If the apothem is \(\displaystyle 6\sqrt3\:cm\) and the question requires us to solve for the length of one of the sides, the problem can be resolved through the use of right triangles and trig functions. 

As long as one angle and one side length is known for a right triangle, trig functions can be used to solve for a mystery side. In the previous image, the side of interest has been labeled as \(\displaystyle x\). Keep in mind that \(\displaystyle x\) is actually half the length of one side of the regular hexagon. In order to solve for \(\displaystyle x\), the first step is to solve for the measure of an internal angle of the hexagon. This angle has been marked in the image. This can be solved for by using:

\(\displaystyle \frac{180^{\circ}(n-2)}{n}\) where \(\displaystyle n\) is the number of sides of the hexagon. In this problem, \(\displaystyle n = 6\).

\(\displaystyle \frac{180^{\circ}(6-2)}{6}\)

\(\displaystyle \frac{180^{\circ}(4)}{6} = 120^{\circ}\)

This measure of \(\displaystyle 120^{\circ}\) is the measure of the entire angle. Keep in mind that the drawn triangle is actually bisecting the internal angle. This means that the angle of interest in the triangle is actually \(\displaystyle 60^{\circ}\).

Now that we have the apothem and one of the angles, we can use trig functions to solve for \(\displaystyle x\). 

Using SOH CAH TOA, we realize that this problem would require us to use the tan function where the ratio would be \(\displaystyle \frac{opp}{adj}\), or for this problem's data,\(\displaystyle \frac{6\sqrt3\:cm}{x}\).

\(\displaystyle tan(60^\circ)=\frac{6\sqrt{3}\:cm}{x}\)

\(\displaystyle x \cdot tan(60^\circ)= 6\sqrt{3}\:cm\)

\(\displaystyle x= \frac{6\sqrt{3}\:cm}{tan(60^\circ)}\)

\(\displaystyle x=6\:cm\)

Now \(\displaystyle 6\:cm\) must be multiplied by \(\displaystyle 2\) because it's the length of half of the total length of one side of the hexagon. Therefore, the final answer is \(\displaystyle 12\:cm\).

Write the perimeter formula for hexagons.

\(\displaystyle P=6s\)

Substitute the perimeter in the formula and solve for the side length.

\(\displaystyle 24a+15b=6s\)

\(\displaystyle s=4a+\frac{15}{6}b=4a+\frac{5}{2}b\)

Write the formula for finding the perimeter of a hexagon.

\(\displaystyle P=6s\)

Substitute the perimeter and solve for the side length.

\(\displaystyle 15=6s\)

\(\displaystyle s=\frac{15}{6}=\frac{5}{2}\)

Write the perimeter formula for a hexagon.

\(\displaystyle P=6s\)

Substitute the perimeter into the formula and solve for the side length.

\(\displaystyle 3+6\sqrt3=6s\)

\(\displaystyle s=\frac{3+6\sqrt3}{6}= \frac{1+2\sqrt3}{2}\)

Write the formula for the perimeter of a hexagon.

\(\displaystyle P=6s\)

Substitute the perimeter and solve for the side.

\(\displaystyle 15a+16b=6s\)

\(\displaystyle s=\frac{15}{6}a+\frac{16}{6}b= \frac{5}{2}a+\frac{8}{3}b\)

This is a regular hexagon; therefore, when we draw in the diagonals, we will create \(\displaystyle 6\) identical equilateral triangles as shown by the figure:

Now, the given line segment also serves as the height to the newly created equilateral triangles. This height also splits an equilateral triangle into two congruent \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles.

Recall that \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles have side lengths that are in the ratio of \(\displaystyle 1:\sqrt3:2\).

Substitute in the given height into the ratio in order to find the length of the base of the \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangle.

\(\displaystyle \frac{\text{height}}{\sqrt3}=\frac{\text{base}}{1}\)

Substitute.

\(\displaystyle \frac{12}{\sqrt3}=\frac{\text{base}}{1}\)

Solve.

\(\displaystyle \text{base}=\frac{12\sqrt3}{3}=4\sqrt3\)

The length of the side of the hexagon is twice the length of the base.

\(\displaystyle \text{side length}=2(\text{base})\)

Substitute in the value of the length of the base to find the side length of the hexagon.

\(\displaystyle \text{side length}=2(4\sqrt3)\)

Solve.

\(\displaystyle \text{side length}=8\sqrt3\)

This is a regular hexagon; therefore, when we draw in the diagonals, we will create \(\displaystyle 6\) identical equilateral triangles as shown by the figure:

Now, the given line segment also serves as the height to the newly created equilateral triangles. This height also splits an equilateral triangle into two congruent \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles.

