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In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the heap of the fodder. If it is to be covered by polythene in rainy se&son then how much minimum polythene sheet is needed? (π = 22/7 and √17.37 = 4.17]

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Given: Height of the heap (h) = 2.1 m, diameter of the base (d) = 7.2 m 

∴ Radius of the base (r) = d/2 = 7.2/2 = 3.6 m

To find: Volume of the heap of the fodder and polythene sheet required

i. Volume of the heap of fodder = (1/3) πr2

= (1/3) x (22/7) x (3.6)2 x 2.1 

= (1/3) x (22/7) x 3.6 x 3.6 x 2.1 

= 1 x 22 x 1.2 x 3.6 x 0.3 

= 28.51 cubic metre

ii. Now, l2 = r2 + h2 

= (3.6)2 + (2.1)2 

= 12.96 + 4.41 

∴ l2 =17.37 

∴ l2 = √17.37 ... [Taking square root on both sides] 

= 4.17 m

iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap

= πrl 

= (22/7) x 3.6 x 4.17 

= 47.18 sq.m 

∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.

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