Difference between revisions of "Heron's Formula"

Revision as of 21:27, 30 June 2006

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

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

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

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

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

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

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

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

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

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

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

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

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

@@ Line 8: / Line 8: @@
 where the [[semi-perimeter]] <math>s=\frac{a+b+c}{2}</math>.
+=== Proof ===
+<math>[ABC]=\frac{ab}{2}\sin C</math>
+<math>=\frac{ab}{2}\sqrt{1-\cos^2 C}</math>
+<math>=\frac{ab}{2}\sqrt{1-(\frac{a^2+b^2-c^2}{2ab})^2}</math>
+<math>=\sqrt{\frac{a^2b^2}{4}(1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2})}</math>
+<math>=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}</math>
+<math>=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}</math>
+<math>=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}</math>
+<math>=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}</math>
+<math>=\sqrt{s(s-a)(s-b)(s-c)}</math>
 === See Also ===
 * [[Brahmagupta's formula]]