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

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

where s is the semiperimeter of the triangle. It is calculated by

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

Heron's formula can also be written as

A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
A=\frac{1}{4}\sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}
A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}
A=\frac{1}{4}\sqrt{4a^2b^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[edit]

Let ΔABC be the triangle with sides a=7, b=4 and c=5. The semiperimeter is   s=\tfrac{1}{2}(a+b+c)=\tfrac{1}{2}(7+4+5)=8 , and the area is

 \begin{align} A &= \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}= \sqrt{8 \cdot (8-7) \cdot (8-4) \cdot (8-5)}\\ &=\sqrt{8 \cdot 1 \cdot 4 \cdot 3}=\sqrt{96}=4\sqrt{6} \approx 9.8 \end{align}

History[edit]

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.[2]

A formula equivalent to Heron's, namely

A=\frac1{2}\sqrt{a^2c^2-\left(\frac{a^2+c^2-b^2}{2}\right)^2}, where a \ge b \ge 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[edit]

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 cosine[edit]

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

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

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

\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

 \begin{align} A & = \frac{1}{2} (\mbox{base}) (\mbox{altitude}) \\ & = \frac{1}{2} ab\sin \gamma \\ & = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\ & = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\ & = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\ & = \sqrt{\frac{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)}{16}} \\ & = \sqrt{\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}\frac{(a + b + c)}{2}} \\ & = \sqrt{\frac{(a + b + c)}{2}\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}} \\ & = \sqrt{s(s-a)(s-b)(s-c)}. \end{align}

The difference of two squares factorization was used in two different steps.

Algebraic proof using the Pythagorean theorem[edit]

Triangle with altitude h cutting base c into d + (c − d).

By the Pythagorean theorem we have a^2=h^2+(c-d)^2 and b^2=h^2+d^2 according to the figure at the right. Subtracting these yields a^2-b^2=c^2-2cd. Thus

d=\frac{-a^2+b^2+c^2}{2c}.

Then we get for the height of the triangle that

 \begin{align} h^2 & = b^2-d^2=\left(\frac{2bc}{2c}\right)^2-\left(\frac{-a^2+b^2+c^2}{2c}\right)^2\\ & = \frac{(2bc-a^2+b^2+c^2)(2bc+a^2-b^2-c^2)}{4c^2}\\ & = \frac{((b+c)^2-a^2)(a^2-(b-c)^2)}{4c^2}\\ & = \frac{(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^2}\\ & = \frac{2(s-a)\cdot 2s\cdot 2(s-c)\cdot 2(s-b)}{4c^2}\\ & = \frac{4s(s-a)(s-b)(s-c)}{c^2} \end{align}

Taking the square root to get the height and inserting the expression into the area formula A=\frac{ch}{2} yields

A=\frac{c}{2}\cdot \frac{2\sqrt{s(s-a)(s-b)(s-c)}}{c}

from which Heron's formula follows.

Trigonometric proof using the Law of cotangents[edit]

Geometrical significance of s-a, s-b, and s-c. See the Law of cotangents for the reasoning behind this.

From the first part of the Law of cotangents proof,[3] we have that the triangle's area is both

 \begin{align} A &= r\big((s-a) + (s-b) + (s-c)\big) = r^2\left(\frac{s-a}{r} + \frac{s-b}{r} + \frac{s-c}{r}\right) \\ &= r^2\left(\cot{\frac{\alpha}{2}} + \cot{\frac{\beta}{2}} + \cot{\frac{\gamma}{2}}\right) \\ \end{align}

and

A = rs

but, since the sum of the half-angles is \tfrac{\pi}{2}, the triple cotangent identity applies, so the first of these is

 \begin{align} A &= r^2\left(\cot{\frac{\alpha}{2}} \cot{\frac{\beta}{2}} \cot{\frac{\gamma}{2}}\right) = r^2\left( \frac{s-a}{r}\cdot \frac{s-b}{r}\cdot \frac{s-c}{r}\right) \\ &= \frac{(s-a)(s-b)(s-c)}{r} \\ \end{align}

Combining the two, we get

A^2 = s(s-a)(s-b)(s-c)

from which the result follows.

Numerical stability[edit]

Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative [4] [5] involves arranging the lengths of the sides so that a \ge b \ge c and computing

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

The brackets in the above formula are required in order to prevent numerical instability in the evaluation.

Other area formulas resembling Heron's formula[edit]

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[6]

A = \frac{4}{3} \sqrt{\sigma (\sigma - m_a)(\sigma - m_b)(\sigma - m_c)}.

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 = (h_a^{-1} + h_b^{-1} + h_c^{-1})/2 we have[7]

A^{-1} = 4 \sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}.

Finally, denoting the semi-sum of the angles' sines as S = [(sin α) + (sin β) + (sin γ)]/2, we have[8]

A = D^{2} \sqrt{S(S-\sin \alpha)(S-\sin \beta)(S-\sin \gamma)}

where D is the diameter of the circumcircle: D=\tfrac{a}{\sin \alpha} = \tfrac{b}{\sin \beta} = \tfrac{c}{\sin \gamma}.

Generalizations[edit]

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,

 A =  \frac{1}{4} \sqrt{- \begin{vmatrix}    0 & a^2 & b^2 & 1 \\ a^2 & 0   & c^2 & 1 \\ b^2 & c^2 & 0   & 1 \\   1 &   1 &   1 & 0 \end{vmatrix} }

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.[9]

Heron-type formula for the volume of a tetrahedron[edit]

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[10]

 \text{volume} = \frac{\sqrt {\,( - a + b + c + d)\,(a - b + c + d)\,(a + b - c + d)\,(a + b + c - d)}}{192\,u\,v\,w}

where

     \begin{align} a & = \sqrt {xYZ} \\ b & = \sqrt {yZX} \\ c & = \sqrt {zXY} \\ d & = \sqrt {xyz} \\ X & = (w - U + v)\,(U + v + w) \\ x & = (U - v + w)\,(v - w + U) \\ Y & = (u - V + w)\,(V + w + u) \\ y & = (V - w + u)\,(w - u + V) \\ Z & = (v - W + u)\,(W + u + v) \\ z & = (W - u + v)\,(u - v + W). \end{align}

See also[edit]

Notes[edit]

  1. ^ "Fórmula de Herón para calcular el área de cualquier triángulo" (in Spanish). Retrieved 30 June 2012. 
  2. ^ Weisstein, Eric W., "Heron's Formula", MathWorld.
  3. ^ The second part of the Law of cotangents proof depends on Heron's formula itself, but this article depends only on the first part.
  4. ^ P. Sterbenz (1973). Floating-Point Computation, Prentice-Hall. 
  5. ^ W. Kahan (24 March 2000). "Miscalculating Area and Angles of a Needle-like Triangle". 
  6. ^ Benyi, Arpad, "A Heron-type formula for the triangle," Mathematical Gazette" 87, July 2003, 324–326.
  7. ^ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.
  8. ^ Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109.
  9. ^ D. P. Robbins, "Areas of Polygons Inscribed in a Circle", Discr. Comput. Geom. 12, 223-236, 1994.
  10. ^ W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?", [1], pp. 16-17.

References[edit]

External links[edit]