If $V$ is the volume of a cuboid of dimensions $a, b, c$ and $S$ is its surface area, then prove that
$\frac{1}{\mathrm{~V}}=\frac{2}{\mathrm{~S}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

Given:

$V$ is the volume of a cuboid of dimensions $a, b, c$ and $S$ is its surface area.

To do:

We have to prove that

$\frac{1}{\mathrm{~V}}=\frac{2}{\mathrm{~S}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$.

Solution:

$V = a \times b \times c$

$=abc$

$S = 2(lb + bc + ca)$

RHS $=\frac{2}{\mathrm{~S}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

$=\frac{2}{\mathrm{~S}}(\frac{b c+c a+a b}{a b c})$

$=\frac{2}{\mathrm{~S}} \times \frac{\mathrm{S}}{2 \mathrm{~V}}$

$=\frac{1}{\mathrm{~V}}$

$=$ LHS.

Hence proved.

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Updated on: 10-Oct-2022

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