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
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