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A tent consists of a frustum of a cone, surmounted by a cone. If the diameters of the upper and lower circular ends of the frustum be 14m and 26m respectively , the height of the frustum be 8m and the slant height of the surmounted conical portion be 12 m, find the area of canvas required to make tent . (Assume that the radii of the upper circular end of the frustum and the base of surmounted conical portion are equal).

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$Given Diameter of the upper end=14m Radius of the upper end(r_{2}) = \frac{d i a m e t e r}{2} = \frac{14}{2} = 7 m Diameter of the lower end=26m Radius of the lower end(r_{1}) = \frac{d i a m e t e r}{2} = \frac{26}{2} = 13 m Height of the frustum(h)=8m Let l be the slant height of the frustum we know that l^{2} = ({r_{1}}^{2} - {r_{2}}^{2}) + h^{2} \Rightarrow l^{2} = (13^{2} - 7^{2}) + 8^{2} \Rightarrow l^{2} = (169 - 49) + 64 \Rightarrow l^{2} = 120 + 64 \Rightarrow l^{2} = 184 \Rightarrow l = 13.56 m Curved surface area of the frustum= π (r_{1} + r_{2}) l = \frac{22}{7} \times (13 + 7) \times 13.56 = \frac{22}{7} \times 20 \times 13.56 = 852.3 m^{2} Diameter of the cone=14m \Rightarrow radius of the cone(r)=7m slant height of the cone(l_{1}) =12m Curved surface area of the cone= π r l_{1} = \frac{22}{7} \times 7 \times 12 = 264 m^{2} Area of the canvas required=curved surface ar ea of frustum + curved surface area of cone = 852.3 + 264 = 1116.3 m^{2}$

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