Question

The surface areas of a sphere and cube are equal. Prove that their volumes are in the ratio $1:\sqrt{\pi /6}.$

Answer 1

Let the radius of the sphere be $r$ and side of cube be $a$.

Since,

Surface area of the sphere $=$ Surface area of cube

$4\pi r^2=6a^2$

$2\pi r^2=3a^2$

$a^2=\frac{2\pi r^2}{3}$

$a=r\sqrt{\frac{2\pi }{3}}$

Therefore,

$\frac{V_1}{V_2}=\frac{\frac{4}{3}\pi \times r^3}{a^3}$

$=\frac{4\pi \times r^3}{3{\left(r\sqrt{\frac{2\pi }{3}}\right)}^3}$

$=\frac{4\pi }{3\left(\frac{2\pi }{3}\sqrt{\frac{2\pi }{3}}\right)}$

$=\frac{2}{\sqrt{\frac{2\pi }{3}}}$

$=\frac{1}{\sqrt{\frac{2\pi }{12}}}$

$=\frac{1}{\sqrt{\frac{\pi }{6}}}$

Hence, the required ratio is $1:\sqrt{\frac{\pi }{6}}$.