Question

If 'V' is the volume of a cuboid of dimensions a $\times$ b $\times$ c and 'S' is its surface area, then prove that:
$\frac{1}{V}=\frac{2}{S}\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]$.

Answer 1

The dimensions of the cuboid are $a,b,c$.

We know that, Volume of the cuboid $V=abc$ and surface area of the cuboid $S=2(ab+bc+ac)$

To prove: $\frac{1}{V}=\frac{2}{S}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]$

Consider LHS, $\frac{1}{V}=\frac{1}{abc}...\left(1\right)$

Consider RHS.

$\frac{2}{S}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]=\frac{2}{2(ab+bc+ac)}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]$

$=\frac{1}{ab+bc+ac}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]$

$=\frac{1}{ab+bc+ac}\left[\frac{ab+bc+ac}{abc}\right]$

$=\frac{1}{abc}$

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

Hence from $\left(1\right)$ and $\left(2\right)$ we get $\frac{1}{V}=\frac{2}{S}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]$