Question

The cores of three cubes are $8cm,6cm$ and $1cm$ respectively. Having melted these cubes a new cube is recastesd. Find the total surface area of the new cube recanted.

Answer 1

Volume of the cube with core of $8cm=(core{)}^3$
$=8^3=512{cm}^3$
Volume of the cube with core $6cm=(core{)}^3$
$=(6{)}^3=216{cm}^3$
Volume of the cube with core $1cm=(\text{core}{)}^3$
$=(1{)}^3=1{cm}^3$
The total volume of three cubes $512+216+1=729{cm}^3$
Having melted these cubes, a new cube is recasted
$\therefore$ The volume of the cube recasted $=729{cm}^3$
$\Rightarrow (\text{core}{)}^3=729$
$\Rightarrow$ core $=\sqrt[3]{3729}$
$={\left(9^3\right)}^{1/3}=9cm$
Total surface area of cuboid recasted $=6(\text{core}{)}^2$
$6\times 9\times 9=486{cm}^2$
Hence the surface area of the new cube $=486{cm}^2$