Question

If v is volume of a cuboid of dimensions a,b,c and 's is its surface area then prove 1/v =2/s [1/a+1/b+1/c]

Answer 1

this is the answer...

Answer 2

Answer:

$\frac{2}{S}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]=\frac{1}{V}$

Step-by-step explanation:

Dimensions of a cuboid:

Length = a units,

Breadth = b units,

Height = h units

$Volume \: of \: the \: cuboid\\(V) = abc\:--(1)$

$Surface \: Area \\(S)=2(ab+bc+ca)\:---(2)$

$Now,\\RHS=\frac{2}{S}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]$

$=\frac{2}{S}[\frac{bc+ac+ab}{abc}]$

$=\frac{2}{2(ab+bc+ca)}[\frac{(ab+bc+ca)}{abc}]$

$=\frac{1}{abc}$

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

Therefore,

$\frac{2}{S}[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]=\frac{1}{V}$

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