Heron's formula

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This article is about calculating the area of a triangle. For calculating a square root, see Heron's method.
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; that is,

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=4, b=13 and c=15. The semiperimeter is   s=\tfrac{1}{2}(a+b+c)=\tfrac{1}{2}(4+13+15)=16 , and the area is


\begin{align}
A &= \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} = \sqrt{16 \cdot (16-4) \cdot (16-13) \cdot (16-15)}\\
&= \sqrt{16 \cdot 12 \cdot 3 \cdot 1} = \sqrt{576} = 24.
\end{align}

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.

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,[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

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 cosines[edit]

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

\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).

The following proof is very similar to one given by Raifaizen.[5] By the Pythagorean theorem we have b^2=h^2+d^2 and a^2=h^2+(c-d)^2 according to the figure at the right. Subtracting these yields a^2-b^2=c^2-2cd. This equation allows to express d in terms of the sides of the triangle:

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

For the height of the triangle we have that h^2 = b^2-d^2. By replacing d with the formula given above and applying the difference of squares identity repeatedly we get


\begin{align}
h^2 & = b^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}

We now apply this result to the formula that calculates the area of a triangle from its height:


\begin{align}
A & = \frac{ch}{2}\\
& = \sqrt{\frac{c^2}{4}\cdot \frac{4s(s-a)(s-b)(s-c)}{c^2}}\\
& = \sqrt{s(s-a)(s-b)(s-c)}
\end{align}

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,[6] 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 [7] [8] 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[9]

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[10]

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[11]

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.[12]

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[13]


\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}

References[edit]

  1. ^ "Fórmula de Herón para calcular el área de cualquier triángulo" (in Spanish). Retrieved 30 June 2012. 
  2. ^ Heath, Thomas L. (1921). A History of Greek Mathematics (Vol II). Oxford University Press. pp. 321–323. 
  3. ^ Weisstein, Eric W., "Heron's Formula", MathWorld.
  4. ^ Niven, Ivan (1981). Maxima and Minima Without Calculus. The Mathematical Association of America. pp. 7–8. 
  5. ^ Raifaizen, Claude H. (1971). "A Simpler Proof of Heron's Formula". Mathematics Magazine 44 (1): 27–28. 
  6. ^ The second part of the Law of cotangents proof depends on Heron's formula itself, but this article depends only on the first part.
  7. ^ P. Sterbenz (1973). Floating-Point Computation, Prentice-Hall. 
  8. ^ W. Kahan (24 March 2000). "Miscalculating Area and Angles of a Needle-like Triangle". 
  9. ^ Benyi, Arpad, "A Heron-type formula for the triangle," Mathematical Gazette" 87, July 2003, 324–326.
  10. ^ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.
  11. ^ Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109.
  12. ^ D. P. Robbins, "Areas of Polygons Inscribed in a Circle", Discr. Comput. Geom. 12, 223-236, 1994.
  13. ^ W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?", [1], pp. 16-17.

External links[edit]