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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)2h113π(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 = πr1l1πr2l2

=227×7×16.55(161)
=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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