Just for the fun of it, I worked out Heron's formula for the area of a triangle in terms of its sides, $A =\sqrt{s(s-a)(s-c)(s-b)}$. I've included it below.
In my (and, I assume, just about everybody's) derivation, the four terms appear as sort of algebraic coincidences, and this made me wonder:
Is there a derivation in which each term has a geometric meaning?
Also, the derivation begins by choosing a particular side, and ends up with the well-known formula which is symmetrical on all the sides. So, that brings to mind this question:
Is there a derivation which is symmetrical in all the sides all the way through?
And those are my questions.
I'm fine with this being a duplicate.
Here is my derivation.
Sides $a, b, c$, altitude $h$ on $c$.
$c$ divided into $p$ and $c-p$.
Equations:
$a^2 = p^2+h^2, b^2 = (c-p)^2+h^2 = c^2-2cp+p^2+h^2 $.
$a^2-b^2 =2cp-c^2$ so $p =\dfrac{a^2-b^2+c^2}{2c} $.
Then
$\begin{array}\\ h^2 &=a^2-p^2\\ &=a^2-\dfrac{(a^2-b^2+c^2)^2}{(2c)^2}\\ &=\dfrac{(2ac)^2-(a^2-b^2+c^2)^2}{(2c)^2}\\ &=\dfrac{(2ac-(a^2-b^2+c^2))(2ac+(a^2-b^2+c^2))}{(2c)^2}\\ &=\dfrac{(b^2-(a^2-2ac+c^2))(-b^2+(a^2+2ac+c^2))}{(2c)^2}\\ &=\dfrac{(b^2-(a-c)^2)((a+c)^2-b^2)}{(2c)^2}\\ &=\dfrac{(b-(a-c))(b+(a-c))((a+c)-b)((a+c)+b)}{(2c)^2}\\ &=\dfrac{(b+c-a)(a+b-c)(a+c-b)(a+b+c)}{(2c)^2}\\ &=\dfrac{(a+b+c-2a)(a+b+c-2c)(a+b+c-2b)(a+b+c)}{(2c)^2}\\ &=\dfrac{(2s-2a)(2s-2c)(2s-2b)2s}{4c^2} \qquad s=\dfrac{a+b+c}{2}\\ &=\dfrac{4(s-a)(s-c)(s-b)s}{c^2}\\ \text{so,} &\text{ since } A = \dfrac{hc}{2}, A^2 = \dfrac{h^2c^2}{4}\\ A^2 &=s(s-a)(s-c)(s-b)\\ A &=\sqrt{s(s-a)(s-c)(s-b)}\\ \end{array} $