This is one of the important question from the cbse board exam point of view because in 2014 cbse board exam this question was asked. Question number 16 from RS Aggarwal book page number 823, exercise 17C, chapter volume and surface area of solid.

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# The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, then find its volume and curved surface area.

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Given data,

height of frustum(h)= 11 cm

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)

Curved surface area of the frustum=π[r2+r1]l

= [22/7]×24π+18π×15082

= [22/7]×42π×1508211

= [22/7]×42×722×1508211

= 42×11.164436

= 468.91 cm²

∴ curved surface area = 468.91 cm²

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