# Heron's formula

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A triangle with sides a, b, and c.

In geometry, Heron's formula (sometimes called Hero's formula) is named after Hero of Alexandria[1] and states that the area of a triangle whose sides have lengths a, b, and c is

${\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}},}$

where s is the semiperimeter of the triangle; that is,

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

Heron's formula can also be written as

${\displaystyle A={\frac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}}$
${\displaystyle A={\frac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}}$
${\displaystyle A={\frac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}}$
${\displaystyle A={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}.}$

Heron's formula is distinguished from other formulas for the area of a triangle, such as half the base times the height or half the modulus of a cross product of two sides, by requiring no arbitrary choice of side as base or vertex as origin.

## Example

Let ΔABC be the triangle with sides a=4, b=13 and c=15. The semiperimeter is   ${\displaystyle s={\tfrac {1}{2}}(a+b+c)={\tfrac {1}{2}}(4+13+15)=16}$ , and the area is

{\displaystyle {\begin{aligned}A&={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {16\cdot (16-4)\cdot (16-13)\cdot (16-15)}}\\&={\sqrt {16\cdot 12\cdot 3\cdot 1}}={\sqrt {576}}=24.\end{aligned}}}

In this example, the side lengths and area are all integers, making it a Heronian triangle. However, Heron's formula works equally well in cases where one or all of these numbers is not an integer.

## History

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c. A.D. 60. It has been suggested that Archimedes knew the formula over two centuries earlier,[2] and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.[3]

A formula equivalent to Heron's, namely

${\displaystyle A={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}}$, where ${\displaystyle a\geq b\geq c}$

was discovered by the Chinese independently of the Greeks. It was published in Shushu Jiuzhang (“Mathematical Treatise in Nine Sections”), written by Qin Jiushao and published in A.D. 1247.

## Proofs

Heron's original proof made use of cyclic quadrilaterals, while other arguments appeal to trigonometry as below, or to the incenter and one excircle of the triangle [2].

### Trigonometric proof using the Law of cosines

A modern proof, which uses algebra and is quite unlike the one provided by Heron (in his book Metrica), follows.[4] Let a, b, c be the sides of the triangle and α, β, γ the angles opposite those sides. We have

${\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}$

by the law of cosines. From this proof get the algebraic statement that

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

The altitude of the triangle on base a has length b·sin γ, and it follows

${\displaystyle b^{2}=h^{2}+d^{2}}$