Question

A tent is made in the form of a frustum of a cone surmounted by another cone.the diameter of the base and top of frustum is 20m and 6m and the height is 24cm. If the height of the tent is 28cm and the radius of the conical part is equal to the radius of the the top of frustum. Find the quantity of canvas area.

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

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

hope it helps you dear