The volume of a conical tent is 1232 m³ and the area of its base is 154 sq. m. Find the length of the canvas required to build the tent, if the canvas is 2 m in width. $$\left( {{\text{Take }}\pi = \frac{{22}}{7}} \right)$$

Solution (By Examveda Team)

$$\eqalign{ & \pi {r^2} = 154 \cr & {r^2} = \frac{{154 \times 7}}{{22}} \cr & {r^2} = 49 \cr & r = \sqrt {49} \cr & r = 7{\text{ m}} \cr & {\text{Also volume}} = 1232 \cr & \frac{1}{3}\pi {r^2}h = 1232 \cr & h = \frac{{1232 \times 3}}{{\pi {r^2}}} \cr & h = \frac{{1232 \times 3}}{{154}} \cr & h = 24{\text{ m}} \cr & {\text{Area of canvas required}} = \pi rl \cr & = \pi r\sqrt {{r^2} + {h^2}} \cr & = \frac{{22}}{7} \times 7 \times \sqrt {{{24}^2} + {7^2}} \cr & = \frac{{22}}{7} \times 7 \times 25 \cr & = 550{\text{ }}{{\text{m}}^2} \cr & {\text{Length}} \times {\text{2}} = 550{\text{ }}{{\text{m}}^2} \cr & {\text{length}}\left( l \right) = \frac{{550}}{2} = 275{\text{ m}} \cr} $$