Question.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Side (a) of cube = 12 cm
Volume of cube $=(a)^{3}=(12 \mathrm{~cm})^{3}=1728 \mathrm{~cm}^{3}$
Let the side of the smaller cube be $a_{1}$.
Volume of 1 smaller cube $=\left(\frac{1728}{8}\right) \mathrm{cm}^{3}=216 \mathrm{~cm}^{3}$
$\left(a_{1}\right)^{3}=216 \mathrm{~cm}^{3}$
$\Rightarrow a_{1}=6 \mathrm{~cm}$
Therefore, the side of the smaller cubes will be 6 cm.
Ratio between surface areas of cubes $=\frac{\text { Surface area of bigger cube }}{\text { Surface area of smaller cube }}$
$=\frac{6 a^{2}}{6 a_{1}^{2}}=\frac{(12)^{2}}{(6)^{2}}$
$=\frac{4}{1}$
Therefore, the ratio between the surface areas of these cubes is 4:1.
Side (a) of cube = 12 cm
Volume of cube $=(a)^{3}=(12 \mathrm{~cm})^{3}=1728 \mathrm{~cm}^{3}$
Let the side of the smaller cube be $a_{1}$.
Volume of 1 smaller cube $=\left(\frac{1728}{8}\right) \mathrm{cm}^{3}=216 \mathrm{~cm}^{3}$
$\left(a_{1}\right)^{3}=216 \mathrm{~cm}^{3}$
$\Rightarrow a_{1}=6 \mathrm{~cm}$
Therefore, the side of the smaller cubes will be 6 cm.
Ratio between surface areas of cubes $=\frac{\text { Surface area of bigger cube }}{\text { Surface area of smaller cube }}$
$=\frac{6 a^{2}}{6 a_{1}^{2}}=\frac{(12)^{2}}{(6)^{2}}$
$=\frac{4}{1}$
Therefore, the ratio between the surface areas of these cubes is 4:1.