Answer :

Let the edges of metal cubes be a_{1}, a_{2} and a_{3}

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

So, let a_{1} = 3x, a_{2} = 4x and a_{3} = 5x

Volume of cube with edge a_{1} = v_{1} = (a_{1})^{3}

⇒v_{1} = (3x)^{3}

∴ v_{1} = 27x^{3} cm^{3}→eqn1

Similarly

Volume of cube with edge a_{2} = v_{2} = (a_{2})^{3}

⇒v_{2} = (4x)^{3}

∴ v_{2} = 64x^{3} cm^{3}→eqn2

Volume of cube with edge a_{3} = v_{3} = (a_{3})^{3}

⇒v_{3} = (5x)^{3}

∴ v_{3} = 125x^{3} cm^{3}→eqn3

Now let the volume of resulting cube be ‘V’ cm^{3}

So, V = v_{1} + v_{2} + v_{3}

⇒ V = 27x^{3} + 64x^{3} + 125x^{3} (from eqn1, eqn2 and eqn3)

∴ V = 216x^{3} cm^{3}→eqn4

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)

⇒ BC^{2} = a^{2} + a^{2} (putting value of BD and CD)

⇒BC^{2} = 2a^{2}

⇒BC = √(2a^{2})

∴ 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} = a^{2} + (a√2)^{2} (putting values of AB, AC and BC)

⇒ 144 × 3 = a^{2} + 2a^{2}

⇒ 144 × 3 = 3a^{2}

⇒ 144 = a^{2}

⇒ a = √144

∴ a = 12 cm

So, volume of the resulting cube = V = a^{3}

⇒ V = 12^{3} (putting value of a)

∴ V = 1728 cm^{3} –eqn5

Equate equation 4 and 5

⇒ 216x^{3} = 1728

⇒ x^{3} = 1728/216

⇒ x^{3} = 8

⇒ x = ∛8

∴ x = 2

So a_{1} = 3x = 3 × 2

⇒ a_{1} = 6 cm

Similarly a_{2} = 4x = 4 × 2

∴ a_{2} = 8 cm

Similarly a_{3} = 5x = 5 × 2

∴ a_{3} = 10 cm

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

Rate this question :