Question

Three Equal Cubes Are Placed Adjacently In A Row Find The Ratio Of The Total Surface Area Of The Resulting Cuboid To That Of The Sum Of The Surface Areas Of Three Cubes.

Answer 1

Let

The side of the cube be $a$.

The height of the cuboid is $a$.

The length of the cuboid is $3a$.

Three cubes placed in row then breadth of cuboid be a

The Sum Of The Surface Areas Of Three Cubes is $6a^2+6a^2+6a^2=18a^2$
The total surface area of the resulting cuboid is $2\left(lb+bh+hl\right)$
$$\begin{gathered} 2\left(lb+bh+hl\right)=\left(3a\times a+a\times a+a\times 3a\right) \\ =2\left(7a^2\right)=14a^2 \end{gathered}$$
So the ratio of The Total Surface Area Of The Resulting Cuboid To That Of The Sum Of The Surface Areas Of Three Cubes is:
$=\frac{14a^2}{18a^2}$
$=\frac{7}{9}$
$=7:9$
The Ratio Of The Total Surface Area Of The Resulting Cuboid To That Of The Sum Of The Surface Areas Of Three Cubes is $7:9$.