Proof of Heron's (Hero's) formula: Triangle ```Name: Unknown Status: N/A Age: N/A Location: N/A Country: N/A 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 Click here to return to the Mathematics Archives

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