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Difference between revisions of "Heron's Formula"
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==Theorem==
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'''Heron's Formula''' (sometimes referred to as '''Hero's formula''') is a [[mathematical formula]] for finding the [[area]] of a [[triangle]] given the three side lengths.
  
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In [[Geometry|geometry]], '''Heron's formula''' (or '''Hero's formula''') gives the [[Area|area]] of a [[Triangle|triangle]] in terms of three side lengths, <math>a</math>, <math>b</math>, <math>c</math>. Letting <math>s</math> be the [[Semi-perimeter|semiperimeter]] of the triangle, <math>s = \frac{1}{2} (a + b + c)</math>, the area <math>A</math> is
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== History ==
  
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<cmath>A=\sqrt{s(s-a)(s-b)(s-c)}</cmath>
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The formula is credited to [[Hero of Alexandria]], and a proof can be found in his book ''Metrica''. Mathematical historian [[Thomas Heath]] 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.
  
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It is named after the first-century engineer [https://en.wikipedia.org/wiki/Heron_of_Alexandria Heron of Alexandria] (or Hero) who proved it in his work ''Metrica'', though it was probably known centuries earlier.
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A formula equivalent to Heron's was discovered by Chinese mathematician [[Qin Jiushao]], published in ''[[Mathematical Treatise in Nine Sections]]'' in 1247:
 +
<cmath>A = \frac{1}{2} \sqrt{a^2 c^2 - (\frac{a^2 + c^2 - b^2}{2}^2)}</cmath>
  
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==History==
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== Statement ==
  
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The formula is credited to [https://en.wikipedia.org/wiki/Hero_of_Alexandria Heron (or Hero) of Alexandria], and a proof can be found in his book ''Metrica''. Mathematical historian [https://en.wikipedia.org/wiki/Thomas_Heath_(classicist) Thomas Heath] suggested that [https://en.wikipedia.org/wiki/Archimedes 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.
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In any triangle with side lengths <math>a</math>, <math>b</math>, <math>c</math>, and [[semiperimeter]] <math>s</math> the area <math>A</math> is equal to
  
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A formula equivalent to Heron's was discovered by Chinese mathematician Qin Jiushao:
+
<cmath>A = \sqrt{s(s-a)(s-b)(s-c)}</cmath>
  
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<cmath>A = \frac{1}{2} \sqrt{a^2 c^2 - (\frac{a^2 + c^2 - b^2}{2}^2)}</cmath> published in ''[https://en.wikipedia.org/wiki/Mathematical_Treatise_in_Nine_Sections Mathematical Treatise in Nine Sections]'' ([https://en.wikipedia.org/wiki/Qin_Jiushao Qin Jiushao], 1247).
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== Proof ==
  
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== Proof ==
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A common formula for area states that
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Using basic [[Trigonometry]], we have
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<cmath>[ABC]=\frac{ab}{2}\sin C</cmath>
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<cmath>[ABC]=\frac{ab}{2}\sin C, </cmath>
 
 
which simplifies to
 
which simplifies to
 
<cmath>[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.</cmath>
 
<cmath>[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.</cmath>
Line 34: Line 34:
 
\end{align*}
 
\end{align*}
  
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==Isosceles Triangle Simplification==
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== Isosceles Triangle Simplification ==
  
 
<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> for all triangles
 
<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> for all triangles
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<math>A=\sqrt{s(s-a)(s-b)(s-b)}</math> simplifies to <math>A=(s-b)\sqrt{s(s-a)}</math>
 
<math>A=\sqrt{s(s-a)(s-b)(s-b)}</math> simplifies to <math>A=(s-b)\sqrt{s(s-a)}</math>
  
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==Square root simplification/modification==
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== Square root simplification/modification ==
 
From <cmath>A=\sqrt{s(s-a)(s-b)(s-c)}</cmath> We can "take out" the <math>1/2</math> in each <math>s</math>, then we have <cmath>A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}</cmath>Using the [[difference of squares]] on the first two and last two factors, we get <cmath>A=\frac{1}{4}\sqrt{(b^2+2bc+c^2-a^2)(a^2-b^2+2bc-c^2)}</cmath>and using the difference of squares again, we get <cmath>A=\frac{1}{4}\sqrt{(2bc)^2-(-a^2+b^2+c^2)^2}</cmath> From this equation (although seemingly not symmetrical), it is much easier to calculate the area of a certain triangle with side lengths with quantities with square roots. One can remember this formula by noticing that when finding the cosine of an angle in a triangle, the formula is <cmath>\cos{A}=\frac{-a^2+b^2+c^2}{2bc}</cmath> and the two terms in the formula are just the [[denominator]] and [[numerator]] of the fraction for <math>\cos{A}</math>, only they're squared. This can also serve as a reason for why the area <math>A</math> is never imaginary. This is equivalent of ending at step <math>4</math> in the proof and distributing.
 
