Question

A tent is made in the form of a frustum of a cone surmounted by another cone, The diameters of the base and the top of the frustum are 20m and 6m respectively, and the height is 24m, If the height of the tent is 28m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required

Answer 1

Let r1 and r2 be radius of the top and bottom of frustum 

Given: 2r1 = 6 m , 2r2 = 20 m  and Height of frustum = 24 m  

So, r1 = 3 m and r2 = 10m  

Therefore,  

Slant height (l) = underoot (24) square + (10-3) square = underoot 576+49 = 25 m  

And radius of cone (r1) = 3m and height  = (28-24) m = 4m 

Slant height (l1) = underoot (3) square + (4) square = underoot 9+16 = 5m 

So,  

Quantity of canvas area = πl (r1+r2) + π r1 l1 

= ( π x 25 (3+10) + π x 3 x 5) 

= (325 π + 15π) 

= 340 π 

= 340 x 3.14 

= 1067.6 m square  

So the quantity of  canvas area is 1067.6 m square

Answer 2

Canvas required 1068. 57 square metre.