Lesson 9-3: Cylinders and Cones

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Lesson 9-3: Cylinders and Cones

Lesson 9-3: Cylinders and Cones Formulas: S.A. = 2πr ( r + h ) V = Cylinders Cylinders are right prisms with circular bases. Therefore, the formulas for prisms can be used for cylinders. Surface Area (SA) = 2B + LA = 2πr ( r + h ) The base area is the area of the circle: The lateral area is the area of the rectangle: 2πrh Volume (V) = Bh = h 2πr h Lesson 9-3: Cylinders and Cones

Lesson 9-3: Cylinders and Cones Example For the cylinder shown, find the lateral area , surface area and volume. 3 cm S.A.= 2•πr2 + 2πr•h L.A.= 2πr•h 4 cm S.A.= 2•π(3)2 + 2π(3)•(4) L.A.= 2π(3)•(4) S.A.= 18π +24π L.A.= 24π sq. cm. S.A.= 42π sq. cm. V = πr2•h V = π(3)2•(4) V = 36π Lesson 9-3: Cylinders and Cones

Lesson 9-3: Cylinders and Cones Formulas: S.A. = π r ( r + l ) V = Cones Cones are right pyramids with a circular base. Therefore, the formulas for pyramids can be used for cones. Lateral Area (LA) = π r l, where l is the slant height. Surface Area (SA) = B + LA = π r (r + l) l h The base area is the area of the circle: Volume (V) = r Notice that the height (h) (altitude), the radius and the slant height create a right triangle. Lesson 9-3: Cylinders and Cones

Lesson 9-3: Cylinders and Cones Example: For the cone shown, find the lateral area surface area and volume. S.A.= πr (r + l ) L.A.= πrl 6 cm 8 S.A.= π•6 (6 + 10) 62 +82 = l 2 10 S.A.= 6π (16) L.A.= π(6)(10) S.A.= 96π sq. cm. L.A.= 60π sq. cm. Note: We must use the Pythagorean theorem to find l. V= 96π cubic cm. Lesson 9-3: Cylinders and Cones