The circumference of one end of a frustum of | vedclass

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(B) Let the radii of the two circular ends be $r_1$ and $r_2$.Given circumferences are $C_1 = 2\pi r_1 = 48\, cm$ and $C_2 = 2\pi r_2 = 34\, cm$.Thus,

Vedclass · Accepted Answer

$1350$

Vedclass · Answer

(B) Let the radii of the two circular ends be $r_1$ and $r_2$.Given circumferences are $C_1 = 2\pi r_1 = 48\, cm$ and $C_2 = 2\pi r_2 = 34\, cm$.Thus,$r_1 = \frac{48}{2\pi}$ and $r_2 = \frac{34}{2\pi}$.The height of the frustum is $h = 10\, cm$.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 (10) \left[ \left(\frac{48}{2\pi}ight)^2 + \left(\frac{34}{2\pi}ight)^2 + \left(\frac{48}{2\pi}ight) \left(\frac{34}{2\pi}ight) ight]$$V = \frac{10\pi}{3} \left[ \frac{2304}{4\pi^2} + \frac{1156}{4\pi^2} + \frac{1632}{4\pi^2} ight]$$V = \frac{10\pi}{3} 	imes \frac{1}{4\pi^2} (2304 + 1156 + 1632)$$V = \frac{10}{12\pi} (5092) = \frac{50920}{12\pi} \approx \frac{50920}{12 	imes 3.14159} \approx 1351.2\, cm^3$.Rounding to the nearest provided option,the correct value is $1350$.

The circumference of one end of a frustum of a right circular cone is $48\, cm$ and of the other end is $34\, cm$. If the height of the frustum is $10\, cm$,its volume (in $cm^3$) is

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