Question

If 'V' is the 'Volume of a Cuboid' of Dimensions a, b c and 'S' is its 'Surface Area' then Prove that $\boxed {\frac{1}{V} = \frac{2}{S} \Big (\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big )}$
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$\mathfrak {\bold {Topic} - Class\: 9 - Volume\: and\: Surface\: Areas\: of\: Solids}$
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Answer 1

Answer:

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

where, V is the volume

a,b,c are the sides

and, S the surface area

please refer to the above attachment

hope this will help you.

Answer 2

Solution :-

Given,

Dimensions of the Cuboid are a, b and c.

Volume of the Cuboid, V = abc and Surface Area of the Cuboid, S = $\bold {2(ab + bc + ac)}$

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To Prove, $\bold {\frac{1}{V} = \frac{2}{S} \Big [\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big ]}$

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Consider LHS to be $\bold {\frac{1}{V} = \frac{1}{abc}}$  . . . (1)

Consider RHS to be,

$\begin {aligned} \bold {\frac{2}{S} \Big [\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big ]} &= \bold {\frac{2}{2(ab + bc + ac)} \Big [\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big ]}\\\\\\& \Rightarrow \frac{1}{ab + bc + ac} \Big [\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big ]\\\\\\& \Rightarrow \frac{1}{ab + bc + ac} \Big [\frac{ab + bc + ac}{abc} \Big ]\\\\\\& = \bold {\frac{1}{abc}} \\\\\\\bold {\frac{2}{S} \Big [\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big ]} & = \bold {\frac{1}{abc}} \end {aligned}$

From . . . (1) and  . . . (2), we get $\bold {\frac{1}{V} = \frac{2}{S} \Big [\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big ]}$

Hence, Proved . . . :)