Basic Derivation of Heron’s Formula
Introduction
This post is a basic derivation of the famous Heron’s formula, which allows calculation of the area of any triangle in terms of its side lengths 
. We refer to the following diagram:
As is convention, let sides be opposite to angles 
. Further, let the altitude length 
and 
. Then, by the Pythagorean Theorem applied to triangles 
and 
, we have the following two equations:
Subtracting the two equations cancels the 
and yields 
. Thus
Plugging 
back into the first equation, we have
We compute the square of the triangle’s area as 
. Plugging in and noting that the 
cancels out, we obtain
by difference of squares. Note that each term in the numerator may itself be simplified by difference of squares, namely
Thus,
![\begin{equation*} A^2 = \frac{(a + b + c)(a - b + c)(a + b - c)(-a + b + c)}{16}][ = (\frac{a + b + c}{2})(\frac{a - b + c}{2})(\frac{a + b - c}{2})(\frac{-a + b + c}{2}) = (s)(s - b)(s - c)(s - a) \end{equation*}](https://i0.wp.com/ignitedminds.blog/wp-content/ql-cache/quicklatex.com-efff5bc9887aae3fd0f2f598d0cacc65_l3.png?resize=640%2C26&ssl=1 "Rendered by QuickLaTeX.com")
where 
. Finally, we arrive at Heron’s Formula:
Although this process may seem a bit tedious, the motivation was clear and the algebra was greatly simplified using the difference of squares factorization.