Question
Three cubes whose edges measure $3 cm, 4 cm$ and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.
✓
Answer
Three cubes of edges $3 cm, 4 cm$ and 5 cm are melted and molded to form a new cube.
i.e., volume of the new cube $=$ sum of the volumes of the three cubes
$=(3)^3+(4)^3+(5)^3 \\
=27+64+125 \\
=216 cm^3$
We have that volume of a cube $=(\text { side })^3$
$\Rightarrow 226=(\text { side })^3$
$\Rightarrow$ side of the new cube $=\sqrt[3]{216}=6 cm$
$\therefore$ Surface area of the new cube $=6 \times(\text { side })^3=6 \times(6)^2=216 cm^2$
i.e., volume of the new cube $=$ sum of the volumes of the three cubes
$=(3)^3+(4)^3+(5)^3 \\
=27+64+125 \\
=216 cm^3$
We have that volume of a cube $=(\text { side })^3$
$\Rightarrow 226=(\text { side })^3$
$\Rightarrow$ side of the new cube $=\sqrt[3]{216}=6 cm$
$\therefore$ Surface area of the new cube $=6 \times(\text { side })^3=6 \times(6)^2=216 cm^2$
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