Answer :

Let the edges of metal cubes be a1, a2 and a3


And it is given that ratio of edges is 3:4:5


So, let a1 = 3x, a2 = 4x and a3 = 5x


Volume of cube with edge a1 = v1 = (a1)3


v1 = (3x)3


v1 = 27x3 cm3eqn1


Similarly


Volume of cube with edge a2 = v2 = (a2)3


v2 = (4x)3


v2 = 64x3 cm3eqn2


Volume of cube with edge a3 = v3 = (a3)3


v3 = (5x)3


v3 = 125x3 cm3eqn3


Now let the volume of resulting cube be ‘V’ cm3


So, V = v1 + v2 + v3


V = 27x3 + 64x3 + 125x3 (from eqn1, eqn2 and eqn3)


V = 216x3 cm3eqn4


It is given that the diagonal of the resulting cube is 12√3 cm



Let the edge of resulting cube be ‘a’ cm


Consider ΔBCD, BDC = 90°


BD = CD = a cm (as they are the edges of cube)



BC2 = a2 + a2 (putting value of BD and CD)


BC2 = 2a2


BC = √(2a2)


BC = a√2 cm


Now consider ΔABC, ABC = 90°


Here, AB = a cm and BC = a√2 cm and AC = 12√3 cm



(12√3)2 = a2 + (a√2)2 (putting values of AB, AC and BC)


144 × 3 = a2 + 2a2


144 × 3 = 3a2


144 = a2


a = √144


a = 12 cm


So, volume of the resulting cube = V = a3


V = 123 (putting value of a)


V = 1728 cm3 –eqn5


Equate equation 4 and 5


216x3 = 1728


x3 = 1728/216


x3 = 8


x = 8


x = 2


So a1 = 3x = 3 × 2


a1 = 6 cm


Similarly a2 = 4x = 4 × 2


a2 = 8 cm


Similarly a3 = 5x = 5 × 2


a3 = 10 cm


The edges of three cubes are 6 cm, 8 cm and 10 cm.


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