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) πr2h
= (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.