Chapter 11 Review

Chapter 11 Review

9th - 10th Grade

26 Qs

quiz-placeholder

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Chapter 11 Review

Chapter 11 Review

Assessment

Quiz

Mathematics

9th - 10th Grade

Medium

Created by

M VT

Used 21+ times

FREE Resource

26 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

What is the volume?

36 cm3

100 cm3

216 m3

None of the above

Answer explanation

V=Bh
B is area of the base. The base is a triangle.
Area of a triangle is 12bh\frac{1}{2}bh .


The base of this triangle is 3 and the height is 4.
1234=6\frac{1}{2}\cdot3\cdot4=6 , so B = 6
What is height (h) in the volume formula? That's the height of the prism itself, which is h=6.

V=Bh
V = (6)(6)
V = 36 cm336\ cm^3

2.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

What is the volume?

141.4 in3

231.4 in3

93.6 in3

None of the above

Answer explanation

V=πr2hV=\pi r^2h
r is the radius, which is r = 3 here.
h is the height, which is h = 5 here.

V=π (3)2(5)V=\pi\ \cdot\left(3\right)^2\left(5\right)
V=π(9)(5)V=\pi\left(9\right)\left(5\right)
V=45πV=45\pi
V=141.3716694V=141.3716694

3.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

What is the volume?

21 mm3

42 mm3

17 mm3

None of the above

Answer explanation

V=13BhV=\frac{1}{3}Bh
B is the area off the base. This pyramid has a square base. What is the area of a square with sides 3mm long? 32=93^2=9
The area is 9, so B = 9.
h is the height of the pyramid itself, which is given as h = 7.

V=13(9)(7)V=\frac{1}{3}\left(9\right)\left(7\right)
V=21V=21

4.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

Find the surface area of the prism.

94

23

105

None of the above

Answer explanation

S.A.=ph+2BS.A.=ph+2B
The base here is rectangular.
The area of a rectangle is l*w.
The length of the base is 3 and the width is 5.
So the area of the base is 3*5, or B = 15.
p is the perimeter of the base (sum of all the side lengths). If the base has 2 sides that measure 3 and 2 sides that measure 5, then p = 3+3+5+5, which is p=16.
The height of the prism is h = 4.

S.A. =(16)(4)+2(15)S.A.\ =\left(16\right)\left(4\right)+2\left(15\right)

S.A.=64+30S.A.=64+30
S.A. = 94S.A.\ =\ 94

5.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

Find the surface area of the cylinder in terms of pi.

54π54\pi

35π35\pi

62π62\pi

None of the above

Answer explanation

S.A.=2πrh + 2πr2S.A.=2\pi rh\ +\ 2\pi r^2
We are given 6, which is the DIAMETER.
62=3\frac{6}{2}=3 , so 3 is the radius, r. r = 3.

The height (h) of the entire cylinder is 6.

S.A.=2π(3)(6)+2π(3)2S.A.=2\pi\left(3\right)\left(6\right)+2\pi\left(3\right)^2
S.A.=2π(18)+ 2π(9)S.A.=2\pi\left(18\right)+\ 2\pi\left(9\right)
S.A. = 36π+18πS.A.\ =\ 36\pi+18\pi
S.A. = 54πS.A.\ =\ 54\pi

6.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

Find the lateral area of the cone in terms of pi.

60π60\pi

35π35\pi

80π80\pi

None of the above

Answer explanation

L.A.=πrlL.A.=\pi rl
'r' is the radius, 6, so r = 6.
'l' is the slant height. We are given both the height and the slant height, but we just need 10, which is our slant height, so l = 10.

L.A.=π(6)(10)L.A.=\pi\left(6\right)\left(10\right)
L.A.=60πL.A.=60\pi

7.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Media Image

Find the surface area of the cone in terms of pi.

144π144\pi

152π152\pi

128π128\pi

None of the above

Answer explanation

S.A.=πrl + πr2S.A.=\pi rl\ +\ \pi r^2
r is the radius, which is given, so r = 8.
'l' is the slant height, which is given, so l = 10.
S.A.=π(8)(10)+π(8)2S.A.=\pi\left(8\right)\left(10\right)+\pi\left(8\right)^2
S.A. = 80π+64πS.A.\ =\ 80\pi+64\pi
S.A.=144πS.A.=144\pi

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