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.
Best Answer
Let $r$ be the radius of the semicircle and $d$ the parallel length. Then, with $2d+2\pi r=400$, the rectangle area is
$$A(r)=2rd=r(400-2\pi r)$$
Set the derivative $A’(r)=400-4\pi r=0$ to obtain the optimal radius $r=\frac{100}{\pi}$ meters and the corresponding straight length $d=100$ meters.