A lune is a plane figure bounded by two circular arcs of unequal radii, i.e., a crescent. (By contrast, a plane
figure bounded by two circular arcs of equal radius is known
as a lens.) For circles of radius 
and 
whose centers are separated by a distance 
, the area of the lune is given by
![A=a^2[tan^(-1)((a^2-b^2+c^2)/(4Delta))+cos^(-1)((b-c)/a)+tan^(-1)((b-c)/(sqrt((a+b-c)(a-b+c))))]
-b^2[tan^(-1)((a^2-b^2-c^2)/(4Delta))+pi/2]+2Delta
=2Delta+a^2sec^(-1)((2ac)/(b^2-a^2-c^2))-b^2sec^(-1)((2bc)/(b^2+c^2-a^2)),](/images/equations/Lune/NumberedEquation1.svg) |
(1)
|
where
 |
(2)
|
is the area of the triangle with side lengths 
, 
,
and 
. The second of these can be obtained
directly by subtracting the areas of the two half-lenses
whose difference producing the colored region above.
In each of the figures above, the area of the lune is equal to the area of the indicated triangle. Hippocrates of Chios squared the above left lune (Dunham 1990, pp. 19-20; Wells 1991, pp. 143-144), as well as two others, in the fifth century BC. Two more squarable lunes were found by T. Clausen in the 19th century (Shenitzer and Steprans 1994; Dunham 1990 attributes these discoveries to Euler in 1771). In the 20th century, N. G. Tschebatorew and A. W. Dorodnow proved that these are the only five squarable lunes (Shenitzer and Steprans 1994).
Hippocrates also proved that, in the figure above, the sum of the areas of the two colored lunes is equal to the area of the triangle (Pappas 1989, pp. 72-73). Individually, the area of the lunes are
 |  | ![1/8[a(2b+api)-2(a^2+b^2)tan^(-1)(a/b)]](/images/equations/Lune/Inline9.svg) |
(3)
|
 |  | ![1/8[a(2b-api)+2(a^2+b^2)tan^(-1)(a/b)],](/images/equations/Lune/Inline12.svg) |
(4)
|
where the lengths of the horizontal and vertical legs of the triangle are 
and 
,
respectively.
