Calculating the area of an irregular polygon Given the length of the sides of an irregular polygon (no coordinates provided) how do you compute the area of the maximum area of the polygon? 
Thanks in advance
 A: The polygon must be cyclic. For details, check this out.
A: Since the very interesting treatments that aelguindy linked to appear to indicate that no formulas are known beyond hexagons, it seems you'll need to determine the area numerically. One efficient approach to doing this might be to solve the condition for the angles to add up to a full circle,
$$\sum_k\arcsin\frac{L_k}{2R}=\pi\;,$$
for $R$ numerically (e.g. using Newton's method). Then the area is
$$
\begin{eqnarray}
A
&=&
\frac{R^2}2\sum_k\sin\left(2\arcsin\frac{L_k}{2R}\right)
\\
&=&
\frac{R}2\sum_kL_k\sqrt{1-\left(\frac{L_k}{2R}\right)^2}\;.
\end{eqnarray}
$$
[Edit as requested:]
Here's how you could solve for $R$ using Newton's method. The general prescription for using Newton's method to solve the equation $f(x)=0$ is
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\;.$$
This is where the tangent to the graph at $x_n$ intersects the $x$ axis. In the present case, we want to solve
$$f(R)=\sum_k\arcsin\frac{L_k}{2R}-\pi=0\;,$$
so we form
$$f'(R)=-\sum_k\frac{L_k}{2R\sqrt{(2R)^2-L_k^2}}$$
and iterate using
$$R_{n+1}=R_n+\frac{\sum_k\arcsin\frac{L_k}{2R_n}-\pi}{\sum_k\frac{L_k}{2R_n\sqrt{(2R_n)^2-L_k^2}}}\;.$$
Now we just need a suitable initial value $R_0$. This must be somewhere between the minimal radius $L_\max/2$ (where $L_\max$ is the maximal side length) and infinity; a reasonable choice might be $L_\max$.
Note that Newton's method isn't guaranteed to converge in general; I think it should work in this case, but I haven't tried it. If it doesn't work, you may have to use a more robust root-finding algorithm.
A: The polygon should be cyclic, meaning that its vertices will lie on the circumference of a circle(assuming the polygon is convex). If the side lengths are given as $x_1, x_2, x_3, x_4,\ldots$ and $s=\frac{p}{2}$ where $p$ is the perimeter then the area is given by something akin to Brahmgupta's formula for $4$ sides or Heron's formula for a triangle. Not sure what the area is given by when the number of sides exceeds $4$.
You can usually find the areas of irregular convex polygons by dividing them into triangles wisely, though. 
A: Following on @aelguindy's comment, this was an area of research by one Prof. David Robbins (who passed away prematurely).  Here is a typical paper from that research effort: he was able to find explicit expressions for the areas of cyclic heptagons and octagons.  Anyone who browses through that paper will understand that this problem is extraordinarily difficult in general.
For amusement, I compared relatively simple results I got from a Putnam exam problem with his result and got agreement, happily.  Here is the problem:

Find the area of a convex octagon inscribed in a circle that has 4
  consecutive sides of length 3 units and 4 consecutive sides of length
  2 units. Give your answer in the form $r+s \sqrt{t}$, where $r$, $s$
  and $t$ are positive integers.

I'll leave it as an exercise for the reader.
A: because we know the sides So the angles formed by these sides at the center of the circle will be in the same ratio as the sides
for example if the sides of the polygon are a,b,c,d and e then the sum of all the angles at the center will be = a*x+b*x+c*x+d*x+e*x and this sum will be equal to 360.Sonow we know all the angles made by the sides at the centre. Now we can find the area of each triangle and sum of these areas is the area of the polygon
