80 likes | 88 Views
Learn how to calculate the volume of different solids such as pyramids, cones, and hexagonal pyramids. Examples and guided practice included.
E N D
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.
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.
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 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.