Question

A diagonal of a cube joins two vertices not in the same face. If the diagonals are$4\sqrt{3}$cm long, what is the volume?

Short answer

The volume is $\frac{8\sqrt{3}}{9}{\text{cm}}^3$when the diagonals are$4\sqrt{3}\text{\hspace{0.17em}cm}$ .

Step-by-step solution

Step 1. Given information.

A diagonal of a cube joins two vertices not in the same face and the diagonals are$4\sqrt{3}\text{cm}$ long.

Step 2. Write the concept.

Since a diagonal of a cube joins two vertices not in the same face.

Then, the factor can be found as:

$\begin{array}{c}k=\frac{4\sqrt{3}}{2}\\ =2\sqrt{3}\text{cm}\end{array}$

Step 3. Determine the volume.

Now, the ratio of the larger cube and smaller cube is cube of k factor,

$\begin{array}{c}\frac{v_l}{v_s}=k^3\\ \frac{{\left(4\sqrt{3}\right)}^2}{v_s}={\left(2\sqrt{3}\right)}^3\text{[simplify]}\\ \frac{64}{v_s}=24\sqrt{3}\text{[multiply]}\end{array}$

Simplify the last expression,

$\begin{array}{c}{v}_s=\frac{64}{24\sqrt{3}}\text{[crossmultiplication]}\\ =\frac{8\sqrt{3}}{9}\text{[simplify]}\end{array}$

Thus, the volume is $\frac{8\sqrt{3}}{9}{\text{\hspace{0.17em}\hspace{0.17em}cm}}^3$when the diagonals are $4\sqrt{3}\text{\hspace{0.17em}\hspace{0.17em}cm}$.