A bucket is in the form of a frustum of a con | vedclass

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(B) Given, volume of the frustum $= 28.49 \, 	ext{L} = 28.49 	imes 1000 \, 	ext{cm}^3 = 28490 \, 	ext{cm}^3$ (since $1 \, 	ext{L} = 1000 \, 	ext

Vedclass · Accepted Answer

$15$

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(B) Given, volume of the frustum $= 28.49 \, 	ext{L} = 28.49 	imes 1000 \, 	ext{cm}^3 = 28490 \, 	ext{cm}^3$ (since $1 \, 	ext{L} = 1000 \, 	ext{cm}^3$).Radius of the top $(r_1) = 28 \, 	ext{cm}$.Radius of the bottom $(r_2) = 21 \, 	ext{cm}$.Let the height of the bucket be $h \, 	ext{cm}$.The volume 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: $28490 = \frac{1}{3} 	imes \frac{22}{7} 	imes h 	imes (28^2 + 21^2 + 28 	imes 21)$.$28490 = \frac{22}{21} 	imes h 	imes (784 + 441 + 588)$.$28490 = \frac{22}{21} 	imes h 	imes 1813$.$h = \frac{28490 	imes 21}{22 	imes 1813}$.$h = \frac{598290}{39886} = 15 \, 	ext{cm}$.

$A$ bucket is in the form of a frustum of a cone and holds $28.490 \, \text{litres}$ of water. The radii of the top and bottom are $28 \, \text{cm}$ and $21 \, \text{cm}$, respectively. Find the height of the bucket. (in $\text{cm}$)

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