Suppose that the original square had side length $x \mathrm{~mm}$. We extend $PQ$ and draw a line through $R$ perpendicular to $PQ$, meeting $PQ$ extended at $T$. $SRTQ$ is a square, since it has three right angles at $S, Q, T$ (which makes it a rectangle) and since $SR=SQ$ (which makes the rectangle a square). Now $RT=SQ=x \mathrm{~mm}$ and $PT=PQ+QT=2x \mathrm{~mm}$. By the Pythagorean Theorem, $PR^{2}=PT^{2}+RT^{2}$ and so $90^{2}=x^{2}+(2x)^{2}$. Therefore, $5x^{2}=8100$ or $x^{2}=1620$. The area of the original square is $x^{2} \mathrm{~mm}^{2}$, which equals $1620 \mathrm{~mm}^{2}$.