Question

Let V be the Volume of the Cuboid and S be the surface area of the Cuboid of dimension
a, b and c then \[\dfrac{1}{V}\]=
(a) \[\dfrac{S}{2}(a+b+c)\]
(b) \[\dfrac{2}{S}(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\]
(c) \[\dfrac{2S}{a+b+c}\]
(d) \[2S(a+b+c)\]

Answer 1

Hint: We use the formula of Volume of a cuboid and the formula of Surface Area of a cuboid and see the possible relations between them to solve the question.

Complete step-by-step answer:
We are given the dimension of the cuboid as a,b and c then Volume of the Cuboid \[V=abc\] and the Surface area of cuboid is \[S=2(ab+bc+ac)\]
Now there are two possible ways to solve these types of questions either we go for creating the formula and relation between V and S or we go by the options and check which one holds correct.
Proceeding by the first way and taking the reciprocal of the Volume we get,
\[\dfrac{1}{V}=\dfrac{1}{abc}\]
As 2 is a non-zero number so multiplying and dividing by 2 in the above equation, we get
\[\dfrac{1}{V}=\dfrac{2}{2(abc)}\]
Again, multiplying and dividing the right-hand side of the above equation by \[(ab+bc+ac)\], we get
\[\dfrac{1}{V}=\dfrac{2(ab+bc+ac)}{2(ab+bc+ac)(abc)}\]
Rearranging the parenthesis, we have
\[\dfrac{1}{V}=\dfrac{2}{2(ab+bc+ac)}\left( \dfrac{ab+bc+ac}{abc} \right)\]
Now because we have Surface area of cuboid is \[S=2(ab+bc+ac)\], then substituting in above equation we get
\[\dfrac{1}{V}=\dfrac{2}{S}\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)\]
Hence \[\dfrac{1}{V}=\dfrac{2}{S}\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)\] is the required answer, which is option (b)

Note: The possibility for the mistake is that you can choose the option (a) as you might get confused with either getting \[\dfrac{2}{S}\]rather than getting \[\dfrac{S}{2}\] because both the terms cancel 2 present in \[S=2(ab+bc+ac)\], which will give incorrect answer.