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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Answer 1

$\huge{\sf{Question:-}}$

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

We know that,

➠$\bold{\sf{Volume\:Of\: Cuboid=L× B× H}}$

➠$\bold{\sf{TSA=2(lb+bh+lh)}}$

$\bold{\sf{Given:-}}$

Length,breadth and height = a,b and c

Total Surface area = S

$\bold{\sf{To\:prove:-}}$

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

$\huge{\sf{Solution:-}}$

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

$\sf{\frac{1}{v}=\frac{2}{S}(\frac{ab+bc+ca}{abc})}$ {$\sf{\red{Taking\:the\:LCM}}$}

Inserting the formula of TSA and Volume,

$\sf{\frac{1}{abc}=\frac{2}{2(ab+bc+ca)}(\frac{ab+bc+ca}{abc})}$

$\sf{\frac{1}{abc}=\frac{1}{abc}}$

$\sf{\frac{1}{V}=\frac{1}{V}}$

${\boxed{\sf{LHS=RHS}}}$

Hence Proved !!