If the radius of the base of the right circular cylinder is reduced by $50 %$ , keeping the same height, what is the ratio of the volume of the reduced cylinder to that of the original?

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Solution

If r be the radius of the original, then $\frac{r}{2}$ is the radius of the reduced cylinder
$∴$ Ratio is $π {(\frac{r}{h})}^{2} = π r^{2} h$
$= 1 : 4$

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If the radius of the base of the right circular cylinder is reduced by 50%, keeping the same height, what is the ratio of the volume of the reduced cylinder to that of the original?

If r be the radius of the original, then r2 is the radius of the reduced cylinder∴ Ratio is π(rh)2=πr2h=1:4

If the radius of the base of the right circular cylinder is reduced by $50 %$ , keeping the same height, what is the ratio of the volume of the reduced cylinder to that of the original?

If r be the radius of the original, then $\frac{r}{2}$ is the radius of the reduced cylinder
$∴$ Ratio is $π {(\frac{r}{h})}^{2} = π r^{2} h$
$= 1 : 4$