A regular hexagon is inscribed in a circle. If the length of the shortest diagonal of the hexagon is 12 cm, find one-sixth of the area (in cm²) of the region lying between the circle and the hexagon.

The correct option is:
$4(2\pi -3\sqrt{3})$

Step-by-Step Explanation:

Let the regular hexagon be inscribed in a circle of radius r. Let the side length of the regular hexagon be a.
For a regular hexagon inscribed in a circle, the side length of the hexagon is equal to the radius of the circle, so:
$a=r$

A regular hexagon has two types of diagonals:
1. Shortest diagonals, which connect vertices that are separated by one vertex (e.g., from vertex 1 to vertex 3).
2. Longest diagonals, which connect opposite vertices (e.g., from vertex 1 to vertex 4), having a length of 2r.

The length of the shortest diagonal of a regular hexagon with side length a is given by the formula:
$d=a\sqrt{3}$
We are given that the length of the shortest diagonal is 12 cm. Therefore:
$a\sqrt{3}=12$

Solving for the side length a (and thus the radius r):
$a=\frac{12}{\sqrt{3}}=4\sqrt{3}\text{cm}$
So, the radius of the circle is:
$r=4\sqrt{3}\text{cm}$

Now, let's find the area of the circle and the area of the regular hexagon.
The area of the circle is:
${\mathrm{Area}}_{\text{circle}}=\pi r^2=\pi {\left(4\sqrt{3}\right)}^2=48\pi {\text{cm}}^2$

The area of a regular hexagon with side length a is composed of 6 equilateral triangles of side length a:
${\mathrm{Area}}_{\text{hexagon}}=6\times \left(\frac{\sqrt{3}}{4}{a}^2\right)$
Substituting $a^2=48$:
${\mathrm{Area}}_{\text{hexagon}}=6\times \frac{\sqrt{3}}{4}$

The total area of the region lying between the circle and the hexagon is the difference between the area of the circle and the area of the hexagon:
${\mathrm{Area}}_{\text{region}}={\mathrm{Area}}_{\text{circle}}-{\mathrm{Area}}_{\text{hexagon}}=48\pi -72\sqrt{3}$

The question asks for one-sixth of this area:
$\frac{1}{6}{\mathrm{Area}}_{\text{region}}=\frac{1}{6}(48\pi -72\sqrt{3})$
Factoring out terms:
$\frac{1}{6}\times 24(2\pi -3\sqrt{3})=4(2\pi -3\sqrt{3})$