Question

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

Answer 1

Given, the dimensions of cuboid are

$l=a$, $b=b$ and $h=c$

Step 1: Find the volume $\mathit{V}$ of the cuboid.

Volume of cuboid $=l\times b\times h$

$V=a\times b\times c$

$V=abc-------\left(i\right)$

Step 2: Find the surface area $\mathit{S}$ of the cuboid.

Surface are of cuboid $=2\left[lb+bh+hl\right]$

$S=2\left[ab+bc+ca\right]-------\left(ii\right)$

Step 3: Divide equation (ii) by (i)

$\Rightarrow \frac{S}{V}=\frac{2\left[ab+bc+ca\right]}{abc}$

$\Rightarrow \frac{S}{V}=2\left[\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}\right]$

$\Rightarrow \frac{S}{V}=2\left[\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right]$

Divide both side by $S$, then we get

$\therefore \frac{1}{V}=\frac{2}{S}\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]$

Hence proved.