Question

Question No: 7
The volume of cube A is equal to the surface area of cube B. If both have integral side lengths, the
minimum surface area (in square units) of cube B will be
Cube A
Cube B
​

Answer 1

The volume of cube A is equal to the surface area of cube B. If both have integral side lengths, the

minimum surface area (in square units) of cube B will be

Answer 2

Answer:

216

Step-by-step explanation:

Let the side length of the cube A is 'a' and side length of cube B is 'b'.

Formula Used:

Volume of cube of side x is $x^3$

Surface are of the cube of the side x is $6x^2$

Given that:

Volume of cube A = Surface area of cube B

$\implies a^3 = 6b^2 \,\,\, s\,.\,t.\,\,\,a,b \in \mathbb{Z^+}$

Consider hit and trail;

$\\For \,\, a=1 , b \notin \mathbb{Z^+}\\For \,\, a=2 , b \notin \mathbb{Z^+}\\\\.\\.\\.\\For \,\, a=6 , we\,\, have\,\, b= 6\,\, \& \,\, b \in \mathbb{Z^+}\\\\\implies 6b^2 = 216$

Minimum surface are of cube B will be 216.

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