If V is the volume of a cuboid of dimensions a, b, c and S is its surface area then prove that

Question:

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

Solution:

Let the length, breadth and height of the cuboid be ab and c, respectively.

∴ Surface area of the cuboid, S = 2(ab + bc + ca

Volume of the cuboid, V = abc

Now,

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

$\Rightarrow \frac{S}{V}=2\left(\frac{a b}{a b c}+\frac{b c}{a b c}+\frac{c a}{a b c}\right)$

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

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

 

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