To follow the argument, a labelled diagram needs to be drawn. The argument is exactly the same as yours, except that everything is given a name.
Let our family of parallelograms have base $b$ and area $A$. So all of our parallelograms have height $h$ where $bh=A$. In particular, the height $h$ is determined.
Note that two of the sides of the parallelogram are given. So to minimize the perimeter, we must minimize the other two sides.
Let $PQRS$ be such a parallelogram, where the vertices $P$, $Q$, $R$, and $S$ are listed in counterclockwise order, and the base is $PQ$. Then the perimeter of the parallelogram is $2(PQ)+2(PS)$. This is $2b+2(NS)$. So to minimize the perimeter, we must minimize $NS$.
Draw a perpendicular from $P$ to the line through $R$ and $S$. Suppose that this perpendicular meets the line through $R$ and $S$ at $N$. Draw a perpendicular from $Q$ to the line through $R$ and $S$. Suppose that this perpendicular meets the line through $R$ and $S$ at $M$.
Note that $PQMN$ is a rectangle with base $PQ$ and height $h$, so it is in our family of parallelograms. Suppose that $N\ne S$. We will show that the rectangle $PQMN$ has perimeter less than the perimeter of $PQRS$. To do this, we need to show that $NS \gt h$.
Note that by the Pythagorean Theorem, $(PS)^2=(PN)^2+(NS)^2=h^2+(NS)^2$. Since $NS\ne 0$, we have $(PN)^2\gt h^2$, that is, $PS \gt h$.
This shows that if $N\ne S$, then $PQRS$ cannot have minimum perimeter in our family. So for minimum perimeter, we must have $N=S$, meaning that we have a rectangle.
Remark: The argument can be greatly shortened. There is only one idea here: If $PQRS$ is not a rectangle, then the side $PS$ is greater than $h$, and therefore our parallelogram has perimeter greater than $2b+2h$, which is the perimeter of the rectangle.
In your case, the area of the triangle can be made to be greater than $72$ if the two remaining sides both have length $14.$ That is, if the triangle is $12-14-14,$ then it is isosceles, so we can find its area pretty easily: Drop an altitude to the $12$-length side, dividing that side into two segments of length $6.$ Then the altitude has length $\sqrt{14^2-6^2} = \sqrt{196-36} = \sqrt{160} = 4\sqrt{10},$ by the Pythagorean theorem, so the area of this triangle is $\dfrac12 (12)(4\sqrt{10}) = 24\sqrt{10},$ which is about $75.89.$
In general, I believe that the area of a triangle given one side and the perimeter is maximized when the two remaining sides have the same length. This should be provable algebraically using Heron's formula.
Best Answer
If I understand the problem correctly, both $s$ and $h$ are fixed, and you want to choose $x$ so that $p(x)=s + \sqrt{x^2 + h^2} + \sqrt{(s-x)^2 + h^2}$ is minimized. Note that $x$ can be negative, in which case one of the base angles is obtuse.
Computing $p'(x)$ and simplifying gives a fraction whose numerator is $$ (x-s)\sqrt{h^2+x^2}+x \sqrt{h^2+(s-x)^2}.$$ Set equal to zero and solve for $x$, giving $x=\frac{s}{2}$. Thus the minimal-perimeter triangle is isosceles on its base $s$.