Question

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

Answer 1

To Prove: $\frac{1}{v}=\frac{2}{s}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

We know that the volume of cuboid $v$ = $abc$

Surface area of cuboid $s$ = $2(ab+bc+ca)$

Now, considering the RHS

we have

$\frac{2}{s}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

putting the value of $s$, we have

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

$=\frac{1}{(ab+bc+ca)}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

$=\frac{1}{(ab+bc+ca)}\left(\frac{ab+bc+ca}{abc}\right)$

$=\frac{1}{abc}$

Using $v$ = $abc$, we have

$=\frac{1}{v}$

Hence, $LHS=RHS=\frac{1}{v}$

Therefore, $\frac{1}{v}=\frac{2}{s}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ proved.