The height and base radius of a cone are each | vedclass

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(D) Let the original radius of the cone be $r$ and the original height be $h$.The original volume of the cone is $V_1 = \frac{1}{3} \pi r^2 h$.Since b

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$8$ times the original.

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(D) Let the original radius of the cone be $r$ and the original height be $h$.The original volume of the cone is $V_1 = \frac{1}{3} \pi r^2 h$.Since both the radius and height are increased by $100 \%$,the new radius $r' = r + 100 \% 	ext{ of } r = 2r$ and the new height $h' = h + 100 \% 	ext{ of } h = 2h$.The new volume of the cone is $V_2 = \frac{1}{3} \pi (r')^2 (h') = \frac{1}{3} \pi (2r)^2 (2h)$.$V_2 = \frac{1}{3} \pi (4r^2) (2h) = 8 	imes (\frac{1}{3} \pi r^2 h) = 8 V_1$.Therefore,the volume of the cone becomes $8$ times the original volume.

The height and base radius of a cone are each increased by $100 \%$. The volume of the cone now becomes

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