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

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

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Solution

a, b, c are the dimenions of a cuboid S is the surface area and V is the volume

$∴ V = a b c a n d S = 2 (a b + b c + c a)$

$R . H . S . = \frac{2}{S} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

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

$= \frac{2}{S} \times \frac{S}{2 V} = \frac{1}{V} = L . H . S .$

$H e n c e, \frac{1}{V} = \frac{2}{S} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

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