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 1

$$\begin{gathered} \mathrm{Three}\mathrm{cubes}\mathrm{of}\mathrm{edges}3\mathrm{cm},4\mathrm{cm}\mathrm{and}5\mathrm{cm}\mathrm{are}\mathrm{melted}\mathrm{and}\mathrm{molded}\mathrm{to}\mathrm{form}\mathrm{a}\mathrm{new}\mathrm{cube}. \\ \mathrm{i}.\mathrm{e}.,\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{cube}=\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{volumes}\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{cubes} \\ =(3{)}^3+(4{)}^3+(5{)}^3 \\ =27+64+125 \\ =216{\mathrm{cm}}^3 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{volume}\mathrm{of}\mathrm{a}\mathrm{cube}=(\mathrm{side}{)}^3 \\ \Rightarrow 216=(\mathrm{side}{)}^3 \\ \Rightarrow \mathrm{Side}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{cube}=\sqrt[3]{216}=6\mathrm{cm} \\ \therefore \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (6{)}^2=216{\mathrm{cm}}^2 \end{gathered}$$