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# A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m, respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of the canvas required to make the tent.

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Assume that the radii of the upper circular end of the frustum and the base of the surmounted conical portion are equal.

This question was asked in 2008 cbse board exam. Question from RS Aggarwal book chapter volume and surface area of solid, page number 823, exercise 17C, problem number 15

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1. For the frustum:
Upper diameter = 14 m
Upper Radius, r = 7 m
Lower diameter = 26 m
Lower Radius, R = 13 m
Given data

Height of the frustum= h = 8 m

Slant height l = √ [h²+[R-r]²]

= √[8²+ [13-7]²]

= √ [64+36]

= √100

l =10 m

For the conical part:

Radius of the base = r = 7 m

Slant height = L =12 m

Total surface area of the tent = Curved area of frustum + Curved area of the cone

=πl[R+r] + πrL

=[22/7]× 10 [13+7]] + [22/7]×7×12

=227200+84

=892.57 m²

∴ area of the canvas required to make the tent is 892.57 m²

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