Chapter 18, Problem 49A

Consider eight one-cubic-centimeter sugar cubes stacked two-by-two to form a single bigger cube. What will be the volume of the combined cube? How does its surface area compare to the total surface area of the eight separate cubes?

To determine

Volume of the combined cube and total surface area in compare to separate cube.

Answer to Problem 49A

Volume of the combined cube is $V_2=8{\text{m}}^3$ and the total surface area is half of the eight separate cubes.

Explanation of Solution

Given:

Number of small cubes, $n=8$

Volume of a small cube, $V_1=1{\text{cm}}^3$

Formula Used:

The volume of cube is, $V=a^3$

Where, $a$ is side of cube.

The total surface area of cube is, $S=6a^2$

Where, $a$ is side of cube.

Calculation:

As volume of a small cube, $V_1=1{\text{cm}}^3$

So, side of small cube is

  $a_1=\sqrt[3]{1}=1\text{cm}$

Let the volume of the combined cube is $V_2$ .

The volume of the combined cube will be equal to the volume of the all small cubes.

So,

  $\begin{array}{l}{V}_2=n{V}_1=8\times 1{\text{cm}}^3\\ \Rightarrow V_2=8{\text{cm}}^3\end{array}$

And side of combined cube is

  $a_2=\sqrt[3]{8}=2\text{cm}$

Now,

The total surface area of small cube is,

  $\begin{array}{l}{S}_1=nS=n\left(6a_1{}^2\right)=8\times 6{\left(1\text{cm}\right)}^2\\ \Rightarrow S_1=48{\text{cm}}^2\end{array}$

While the total surface area of the combined cube is,

  $\begin{array}{l}{S}_2=6a_2{}^2=6\times {\left(2\text{cm}\right)}^2\\ \Rightarrow S_2=24{\text{cm}}^2\end{array}$

So, $S_2=\frac{1}{2}{S}_1$

Conclusion:

Therefore, the volume of the combined cube is $V_2=8{\text{m}}^3$ and the total surface area is half of the eight separate cubes.