Derivations of Heron's FormulaDate: 11/24/98 at 00:52:32 From: Hath Subject: Heron's theory I understand how to use Heron's Theory, but how exactly is it derived? Date: 11/24/98 at 12:07:31 From: Doctor Rob Subject: Re: Heron's theory Here is one derivation: Proof of Hero's Formula http://mathforum.org/library/drmath/view/54957.html Another proof uses these three formulas: K = (1/2)*a*b*sin(C) c^2 = a^2 + b^2 - 2*a*b*cos(C) sin^2(C) + cos^2(C) = 1 The first uses the definition of the sine as the side opposite over the hypotenuse in the right triangle whose hypotenuse is side a, one leg is part of side b, and the other leg is the altitude h to side b. The second is the Law of Cosines, and the third is one of the simplest identities from trigonometry. Another proof uses the Pythagorean Theorem instead of the trigonometric functions sine and cosine. Drop from B a perpendicular to side b. Let its length be h. Let the side b be divided by it into parts of lengths x and b - x. Then using the Pythagorean Theorem on the two small triangles, and the usual formula for area of a triangle: a^2 = h^2 + x^2 c^2 = h^2 + (b-x)^2 K = b*h/2 Then by subtracting the second equation from the first: a^2 - c^2 = 2*b*x - b^2 x = (a^2+b^2-c^2)/(2*b) Substituting this back into the first equation: h^2 = a^2 - (a^2+b^2-c^2)^2/(4*b^2) Now using the third equation: K^2 = b^2*h^2/4 = (b^2/4)*[a^2 - (a^2+b^2-c^2)^2/(4*b^2)] = a^2*b^2/4 - (a^2+b^2-c^2)^2/16 = (4*a^2*b^2-a^4-2*a^2*b*2-b^4+2*b^2*c^2-c^4+2*a^2*c^2)/16 = -(a^4+b^4+c^4-2*b^2*c^2-2*a^2*c^2-2*a^2*b^2)/16 = -(c^4 + [-2*b^2-2*a^2]*c^2 + [a^4-2*a^2*b^2+b^4])/16 We factor this by completing the square on c^2, then using the difference of two squares three times, then a + b + c = 2*s: K^2 = -(c^4 + [-2*b^2-2*a^2]*c^2 + [a^2+b^2]^2 - 4*a^2*b^2)/16 = -([c^2-a^2-b^2]^2-[2*a*b]^2)/16 = -([c^2-a^2-2*a*b-b^2]*[c^2-a^2+2*a*b-b^2])/16 = -(c^2-[a+b]^2)^2*(c^2-[a-b]^2)/16 = -(c+[a+b])*(c-[a+b])*(c+[a-b])*(c-[a-b])/16 = (a+b+c)*(a+b-c)*(a-b+c)*(-a+b+c)/16 = (2*s)*(2*s-2*c)*(2*s-2*b)*(2*s-2*a)/16 = s*(s-a)*(s-b)*(s-c) K = sqrt[s*(s-a)*(s-b)*(s-c)] - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/