Question

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

$\sf \: \dfrac{1}{v} = \dfrac{2}{s} \bigg( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \bigg)$
Class 8 Maths Question .​

Answer 1

GIVEN :-

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

$\sf \: \dfrac{1}{V} = \dfrac{2}{S} \bigg( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \bigg)$

PROOF :-

Taking RHS,

$\implies \: \sf \: \dfrac{2}{S} \bigg( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \bigg)$

$\implies \: \sf \: \dfrac{2}{S} \bigg( \dfrac{bc + ca + ab}{abc} \bigg)$

$\implies \: \sf \: \dfrac{2(bc + ca + ab)}{S \times abc}$

Now as we know that S of cuboid = 2(lb + bh + lh) on substituting the values we get,

$\implies \: \sf \: \dfrac{ \cancel{2(bc + ca + ab)}}{ \cancel{2(bc + ca + ab) }\times abc}$

$\implies \sf \: \dfrac{1}{abc} = \dfrac{1}{V}$

Hence Proved .

Answer 2

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Question:-

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

$\sf \frac{1}{v} = \frac{2}{s} ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} )$

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Given:-

• The dimensions of cuboid are a,b and c

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Solution:-

We know that,

• Volume of the cuboid V = abc

• surface area of cuboid S = 2(ab + bc + ac)

$\sf \frac{1}{v} = \frac{2}{s} ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} )$

By Taking RHS,

$\sf\green\implies \: \frac{2}{s} ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} )$

$\sf\pink\implies \: \frac{2}{s} ( \frac{bc + ca + ab)}{abc} )$

$\sf\red\implies \: \frac{2(bc + ca + ab)}{s \times abc}$

we know,

That Surface (S) of cuboid = 2(lb + bh + lh)

On substituting Values,

$\sf\implies \cancel\frac{2(bc +ca + ab)}{2(bc + ca + ab) \times abc}$

$\star{\boxed{\sf{ \frac{1}{abc} = \frac{1}{v} }}}$

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