In a field, dry fodder for the cattle is heaped in a conical shape. Th

1.	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 = \(\frac{22}7\) and \(\sqrt{17.37}\) = 4.17 ]
Answer» 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.

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 = \(\frac{22}7\) and \(\sqrt{17.37}\) = 4.17 ]

Answer»

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.