Difference between revisions of "Heron's Formula"

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 $6$ .

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.

@@ Line 16: / Line 16: @@
 <math>=\frac{ab}{2}\sqrt{1-\cos^2 C}</math>
-<math>=\frac{ab}{3}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}</math>
+<math>=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}</math>
 <math>=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}</math>
@@ Line 30: / Line 30: @@
 <math>=\sqrt{s(s-a)(s-b)(s-c)}</math>
-== Proof 2 ==
+==Isosceles Triangle Simplification==
-Imagine a triangle with an altitude dropped from vertex A.
-Now the side of "a" is split into two sections.
-Label the right section "x" and the left section "a-x".
-Now we can prove Heron's formula.
+<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> for all triangles
-<math>b^2 - (a-x)^2 = h^2</math>
+<math>b=c</math> for all isosceles triangles
-<math>b^2 - a^2 + 2ax - x^2 = h^2</math>
-<math>c^2 - h^2 = h^2</math>
+<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>
-<math>b^2 - a^2 + 2ax - x^2 = c^2 - x^2</math>
+==Example==
+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>.
-<math>2ax = c^2 - b^2 + a^2</math>
+Then you have <math>6-3=3</math>, <math>6-4=2</math>, <math>6-5=1</math>.
-<math>x = \frac{c^2 - b^2 + a^2}{2a}</math>
+<math>1+2+3=6.</math>
+<math> 6\cdot 6 = 36</math>
-<math>h^2 = c^2 - x^2 = (c-x) (c+x)</math>
+The square root of <math>36</math> is <math>6</math>. The area of your triangle is <math>6</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>[ABC] = \sqrt(s(s - a)(s - b)(s - c))</math>
 == See Also ==

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