Find the area of the regular polygon described given the side-length I am being asked to calculate the area of a triangle with a side-length of 15.5 inches.  
The formula for calculating a regular polygon's area is 1/2Pa
Where P is the perimeter of the polygon, and a is the apothem.  I am completely lost.
 A: Hint:  If you cut the equilateral triangle into two by bisecting an angle, you make two right triangles.  Can you identify the base and height?
Added:  Look at the right triangles and try to find x
A: It is best not to use a general formula that one has not great control over, when the problem is a very concrete one.
However, one can compute the apothem. Recall the apothem is the (length of) a perpendicular from the centre of the equilateral triangle to any one of the sides. So let our equilateral triangle be $ABC$, and let $P$ be the centre. Draw the line through $P$ which is perpendicular to $AB$. Suppose it meets $AB$ at $Q$. The apothem is the length of $PQ$.
Since $\angle CAB=60^\circ$, by symmetry we have $\angle PAQ=30^\circ$. Note that $AQ=15.5/2$.  So if $x$ is the apothem, then 
$$\frac{x}{(15.5/2)}=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$$
 It follows that
$$x=\frac{15.5}{2}\cdot \frac{1}{\sqrt{3}}.$$
If you do not know the "special angles" fact that $\tan(30^\circ)=\frac{1}{\sqrt{3}}$, you can use the calculator to evaluate $\tan(30^\circ)$ to high precision. 
