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

We have,

Volume of the cuboid

Surface area of the cuboid

Dimensions of the cuboid

We need to prove,

We know that,

And

Hence,

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

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If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that 1V=2S 1a+1b+1c

We have, Volume of the cuboid Surface area of the cuboid Dimensions of the cuboid We need to prove, We know that, And Hence,

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

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

We have,

Volume of the cuboid

Surface area of the cuboid

Dimensions of the cuboid

We need to prove,

We know that,

And

Hence,