Question

If $V$ is the volume of cuboid whose length is ${}^{\prime }{a}^{\prime }$, breadth is $b$ and height is $c$ and $S$ is its surface area then prove that-
$\frac{1}{V}=\frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

Answer 1

As we know that the volume of a cuboid V = $abc$

and the surface area of the cuboid S = $2ab+2bc+2ca$

$\frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{2}{S}\left(\frac{ab+bc+ca}{abc}\right)$

As we know that S = $2(ab+bc+ca)$

$\frac{S}{2}=ab+bc+ca$

$\frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{2}{S}\times \frac{S}{2}\left(\frac{1}{abc}\right)$

$=\left(\frac{1}{abc}\right)=\frac{1}{V}$

Hence, proved