Question

A solid cube is cut into two cuboids of equal volumes. The ratio of the total surface area of the given cube and that of one of the cuboids is

• A. $6a^2:4a^2$
• B. $4a^2:6a^2$
• C. $3:4$
• D. $2:3$

Answer 1

The correct option is A $6a^2:4a^2$

Let the edge of the solid cube = a
Since, cube is cut into two cuboids of equal volumes.
$\therefore$ length = a, breadth = a, height $=\frac{a}{2}$
Total surface area of cube = $6a^2$ sq. unit.
Total surface area of one cuboid
$=2(a\times a+a\times \frac{a}{2}+\frac{a}{2}\times a)=4a^{2}sq.unit$
Required ratio $=6a^2:4a^2=3:2$