Hero's formula for the area of a triangle is exceedingly neat-o: it gives a simple expression for the area in terms of the sides. But surely nothing similar can exist for a quadrilateral (a shape with 4 sides)?

Well, obviously not. After all, you can have a convex and a concave quadrilateral with the same sides. Their areas will be different, while the sides the same. So we can't hope for a formula for all quadrilaterals.

Suprisingly amazingly stunningly, though, we still have an extension of Hero's formula!

Let a,b,c,d be the sides of a quadrilateral that can be inscribed in a circle. Denote half the perimeter (the semi circumference?) by s=(a+b+c+d)/2. Then the quadrilateral's area is sqrt((s-a)(s-b)(s-c)(s-d)).

It's easy to verify the result is correct for rectangles (which are indeed bounded in circles).

Interestingly, Hero's formula for triangles is a special case of this formula. Think of a triangle as having 4 sides, one of length 0. All triangles can be inscribed in a circle, so this formula applies to any triangle, taking e.g. d=0.

Log in or register to write something here or to contact authors.