EXAMPLE 1

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# EXAMPLE 1 - PowerPoint PPT Presentation

1. a. V = Bh. 3. 1. 1. = ( 4 6)(9). 3. 2. EXAMPLE 1. Find the volume of a solid. Find the volume of the solid. = 36 m 3. b. 1. V = Bh. 3. 1. = (π r 2 ) h. 3. 1. = (π 2.2 2 )(4.5). 3. EXAMPLE 1. Find the volume of a solid. = 7.26π.

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1

a.

V = Bh

3

1

1

= ( 4 6)(9)

3

2

EXAMPLE 1

Find the volume of a solid

Find the volume of the solid.

= 36 m3

b.

1

V = Bh

3

1

= (πr2)h

3

1

= (π 2.22)(4.5)

3

EXAMPLE 1

Find the volume of a solid

= 7.26π

≈ 22.81 cm3

Originally, the pyramid had height 144 meters and volume 2,226,450 cubic meters. Find the side length of the square base.

1

V = bh

3

1

2,226,450 = (x2)(144)

3

EXAMPLE 2

Use volume of a pyramid

ALGEBRA

SOLUTION

Write formula.

Substitute.

Originally, the side length of the base was about 215meters.

EXAMPLE 2

Use volume of a pyramid

6,679,350 = 144x2

Multiply each side by 3.

46,384 ≈ x2

Divide each side by 144.

215 ≈ x

Find the positive square

root.

Volume isv = bh

1

1

1

= (41.57)(11)

v= bh

3

3

3

for Examples 1 and 2

GUIDED PRACTICE

Find the volume of the solid. Round your answer to two decimal places, if necessary.

1. Hexagonal pyramid

SOLUTION

Area of a hexagon of base 4 is 41.57

= 152.42 yd3

2. Right cone

Value of a cone is v = bh

1

3

for Examples 1 and 2

GUIDED PRACTICE

SOLUTION

First find by Pythagorean method

h = (82) (5)2

v = bh

= (π 52)(6.24)

1

1

3

3

for Examples 1 and 2

GUIDED PRACTICE

Substitute.

= 6.24

Simplify

Write formula.

Substitute.

= 163.49m3

Simplify

1

V = bh

3

1

1350π = (π182)h

3

The Height of the cone is 12.5m

for Examples 1 and 2

GUIDED PRACTICE

3. The volume of a right cone is 1350πcubic meters and the radius is 18meters. Find the height of the cone.

SOLUTION

Write formula.

Substitute.

4050π = π(18)2h

Multiply each side by 3.

12.5 = h

Divide each side by 324 π.