Derivations of Heron's Formula

Date: 11/24/98 at 00:52:32
From: Hath
Subject: Heron's theory

I understand how to use Heron's Theory, but how exactly is it derived?

Date: 11/24/98 at 12:07:31
From: Doctor Rob
Subject: Re: Heron's theory

Here is one derivation:

   Proof of Hero's Formula
   http://mathforum.org/library/drmath/view/54957.html   

Another proof uses these three formulas:

   K = (1/2)*a*b*sin(C)
   c^2 = a^2 + b^2 - 2*a*b*cos(C)
   sin^2(C) + cos^2(C) = 1

The first uses the definition of the sine as the side opposite over the 
hypotenuse in the right triangle whose hypotenuse is side a, one leg is 
part of side b, and the other leg is the altitude h to side b. The 
second is the Law of Cosines, and the third is one of the simplest
identities from trigonometry.

Another proof uses the Pythagorean Theorem instead of the trigonometric
functions sine and cosine. Drop from B a perpendicular to side b. Let
its length be h. Let the side b be divided by it into parts of lengths
x and b - x. Then using the Pythagorean Theorem on the two small
triangles, and the usual formula for area of a triangle:

   a^2 = h^2 + x^2
   c^2 = h^2 + (b-x)^2
     K = b*h/2

Then by subtracting the second equation from the first:

   a^2 - c^2 = 2*b*x - b^2
           x = (a^2+b^2-c^2)/(2*b)

Substituting this back into the first equation:

   h^2 = a^2 - (a^2+b^2-c^2)^2/(4*b^2)

Now using the third equation:

   K^2 = b^2*h^2/4
       = (b^2/4)*[a^2 - (a^2+b^2-c^2)^2/(4*b^2)]
       = a^2*b^2/4 - (a^2+b^2-c^2)^2/16
       = (4*a^2*b^2-a^4-2*a^2*b*2-b^4+2*b^2*c^2-c^4+2*a^2*c^2)/16
       = -(a^4+b^4+c^4-2*b^2*c^2-2*a^2*c^2-2*a^2*b^2)/16
       = -(c^4 + [-2*b^2-2*a^2]*c^2 + [a^4-2*a^2*b^2+b^4])/16

We factor this by completing the square on c^2, then using the 
difference of two squares three times, then a + b + c = 2*s:

   K^2 = -(c^4 + [-2*b^2-2*a^2]*c^2 + [a^2+b^2]^2 - 4*a^2*b^2)/16
       = -([c^2-a^2-b^2]^2-[2*a*b]^2)/16
       = -([c^2-a^2-2*a*b-b^2]*[c^2-a^2+2*a*b-b^2])/16
       = -(c^2-[a+b]^2)^2*(c^2-[a-b]^2)/16
       = -(c+[a+b])*(c-[a+b])*(c+[a-b])*(c-[a-b])/16
       = (a+b+c)*(a+b-c)*(a-b+c)*(-a+b+c)/16
       = (2*s)*(2*s-2*c)*(2*s-2*b)*(2*s-2*a)/16
       = s*(s-a)*(s-b)*(s-c)
     K = sqrt[s*(s-a)*(s-b)*(s-c)]

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/