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, 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.
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. 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
Heron's formula as given above is numerically unstable for triangles with a very small angle. 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 formulas resembling Heron's formula
Three other area formulas have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides a, b, and c respectively as ma, mb, and mc and their semi-sum (ma + mb + mc)/2 as σ, we have
Next, denoting the altitudes from sides a, b, and c respectively as ha, hb, and hc, and denoting the semi-sum of the reciprocals of the altitudes as we have
Finally, denoting the semi-sum of the angles' sines as S = [(sin α) + (sin β) + (sin γ)]/2, we have