Question

Three cubes with sides in the ratio $3:4:5$ are melted to form a single cube whose diagonal is $12\sqrt{3}cm$. Identify the sides of the cubes

• A. $3cm,4cm,5cm$
• B. $6cm,8cm,10cm$
• C. $9cm,12cm,15cm$
• D. None of these

Answer 1

Answer: (B)

$6cm,8cm,10cm$

Sides of $3$ cubes are in the ratio $3:4:5$
Let side of Ist cube$=3x$
Side of IInd cube$=4x$
Side of IIIrd cube$=5x$
Diagonal of another cube$=12\sqrt{3}㎝$
Let side of another cube$=y$
$\therefore \sqrt{y^2+y^2+y^2}=12\sqrt{3}$
$\sqrt{3}y=12\sqrt{3}$
$y=12㎝$
Now, three cubes of sides $3x,4x$ & $5x$ are melted to form a single cube of side $12㎝$
$\therefore$ Volume of 3 cubes=Volume of bigger cube
${\left(3x\right)}^3+{\left(4x\right)}^3+{\left(5x\right)}^3={\left(12\right)}^3$
$27x^3+64x^3+125x^3=1728$
$216x^3=1728$
$x^3=\frac{1728}{216}$
$x={\left[\frac{1728}{216}\right]}^{1/3}$
$x=\frac{12}{6}=2$
$\therefore$ Sides of the cubes are $3\times 2cm,4\times 2cm\&5\times 2㎝$
$=6㎝,8㎝10㎝$