Question

Q-The perimeters of the two circular ends of the frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm.
find curved surface area

Answer 1

Given height of frustum(h)= 11 cm

Let the radius, r1 and radius, r2 be the radii of the circular ends of the frustum and h be its height.

perimeter = 36 cm (given)

2 π r1=36

r1=36/2π

r1=18/π

perimeter=48 cm

2 π r2=48

r2= 24/π

Volume of Frustum (V)= 1/3 π {(r1)²+ (r2)² + (r1r2)} h

V= 1/3 x π x h {(18/π)²+(24/π)²+(18/π x 24/π)}

V= 1/3 x π x 11 {324/π²+576/π²+432/π²}

V= 1/3 x π x 11x 1/π² (324+576+432)

V= 1/3 x π x 11x 1/π² (1332)

V= 11/3x 1/π (1332)

V= 11/3 x 7/22 (1332)

V= (11 x 7× 1332)/ (22 ×3)

V= 7 x 222

V= 1554 cm³

Hence, the volume of frustum is 154 cm³

HOPE THIS WILL HELP YOU...

Answer 2

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Let the radius, r1 and radius, r2 be the radii of the circular ends of the frustum and h be its height.

perimeter = 36 cm (given)

2 π r1=36

r1=36/2π

r1=18/π

perimeter=48 cm

2 π r2=48

r2= 24/π

Volume of Frustum (V)= 1/3 π {(r1)²+ (r2)² + (r1r2)} h

V= 1/3 x π x h {(18/π)²+(24/π)²+(18/π x 24/π)}

V= 1/3 x π x 11 {324/π²+576/π²+432/π²}

V= 1/3 x π x 11x 1/π² (324+576+432)

V= 1/3 x π x 11x 1/π² (1332)

V= 11/3x 1/π (1332)

V= 11/3 x 7/22 (1332)

V= (11 x 7× 1332)/ (22 ×3)

V= 7 x 222

V= 1554 cm³

Hence, the volume of frustum is 154 cm³

csa =22/7×11(36/2×22/7+18/22/7)

=396 cm.