This pdf shows the formula of Volume of a cylinder and cone with proper explanation and examples that are step by step solved.
1. Lesson 9-3 and Cones Lesson 9-3: Cylinders and Cones 1
2. Formulas: S.A. = 2πr ( r + h ) Cylinders V= r h 2 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 ) 2 The base area is the area of the circle: r The lateral area is the area of the rectangle: 2πrh 2 Volume (V) = Bh = r h 2πr h h Lesson 9-3: Cylinders and Cones 2
3. Example For the cylinder shown, find the lateral area , surface area and 3 cm L.A.= 2πr•h S.A.= 2•πr 2 + 2πr•h 4 cm L.A.= 2π(3)•(4) S.A.= 2•π(3)2 + 2π(3)•(4) L.A.= 24π sq. cm. S.A.= 18π +24π S.A.= 42π sq. cm. V = πr2•h V = π(3)2•(4) V = 36π Lesson 9-3: Cylinders and Cones 3
4. Formulas: S.A. = π r ( r + l ) 1 2 Cones V= 3 r h 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) h l The base area is the area of the circle: r2 1 1 2 Volume (V) = Bh r h r 3 3 Notice that the height (h) (altitude), the radius and the slant height create a right triangle. Lesson 9-3: Cylinders and Cones 4
5. Example: For the cone shown, find the lateral area surface area and volume. S.A.= πr (r + l ) L.A.= πrl S.A.= π•6 (6 + 10) 62 +82 = l 2 S.A.= 6π (16) 8 10 L.A.= π(6)(10) S.A.= 96π sq. cm. L.A.= 60π sq. cm. 1 2 V r h 6 cm Note: We must use the 3 Pythagorean theorem 1 to find l. V 62 8 V= 96π cubic cm. 3 Lesson 9-3: Cylinders and Cones 5