The volume of a frustum cone is 560 $m^{3}$ , whose base and upper area of a circle is 90 and 10 $m^{2}$ . Find the height of the cone.

40 cm

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34 cm

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24 cm

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54 cm

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Solution

The correct option is C 24 cm
Volume of a frustum cone = $\frac{1}{3} (A_{1} + A_{2} + \sqrt{A_{1} A_{2}}) h$
560 = $\frac{1}{3} (90 + 10 + \sqrt{40 \times 10}) \times h$
560 = $\frac{1}{3} (50 + \sqrt{400}) \times h$
560 = $\frac{1}{3} (50 + 20) \times h$
$560 \times 3$ = $70 \times h$
h = 1,680/70 = 24 $c m$

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