 # Three equal cubes are placed adjacently in a row.

Question:

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.

Solution:

Suppose that the side of the cube $=\mathrm{x} \mathrm{cm}$

Surface area of the cube $=6 \times(\text { side })^{2}=6 \times \mathrm{x}^{2}=6 \mathrm{x}^{2} \mathrm{~cm}^{2}$

i.e., the sum of the surface areas of three such cubes $=6 \mathrm{x}^{2}+6 \mathrm{x}^{2}+6 \mathrm{x}^{2}=18 \mathrm{x}^{2} \mathrm{~cm}^{2}$

Now, these three cubes area placed together to form a cuboid.

Then the length of the new cuboid will be 3 times the edge of the cube $=3 \times \mathrm{x}=3 \mathrm{x} \mathrm{cm}$

Breadth of the cuboid $=\mathrm{x} \mathrm{cm}$

Height of the cuboid $=\mathrm{x} \mathrm{cm}$

$\therefore$ Total surface area of the cuboid $=2 \times($ length $\times$ breadth $+$ breadth $\times$ height $+$ length $\times$ height $)$

$=2 \times(3 \mathrm{x} \times \mathrm{x}+\mathrm{x} \times \mathrm{x}+3 \mathrm{x} \times \mathrm{x})$

$=2 \times\left(3 \mathrm{x}^{2}+\mathrm{x}^{2}+3 \mathrm{x}^{2}\right)$

$=2 \times\left(7 \mathrm{x}^{2}\right)$

$=14 \mathrm{x}^{2} \mathrm{~cm}$

i.e., the ratio of the total surface area cuboid to the sum of the surface areas of the three cubes $=14 x^{2} \mathrm{~cm}^{2}: 18 x^{2} \mathrm{~cm}^{2}$

$=7: 9$

Hence, the ratio is $7: 9$.