Closed form for the area of a convex cyclic n-gon, given the set of edge lengths Let's say we are given a set of positive reals, and we're told that these are the edges of a convex cyclic $n$-gon, and we must compute it's area. 
For $n = 3$ there is the famous Heron's formula:
$$A = {1\over 4}\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2}$$
For $n = 4$ there is Brahmagupta's formula:
$$A = {1\over 4}\sqrt{(a^2 + b^2 + c^2 + d^2)^2 + 8abcd - 2(a^4 + b^4 + c^4 + d^4)}$$
Is there a generalization for arbitrary $n$? Does it even exist?
 A: This paper (Cyclic Polygons with Rational Sides and Area by Buchholz and Macdougall) mentiones explicit formulae for $n\leqslant 6$, and it seems it would be difficult (or at least nobody seems to have done it up to now) to explicitly construct a formula for arbitrary $n$.
But note that you can divide any cyclic $n$-gon into cyclic triangles, and apply Herons formula on these, but this requires you to calculate the chords which form the new sides of the triangles.
You might also be intersted in this: Calculate the area of an irregular cyclic convex polygon
And this: Areas of Polygons Inscribed in a Circle by
David P. Robbins 
A: In case the links go down (shout-out to Martin Buettner), at least for the pentagon we can triangulate it trigonometrically.  But beware: it gets a bit complicated.  What we see for the pentagon heralds the complexity of the problem for larger numbers of sides.
Let $ABCDE$ be the pentagon, with sides $AB=a, BC=b, CD=c, DE=d, EA=e$.  Form the diagonals $AC=x, AD=y$.  Now, apply the Law of Cosines to triangles $ABC$ and $ACD$.  We get:
$\cos(ABC)=\frac{a^2+b^2-x^2}{2ab}$
$\cos(CDA)=\frac{c^2+y^2-x^2}{2cy}$
The angles on the left side of the equations must be supplementary if quarilateral $ABCD$ is to be cyclic.  So the left sides sum to zero, and therefore the right sides do the same.  Form this sum and isolate the first power of $y$  leaving $y^2$ in the "solved" expression.  This gives:
$y=(\frac{ab}{c})(\frac{c^2+y^2-x^2}{x^2-a^2-b^2}) ... Eq. 1$
We now do the same thing using triangles $CDE$ and $ACD$, focusing on a different angle of the latter triangle.
$\cos(CDE)=\frac{d^2+e^2-y^2}{2de}$
$\cos(ACD)=\frac{c^2+x^2-y^2}{2cx}$
Again the angles are supplementary, their cosines add up to zero and so do the right sides of the equations above.  This time solve the sum for $y^2$; do not take the square root:
$y^2=\frac{cx(d^2+e^2)+de(c^2+x^2)}{cx+de} ... Eq. 2$
Now comes the good part.  Take the right side of Eq. 2 and substitute that for $y^2$ in the right side of Eq. 1.  Square the resulting equation and match the resulting expression for $y^2$ with the right side of Eq. 2.  You now have a rational equation for $x$, which can be turned to a (complicated) polynomial equation.  That equation turns out to be degree 7:
$x^7(cde)+$
$x^6(c^2d^2+c^2e^2+d^2e^2-a^2b^2)+$
$x^5(cde)[c^2+d^2+e^2-2(a^2+b^2)]+$
$x^4[c^2d^2e^2-2(a^2+b^2)(c^2d^2+c^2e^2+d^2e^2)+2a^2b^2(c^2+d^2+e^2)]+$
$x^3(cde)[-2(a^2+b^2)(c^2+d^2+e^2)+(a^4+6a^2b^2+b^4)^2]+$
$x^2[-2(a^2+b^2)c^2d^2e^2+(a^2+b^2)^2(c^2d^2+c^2e^2+d^2e^2)-a^2b^2(c^2+d^2+e^2)^2]+$
$x(cde)(a^2-b^2)^2(c^2+d^2+e^2)+$
$(a^2-b^2)^2c^2d^2e^2=0,$
and (unless the pentagon is "rigged") we expect it to be irreducible.  So given a set of side lengths we cannot generally construct a cyclic pentagon -- even if we are allowed a marked ruler!
For each root obtained (in general, numerically) for $x$, we find $y^2$ using Eq. 2 and then $y$ with Eq. 1.  Of course $y$ from Eq. 1 must be a square root of $y^2$ from Eq. 2, but Eq. 1 also determines a specific sign.  At least for side lengths where a pentagon can be drawn in real space, there is one root where both $x$ and $y$ are positive.  That is your convex cyclic pentagon.  The other roots correspond to various "crossed" pentagon configurations -- the pentagram, the "crystal ball" pentagon where a quadrilateral sits on top of a triangle, and the "vampire" pentagon where three triangles look like fangs. The reversed sign of $x$ incusive-or $y$  in these roots come from sides crossing each other so they're effectively reversed relative to the $x$ or $y$ diagonal.
Now that you have $x$ and $y$, you can use Heron's Formula on each triangle and add them up to get the area of the convex pentagon.  For the crossed pentagon roots you get a net or signed area by properly giving each triangle a $+$ sign (if the triangle has zero or two "negative" diagonals) or $-$ sign (one "negative" diagonal) and then doing the addition.
Yes, it's complicated.  But, actually, it's simpler than a direct area calculation, and you learn more about the geometry of the pentagon.
A: This is discussed in this very nice 2004 paper by Igor Pak.
A: Yes there is, but only for cycling n-gons. Also it's not an exact formula, but a "quite good" approximation. Its goodness has to be determined. 
What I am going to write down comes from an ArXiV Article I read many months ago, hence I just remember the formula without the names of the scientists, but a Google research like "generalization go Brahmagupta" shall work. 
Anyway it's not an obscure lost hidden forgotten formula so I guess the internet gots it within. Do your researches.
The formula is given by:
$$A = \sqrt{s^{4-n}(s-x_1)(s-x_2)\ldots (s-x_n)}$$
Where
$$s = \frac{1}{2}(x_1+ x_2 + \ldots +x_n)$$
and $x_i$ are the sides of the polygon, whilst $n$ is the number of sides.
For the case of regular polygons this formula slightly overestimates the true area.  For a regular pentagon the overestimate is about 1.3%.  For higher regular polygons it grows to a maximum of about 15.6%.  Asymptotic analysis of this formula versus the exact one for large $n$ shows a limiting ratio of
$\pi/e = 1.155727... $
for this formula divided by the exact area of a regular $n$-gon.
