Difference between revisions of "Heron's Formula"

(Example)
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Then you have 6-3=3, 6-4=2, 6-5=1.
 
Then you have 6-3=3, 6-4=2, 6-5=1.
 
1+2+3= 6
 
1+2+3= 6
<math> 6*6 = 36</math>
+
<math> 6\cdot 6 = 36</math>
 
The square root of 36 is 6. The area of your triangle is 6.
 
The square root of 36 is 6. The area of your triangle is 6.
  

Revision as of 19:47, 27 May 2020

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. 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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