Three cubes of a metal whose edges are in the ratio 3 : 4 : 5

Solution:

Let the edge of the metal cubes be $3 x, 4 x$ and $5 x$.

Let the edge of the single cube be $a$.

As,

Diagonal of the single cube $=12 \sqrt{3} \mathrm{~cm}$

$\Rightarrow a \sqrt{3}=12 \sqrt{3}$

$\Rightarrow a=12 \mathrm{~cm}$

Now,

Volume of the single cube $=$ Sum of the volumes of the metallic cubes

$\Rightarrow a^{3}=(3 x)^{3}+(4 x)^{3}+(5 x)^{3}$

$\Rightarrow 12^{3}=27 x^{3}+64 x^{3}+125 x^{3}$

$\Rightarrow 1728=216 x^{3}$

$\Rightarrow x^{3}=\frac{1728}{216}$

$\Rightarrow x^{3}=8$

$\Rightarrow x=\sqrt[3]{8}$

$\Rightarrow x=2$

So, the egdes of the cubes are $3 \times 2=6 \mathrm{~cm}, 4 \times 2=8 \mathrm{~cm}$ and $5 \times 2=10 \mathrm{~cm}$.

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm