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EXAMPLE 1

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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EXAMPLE 1

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  1. 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

  2. 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

  3. 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.

  4. ANSWER 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.

  5. 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

  6. 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

  7. 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

  8. 1 V = bh 3 1 1350π = (π182)h 3 ANSWER 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.

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