Question

Three cubes of a metal whose edges are in the ratios 3 : 4 : 5 are melted and converted into a single cube whose diagonal is $12\sqrt{3}$ cm . Find the edges of the three cubes.

Answer 1

The three cubes of metal are in the ratio 3 : 4 : 5.
Let the edges of the cubes be 3x, 4x and 5x.
Volume of the three cubes will be
$$\begin{gathered} V_1={\left(3x\right)}^3 \\ {V}_2={\left(4x\right)}^3 \\ {V}_3={\left(5x\right)}^3 \end{gathered}$$
Diagonal of the single cube = $12\sqrt{3}\mathrm{cm}$
We know diagonal of the cube = $a\sqrt{3}=12\sqrt{3}$
Hence, the side of the cube = 12 cm
Volume of the bigger cube $V_b={\left(12\right)}^3$
Volume of the three cubes = Volume of the single
$$\begin{gathered} {\left(3x\right)}^3+{\left(4x\right)}^3+{\left(5x\right)}^3={\left(12\right)}^3 \\ \Rightarrow 27x^3+64x^3+125x^3=1728 \\ \Rightarrow 216x^3=1728 \\ \Rightarrow x^3=\frac{1728}{216}=8 \\ \Rightarrow x=2 \end{gathered}$$
Hence, the edges of the three cubes will be $3\times \left(2\right),4\times \left(2\right),5\times \left(2\right)=6,8,10$ cm