I have been searching all over the internet looking for the general formula for the area of a lune and have only found this one on the Wolfram MathWorld page:
I emailed the author asking for a derivation and received no response. Does anyone know how this is derived geometrically? Does anyone know a simpler formula that is derived geometrically?
Based on the clarifications from the OP, the following geometric derivation is proposed. In all of the following, $\pm$ or $\mp$ signs are used; the top sign corresponds to crescent-shaped lunes where the chord joining the endpoints lies outside the crescent, the bottom signs correspond to convex lenses where this chord passes inside the area.
The lune may be regarded as a superposition of two sectors and a triangle. We then have the area formula
$S(\text{lune})=S(\text{sector 1})\mp S(\text{sector 2})\pm 2S(\text{triangle})$
Sector 1 = sector of radius $a$, corresponding to the convex arc in a crescent lune.
Sector 2 = sector of larger radius $b$, corresponding to the concave arc in a crescent lune.
Triangle = triangle with vertices at the centers and at either vertex of the lune
We render these component areas separately.
Triangle area
With the radii $a,b$ and the center-to-center distance $c$ given, the triangle area is rendered by Heron's Formula. This area is to be taken twice since two mirror-image triangles appear in the superposition.
$2S(\text{triangle})=\frac12\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$
Sector 1 area
The areas of Sector 1 is rendered as
$S(\text{sector 1})=a^2\theta_1$
where $\theta_1$ measures half the sector arc and is given by the Law of Cosines:
$\cos(\theta_1)=\mp\dfrac{a^2+c^2-b^2}{2ac}$
The sign comes from the fact that for a crescent lune we must take the exterior angle of the triangle at the center of this sector. The inverse cosine will provide a unique value for $\theta_1$ between $0°$ and $180°$.
Sector 2 area
The area of sector 2 is rendered by a method similar to that used for sector 1, except here the interior angle is always used so no sign is needed in the cosine term.
$S(\text{sector 2})=b^2\theta_2$
$\cos(\theta_2)=\dfrac{b^2+c^2-a^2}{2bc}$
So the lune area is given by
$S(\text{lune})=a^2\cos^{-1}\left(\mp\dfrac{a^2+c^2-b^2}{2ac}\right)\mp b^2\cos^{-1}\left(\dfrac{b^2+c^2-a^2}{2bc}\right)\pm \frac12\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}.$