# What is Heron's formula?

Jan 12, 2015

Heron's formula allows you to evaluate the area of a triangle knowing the length of its three sides.
The area $A$ of a triangle with sides of lengths $a , b$ and $c$ is given by:

A=sqrt(sp×(sp-a)×(sp-b)×(sp-c))

Where $s p$ is the semiperimeter:

$s p = \frac{a + b + c}{2}$

For example; consider the triangle:

The area of this triangle is A=(base×height)/2
So: A=(4×3)/2=6
Using Heron's formula:
$s p = \frac{3 + 4 + 5}{2} = 6$
And:
A=sqrt(6×(6-5)×(6-4)×(6-3))=6

The demonstration of Heron's formula can be found in textbooks of geometry or maths or in many websites. If you need it have a look at:
http://en.m.wikipedia.org/wiki/Heron%27s_formula

Jun 15, 2018

Heron's Formula is usually the worst choice for finding the area of a triangle.

#### Explanation:

Alternatives:

Area $S$ of a triangle with sides $a , b , c$

$16 {S}^{2} = \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right)$

Area $S$ of a triangle with squared sides $A , B , C$

$16 {S}^{2} = 4 A B - {\left(C - A - B\right)}^{2} = {\left(A + B + C\right)}^{2} - 2 \left({A}^{2} + {B}^{2} + {C}^{2}\right)$

Area of a triangle with vertices $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \left({x}_{3} , {y}_{3}\right)$

$S = \frac{1}{2} | \left({x}_{1} - {x}_{3}\right) \left({y}_{2} - {y}_{3}\right) - \left({x}_{2} - {x}_{3}\right) \left({y}_{1} - {y}_{3}\right) | = \frac{1}{2} | {x}_{1} {y}_{2} - {x}_{2} {y}_{1} + {x}_{2} {y}_{3} - {x}_{3} {y}_{2} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} |$

Oh yeah, Heron's Formula is

$S = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s = \frac{1}{2} \left(a + b + c\right)$