Heron's formula

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A triangle with sides a, b, and c.
A triangle with sides a, b, and c.

In geometry, Heron's formula (also called Hero's formula) states that the area (A) of a triangle whose sides have lengths a, b and c is

A = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\,

where s is the semiperimeter of the triangle:

s=\frac{a+b+c}{2}.

Heron's formula can also be written as

A={\ \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)\,}\ \over 4}.\,

Contents

The formula is credited to Heron of Alexandria, and a proof can be found in his book, Metrica, written c. 60 AD. It has been suggested that Archimedes knew the formula, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that it predates the reference given in the work. [1]

A modern proof, which uses algebra and trigonometry and is quite unlike the one provided by Heron, follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have

\cos(C) = \frac{a^2+b^2-c^2}{2ab}

by the law of cosines. From this we get with some algebra

\sin(C) = \sqrt{1-\cos^2(C)} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.

The altitude of the triangle on base a has length bsin(C), and it follows

S\, = \frac{1}{2} (\mbox{base}) (\mbox{altitude})
= \frac{1}{2} ab\sin(C)
= \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2}
= \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))}
= \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)}
= \frac{1}{4}\sqrt{(c -(a -b))((c +(a -b))((a +b) -c))((a +b) +c)}
= \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.

The difference of two squares factorization was used in two different steps.

Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative[2] involves arranging the lengths of the sides so that: abc and computing

S = \frac{1}{4}\sqrt{(a+(b+c)) (c-(a-b)) (c+(a-b)) (a+(b-c))}

The brackets in the above formula are required in order to prevent numerical instability in the evaluation.

\frac{1}{4} \sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}

Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral; both of which are special cases of Bretschneider's formula for the area of a quadrilateral.

Expressing Heron's formula with a determinant in terms of the squares of the distances between the three given vertices,

S =  \frac{1}{4} \sqrt{ \begin{vmatrix}    0 & a^2 & b^2 & 1 \\ a^2 & 0   & c^2 & 1 \\ b^2 & c^2 & 0   & 1 \\   1 &   1 &   1 & 0 \end{vmatrix} }

illustrates its similarity to Tartaglia's formula for the volume of a three-simplex.

  1. ^ http://mathworld.wolfram.com/HeronsFormula.html
  2. ^ http://http.cs.berkeley.edu/~wkahan/Triangle.pdf
  • Heath, Thomas L. (1921). A History of Greek Mathematics (Vol II). Oxford University Press, 321-323. 

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