In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first.
where s is the semi-perimeter of the triangle; that is,
Heron's formula can also be written as
Let △ABC be the triangle with sides a = 4, b = 13 and c = 15. This triangle’s semiperimeter is
s = 1/2(a + b + c) = 1/2(4 + 13 + 15) = 16, and the area is
The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book Metrica, written around AD 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,
Heron's original proof made use of cyclic quadrilaterals. Other arguments appeal to trigonometry as below, or to the incenter and one excircle of the triangle, or to De Gua's theorem (for the particular case of acute triangles).
Trigonometric proof using the law of cosines
A modern proof, which uses algebra and is quite different from 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. Applying the law of cosines we get
From this proof, we get the algebraic statement that
The altitude of the triangle on base a has length b sin γ, and it follows
The difference of two squares factorization was used in two different steps.
Algebraic proof using the Pythagorean theorem
The following proof is very similar to one given by Raifaizen. By the Pythagorean theorem we have b2 = h2 + d2 and a2 = h2 + (c − d)2 according to the figure at the right. Subtracting these yields a2 − b2 = c2 − 2cd. This equation allows us to express d in terms of the sides of the triangle:
For the height of the triangle we have that h2 = b2 − d2. By replacing d 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:
Trigonometric proof using the law of cotangents
and A = rs, but, since the sum of the half-angles is π/2, the triple cotangent identity applies, so the first of these is
Combining the two, we get
from which the result follows.
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 a ��� b ≥ c 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 a, b, and c respectively as ma, mb, and mc and their semi-sum 1/2(ma + mb + mc) 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 H = 1/2(h−1
a + h−1
b + h−1
c) we have
Finally, denoting the semi-sum of the angles' sines as S = 1/2(sin α + sin β + sin γ), we have
where D is the diameter of the circumcircle: D = a/sin α = b/sin β = c/sin γ.
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.
Heron-type formula for the volume of a tetrahedron
If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then
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