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

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## Given, the dimensions of cuboid are$l=a$, $b=b$ and $h=c$Step 1: Find the volume $\mathbit{V}$ of the cuboid.Volume of cuboid $=l×b×h$$V=a×b×c$$V=abc-------\left(i\right)$ Step 2: Find the surface area $\mathbit{S}$ of the cuboid.Surface are of cuboid $=2\left[lb+bh+hl\right]$$S=2\left[ab+bc+ca\right]-------\left(ii\right)$ Step 3: Divide equation (ii) by (i)$⇒\frac{S}{V}=\frac{2\left[ab+bc+ca\right]}{abc}$$⇒\frac{S}{V}=2\left[\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}\right]$$⇒\frac{S}{V}=2\left[\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right]$Divide both side by $S$, then we get$\therefore \frac{1}{V}=\frac{2}{S}\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]$Hence proved.  Suggest Corrections  0      Similar questions
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