Difference between revisions of "Heron's Formula"
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<math>=\sqrt{s(s-a)(s-b)(s-c)}</math> | <math>=\sqrt{s(s-a)(s-b)(s-c)}</math> | ||
− | == Proof == | + | == Proof 2 == |
+ | |||
+ | \includegraphics{Heron.png} | ||
+ | |||
<math>b^2 - (a-x)^2 = h^2</math> | <math>b^2 - (a-x)^2 = h^2</math> | ||
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<math>2ax = c^2 - b^2 + a^2</math> | <math>2ax = c^2 - b^2 + a^2</math> | ||
− | <math>x = (c^2 - b^2 + a^2)/2a</math> | + | <math>x = \frac{c^2 - b^2 + a^2}{2a}</math> |
+ | |||
+ | <math>h^2 = c^2 - x^2 = (c-x) (c+x)</math> | ||
+ | <math> = (c - \frac{c^2 - b^2 + a^2}{2a}) (c + \frac{c^2 - b^2 + a^2}{2a})</math> | ||
+ | <math> = \frac{-(2ac + c^2 + a^2) + b^2}{2a} * \frac{(2ac + c^2 + a^2) - b^2}{2a}</math> | ||
+ | <math> = \frac{b^2 - (a^2 - 2ac + c^2)}{2a}</math> | ||
+ | <math> = \frac{b^2 - (a - c)^2}{2a} * \frac{(a + c)^2 - b^2}{2a}</math> | ||
+ | <math> = \frac{(b + a - c)(b - a + c)}{2a} * \frac{(a + c -b)(a + b + c)}{2a}</math> | ||
+ | <math> = \frac{(2s - 2c)(2s - 2a)(2s - 2b)(2s)}{4a^2}</math> | ||
+ | |||
+ | <math>h = \frac{4\sqrt(s(s - a)(s - b)(s - c))}{2a}</math> | ||
+ | <math> = \frac{2\sqrt(s(s - a)(s - b)(s - c))}{a}</math> | ||
+ | |||
+ | <math>[ABC] = \frac{1}{2} * a * h</math> | ||
+ | <math>Substitute h with the equation and you get</math> | ||
+ | |||
+ | $[ABC] = \sqrt(s(s - a)(s - b)(s - c)) | ||
== See Also == | == See Also == |
Revision as of 19:04, 8 February 2016
Heron's Formula (sometimes called Hero's formula) is a formula for finding the area of a triangle given only the three side lengths.
Contents
[hide]Theorem
For any triangle with side lengths , the area can be found using the following formula:
where the semi-perimeter .
Proof
Proof 2
\includegraphics{Heron.png}
$[ABC] = \sqrt(s(s - a)(s - b)(s - c))
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.