Question

Two identical cubes each of volume $64{\mathrm{cm}}^3$ are joined together end to end. What is the surface area of the resulting cuboid?

Answer 1

Calculating the surface area of the resulting cuboid:

Cube:

A three-dimensional figure which has six equal square faces is called a cube.

Cuboid:

A three-dimensional figure which has six rectangular faces is called a cuboid.

The side length of the cube is $\mathrm{a}$.

Each faces in the cube is square.

Area of one squared surface$={\mathrm{a}}^2$

Surface area of a cube$=6{\mathrm{a}}^2$

Since the two identical cubes joined together, it forms a cuboid

Calculating the surface area of the resulting cuboid:$$\begin{gathered} \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}=2(\mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{cube})-2(\mathrm{Area}\mathrm{of}\mathrm{one}\mathrm{squared}\mathrm{surface}) \\ \Rightarrow \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}=2\left(6{\mathrm{a}}^2\right)-2\left({\mathrm{a}}^2\right) \\ \Rightarrow \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}=12{\mathrm{a}}^2-2{\mathrm{a}}^2 \\ \Rightarrow \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}=10{\mathrm{a}}^2 \end{gathered}$$

Surface area of the resulting cuboid is $10{\mathrm{a}}^2...\left(1\right)$

The volume of each cube is given as $64{\mathrm{cm}}^3$.

Volume of cube

$\begin{array}{rcl}\mathrm{Volumeof}\mathrm{cube}& =& {\mathrm{a}}^3\\ & \Rightarrow & {\mathrm{a}}^3=64\\ & \Rightarrow & \mathrm{a}=\sqrt[3]{64}\\ & \Rightarrow & \mathrm{a}=\sqrt[3]{4\times 4\times 4}\\ & \Rightarrow & \mathrm{a}=4\end{array}$

Substitute the value of $\mathit{a}$ in $\left(1\right)$:

$$\begin{gathered} \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}=10\left(4^2\right) \\ \Rightarrow \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}=10\times 16 \\ \Rightarrow \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{resulting}\mathrm{cuboid}=160 \end{gathered}$$

Hence, the area of the resulting cuboid is $\mathbf{160}\mathbf{}{\mathbf{cm}}^{\mathbf{2}}$.