Question

Two cubes, each of volume 512 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Answer 1

$$\begin{gathered} \mathrm{Two}\mathrm{cubes}\mathrm{each}\mathrm{of}\mathrm{volume}512{\mathrm{cm}}^3\mathrm{are}\mathrm{joined}\mathrm{end}\mathrm{to}\mathrm{end}. \\ \mathrm{Now},\mathrm{volume}\mathrm{of}\mathrm{a}\mathrm{cube}=(\mathrm{side}{)}^3 \\ \Rightarrow 512=(\mathrm{side}{)}^3 \\ \Rightarrow \mathrm{Side}\mathrm{of}\mathrm{the}\mathrm{cube}=\sqrt[3]{512}=8\mathrm{cm} \\ \mathrm{If}\mathrm{the}\mathrm{cubes}\mathrm{area}\mathrm{joined}\mathrm{side}\mathrm{by}\mathrm{side},\mathrm{then}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}\mathrm{is}2\times 8\mathrm{cm}=16\mathrm{cm}. \\ \mathrm{Breadth}=8\mathrm{cm} \\ \mathrm{Height}=8\mathrm{cm} \\ \therefore \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (16\times 8+8\times 8+16\times 8) \\ =2\times (128+64+128) \\ =640{\mathrm{cm}}^2 \end{gathered}$$