EXAMPLE 1 Find the volume of a solid Find the volume of the solid. V = Bh 1 3 = 36 m 3 a.a. = ( 4 6)(9) 1 3 1 2.

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EXAMPLE 1 Find the volume of a solid Find the volume of the solid. V = Bh 1 3 = 36 m 3 a.a. = ( 4 6)(9)

EXAMPLE 1 Find the volume of a solid V = Bh 1 3 = (πr 2 )h 1 3 = 7.26π ≈ cm 3 b.b. = (π )(4.5) 1 3

EXAMPLE 2 Use volume of a pyramid ALGEBRA Originally, the pyramid had height 144 meters and volume 2,226,450 cubic meters. Find the side length of the square base. SOLUTION V = bh 1 3 Write formula. 2,226,450 = (x 2 )(144) 1 3 Substitute.

EXAMPLE 2 Use volume of a pyramid 6,679,350 = 144x 2 Multiply each side by 3. 46,384 ≈ x 2 Divide each side by ≈ x Find the positive square root. Originally, the side length of the base was about 215 meters. ANSWER

GUIDED PRACTICE for Examples 1 and 2 Find the volume of the solid. Round your answer to two decimal places, if necessary. 1. Hexagonal pyramid SOLUTION Volume is v = bh 1 3 Area of a hexagon of base 4 is v = bh 1 3 = (41.57)(11) 1 3 = yd 3

GUIDED PRACTICE for Examples 1 and 2 2. Right cone SOLUTION Value of a cone is v = bh 1 3 First find by Pythagorean method

GUIDED PRACTICE for Examples 1 and 2 v = bh 1 3 Write formula. Substitute. h = (8 2 ) (5) 2 – = 6.24 = (π 5 2 )(6.24) 1 3 = m 3 Simplify Substitute. Simplify

GUIDED PRACTICE for Examples 1 and 2 3. The volume of a right cone is 1350π cubic meters and the radius is 18 meters. Find the height of the cone. SOLUTION V = bh π = (π18 2 )h π = π(18) 2 h 12.5 = h Write formula. Substitute. Multiply each side by 3. Divide each side by 324 π. The Height of the cone is 12.5 mANSWER