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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

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

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

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

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

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