# Volume of Cylinders and Cones

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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