If V is the volume of a cubiod 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

$∵$ a,b,c are the dimensions of a cubiod
$∴$ Volume (V) = abc
Surface area (S) = 2(ab+bc+ca)
Now
LHS = $\frac{2}{S} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
= $\frac{2}{2 (a b + b c + c a)} \times \frac{b c + c a + a b}{a b c}$
= $\frac{1}{a b c} = \frac{1}{V}$ = RHS
Hence proved

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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 that

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If V is the volume of a cubiod of dimensions a, b, c and S is its surface area, then prove that 1V=2S(1a+1b+1c)

∵ a,b,c are the dimensions of a cubiod ∴ Volume (V) = abc Surface area (S) = 2(ab+bc+ca) Now LHS = 2S(1a+1b+1c) = 22(ab+bc+ca)×bc+ca+ababc = 1abc=1V= RHS Hence proved

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

$∵$ a,b,c are the dimensions of a cubiod
$∴$ Volume (V) = abc
Surface area (S) = 2(ab+bc+ca)
Now
LHS = $\frac{2}{S} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
= $\frac{2}{2 (a b + b c + c a)} \times \frac{b c + c a + a b}{a b c}$
= $\frac{1}{a b c} = \frac{1}{V}$ = RHS
Hence proved