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Difference between revisions of "Heron's Formula"

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'''Heron's Formula''' (sometimes called Hero's formula) is a [[mathematical formula | formula]] for finding the [[area]] of a [[triangle]] given only the three side lengths.
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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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== Theorem ==
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== History ==
  
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For any triangle with side lengths <math>{a}, {b}, {c}</math>, the area <math>{A}</math> can be found using the following formula:
+
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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<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math>
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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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where the [[semi-perimeter]] <math>s=\frac{a+b+c}{2}</math>.
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== Statement ==
  
 +
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
 +
 +
<cmath>A = \sqrt{s(s-a)(s-b)(s-c)}</cmath>
  
 
== Proof ==
 
== Proof ==
  
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<math>[ABC]=\frac{ab}{2}\sin C</math>
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A common formula for area states that
 +
<cmath>[ABC]=\frac{ab}{2}\sin C</cmath>
 +
which simplifies to
 +
<cmath>[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.</cmath>
  
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<math>=\frac{ab}{2}\sqrt{1-\cos^2 C}</math>
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The [[Law of Cosines]] states that in triangle <math>ABC</math>, <math>c^2 = a^2 + b^2 - 2ab\cos C</math>, which can be written as <math>\cos C = \frac{a^2 + b^2 - c^2}{2ab}</math>. Thus,
 +
<math>[ABC]=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}.</math>
  
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<math>=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}</math>
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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}} \\
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&= \sqrt{s(s - a)(s - b)(s - c)}
 +
\end{align*}
  
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<math>=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}</math>
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== Isosceles Triangle Simplification ==
 +
 
 +
<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> for all triangles
  
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<math>=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}</math>
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<math>b=c</math> for all isosceles triangles
  
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<math>=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}</math>
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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>
  
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<math>=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}</math>
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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.
  
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<math>=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}</math>
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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.
  
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<math>=\sqrt{s(s-a)(s-b)(s-c)}</math>
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== 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.
  
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==Isosceles Triangle Simplification==
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== Problems ==
  
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<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> for all triangles
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=== Introductory ===
  
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<math>b=c</math> for all isosceles triangles
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=== Intermediate ===
  
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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>\blacksquare</math>
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=== Olympiad ===
  
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==Example==
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{{problem}}
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Let's say that you have a right triangle with the sides 3,4, and 5. Your semi- perimeter would be 6.
 
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Then you have 6-3=3, 6-4=2, 6-5=1.
 
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1+2+3= 6
 
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<math> 6\cdot 6 = 36</math>
 
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The square root of 36 is 6. The area of your triangle is 6.
 
  
 
== 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]
 
−
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 20:11, 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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