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Question

# The radii of the ends of a frustum of a cone 45cm high are 28 cm and 7cm. Find its volume, the curved surface area, and the total surface area (take π=227).

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Solution

## The frustum can be viewed as a difference of two right circular cones OAB and OCD . let the height of the cone OAB be h1 and its slant height l1 i.e., OP=h1 and OA=OB=l1. Let h2 be the height of cone OCD and l2 its Slant height . We have r1=28cm , r2=7cm and the height of frustum (h) =45cm h1=45+h2 ----(i) Since the triangles OPB and OQD are similar , we have h1h2=287=41−−−−−(ii) From (i) and (ii) h2=15cm and h1= 60 cm. now, the volume of the frustum = volume of the cone OAB – volume of the cone OCD =13π(r1)2h1−13π(r2)2h2 =[13(227)(28)2(60)−13(227)(7)2(15)] =48510cm2 The respective slant height l2and l1 of the cones OCD and OAB are given by l2=√(7)2+(15)2=16.55cm l1=√(28)2+(60)2=4√(7)2+(15)2=66.20cm Thus, the curved surface area of the frustum = πr1l1−πr2l2 =227×7×16.55(16−1) =5461.5cm2 Now, the total surface of the frustum = curved surface area + π(r1)2+π(r2)2 =5461.5+227(28)(28)+227(7)(7) = 5461.5+2464+154 =8079.5cm2

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