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

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

Heron's formula can also be written as

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.


Let ΔABC be the triangle with sides a=4, b=13 and c=15. The semiperimeter is    , and the area is

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.


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

, where

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.


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

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

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