TITLE: Area formula for cyclic pentagon?
QUESTION [8 upvotes]: For a triangle, we have 
$$Area=\sqrt{p(p-a)(p-b)(p-c)},\qquad p=\frac{a+b+c}{2}.$$
For a cyclic quadrilateral, we have
$$Area=\sqrt{(p-a)(p-b)(p-c)(p-d)},\qquad p=\frac{a+b+c+d}{2}.$$
Is there a similar formula for a cyclic pentagon?

REPLY [5 votes]: Hint: The generalization of Heron formula to cyclic pentagon and hexagon can be found here: 
Areas of polygons inscribed in a circle
Also the more recent paper for general polygons:
On the Areas of Cyclic and Semicyclic Polygons