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 that one of the cuboids.

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

Answer 1

The correct option is D

$3:2$

Step 1: Given data

A given cube is cut into two cuboids of equal volume.

Let, edge of given cube be $a$

$\frac{Totalsurfaceareaofcube}{Totalsurfaceareaofacuboid}=?$

Step 2: Using the formula

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

Total surface area of cuboid$=2(lb+bh+hl)$

Step 3: Determining the required ratio

As per the given data, since the cube is cut into two cuboids of same volume.

Thus, dimensions of cuboid are given as,

$l=a;b=\frac{a}{2};h=a$

$$\begin{gathered} \frac{Totalsurfaceareaofcube}{Totalsurfaceareaofacuboid}=\frac{6a^2}{2(a\times {\displaystyle \frac{a}{2}}+{\displaystyle \frac{a}{2}}\times a+a\times a)} \\ =\frac{6a^2}{2({\displaystyle \frac{a^2}{2}}+{\displaystyle \frac{a^2}{2}}+a^2)} \\ =\frac{6a^2}{2\left({\displaystyle \frac{a^2+a^2+2a^2}{2}}\right)} \\ =\frac{6a^2}{4a^2} \\ =\frac{3}{2} \end{gathered}$$

Thus, the required ratio is $3:2$

Hence, option D is correct.