Three metallic cubes of side $3 c m, 4 c m$ and $5 c m$ respectively are melted and are recast into a single cube. Find the total surface area of the new cube.

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Solution

Volume of new cube $=$ Sum of volume of three cubes

$= 3^{2} + 4^{2} + 5^{2}$

$= 216 c m^{3}$

i.e., $(e d g e)^{3} = 216 c m^{3}$

So, Edge of the single cube $= 6 c m$

And, Surface area of the cube $= 6 (e d g e)^{2} = 6 \times 6^{2} = 216 c m^{2}$

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Three metallic cubes of side 3 cm,4 cm and 5 cm respectively are melted and are recast into a single cube. Find the total surface area of the new cube.

Volume of new cube = Sum of volume of three cubes =32+42+52 =216 cm3 i.e., (edge)3=216 cm3 So, Edge of the single cube =6 cm And, Surface area of the cube =6(edge)2=6×62=216 cm2

Three metallic cubes of side $3 c m, 4 c m$ and $5 c m$ respectively are melted and are recast into a single cube. Find the total surface area of the new cube.

Volume of new cube $=$ Sum of volume of three cubes

$= 3^{2} + 4^{2} + 5^{2}$

$= 216 c m^{3}$

i.e., $(e d g e)^{3} = 216 c m^{3}$

So, Edge of the single cube $= 6 c m$

And, Surface area of the cube $= 6 (e d g e)^{2} = 6 \times 6^{2} = 216 c m^{2}$