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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. Hero's Formula is not easy to come up with. Here's what happens when I try to derive it from scratch:
h^2 = a^2-x^2 [Pythagorean Theorem]
h^2 = b^2-c^2+2cx-x^2 [Pythagorean Theorem]
2cx = a^2-b^2+c^2
x = (a^2-b^2+c^2)/2c
h^2 = a^2 - [(a^2-b^2+c^2)/2c]^2
A = hc/2
A = sqr[4a^2c^2-(a^2-b^2+c^2)^2]/4 [this is where I stopped, when I was young]
A = sqr(2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4)/4
This looks like a good formula. 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)=-a^4+0a^3b+2a^2b^2+0ab^3+0a^3c+2a^2c^2+0ac^3-b^4+0b^3c+2b^2c^2+0bc^3-c^4
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. 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. But, I stopped short of Hero's Formula, saying that it could be modified to look like my formula. I have actually done that, in the past, so we can assume that it is not difficult. So, I have not finished my derivation. I hope to tidy that up, eventually.
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.