Question

A cube of side 4 cm is cut into 1 cm cubes. What is the ratio of the surface areas of the original cube and cut-out cubes ?

Answer 1

Volume of the original cube having side of length $4cm=(4{)}^3=64c{m}^3$
[$\because$ volume of cube with side $a=a^3$]
Volume of the cut-out cubes with side of length $1cm=1c{m}^3$
$\therefore$ Number of cut-out cubes $=\frac{Volumeoftheoriginalcube}{Volumeofasmallercube}=\frac{64}{1}=64$
Now, surface area of cut-out cubes $=64\times 6\times (1{)}^{2}c{m}^2$
[$\therefore$ surface area of cube with side $a=6a^2$]
and surface area of the original cube $=6\times 4^{2}c{m}^2$
$\therefore$ The required ratio of surface areas of the original cube and cut-out cubes $=\frac{6\times 4^2}{64\times 6\times 1^2}=1:4$