Recall that \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles have side lengths that are in the ratio of \(\displaystyle 1:\sqrt3:2\).

Substitute in the given height into the ratio in order to find the length of the base of the \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangle.

\(\displaystyle \frac{\text{height}}{\sqrt3}=\frac{\text{base}}{1}\)

Substitute.

\(\displaystyle \frac{16}{\sqrt3}=\frac{\text{base}}{1}\)

Solve.

\(\displaystyle \text{base}=\frac{16\sqrt3}{3}\)

The length of the side of the hexagon is twice the length of the base.

\(\displaystyle \text{side length}=2(\text{base})\)

Substitute in the value of the length of the base to find the side length of the hexagon.

\(\displaystyle \text{side length}=2\left(\frac{16\sqrt3}{3}\right)\)

Solve.

\(\displaystyle \text{side length}=\frac{32\sqrt3}{3}\)

This is a regular hexagon; therefore, when we draw in the diagonals, we will create \(\displaystyle 6\) identical equilateral triangles as shown by the figure:

Now, the given line segment also serves as the height to the newly created equilateral triangles. This height also splits an equilateral triangle into two congruent \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles.

Recall that \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles have side lengths that are in the ratio of \(\displaystyle 1:\sqrt3:2\).

Substitute in the given height into the ratio in order to find the length of the base of the \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangle.

\(\displaystyle \frac{\text{height}}{\sqrt3}=\frac{\text{base}}{1}\)

Substitute.

\(\displaystyle \frac{22}{\sqrt3}=\frac{\text{base}}{1}\)

Solve.

\(\displaystyle \text{base}=\frac{22\sqrt3}{3}\)

The length of the side of the hexagon is twice the length of the base.

\(\displaystyle \text{side length}=2(\text{base})\)

Substitute in the value of the length of the base to find the side length of the hexagon.

\(\displaystyle \text{side length}=2\left(\frac{22\sqrt3}{3}\right)\)

Solve.

\(\displaystyle \text{side length}=\frac{44\sqrt3}{3}\)

This is a regular hexagon; therefore, when we draw in the diagonals, we will create \(\displaystyle 6\) identical equilateral triangles as shown by the figure:

Now, the given line segment also serves as the height to the newly created equilateral triangles. This height also splits an equilateral triangle into two congruent \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles.

Recall that \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles have side lengths that are in the ratio of \(\displaystyle 1:\sqrt3:2\).

Substitute in the given height into the ratio in order to find the length of the base of the \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangle.

\(\displaystyle \frac{\text{height}}{\sqrt3}=\frac{\text{base}}{1}\)

Substitute.

\(\displaystyle \frac{14}{\sqrt3}=\frac{\text{base}}{1}\)

Solve.

\(\displaystyle \text{base}=\frac{14\sqrt3}{3}\)

The length of the side of the hexagon is twice the length of the base.

\(\displaystyle \text{side length}=2(\text{base})\)

Substitute in the value of the length of the base to find the side length of the hexagon.

\(\displaystyle \text{side length}=2\left(\frac{14\sqrt3}{3}\right)\)

Solve.

\(\displaystyle \text{side length}=\frac{28\sqrt3}{3}\)

This is a regular hexagon; therefore, when we draw in the diagonals, we will create \(\displaystyle 6\) identical equilateral triangles as shown by the figure:

Now, the given line segment also serves as the height to the newly created equilateral triangles. This height also splits an equilateral triangle into two congruent \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles.

Recall that \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangles have side lengths that are in the ratio of \(\displaystyle 1:\sqrt3:2\).

Substitute in the given height into the ratio in order to find the length of the base of the \(\displaystyle 30^\circ-60^\circ-90^\circ\) triangle.

\(\displaystyle \frac{\text{height}}{\sqrt3}=\frac{\text{base}}{1}\)

Substitute.

\(\displaystyle \frac{18}{\sqrt3}=\frac{\text{base}}{1}\)

Solve.

\(\displaystyle \text{base}=\frac{18\sqrt3}{3}=6\sqrt3\)

The length of the side of the hexagon is twice the length of the base.

\(\displaystyle \text{side length}=2(\text{base})\)

Substitute in the value of the length of the base to find the side length of the hexagon.

\(\displaystyle \text{side length}=2(6\sqrt3)\)

Solve.

\(\displaystyle \text{side length}=12\sqrt3\)