Question

A metal cube of edge $\mathbf{12}\mathbf{}\mathit{c}\mathit{m}$ is melted and formed into three smaller cubes. If the edges of the two smaller cubes are $\mathbf{6}\mathbf{}\mathit{c}\mathit{m}$ and $\mathbf{8}\mathbf{}\mathit{c}\mathit{m}$, find the edge of the third smaller cube.

Answer 1

Step 1: Find the edge of the third cube.

Given,

The length $\left(l\right)$ of edge of a metal cube $l=12cm$.

The length of edges of the two smaller cubes are $l_1=6cm$ and $l_2=8cm$.

Let us consider the length of edge of the third smaller cube be $l_3=xcm$.

Volume of large cube $V=V_1+V_2+V_3$ $\left(\because VolumeofcubeV=l^3\right)$

$l^3=l_1^3+l_2^3+l_3^3$

${12}^3=6^3+8^3+x^3$

$1728=216+512+x^3$

$1728=728+x^3$

$x^3=1728-728$

$x^3=1000$

$x^3={10}^3$ $\left(\because {10}^3=1000\right)$

$\therefore x=10cm$

Hence, the edge of the third smaller cube are $10cm$