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

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Solution

Given dimensions of cuboid are a, b and c.
Therefore, volume of cuboid, V = abc → (1)
Surface area of cuboid, S = 2(ab + bc + ca) → (2)
Now divide (2) with (1), we get
$s over v equals fraction numerator 2 open parentheses a b plus b c plus c a close parentheses over denominator a b c end fraction space space space space equals 2 open square brackets fraction numerator a b over denominator a b c end fraction plus fraction numerator b c over denominator a b c end fraction plus fraction numerator c a over denominator a b c end fraction close square brackets space space space space space equals 2 open square brackets 1 over c plus 1 over a plus 1 over b close square brackets h e n c e space 1 over v equals 2 over s open square brackets 1 over c plus 1 over a plus 1 over b close square brackets$

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

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The solution for the equations $c_{1} x + b_{1} y + a_{1} = 0$ and $c_{2} x + b_{2} y + a_{2} = 0$ is

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Q. If

V

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^{'} a^{'}

, breadth is

b

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S

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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 that1V=2S(1a+1b+1c).

Given dimensions of cuboid are a, b and c. Therefore, volume of cuboid, V = abc → (1) Surface area of cuboid, S = 2(ab + bc + ca) → (2) Now divide (2) with (1), we get

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