Three cubes of metal whose edges are in the ratio 3:4:5 are melted down in to a single cube whose diagonal is `12(sqrt(3))` cm. Find the edges of the three cubes.

Answer

84 `pi cm^(2)`,12 `pi m^(3)`

Explanation

The ratio in the edges =3:4:5
Let edges be 3x,4xand 5x respectively.
Volumes of three cubes will be `27x^(3),64x^(3)` and `125x^(3)`in `cm^(3)` respectively.
Now sum of volumes of these three cubes `=27x^(2),64x^(3)`and `125 x^(3)`=`216x^(3)cm^(3)`
Let edge of new cube be a cm.
Diagonal of new cube=`asqrt(3)cm`
`A(sqrt(3))=12(sqrt(12))`
volume of new cube =`(12^(3))=1728 cm^(3)`
Now by the given condition
`216x^(3)=1728`
`x^(3)=8`
x=2
Edge of I cube =`3xx2=6cm`
Edge of II =`4xx2=8 cm`
Edge of III cube =5`xx`2=10 cm