Volume of Prisms Pyramids Cylinders and Cones LEARNING

Volume of Prisms, Pyramids, Cylinders and Cones LEARNING TARGET: I can explain the difference between the volume of pyramids, cones, and prisms and cylinders.

I. Finding the volume of a Prism V = Bh V = lwh Area of the base (rectangular) Height (h) V= (½ bh) h (triangular) Area of Base (B) Height of triangle Height of prism

Ex. 1: Find the Volume of the Prism V = Bh and rectangular so V = lwh 10 in 5 in = (3 in • 5 in)(10 in) = (15 in 2)(10 in) 3 in = 150 in 3

Ex. 2: Find the volume of the following prism. Triangle 29 m V = Bh a = ½bh • h 40 m a 2 20 m Height of the base: + b 2 = c 2 = ½(20 m)__ 21 • (40 m) a 2 + 202 = 292 = 210 m 2 • 40 m a 2 + 400 = 841 = 8400 m 3 a 2 = 441 b = 21

Ex. 3: Yet another prism! Find the volume. 8 10 in 8 in h 4 V = Bh a 2 + b 2 = c 2 = ½bh • h 42 + h 2 = 82 = ½(8 in) 6. 9 __ • (10 in) 16 + h 2 = 64 = (27. 7 in 2) • (10 in) = 277 in 3 h 2 = 48 h = 6. 9

II. Volume of a Cylinder r h

Ex. 4: Find the volume of the following cylinder. 16 ft 9 ft

What have we learned? ? ? Volume of Prisms & Cylinders r V = (area of base) h h 10 in 5 in 3 in

I. Volume of a Pyramid Vp = ⅓(area of base) h Area of the Base A = l • w A = ½bh Height of the pyramid, not to be confused with the slant height (l) h

Ex. 1: Volume of a right Pyramid Find the volume of a square pyramid with base edges of 15 cm & a height of 22 cm. Square V = ⅓(area of base)h = (⅓)b • h 22 cm 15 cm = (⅓)15 • 22 = (⅓)4950 = 1650 cm 3

Ex. 2: Another square pyramid Find the area of a square pyramid w/ base edges 16 ft long & a slant height 17 ft. V = (⅓)Bh a 2 b 2 c 2 + = h 2 + 82 = 172 h 2 + 64 = 289 h 2 = 225 h = 15 = (⅓)l • w • h h 17 ft 8 ft 16 ft = (⅓)16 • ___ 15 = (⅓)3840 16 ft = 1280 ft 3

II. Volume of a Cone 2 h Area of the Base r Height of the cone, not to be confused with the slant height (l)

Ex. 3: Find the volume of the following right cone w/ a diameter of 6 in. Circle 11 in 3 in

Ex. 5: Solve for the missing variable. The following cone has a volume of 110π. What is its radius. V = ⅓(πr 2)h 110π = (⅓)πr 2(10) 110 = (⅓)r 2(10) 10 cm 11 = (⅓)r 2 33 = r 2 r r = √(33) = 5. 7 cm

What have we learned? ? ? Volume of Cones & Pyramids V = ⅓ (area of base)h h h r