From <cmath>A=\sqrt{s(s-a)(s-b)(s-c)}</cmath> We can "take out" the <math>1/2</math> in each <math>s</math>, then we have <cmath>A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}</cmath>Using the [[difference of squares]] on the first two and last two factors, we get <cmath>A=\frac{1}{4}\sqrt{(b^2+2bc+c^2-a^2)(a^2-b^2+2bc-c^2)}</cmath>and using the difference of squares again, we get <cmath>A=\frac{1}{4}\sqrt{(2bc)^2-(-a^2+b^2+c^2)^2}</cmath> From this equation (although seemingly not symmetrical), it is much easier to calculate the area of a certain triangle with side lengths with quantities with square roots. One can remember this formula by noticing that when finding the cosine of an angle in a triangle, the formula is <cmath>\cos{A}=\frac{-a^2+b^2+c^2}{2bc}</cmath> and the two terms in the formula are just the [[denominator]] and [[numerator]] of the fraction for <math>\cos{A}</math>, only they're squared. This can also serve as a reason for why the area <math>A</math> is never imaginary. This is equivalent of ending at step <math>4</math> in the proof and distributing.
  
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===Note===
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=== Note ===
 
Replacing <math>-a^2+b^2+c^2</math> as <math>2bc\cos{A}</math>, the area simplifies down to <math>\frac{1}{4}\sqrt{(2bc\sin{A})^2}</math>, or <math>\frac{1}{4}\cdot2bc\sin{A}</math>, or <math>\frac{1}{2}bc\sin{A}</math>, another common area formula for the triangle.
 
Replacing <math>-a^2+b^2+c^2</math> as <math>2bc\cos{A}</math>, the area simplifies down to <math>\frac{1}{4}\sqrt{(2bc\sin{A})^2}</math>, or <math>\frac{1}{4}\cdot2bc\sin{A}</math>, or <math>\frac{1}{2}bc\sin{A}</math>, another common area formula for the triangle.
  
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==Example==
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== Note ==
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In general, it is a good advice '''not''' to use Heron's formula in computer programs whenever we can avoid it. For example, whenever vertex coordinates are known, vector product is a much better alternative. Main reasons:
 +
* Computing the square root is much slower than multiplication.
 +
* For triangles with area close to zero Heron's formula computed using floating point variables suffers from precision problems.
  
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Let <math>\triangle ABC</math> be the triangle with sides <math>a = 3</math>, <math>b = 4</math>, and <math>c = 5</math>. This triangle's semiperimeter is
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== Problems ==
  
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\begin{align*}
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=== Introductory ===
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s &= \frac{1}{2} (a + b + c) \\
 
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&= \frac{1}{2} (3 + 4 + 5) \\
 
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&= 16
 
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\end{align*}
 
  
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therefore <math>s - a = 13</math>, <math>s - b = 12</math>, <math>s - c = 11</math>, and the area is
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=== Intermediate ===
  
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\begin{align*}
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=== Olympiad ===
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A &= \sqrt{s(s - a)(s - b)(s - c)} \\
 
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&= \sqrt{13 \times 12 \times 11} \\
 
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&= 24
 
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\end{align*}
 
  
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In this example, the triangle's side lengths and area are [[Integer|integers]], making it a [https://en.wikipedia.org/wiki/Heronian_triangle Heronian triangle]. However, Heron's formula works equally well when the side lengths are [[Real number|real numbers]]. As long as they obey the strict [[Triangle Inequality|triangle inequality]], they define a triangle in the [https://en.wikipedia.org/wiki/Euclidean_plane Euclidean plane] whose area is a positive real number.
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{{problem}}
  
 
== See Also ==
 
== See Also ==
 +
 
* [[Brahmagupta's formula]]
 
* [[Brahmagupta's formula]]
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* [[Geometry]]
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* [[Triangle]]
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* [[Area]]
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== External Links ==
 
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* [http://www.scriptspedia.org/Heron%27s_Formula Heron's formula implementations in C++, Java and PHP]
 
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* [http://www.artofproblemsolving.com/Resources/Papers/Heron.pdf Proof of Heron's Formula Using Complex Numbers]
 
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* [https://en.wikipedia.org/wiki/Heron's_formula Heron's Formula - Wikipedia]
 
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In general, it is a good advice <b>not</b> to use Heron's formula in computer programs whenever we can avoid it. For example, whenever vertex coordinates are known, vector product is a much better alternative. Main reasons:
 
−
* Computing the square root is much slower than multiplication.
 
