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A bucket open at the top is of the form of a frustum of a cone. The diameters of its upper and lower circular ends are 40cm and 20cm respectively. If the total of 17600cm^3 of water can be filled in the bucket, find its total surface area.

$Given Diameter of upper end(R)=40cm radius = \frac{d i a m e t e r}{2} = \frac{40}{2} = 20 c m Diameter of lower end(r)=20cm radius = \frac{d i a m e t e r}{2} = \frac{20}{2} = 10 c m {Volume of the frustum=17600cm}^{3} let h be the height of the frustum we know that Volume of the frustum= \frac{π h}{3} (R^{2} + r^{2} + R r) \Rightarrow 17600 = \frac{22}{7} \times \frac{1}{3} \times h (20^{2} + 10^{2} + 20 \times 10) \Rightarrow 17600 = \frac{22}{7} \times \frac{h}{3} (400 + 100 + 200) \Rightarrow 17600 = \frac{22}{7} \times \frac{h}{3} (700) \Rightarrow h = \frac{176 \times 3}{22} \Rightarrow h = 24 c m we know that slant height(l) = \sqrt{h^{2} + (R - r)^{2}} \Rightarrow l = \sqrt{24^{2} + (20 - 10)^{2}} \Rightarrow l = \sqrt{576 + (10)^{2}} \Rightarrow l = \sqrt{576 + 100} \Rightarrow l = \sqrt{676} \Rightarrow l = 26 c m we know that Total surface area of frustum of cone= (area of the base) + (area of the top) + (lateral surface area) = π [R^{2} + r^{2} + l (R + r)] = \frac{22}{7} [(20)^{2} + (10)^{2} + 26 (20 + 10)] = \frac{22}{7} [400 + 100 + 26 (30)] = \frac{22}{7} [500 + 780] = \frac{22}{7} (1280) = 4022.8 c m^{2}$

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