Question

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

Answer 1

\[\text { It is given that V is the volume of a cuboid of length = a, breadth = b and height = c . Also, S is surface area of cuboid . } \]

\[\text { Then, V = a } \times b \times c\]

\[\text { Surface area of the cuboid } = 2 \times \text { (length } \times \text { breadth + breadth  }\times \text { height + length } \times \text { height) }\]

\[ \Rightarrow S = 2 \times (a \times b + b \times c + a \times c)\]

\[\text { Let us take the right - hand side of the equation to be proven } . \]

\[\frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) = \frac{2}{2 \times (a \times b + b \times c + a \times c)} \times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]

\[=\frac{1}{(a \times b + b \times c + a \times c)} \times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]

\[\text { Now, multiplying the numerator and the denominator with a } \times b \times c, \text { we get: } \]

\[\frac{1}{(a \times b + b \times c + a \times c)} \times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \times \frac{a \times b \times c}{a \times b \times c}\]

\[=\frac{1}{(a \times b + b \times c + a \times c)} \times (\frac{a \times b \times c}{a}+\frac{a \times b \times c}{b}+\frac{a \times b \times c}{c}) \times \frac{1}{a \times b \times c}\]

\[=\frac{1}{(a \times b + b \times c + a \times c)} \times (b\times c+a\times c+a\times b) \times \frac{1}{a \times b \times c}\]

\[=\frac{1}{(a \times b + b \times c + a \times c)}\times(a\times b+b\times c+a\times c) \times \frac{1}{a \times b \times c}\]

\[=\frac{1}{a \times b \times c}\]

\[=\frac{1}{V}\]

\[ \therefore \frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) = \frac{1}{V}\]