Difference between revisions of "Heron's Formula"

 
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Sometimes called "hero's formula", it states that for any triangle with side lengths a, b, and c,
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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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== Theorem ==
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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:
  
 
<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math>
 
<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math>
where
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<math>s=\frac{a+b+c}{2}</math>
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where the [[semi-perimeter]] <math>s=\frac{a+b+c}{2}</math>.
and
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A=area
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== Proof ==
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<math>[ABC]=\frac{ab}{2}\sin C</math>
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<math>=\frac{ab}{2}\sqrt{1-\cos^2 C}</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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<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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<math>=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}</math>
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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>=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}</math>
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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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<math>=\sqrt{s(s-a)(s-b)(s-c)}</math>
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==Isosceles Triangle Simplification==
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<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> for all triangles
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<math>b=c</math> for all isosceles 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>\blacksquare</math>
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==Example==
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Let's say that you have a right triangle with the sides <math>3</math> ,<math>4</math> , and <math>5</math>. Your semi- perimeter would be <math>6</math> since <math>(3+4+5)</math> ÷ <math>2</math> is <math>6</math>.
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Then you have <math>6-3=3</math>, <math>6-4=2</math>, <math>6-5=1</math>.
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<math>1+2+3=6.</math>
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<math> 6\cdot 6 = 36</math>
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The square root of <math>36</math> is <math>6</math>. The area of your triangle is <math>6</math>.
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== See Also ==
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* [[Brahmagupta's formula]]
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* [[Geometry]]
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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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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:
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* Computing the square root is much slower than multiplication.
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* For triangles with area close to zero Heron's formula computed using floating point variables suffers from precision problems.
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[[Category:Geometry]]
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[[Category:Theorems]]

Latest revision as of 22:49, 1 February 2021

Heron's Formula (sometimes called Hero's formula) is a formula for finding the area of a triangle given only the three side lengths.

Theorem

For any triangle with side lengths ${a}, {b}, {c}$, the area ${A}$ can be found using the following formula:

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

where the semi-perimeter $s=\frac{a+b+c}{2}$.


Proof

$[ABC]=\frac{ab}{2}\sin C$

$=\frac{ab}{2}\sqrt{1-\cos^2 C}$

$=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$

$=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}$

$=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$

$=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$

$=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$

$=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$

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

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

Example

Let's say that you have a right triangle with the sides $3$ ,$4$ , and $5$. Your semi- perimeter would be $6$ since $(3+4+5)$ ÷ $2$ is $6$. Then you have $6-3=3$, $6-4=2$, $6-5=1$. $1+2+3=6.$ $6\cdot 6 = 36$ The square root of $36$ is $6$. The area of your triangle is $6$.

See Also

External Links

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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