Proof of Hero's formulaDate: 09/08/97 at 13:25:23 From: Russ Boyd Subject: Hero's formula Could you tell me where to find a proof of Hero's formula or help on how to derive it? I am a teacher and my class is working on it. Date: 09/08/97 at 16:05:14 From: Doctor Wilkinson Subject: Re: Hero's formula The book "Journey through Genius" has a proof very much like Hero's. You can derive it by brute force from the law of cosines. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/08/97 at 16:35:37 From: Doctor Anthony Subject: Re: Hero's formula We use the formula area = (1/2)bc.sin(A) And sin(A) = 2 sin(A/2).cos(A/2) so area = bc.sin(A/2).cos(A/2) Next we find expressions for cos(A/2) and sin(A/2) in terms of s, a, b, c, where a, b, c are the sides of the triangle and s is the semi-perimeter. So s = (a+b+c)/2 b^2 + c^2 - a^2 cos(A) = 2cos^2(A/2) - 1 = ---------------- 2bc b^2 + c^2 - a^2 2cos^2(A/2) = 1 + ------------------ 2bc (b+c)^2 - a^2 (b+c-a)(b+c+a) = -------------- = -------------- 2bc 2bc (2s-2a)2s = ----------- 2bc s(s-a) cos^2(A/2) = --------- bc b^2 + c^2 - a^2 Next cos(A) = 1 - 2 sin^2(A/2) = ---------------- 2bc b^2 + c^2 - a^2 2sin^2(A/2) = 1 - ---------------- 2bc a^2 - (b-c)^2 (a-b+c)(a+b-c) = -------------- = -------------- 2bc 2bc (2s-2b)(2s-2c) = -------------- 2bc (s-b)(s-c) sin^2(A/2) = ------------ bc We can now return to the formula for the area of the triangle: (1/2)bc.sin(A) = bc.sin(A/2).cos(A/2) bc.sqrt[s(s-a)(s-b(s-c)] = ------------------------ bc = sqrt[s(s-a)(s-b)(s-c)] -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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