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. $4{\mathrm{a}}^2:6{\mathrm{a}}^2$
• C. $3:4$
• D. $2:3$

Answer 1

The correct option is A

$6a^2:4a^2$

Step 1: Given data

Let the edge of the solid cube $=\mathrm{a}$

Since the cube is cut into two cuboids of equal volumes.

Only the height of the cuboid will be half of the height of the cube and other dimensions will be the same as that of the cube.
$\therefore$Length$=\mathrm{a}$, breadth$=\mathrm{a}$, height$=\frac{a}{2}$
Step 2: Calculating ratio of the total surface area of the given cube and that of one of the cuboids

The total surface area of the cube total$=6{\left(side\right)}^2$

So, The total surface area of a cube$=6a^2$

The total surface area of a cuboid $=2(lb+bh+hl)$

Thus, the total surface area of one of the two cuboids$=2\left(a\times a+a\times \frac{a}{2}+\frac{a}{2}\times a\right)=4{\mathrm{a}}^2$

Required ratio $=\frac{Totalsurfaceareaofacube}{Totalsurfaceareaofacuboid}=\frac{6a^2}{4a^2}=\frac{3}{2}$

Hence, the ratio of the total surface area of the given cube and that of one of the cuboids will be $3:2$

Hence, option A is correct.