Q. 44.3( 6 Votes )

# Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. Find the edge of the new cube formed.

Answer :

Given,

Edge of the first cube = 6 cm

Volume of the first cube = a^{3} = (6)^{3} cm

Edge of the second cube = 8 cm

Volume of the second cube = a^{3} = (8)^{3} cm

Edge of the third cube = 10 cm

Volume of the third cube = a^{3} = (10)^{3} cm

So,

Volume of the formed cube = Volume of the First + Second + Third Cube

Volume of the formed cube = v1 + v2 + v3

= 6^{3} + 8^{3} + 10^{3}

= 216 + 512 + 1000

= 1728

Now volume of new cube = a^{3}= 1728

Edge of new cube = a = ∛1728

a = 12

Therefore surface area of new cube = 6a^{2}

= 6 × 12^{2}

= 6 × 12 × 12

= 864 cm^{2}

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