Rationale for a convention: Why use the semiperimeter in Heron's formula? 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:
$$
\sqrt{s(s-a)(s-b)(s-c)}\text{ where }s=(a+b+c)/2\text{ is the semiperimeter}
$$
$$
\phantom{\frac{\vert}{}}\sqrt{s(s-a)(s-b)(s-c)}\text{ where $s$ is the semiperimeter}
$$
$$
\sqrt{s(s-a)(s-b)(s-c)}\text{ where }s=(a+b+c)/2
$$
$$
\frac 1 4 \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}
$$

The reason I think is simplicity:
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
Is much more simple than:
$$A=\frac{1}{4}\sqrt{(a+b+c)(b+c−a)(c+a−b)(a+b−c)}$$
I see in the comments that you say we have to add $\text{where } 2s=a+b+c$, in the standard form, and so the length of the two forms become the same. However, this is a useless argument. Continuing with the logic we must add, $\text{where a,b,c  are lengths of sides of the triangle}.$ This, everyone will agree that is pointless, and will unnecessarily increase the length and apparent complexity of the formula.
The same goes with $s$ here, we do need to add $s$ is this and this. Anyone studying geometry will immediately recognize that $s$ stands for the semi-perimeter. For example, we would never recognize in a geometry question, $p$ as the perimeter, however when we see $s$, the first thought that comes is the semi-perimeter. Hence, I think, its a bit stubborn to add that part as you have done in the postscript, and then compare lengths. 
And this said, the point is not on length but on how simple the formula is. The better form must be the one that allows more easily to do what the formula is about, that is calculating the area. Let us take an example with a $13, 14, 15$ triangle, and calculate using both forms as I would have done myself with a pen and paper:
$$
\begin{array}{c|lcr}
\text{Form 1} & \text{     Form 2} \\
\hline
{ s=\frac{13+14+15}{2}=21 \\
  s - a=21 - 13=8 \\
  s - b=21 - 14=7 \\
  s - c=21 - 15=6 \\
  \text{So, } A=\sqrt{21\cdot 7 \cdot 8 \cdot 6}=\sqrt{3 \cdot (7 \cdot 7) \cdot 2 \cdot 4  \cdot 2 \cdot  3} \\
   =7\cdot 3\cdot 2\cdot 2=84 } & {a + b + c =13+14+15=42 \\
 a + b - c =13+14-15=27-15=12 \\
 a + c - b =13+15-14=28-14=14 \\
 b + c - a =14+15-13=28-16=16 \\
 \text{So, } A=\frac{1}{4}\sqrt{42\cdot 12 \cdot 14 \cdot 16}\\=\frac{1}{4}\sqrt{2 \cdot 21 \cdot 2 \cdot 6 \cdot 2 \cdot 7 \cdot 2 \cdot 8} \\=\sqrt{21 \cdot 6 \cdot 7 \cdot 8}=7\cdot 3\cdot 2\cdot 2=84}\\ 
\end{array}
$$
Even with small dimensions, we can see that the first form has an advantage. We can calculate $s$ easily and then $s-a$, $s-b$, .. as its just a subtraction of two terms. However, calculating $a + b - c$, $a + c - b$ .. are three terms calculation are very unrelated to each other. This becomes clearer when the sides are larger and variable. Thus, we can see that the standard form has advantages, its easier to use, write and state. There is one more important aspect. 
There is an important area formula in terms of $s$, using the inradius, $A =rs$, where r is the inradius. If we draw the incircle, and the sides divided by the points of tangency can be conveniently expressed in terms of $s$. There are many examples where properties of triangles could be reprsented in terms of $s$, and without using $s$, it would be very untidy. As an example, the standard form allows a convenient expression of the inradius, whle the other form would give a very untidy expression:
$$r=\frac{A}{s}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$    
Generalizations of the Herons formula, like the Brahmagupta's formula, or even the Bretschneider's formula, which gives us the area of a general convex quadrilateral:
$$A=\sqrt {(s-a)(s-b)(s-c)(s-d) - abcd  \cdot \cos^2 \left(\frac{\alpha + \gamma}{2}\right)}$$
This is a very convenient and simple expression. However, if we replace $s$ by $\frac{a+b+c+d}{2}$, and rewrite the formula without introducing $s$:
$$A = \frac{1}{4}\sqrt{(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d) - 16abcd \cos^2 \left(\frac{\alpha + \gamma}{2}\right)}$$   
This is an extremely unwieldy formula and unmanageable for general use. Thus, we can see that writing out the formula in full is a bad idea, which is apparent as we generalize the formula. On the other hand, the standard form, is easy to state and work with and more useful, and hence I think the preference over other forms.
