A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. The area of the canvas required for the tent is :

Solution(By Examveda Team)

Radius, r = 12 m
Height of conical part, h = (16 - 11) m = 5 m
Slant height of conical part,
$$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\, = \sqrt {{{\left( {12} \right)}^2} + {{\left( 5 \right)}^2}} \cr & \,\,\,\, = \sqrt {169} \cr & \,\,\,\, = 13\,m \cr} $$
Height of cylindrical part, H = 11 m
Area of canvas required :
= Curved surface area of cylinder + Curved surface area of cone
$$\eqalign{ & = 2\pi rH + \pi rl \cr & = \left[ {\frac{{22}}{7}\left( {2 \times 12 \times 11 + 12 \times 13} \right)} \right]{{\text{m}}^2} \cr & = \left[ {\frac{{22}}{7}\left( {264 + 156} \right)} \right]{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 420} \right){{\text{m}}^2} \cr & = 1320\,{{\text{m}}^2} \cr} $$