Consider the following extension of Heron's Formula. For a cyclic $n$-gon $C$ with side lengths $x_1, x_2, \dots, x_n$ and semi-perimeter $P = \frac{1}{2} \left( x_1 + x_2 + \dots + x_n\right)$ define: $$ M = \sqrt{P^{4-n} (P-x_1)(P-x_2) \dots (P-x_n)} $$
After some experimentation in GeoGebra, it turns out that $M$ is pretty close (but not equal) to the usual area for cyclic $n$-gons. This suggests that $M$ is less than a constant multiple of area $A$. Does anyone have an idea of how to prove this?
To get a sense of what might happen, for regular $n$-gons $R_n$ it turns out that: $\displaystyle \lim_{n \rightarrow \infty} M(R_n)/A(R_n) = \pi/e$.
Observe that $M$ is a homogeneous function $M(\lambda C) = \lambda^2 M(C)$. Area is also homogenous in the same degree $A(\lambda C) = \lambda^2 A(C)$. Thus, $M/A$ is scale invariant. We can pick a scale to work. Can anyone show that $M$ is bounded on cyclic polygons of area one?
Not a full answer, but some assorted observations:
When $n=4$, the formula is exact; this is Brahmagupta's formula.
For a given cyclic ordering of side lengths in a polygon, the area is maximized by taking the polygon to be cyclic (which uniquely specifies it up to congruence), so lower bounds for the ratio $M/A$ in the cyclic case are actually lower bounds for all polygons.
There are, for each $n$, cyclic polygons with $M$ arbitrarily close to $A$; simply take any triangle with side lengths $a,b,c$ and copy one of its vertices $n-3$ times. (To avoid degenerate cases, just take the neighboring points to be extremely close to one of the vertices - the limits work out fine.) Then the area is that of the original triangle, and the Heron-like formula is $$\sqrt{s^{4-n}(s-a)(s-b)(s-c)(s-0)(s-0)\ldots(s-0)}=\sqrt{s(s-a)(s-b)(s-c)}$$ In fact, this same process shows that the interval of possible ratios for a given $n$ is a subset of the possible ratios for higher $n$; in particular, a bound for $n$-gons implies the same bound must hold for all $k$-gons with $k<n$.
In all experimental data I have collected thus far (tens of thousands of randomly-generated cyclic polygons), $M/A$ appears to be bounded below by $1$ and above by $\pi/e$. For large random cyclic polygons (convex hull of many independently chosen random points on the circle), "typical" such polygons have a ratio very close to $\pi/e$. This follows from the fact that a generic large polygon has $M\approx s^2/e, A\approx s^2/\pi$, with each approximate equality being a $\le$ inequality in general.
For all $n$, the regular $n$-gon appears to maximize the ratio $M/A$, though this is based on experimental data only. (I suspect that proving this is the best route to showing the upper bound of $\pi/e$.)
Some plots of the ratio $M/A$ for different numbers of independently chosen points on the circle are shown below.
$n=5$, $100000$ samples:
(The regular pentagon has $M/A \approx 1.013$.)
$n=6$, $100000$ samples:
$n=10$, $10000$ samples:
$n=100$, $10000$ samples:
($M/A=1$ is of course possible here, it's just vanishingly unlikely.)