TITLE: Rationale for a convention: Why use the semiperimeter in Heron's formula?
QUESTION [11 upvotes]: Heron's formula says that the area of a triangle whose sides have lengths $a, b, c$ is $\sqrt{s(s-a)(s-b)(s-c)}$ where $s=(a+b+c)/2$ is the semiperimeter.  It can also be stated by saying that the area is $\frac14\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$.  Is there a substantial reason to prefer the first form, using the semiperimeter, over the second?  It seems as if that's the only form I've seen in published sources.
(Either way, it's the simplest function of $a$, $b$, $c$ that is 2nd-degree homogeneous and is equal to $0$ whenever the three vertices are on a common line or at a common point.)
PS inspired by comments:
Let's compare lengths:
$$
\sqrt{s(s-a)(s-b)(s-c)}\text{ where }s=(a+b+c)/2\text{ is the semiperimeter}
$$
$$
\phantom{\frac{\vert}{}}\sqrt{s(s-a)(s-b)(s-c)}\text{ where $s$ is the semiperimeter}
$$
$$
\sqrt{s(s-a)(s-b)(s-c)}\text{ where }s=(a+b+c)/2
$$
$$
\frac 1 4 \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}
$$

REPLY [2 votes]: You need to see the proof of that formula. If the proof has not to do directly with the concept of semiperimeter, than it means that two equivalent ways writting that formula are of the same power (substantially speaking). One of the proves is in this link here: http://jwilson.coe.uga.edu/emt725/Heron/HeronProofAlg.html and it has to do nothing with the concept of perimeter, so I can say that it doesn't have a substantial difference on the way you write the Heron's formula.