
56 𝑖𝑛.3
441 𝑖𝑛.3
43%
36%
14%
7%
A.
B.
C.
D.
27
11-5
1 2
𝑉 = 𝑟 ℎ
3
1
2
𝑉 =  7 (27)
3
3
𝑉 = 441 𝑖𝑛.
11-6
28
14. Find the surface area of the sphere
with the given dimension. Leave your
answer in terms of .
A. 2,450 m2
B. 19,600 m2
radius of 70 m
C. 9,800 m2
D. 4,900 m2
57%
36%
7%
0%
A.
B.
C.
D.
29
radius of 70 m
2
4𝜋𝑟
𝑆𝐴 =
2
𝑆𝐴 = 4𝜋(70)
2
𝑆𝐴 = 19600𝜋 𝑚
11-6
11-6
30
15. Find the surface area of the sphere
with a circumference of 67 cm. Round
to the nearest tenth. A. 1428.9 cm2
B.
C.
D.
36%
29%
21%
14%
A.
B.
C.
D.
714.4 cm2
10.7 cm2
113.7 cm2
31
surface area of the sphere with a
circumference of 67 cm
𝐶 = 2𝜋𝑟
67 = 2𝜋𝑟
67
𝑟=
2𝜋
𝑆𝐴 = 4𝜋𝑟 2
2
67
𝑆𝐴 = 4𝜋
2𝜋
𝑆𝐴 = 1428.893079
𝑆𝐴 = 1428.9 𝑐𝑚2
11-6
11-6
32
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
A.
B.
4 mm
C.
D.
57%
21%
14%
7%
A.
B.
C.
D.
33
4 mm
11-6
4 3
𝑉 = 𝜋𝑟
3
4
3
𝑉 = 𝜋(4)
3
𝑉 = 268.0825731
3
𝑉 ≈ 268𝑚𝑚
34
11-6
17. The volume of a sphere is 2410 m3.
What is the surface area of the sphere to
the nearest tenth?
A. 30,285 m2
B. 153.1 m2
C. 1864.6 m2
D. 932.3 m2
36%
36%
21%
7%
A.
B.
C.
D.
35
4 3
𝑉 = 𝜋𝑟
3
4 3
2410𝜋 = 𝜋𝑟
3
3
4 33
2410 = 𝑟
4
3 4
1807.5 = 𝑟 3
3
𝑟 = 1807.5
𝑆𝐴 = 4𝜋𝑟 2
3
𝑆𝐴 = 4𝜋 1807.5 2
𝑆𝐴 = 1864.641739
𝑆𝐴 ≈ 1864.6 𝑚2
11-6
11-7
36
18. Are the two figures similar? If so, give the
similarity ratio of the smaller figure to the
1
larger figure.
57%
21%
14%
7%
A.
B.
C.
D.
A.
Yes,
B.
Yes,
C.
Yes,
D.
no
1.6
1
1.8
1
2.8
11-7
37
Yes;
3
5.4
=
2.6
4.68
=
5
9
11-7
38
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? A. 3.78 lb
B.
C.
64%
D.
14%
14%
7%
A.
B.
C.
D.
34.02 Ib
1.26 lb
11.34 lb
39
𝑎 1
=
𝑏 3
3
𝑎
1
.42
=
=
3
𝑏
27
𝑥
𝑥 = 11.34 𝑙𝑏
11-7
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?
3
A.
B.
C.
29%
29%
29%
14%
A.
B.
C.
D.
D.
338 yd
1,743 yd3
1,009 yd3
876 yd3
𝑎2
𝑏2
𝑎
𝑏
41
=
=
385
1149
385
1149
385
1149
𝑥=
3
3
𝑥
=
1743
1743 ∗
385
1149
3
3
x= 338.0714049  338 yd3
11-7
42
Chapter 11 Review
How to study:
 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
𝑭+𝑽= 𝑬+𝟐
Prism:
𝑳𝑨 = 𝒑𝒉
𝑺𝑨 = 𝒑𝒉 + 𝟐𝑩
𝑽 = 𝑩𝒉
Cylinder:
𝑳𝑨 = 𝟐𝝅𝒓𝒉
𝑺𝑨 = 𝟐𝝅𝒓𝒉 + 𝟐𝝅𝒓𝟐
𝑽 = 𝝅𝒓𝟐 𝒉
Pyramid:
𝟏
𝑳𝑨 = 𝒑𝒍
𝟐
𝟏
𝑺𝑨 = 𝒑𝒍 + 𝑩
𝟐
𝟏
𝑽 = 𝑩𝒉
𝟑
Cone:
𝑳𝑨 = 𝝅𝒓𝒍
𝑺𝑨 = 𝝅𝒓𝒍 + 𝝅𝒓𝟐
𝟏
𝑽 = 𝝅𝒓𝟐 𝒉
𝟑
Sphere:
𝑪 = 𝟐𝝅𝒓
𝑺𝑨 = 𝟒𝝅𝒓𝟐
𝟒
𝑽 = 𝝅𝒓𝟑
𝟑
Similarity ratio:
𝒂
𝒃
Area:
Volume:
𝒂𝟐
𝒃𝟐
𝒂𝟑
𝒃𝟑