If V is the volume of a cuboid of dimensions a × b × c and A is its surface area, then $\frac{A}{V}$ is

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Solution

We know, surface area of a cuboid of dimensions l×b×h, A = 2(lb + bh + hl)
So, surface area of a cuboid of dimensions a×b×c = 2(ab + bc + ca)
And, volume of the cuboid, V = abc
$∴ \frac{A}{V} = \frac{2 (a b + b c + c a)}{a b c} = 2 (\frac{1}{c} + \frac{1}{a} + \frac{1}{b})$

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If V is the volume of a cuboid of dimensions a × b × c and A is its surface area, then AV is

We know, surface area of a cuboid of dimensions l×b×h, A = 2(lb + bh + hl) So, surface area of a cuboid of dimensions a×b×c = 2(ab + bc + ca) And, volume of the cuboid, V = abc ∴ AV=2(ab+bc+ca)abc=2(1c+1a+1b)

If V is the volume of a cuboid of dimensions a × b × c and A is its surface area, then $\frac{A}{V}$ is

We know, surface area of a cuboid of dimensions l×b×h, A = 2(lb + bh + hl)
So, surface area of a cuboid of dimensions a×b×c = 2(ab + bc + ca)
And, volume of the cuboid, V = abc
$∴ \frac{A}{V} = \frac{2 (a b + b c + c a)}{a b c} = 2 (\frac{1}{c} + \frac{1}{a} + \frac{1}{b})$