You can redraw the figure given to notice the little equilateral triangle that is formed within the hexagon. Since a hexagon can have the \(\displaystyle 360\) degrees of its internal rotation divided up evenly, the central angle is \(\displaystyle 60\) degrees. The two angles formed with the sides also are \(\displaystyle 60\) degrees. Thus, you could draw:

Now, the \(\displaystyle 4\sqrt{3}\) is located on the side that is the same as \(\displaystyle \sqrt{3}\) on your standard \(\displaystyle 30-60-90\) triangle. The base of the little triangle formed here is \(\displaystyle 1\) on the standard triangle. Let's call our unknown value \(\displaystyle x\).

We know, then, that:

\(\displaystyle \frac{4\sqrt{3}}{\sqrt{3}}=\frac{x}{1}\)

Or, \(\displaystyle x=4\)

Now, this is only half of the size of the hexagon's side. Therefore, the full side length is \(\displaystyle 4*2=8\).

Since this is a regular hexagon, all of the sides are of equal length.  This means that your total perimeter is \(\displaystyle 8*6\) or \(\displaystyle 48\).

The area of a regular hexagon is defined by the equation:

\(\displaystyle \frac{3\sqrt{3}}{2} * s^2\), where \(\displaystyle s\) is the length of a side.

This is derived from the fact that the regular hexagon can be split up into \(\displaystyle 6\) little equilateral triangles, each having an area of 

\(\displaystyle \frac{\sqrt{3}}{4}*s^2\)

To visualize this, consider the drawing:

Each triangle formed like this will be equilateral.  It is easiest to remember this relationship and memorize the general area equation for equilateral triangles. (It is useful in many venues!)

So, for your data, you know:

\(\displaystyle 315.375\sqrt{3}=\frac{3\sqrt{3}}{2} * s^2\)

Solving for \(\displaystyle s\), you get:

\(\displaystyle s^2 = 210.25\)

This means that \(\displaystyle s=14.5\)

Therefore, the perimeter of the figure is equal to \(\displaystyle 6*s\) or \(\displaystyle 87\:units\).

There are 6 sides in a hexagon.  

Therefore, given a side length of 16, the perimeter is:

\(\displaystyle P=6s=6 \cdot 16=96\)

Write the formula for the perimeter of a hexagon.  

\(\displaystyle P=6s\)

Substitute the given length.

\(\displaystyle P=6s = 6\left(\frac{x}{3}\right)=2x\)

To begin, calculate the side length of the hexagon. Since it is regular, its sides are of equal length. This means that a given side is \(\displaystyle \frac{105}{6}\) or \(\displaystyle 17.5\) in length. Now, consider your figure like this: 

The little triangle at the top forms an equilateral triangle. This means that all of its sides are \(\displaystyle 17.5\). You could form six of these triangles in your figure if you desired. This means that the long diagonal is really just \(\displaystyle 2*17.5\) or \(\displaystyle 35\).

You could redraw your figure as follows.  Notice that this kind of figure makes an equilateral triangle within the hexagon.  This allows you to create a useful \(\displaystyle 30-60-90\) triangle.

The \(\displaystyle 7\) in the figure corresponds to \(\displaystyle \sqrt{3}\) in a reference \(\displaystyle 30-60-90\) triangle. The hypotenuse is \(\displaystyle 2\) in the reference triangle. 

Therefore, we can say:

\(\displaystyle \frac{7}{\sqrt{3}}=\frac{h}{2}\)

Solve for \(\displaystyle h\):

\(\displaystyle h=\frac{14}{\sqrt{3}}\)

Rationalize the denominator:

\(\displaystyle h=\frac{14}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{14\sqrt{3}}{3}\)

Now, the diagonal of a regular hexagon is actually just double the length of this hypotenuse. (You could draw another equilateral triangle on the bottom and duplicate this same calculation set—if you wanted to spend extra time without need!) Thus, the length of the diagonal is:

\(\displaystyle \frac{28\sqrt{3}}{3}\)

Use the formula for perimeter to solve for the length of a side of the regular hexagon:

\(\displaystyle P=6s\)

Where \(\displaystyle P\) is perimeter and \(\displaystyle s\) is the length of a side.

In this case:

\(\displaystyle 112.8=6s\)

\(\displaystyle s=\frac{112.8}{6}=18.8\)

Use the formula for perimeter to solve for the side length:

\(\displaystyle P=6s\)

\(\displaystyle 85.8=6s\)

\(\displaystyle s=14.3\)

Use the formula for perimeter to solve for the side length:

\(\displaystyle P=6s\)

\(\displaystyle 114=6s\)

\(\displaystyle s=19\)

Use the formula for perimeter to solve for the side length:

\(\displaystyle P=6s\)

\(\displaystyle 54=6s\)

\(\displaystyle s=9\)