Question

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

• A. 2 : 1
• B. 1 : 2
• C. 2 : 3
• D. 3 : 2

Answer 1

The correct option is D 3 : 2

Let the edge of the solid cube be $a$ cm. Since the cube is cut into two cuboids of equal volumes. Dimensions of each of the cuboid will be $a$ cm x $a$ cm x $\frac{a}{2}$ cm.
Total surface area of one cuboid
$=2(lb+bh+hl)$
$=2(a\times a+a\times \frac{a}{2}+\frac{a}{2}\times a)$
$=2(a^2+\frac{a^2}{2}+\frac{a^2}{2})$
$=2\left(2a^2\right)=4a^2{\text{cm}}^2$
Total surface area of cube = $6(side{)}^2$
= $6a^2$
Ratio $=\frac{\text{TSA of cube}}{\text{TSA of cuboid}}$
$=\frac{6a^2}{4a^2}=\frac{3}{2}$
Hence the ratio of the total surface area of the given cube and one of the cuboid is 3 :2.