Transcript Geometry - BakerMath.org

Geometry
Volume of Prisms and Cylinders
Goals
Find the volume of prisms.
 Find the volume of cylinders.
 Solve problems using volume.

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Volume
The number of cubic units contained in a
solid.
 Measured in cubic units.
 Basic Formula:

V = Bh

B = area of the base, h = height
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Cubic Unit
V = s3
V = 1 cu. unit
s
1
1
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1
s
s
Cavalieri’s Principle
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Prism: V = Bh
B
B
h
B
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h
h
Cylinder: V =
r
B
h
V = Bh
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2
r h
h
Example 1
Find the volume.
Triangular Prism
V = Bh
8
Base = 40
3
10
Abase = ½ (10)(8) = 40
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V = 40(3) = 120
Example 2
Find the volume.
V = Bh
12
10
The base is a ?
Hexagon
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12
Example 2 Solution
12
6 3
?
12
?
?
6
10
A
1
2

1
2
ap
6 3  72
 216 3
 374.1
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Example 2 Solution
V = Bh
12
374.1
10
V = (374.1)(10)
V  3741
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Example 3
A soda can measures 4.5 inches
high and the diameter is 2.5 inches.
Find the approximate volume.
V = r2h
(The diameter is 2.5 in.
The radius is 2.5 ÷ 2
inches.)
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V = (1.252)(4.5)
V  22 in3
Example 4
A wedding cake has three layers.
The top cake has a diameter of 8
inches, and is 3 inches deep.
The middle cake is 12 inches in
diameter, and is 4 inches deep.
The bottom cake is 14 inches in
diameter and is 6 inches deep.
Find the volume of the entire
cake, ignoring the icing.
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Example 4 Solution
VTop = (42)(3) = 48  150.8 in3
VMid = (62)(4) = 144  452.4 in3
VBot =
(72)(6)
= 294  923.6
8
in3
486  1526.8
in3
12
r=4
3
r=6
4
14
6
r=7
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Example 5
48
44
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A manufacturer of concrete
sewer pipe makes a pipe
segment that has an
outside diameter (o.d.) of
48 inches, an inside
diameter (i.d.) of 44 inches,
and a length of 52 inches.
Determine the volume of
concrete needed to make
52 one pipe segment.
Example 5 Solution
Strategy:
Find the area of the ring at
the top, which is the area
of the base, B, and multiply
by the height.
View of the Base
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Example 5 Solution
48
44
Strategy:
Find the area of the ring at
the top, which is the area
of the base, B, and multiply
by the height.
Area of Outer Circle:
Aout = (242) = 576
52 Area of Inner Circle:
Ain = (222) = 484
Area of Base (Ring):
ABase = 576 - 484 = 92
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V = Bh
Example 5 Solution
48
ABase = B = 92
44
V = (92)(52)
V = 4784
52
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V  15,029.4 in3
Example 5 Alternate Solution
48
Vouter = (242)(52)
44
Vouter = 94,096.98
Vinner = (222)(52)
52
Vinner = 79,067.60
V = Vouter – Vinner
V  15,029.4 in3
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Example 6
4
L
5
A metal bar has a volume of 2400 cm3. The
sides of the base measure 4 cm by 5 cm.
Determine the length of the bar.
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Example 6 Solution
4
5
L
 Method
V
1
= Bh
 B = 4  5 = 20
 2400 = 20h
 h = 120 cm
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 Method
2
V = L  W  H
 2400 = L  4  5
 2400 = 20L
 L = 120 cm
V r h
2
Example 7
Diameter = 7
115   r  3
2
3 in
V = 115 in3
A 3-inch tall can has a
volume of 115 cubic inches.
Find the diameter of the can.
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115  9.42 r
115
2
r
9.42
2
12.21  r
r  12.21
r  3.5
2
Summary

The volumes of prisms and cylinders are
essentially the same:
V = Bh
&
V=
2
r h
where B is the area of the base, h is the
height of the prism or cylinder.
 Use what you already know about area of
polygons and circles for B.

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r
B
h
V = Bh
Add these to your formula sheet.
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h
V = r2h
Which
Holds
More?
This one!
2.3 in
4 in
3.2 in
4.5 in
1.6 in
2
V  (3.2)(1.6)(4)
 20.48
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 2.3 
V  
  4.5
 2 
 18.7
What would the height of cylinder 2
have to be to have the same
r=3
volume as cylinder 1?
r=4
#2
#1
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8
h
Solution
r=4
#1
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8
 
V  4 8
2
 128
Solution
 
r=3
128   3 h
2
128
h
9
h  14.2
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#2
h
Practice Problems
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