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© Copyright 1998, Jim Loy
Hero's (or Heron's) Formula is used to calculate the area of a triangle, given the lengths of the three sides:
A = sqr[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2
Here I use sqr() for square root, rather than draw the square root sign with a graphic file. The variable s is called the semi-perimeter. Hero's Formula is not easy to come up with. Here's what happens when I try to derive it from scratch:
h² = a²-x² [Pythagorean Theorem]
h² = b²-c²+2cx-x² [Pythagorean Theorem]
2cx = a²-b²+c²
x = (a²-b²+c²)/2c
h² = a² - [(a²-b²+c²)/2c]²
A = hc/2
A = sqr[4a²c²-(a²-b²+c²)²]/4 [this is where I stopped, when I was young]
A = sqr(2a²b²+2a²c²+2b²c²-a^4-b^4-c^4)/4
This looks like a good formula. Here I am having difficulty with the notation, as superscripts are apparently not allowed in "preformatted" HTML text. Since I know where I am headed (Hero's Formula), I figure that the quantity under the square root sign is equal to (a+b+c)(-a+b+c)(a-b+c)(a+b-c). Where did I get that? If the perimeter is p, then (a+b+c)(-a+b+c)(a-b+c)(a+b-c) = p(p-2a)(p-2b)(p-2c), and using s for the semiperimeter, (a+b+c)(-a+b+c)(a-b+c)(a+b-c) = 16s(s-a)(s-b)(s-c). Let's multiply (a+b+c)(-a+b+c)(a-b+c)(a+b-c) out, and see if it is the same as the quantity under the square root sign in my formula:
(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=-a4+0a3b+2a2b2+0ab3+0a3c+2a2c2+0ac3-b4+0b3c+2b2c2+0bc3-c4
That seems to be the same. So I can express my formula as: A=sqr((a+b+c)(-a+b+c)(a-b+c)(a+b-c))/4, which is much more symmetric. I can now substitute s=(a+b+c)/2 into this, and we get Hero's Formula: A=sqr[s(s-a)(s-b)(s-c)].
The above mathematics is a derivation, not a proof. I start from basic principles, and derive my equations. My aim was to derive Hero's Formula, as I was amazed that anyone could come up with that elegant formula from basic principles.
Above, I mentioned that I did this derivation when I was young. I stopped at the asymmetrical formula above. I should have suspected that a symmetrical formula was possible. But I was actually deriving a formula for the area of a trapezoid (which I had divided into two right triangles and a rectangle). There was no implied symmetry there.