jmc

Let $AS=x$ and $SD=y$.

Since $△SAP$ and $△SDR$ are isosceles, then $AP=x$ and $DR=y$.

Since there are two pairs of identical triangles, then $BP=BQ=y$ and $CQ=CR=x$. $△SDR$ is right-angled (since $ABCD$ is a square) and isosceles, so its area (and hence the area of $△BPQ$) is $\frac{1}{2}{y}^2$.

Similarly, the area of each of $△SAP$ and $△QCR$ is $\frac{1}{2}{x}^2$.

Therefore, the total area of the four triangles is $2(\frac{1}{2}{x}^2)+2(\frac{1}{2}{y}^2)=x^2+y^2$, so $x^2+y^2=200$.

Now, by the Pythagorean Theorem, used first in $△PRS$, then in $△SAP$ and $△SDR$, $$\begin{array}{rl}P{R}^2& =P{S}^2+S{R}^2\\ & =(S{A}^2+A{P}^2)+(S{D}^2+D{R}^2)\\ & =2x^2+2y^2\\ & =2(200)\\ & =400\end{array}$$ so $PR=\boxed{20}$ m.