If the radius of a cone is doubled and its height is reduced to half, how does the volume change?

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Accepted Answer

1. Use the cone volume formula: $V=\frac{1}{3}\pi r^2 h$.
2. New radius $r'=2r$ so $r'^2=4r^2$. New height $h'=h/2$.
3. New volume: $V'=\frac{1}{3}\pi\cdot4r^2\cdot\frac{h}{2}=2\cdot\frac{1}{3}\pi r^2 h=2V$.
4. So the volume becomes double the original.

FINAL_ANSWER: b

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