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

Given: Height of the heap (h) = 2.1 m. 

diameter of the base (d) = 7.2 m 

∴Radius of the base (r) = d/2 =\(\frac{7.2}2\) = 3.6 m 

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

Solution:

i. Volume of the heap of fodder = 1/3πr²h 

=1/3 x \(\frac{22}7\) x (3.6)² x 2.1 

= 1/3 x \(\frac{22}7\) 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 =\(\sqrt{17.37}\) .. .[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 

= \(\frac{22}7\) 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.