Two solid right circular cones have the same height. The radii of their bases are $r_{1}$ and $r_{2}$ . They are melted and recast into a cylinder of same height. Show that the radius of the base of the cylinder is $\sqrt{\frac{r_{1}^{2} + r_{2}^{2}}{3}}$

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Solution

Let $h$ be the height of two given cones of base radii $r_{1}$ and $r_{2}$ respectively.
Further, let $R$ be the radius of the cylinder.
It is given that the cylinder is also of height $h$ .
It is also given that the cylinder is casted out of both cones.
Therefore, volume of the cylinder = sum of the volumes of two cones

We know that, volume of a cylinder $= π r^{2} h$ and volume of a cone $= \frac{1}{3} π r^{2} h$
$∴ π R^{2} h = \frac{1}{3} π r_{1}^{2} h + \frac{1}{3} π r_{2}^{2} h$

$\Rightarrow π R^{2} h = \frac{1}{3} π h (r_{1}^{2} + r_{2}^{2})$

$\Rightarrow R^{2} = \frac{1}{3} (r_{1}^{2} + r_{2}^{2})$

$\Rightarrow R = \sqrt{\frac{(r_{1}^{2} + r_{2}^{2})}{3}}$ $[H e n c e p r o v e d]$

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Two solid right circular cones have the same height. The radii of their bases are $r_{1}$ and $r_{2}$ . They are melted and recast into a cylinder of same height. Show that the radius of the base of the cylinder is $\sqrt{\frac{r_{1}^{2} + r_{2}^{2}}{3}}$