Ask A Scientist Mathematics Archive

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?
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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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