Heron's formula

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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

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

1 History
2 Proof
3 Numerical stability
4 Multiplied out form of Heron's formula
5 Generalizations
6 See also
7 References
8 External links

[edit] History

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]

[edit] Proof

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.

[edit] Numerical stability

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: a ≥ b ≥ c and computing

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