How many times the volume of a cone decreases if the radius of the base is reduced by half?

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Solution

Volume of cone $= \frac{1}{3} π r^{2} h = v$
When, radius is reduced by half.
$V_{1} = \frac{1}{3} π (\frac{r}{2})^{2} h = \frac{1}{3} \frac{π r^{2} h}{4} V_{1} = \frac{v}{4}$

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How many times the volume of a cone decreases if the radius of the base is reduced by half?

Volume of cone =13πr2h=v When, radius is reduced by half. V1=13π(r2)2h=13πr2h4V1=v4

Volume of cone $= \frac{1}{3} π r^{2} h = v$
When, radius is reduced by half.
$V_{1} = \frac{1}{3} π (\frac{r}{2})^{2} h = \frac{1}{3} \frac{π r^{2} h}{4} V_{1} = \frac{v}{4}$