A 22.5 m high tent is in the shape of a frustum of a cone surmounted by a hemisphere. If the diameters of the upper and the lower circular ends of the frustum are 21 m and 39 m, respectively, then find the area of cloth (in m²) used to make the tent (ignoring the wastage). $$\left( {{\text{Use }}\pi = \frac{{22}}{7}} \right)$$

Solution (By Examveda Team)

$$\eqalign{ & r = \frac{{21}}{2} \cr & R = \frac{{39}}{2} \cr & {\text{Curved serface area of Hemisphere}} \cr & = 2\pi {r^2} \cr & = 2 \times \frac{{22}}{7} \times \frac{{21 \times 21}}{{2 \times 2}} \cr & = 693\,{\text{c}}{{\text{m}}^2} \cr & H = \left( {22.5 - 10.5} \right) = 12{\text{ cm}} \cr & {L^2} = {H^2} + {\left( {R - r} \right)^2} \cr & {L^2} = {12^2} + {\left( {\frac{{39}}{2} - \frac{{21}}{2}} \right)^2} \cr & {L^2} = 144 + 81 \cr & L = \sqrt {225} \cr & L = 15{\text{ cm}} \cr & {\text{Curved serface area of Frustum of cone}} \cr & = \pi l\left( {R + r} \right) \cr & = \frac{{22}}{7} \times 15 \times \left[ {\left( {\frac{{39}}{2}} \right) + \left( {\frac{{21}}{2}} \right)} \right] \cr & = \frac{{22}}{7} \times 15 \times 30 \cr & = \frac{{9900}}{7}{\text{ c}}{{\text{m}}^2} \cr & = 1414\frac{2}{7}{\text{ c}}{{\text{m}}^2} \cr & \therefore {\text{Total area of cloth}} \cr & = \left( {693 + 1414\frac{2}{7}} \right){\text{c}}{{\text{m}}^2} \cr & = 2107\frac{2}{7}{\text{ c}}{{\text{m}}^2} \cr} $$