Geometry

30
Geometry Volume of Prisms and Cylinders

description

Geometry. Volume of Prisms and Cylinders. Goals. Find the volume of prisms. Find the volume of cylinders. Solve problems using volume. 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. Cubic Unit. - PowerPoint PPT Presentation

Transcript of Geometry

Geometry

Volume of Prisms and Cylinders

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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

11

1

V = 1 cu. unit

ss

s

V = s3

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Cavalieri’s Principle

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B

BB

hh

h

Prism: V = Bh

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Cylinder: V = r2h

B

h

r

h

V = Bh

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Example 1 Find the volume.

10

8

3

Triangular Prism

V = Bh

Base = 40

V = 40(3) = 120

Abase = ½ (10)(8) = 40

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Example 2 Find the volume.

12

10

V = Bh

The base is a ?

Hexagon

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Example 2 Solution

12

10

12

?12

?

?

6

12

12 6 3 72

216 3

374.1

A ap

6 3

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Example 2 Solution

12

10

374.1

V = Bh

V = (374.1)(10)

V 3741

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Example 3A soda can measures 4.5 inches high and the diameter is 2.5 inches. Find the approximate volume.

V = r2h

V = (1.252)(4.5)

V 22 in3

(The diameter is 2.5 in. The radius is 2.5 ÷ 2 inches.)

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Example 4A 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

8

12

14

3

4

6

r = 4

r = 6

r = 7

VTop = (42)(3) = 48 150.8 in3

VMid = (62)(4) = 144 452.4 in3

VBot = (72)(6) = 294 923.6 in3

486 1526.8 in3

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Example 5A 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 one pipe segment.

44

48

52

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Example 5 SolutionStrategy:

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 SolutionStrategy:

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

Area of Inner Circle:

Ain = (222) = 484

Area of Base (Ring):

ABase = 576 - 484 = 92

44

48

52

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Example 5 Solution V = Bh

ABase = B = 92

V = (92)(52)

V = 4784

V 15,029.4 in3

44

48

52

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Example 5 Alternate Solution

Vouter = (242)(52)

Vouter = 94,096.98

Vinner = (222)(52)

Vinner = 79,067.60

V = Vouter – Vinner

V 15,029.4 in3

44

48

52

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Example 6

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.

4

5L

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Example 6 Solution

Method 1 V = Bh B = 4 5 = 20 2400 = 20h h = 120 cm

Method 2 V = L W H 2400 = L 4

5 2400 = 20L L = 120 cm

4

5L

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Example 7

3 in

V = 115 in3

A 3-inch tall can has a volume of 115 cubic inches. Find the diameter of the can.

2

2

2

2

2

115 3

115 9.42

115

9.42

12.21

12.21

3.5

V r h

r

r

r

r

r

r

Diameter = 7

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Summary

The volumes of prisms and cylinders are essentially the same:

V = Bh & V = r2h 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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V = Bh V = r2h

B

h h

r

Add these to your formula sheet.

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Which Holds More?

(3.2)(1.6)(4)

20.48

V

3.2 in 1.6 in

4 in 4.5 in

2.3 in

This one!

2

2.34.5

2

18.7

V

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What would the height of cylinder 2 have to be to have the same volume as cylinder 1?

r = 4

h

r = 3

8#1#2

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Solution

24 8

128

V

r = 4

8#1

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Solution

h

r = 3

#2

2128 3

128

914.2

h

h

h

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Practice Problems