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Date: Around 1993


Question:
I am trying to find a complete proof of a formula that is either Heron's formula or Hero's formula (I have seen it named both ways). The formula computes the area of a triangle when only the lengths of the sides are known. The formula: If a, b, and c are sides of a triangle and s = (1/2)*(a+b+c) (s is half of the perimeter) then the area of the triangle is the square root of s * (s-a) * (s-b) * (s-c). The texts that have this formula which I have been using either offer it with no proof, or say that the proof is too long to be printed in that text. Can you refer me to a book that has this proof, or find it?



Replies:
Given a triangle with sides a,b,c, semiperimeter s, and area A, show that A^2 = s(s-a)(s-b)(s-c). Solution: Drop an altitude (of length h) to the side of length c. Then A = (1/2)ch, so A^2 = c^2 h^2 / 4. Use the Pythagorean Theorem to obtain the following system:
(1) x^2 + h^2 = a^2
(2) y^2 + h^2 = b^2 (3) x + y = c

Substitute y = c - x into (2) and simplify. Then subtract the result from (1). You will find that 2cx = a^2 - b^2 + c^2.

From (1),
4c^2 h^2 = 4a^2 c^2 - 4c^2 x^2
= (2ac + 2cx) (2ac - 2cx)
= (2ac + a^2 - b^2 + c^2)(2ac - a^2 + b^2 -c^2)

From (1),
4c^2 h^2 = 4a^2 c^2 - 4c^2 x^2
= (2ac + 2cx) (2ac - 2cx)
= (2ac + a^2 - b^2 + c^2)(2ac - a^2 + b^2 -c^2)
= ((a+c)^2 - b^2) (b^2 - (a-c)^2)
= (a+c+b)(a+c-b)(b+a-c)(b-a+c)
= (2s)(2s-2b)(2s-2c)(2s-2a)
= 16s(s-a)(s-b)(s-c)

Chris Baker



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