−
* For triangles with area close to zero Heron's formula computed using floating point variables suffers from precision problems.
 
  
 
[[Category:Geometry]]
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]
 +
{{stub}}

Latest revision as of 19:11, 24 February 2025

Heron's Formula (sometimes referred to as Hero's formula) is a mathematical formula for finding the area of a triangle given the three side lengths.

History

The formula is credited to Hero of Alexandria, and a proof can be found in his book Metrica. Mathematical historian Thomas Heath 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 was discovered by Chinese mathematician Qin Jiushao, published in Mathematical Treatise in Nine Sections in 1247: \[A = \frac{1}{2} \sqrt{a^2 c^2 - (\frac{a^2 + c^2 - b^2}{2}^2)}\]

Statement

In any triangle with side lengths $a$, $b$, $c$, and semiperimeter $s$ the area $A$ is equal to

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

Proof

A common formula for area states that \[[ABC]=\frac{ab}{2}\sin C\] which simplifies to \[[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.\]

The Law of Cosines states that in triangle $ABC$, $c^2 = a^2 + b^2 - 2ab\cos C$, which can be written as $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Thus, $[ABC]=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}.$

Now, we can simplify: \begin{align*} [ABC] &= \sqrt{\frac{a^2b^2}{4} \left( 1 - \frac{(a^2 + b^2 - c^2)^2}{4a^2b^2} \right)} \\ &= \sqrt{\frac{4a^2b^2 - (a^2 + b^2 - c^2)^2}{16}} \\ &= \sqrt{\frac{(2ab + a^2 + b^2 - c^2)(2ab - a^2 - b^2 + c^2)}{16}} \\ &= \sqrt{\frac{((a + b)^2 - c^2)((a - b)^2 - c^2)}{16}} \\ &= \sqrt{\frac{(a + b + c)(a + b - c)(b + c - a)(a + c - b)}{16}} \\ &= \sqrt{s(s - a)(s - b)(s - c)} \end{align*}

Isosceles Triangle Simplification

$A=\sqrt{s(s-a)(s-b)(s-c)}$ for all triangles

$b=c$ for all isosceles triangles

$A=\sqrt{s(s-a)(s-b)(s-b)}$ simplifies to $A=(s-b)\sqrt{s(s-a)}$

Square root simplification/modification

From \[A=\sqrt{s(s-a)(s-b)(s-c)}\] We can "take out" the $1/2$ in each $s$, then we have \[A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\]Using the difference of squares on the first two and last two factors, we get \[A=\frac{1}{4}\sqrt{(b^2+2bc+c^2-a^2)(a^2-b^2+2bc-c^2)}\]and using the difference of squares again, we get \[A=\frac{1}{4}\sqrt{(2bc)^2-(-a^2+b^2+c^2)^2}\] From this equation (although seemingly not symmetrical), it is much easier to calculate the area of a certain triangle with side lengths with quantities with square roots. One can remember this formula by noticing that when finding the cosine of an angle in a triangle, the formula is \[\cos{A}=\frac{-a^2+b^2+c^2}{2bc}\] and the two terms in the formula are just the denominator and numerator of the fraction for $\cos{A}$, only they're squared. This can also serve as a reason for why the area $A$ is never imaginary. This is equivalent of ending at step $4$ in the proof and distributing.

Note

Replacing $-a^2+b^2+c^2$ as $2bc\cos{A}$, the area simplifies down to $\frac{1}{4}\sqrt{(2bc\sin{A})^2}$, or $\frac{1}{4}\cdot2bc\sin{A}$, or $\frac{1}{2}bc\sin{A}$, another common area formula for the triangle.

Note

In general, it is a good advice not to use Heron's formula in computer programs whenever we can avoid it. For example, whenever vertex coordinates are known, vector product is a much better alternative. Main reasons:

  • Computing the square root is much slower than multiplication.
  • For triangles with area close to zero Heron's formula computed using floating point variables suffers from precision problems.

Problems

Introductory

Intermediate

Olympiad

This problem has not been edited in. Help us out by adding it.

See Also

This article is a stub. Help us out by expanding it.

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