Question

Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a single cube whose diagonal is 12 root ( 3 ) cm. Find the edges of the three cubes .

Answer 1

Ratio of the lengths of the edges of the cubes = 3:4:5
Let the edges of the cubes be 3x, 4x and 5x
Volumes of the cubes = (3x)3 cu units, (4x)3 cu units, (5x)3 cu units
                                = 27x3 cu units, 64x3 cu units, 125x3 cu units
Total volume = (27x3 +  64x3 + 125x3)
                   = 216 x3 cu units
Diagonal of the new cube formed = 15√3 
Let the edge of the new cube formed = 's'  units
Diagonal = s√3 
⇒ s√3 = 15√3
⇒ s = 15 units
Volume of the new cube formed = (15)3 cu units = 3375 cu units
⇒ 216 x3 = 3375
⇒ x3 = (3375 / 216)
⇒ x = 15/6 = 5/2 = 2.5
Therefore, the edges of the cubes are (3 x 2.5) i.e. 7.5 units, (4 x 2.5) i.e. 1 units, (5 x 2.5) i.e. 1.25 units.
                   

Answer 2

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Let the edges of three cubes (in cm) be 3x, 4x and 5x respectively. Let a be the side of the new cube so formed after melting.

Volume of new cube = Sum of the volumes of three smaller cubes

$\sf{{a}^{3} = {(3x)}^{3} + {(4x)}^{3} + {(5x)}^{3} = {216x}^{3} = {(6x)}^{3} }$

$\therefore$ $\sf{a = 6x}$

Diagonal of the new cube = 12√3cm

$\longrightarrow$$\sf{ \sqrt{ {a}^{2} + {a}^{2} + {a}^{2} } = 12 \sqrt{3}}$

$\longrightarrow$$\sf{a \sqrt{3} = 12 \sqrt{3}}$

$\longrightarrow$$\sf{a = 12cm}$

$\longrightarrow$$\sf{6x = 12cm}$

$\therefore$ $\sf{x = 2cm}$

Hence, the ages of the three cubes are 6cm, 8cm and 10cm respectively.