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 side
s. 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.