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?

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.
