The radii of a bucket in the shape of a frust | vedclass

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(B) The volume $V$ of a frustum of a cone is given by the formula $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$.Given: $R = 28 \, cm$, $r = 21 \, cm$, and

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$15$

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(B) The volume $V$ of a frustum of a cone is given by the formula $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$.Given: $R = 28 \, cm$, $r = 21 \, cm$, and $V = 28.490 \, 	ext{litres} = 28490 \, cm^3$ (since $1 \, 	ext{litre} = 1000 \, cm^3$).Substituting the values into the formula:$28490 = \frac{1}{3} 	imes \frac{22}{7} 	imes h 	imes (28^2 + 21^2 + 28 	imes 21)$.$28490 = \frac{22}{21} 	imes h 	imes (784 + 441 + 588)$.$28490 = \frac{22}{21} 	imes h 	imes (1813)$.$28490 = \frac{22 	imes 1813}{21} 	imes h$.$28490 = \frac{39886}{21} 	imes h$.$h = \frac{28490 	imes 21}{39886} = \frac{598290}{39886} = 15 \, cm$.

The radii of a bucket in the shape of a frustum of a cone are $28 \, cm$ and $21 \, cm$. If its capacity is $28.490 \, \text{litres}$, find its height in $cm$.

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