Heron's formula states that the area of a triangle whose sides have lengths,, and is where is the semi-perimeter of the triangle; that is, Heron's formula can also be written as
Example
Let be the triangle with sides, and. The semiperimeter is, and the area is In this example, the side lengths and area are 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 of Alexandria, and a proof can be found in his book, Metrica, written CE 60. It has been suggested that Archimedes knew the formula over two centuries earlier, 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. A formula equivalent to Heron's, namely was discovered by the Chinese independently of the Greeks. It was published in Mathematical Treatise in Nine Sections.
A modern proof, which uses algebra and is quite different from the one provided by Heron, follows. Let,, be the sides of the triangle and,, the angles opposite those sides. Applying the law of cosines we get From this proof, we get the algebraic statement that The altitude of the triangle on base has length, and it follows The difference of two squares factorization was used in two different steps.
The following proof is very similar to one given by Raifaizen. By the Pythagorean theorem we have and according to the figure at the right. Subtracting these yields. This equation allows us to express in terms of the sides of the triangle: For the height of the triangle we have that. By replacing with the formula given above and applying the difference of squares identity we get We now apply this result to the formula that calculates the area of a triangle from its height:
From the first part of the Law of cotangents proof, we have that the triangle's area is both and, but, since the sum of the half-angles is, the triple cotangent identity applies, so the first of these is Combining the two, we get from which the result follows.
Numerical stability
Heron's formula as given above is numerically unstable for triangles with a very small angle when using floating point arithmetic. A stable alternative involves arranging the lengths of the sides so that and computing The brackets in the above formula are required in order to prevent numerical instability in the evaluation.
Other area formulae resembling Heron's formula
Three other area formulae have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides,, and respectively as,, and and their semi-sum as, we have Next, denoting the altitudes from sides,, and respectively as,, and, and denoting the semi-sum of the reciprocals of the altitudes as we have Finally, denoting the semi-sum of the angles' sines as, we have where is the diameter of the circumcircle:.
Generalizations
Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special cases of Bretschneider's formula for the area of a quadrilateral. Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero. Heron's formula is also a special case of the formula for the area of a trapezoid or trapezium based only on its sides. Heron's formula is obtained by setting the smaller parallel side to zero. Expressing Heron's formula with a Cayley–Menger determinant in terms of the squares of the distances between the three given vertices, illustrates its similarity to Tartaglia's formula for the volume of a three-simplex. Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by David P. Robbins.
Heron-type formula for the volume of a tetrahedron
If,,,,, are lengths of edges of the tetrahedron, then where
