Question

Three cubes of metal whose edges are in the ratio 3 : 4 : 5, are melted down into a single cube whose diagonals is $12\sqrt{3}cm.$ Find the edges of three cubes.

Answer 1

$\text{Ratio in the sides of three cubes}=3:4:5$

$\text{Let side of first cube}=3x$

$\text{and side of second cube}=4x$

$\text{and side of third cube}=5x$

$\therefore \text{Sum of volume of three cubes}$

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

$=27x^3+64x^3+125x^3=216x^3$

$\therefore \text{Volume of the cube formed by melting these three cubes}=216x^3$

$\therefore Side=\sqrt[3]{3216x^3}=6x$

$\text{Now diagonal}=\sqrt{3}\times side=\sqrt{3}\times 6x$

$=6\sqrt{3}x$

$\therefore 12\sqrt{3}=6\sqrt{3}x$

$\Rightarrow x=\frac{12\sqrt{3}}{6\sqrt{3}}=2cm$

$\therefore \text{Side of first cube}=3x=3\times 2=6cm$

$\text{Side of second cube}=4x=4\times 2=8cm$

$\text{and side of third cube}=5x=5\times 2=10cm$