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If the areas of circular bases of a frustum of a cone are $4 c m^{2}$ and $9 c m^{2}$ respectively and the height of the frustum is 12 cm, then its volume is

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48 c m^{3}

76 c m^{3}

42 c m^{3}

44 c m^{3}

answer is C.

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Detailed Solution

Given

π R^{2} = 9, π r^{2} = 4, h = 12 c m

Volume of frustum $= \frac{π h}{3} (R^{2} + r^{2} + R r)$

$= \frac{π \times 12}{3} (\frac{9}{11} + \frac{4}{11} + \frac{3 \times 2}{π})$

$= 4 π (\frac{19}{π})$

$= 76 c m^{3}$

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