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A Simple Derivation of Heron’s Formula

By Mark Levi

Heron’s formula gives the area \(A\) of a triangle with sides \(a, b, c\):

\[A =  \sqrt{s(s-a)(s-b)(s-c) } ,\tag1\]

where \(s= \frac{1}{2} (a+b+c)\) is the semiperimeter. Most proofs hide the simple reason for this result.  This reason is twofold:

  1. An observation that \(A ^2\) is a polynomial of degree \(4\) in \(a, b, c\). 
  2. \(A ^2 =0\) if the triangle degenerates into a point or a segment, i.e., if \(a+b+c =0\) or if \(a+b-c\) or any of its cyclic permutations vanish.

Taking \((1)\) for granted for the moment, \((2)\) implies that 

\[A ^2 =  k (a+b+c)(a+b-c)(b+c-a)(c+a-b),\tag2\]

where the unknown constant \(k\) is independent of \(a, b, c\). To find \(k\), we apply (2) to a right triangle with sides \(1, 1, \sqrt{ 2 }\), thus obtaining

\[\biggl( \frac{1}{2} \biggl)   ^2 =  k (2+ \sqrt{ 2 }) (2- \sqrt{ 2 })\sqrt{ 2 }\sqrt{ 2 },\]

or \(k = 1/2^4\). With this value, \((2)\) becomes Heron’s formula \((1)\).

To justify \((1)\), we write 

\[A ^2 = a^2 b^2  \sin ^2  \theta   = a^2 b^2-a^2 b^2 \cos ^2 \theta,\]

where by the theorem of cosines \(4 a^2 b^2 \cos ^2 \theta =(c ^2 - a ^2  - b ^2 )  ^2\). Observation \((2)\) allowed us to avoid the algebra of factoring \(A ^2\). 

Mark Levi ([email protected]) is a professor of mathematics at the Pennsylvania State University.

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