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(D) For the frustum of a cone,the circumference of the base is $C_1 = 2 \pi r_1 = 48 \, cm$,so $r_1 = \frac{24}{\pi} \, cm$.The circumference of the t

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$1554$

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(D) For the frustum of a cone,the circumference of the base is $C_1 = 2 \pi r_1 = 48 \, cm$,so $r_1 = \frac{24}{\pi} \, cm$.The circumference of the top is $C_2 = 2 \pi r_2 = 36 \, cm$,so $r_2 = \frac{18}{\pi} \, cm$.The height of the frustum is $h = 11 \, cm$.The volume $V$ of a frustum of a cone is given by $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.Substituting the values: $V = \frac{1}{3} \pi (11) \left[ (\frac{24}{\pi})^2 + (\frac{18}{\pi})^2 + (\frac{24}{\pi})(\frac{18}{\pi}) ight]$.$V = \frac{11 \pi}{3} \left[ \frac{576}{\pi^2} + \frac{324}{\pi^2} + \frac{432}{\pi^2} ight]$.$V = \frac{11 \pi}{3} \left[ \frac{1332}{\pi^2} ight] = \frac{11 	imes 1332}{3 \pi} = \frac{11 	imes 444}{\pi} = \frac{4884}{\pi}$.Using $\pi \approx \frac{22}{7}$,$V = \frac{4884 	imes 7}{22} = 222 	imes 7 = 1554 \, cm^3$.

For a given frustum of a cone,the circumference of the base is $48 \, cm$ and the circumference of the top is $36 \, cm$. If its height is $11 \, cm$,find its volume (in $cm^3$).

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