The height of a right circular cone and the r | vedclass

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(B) Let the height of the original cone be $H = 9 \, cm$ and the radius of its base be $R = 3 \, cm$.Let the radius of the upper circular surface of t

Vedclass · Accepted Answer

$\sqrt[3]{13} \, cm$

Vedclass · Answer

(B) Let the height of the original cone be $H = 9 \, cm$ and the radius of its base be $R = 3 \, cm$.Let the radius of the upper circular surface of the frustum be $r$ and the height of the small cone cut from the top be $h'$.By similar triangles,the ratio of the radius to the height is constant: $\frac{r}{h'} = \frac{R}{H} = \frac{3}{9} = \frac{1}{3}$,so $h' = 3r$.The volume of the original cone is $V_{total} = \frac{1}{3} \pi R^2 H = \frac{1}{3} 	imes \frac{22}{7} 	imes 3^2 	imes 9 = \frac{22 	imes 27}{7} = \frac{594}{7} \, cm^3$.The volume of the small cone at the top is $V_{small} = \frac{1}{3} \pi r^2 h' = \frac{1}{3} \pi r^2 (3r) = \pi r^3$.The volume of the frustum is $V_{frustum} = V_{total} - V_{small} = 44 \, cm^3$.Substituting the values: $\frac{594}{7} - \frac{22}{7} r^3 = 44$.Multiply by $7$: $594 - 22r^3 = 308$.$22r^3 = 594 - 308 = 286$.$r^3 = \frac{286}{22} = 13$.Therefore,$r = \sqrt[3]{13} \, cm$.

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