, 2014

Suppose that the original square had side length $x$ mm. We extend $PQ$ and draw a line through $R$ perpendicular to $PQ$, meeting $PQ$ extended at $T$. [[IMAGE0]] $SRTQ$ 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 $RT=SQ = x$ mm and $PT=PQ+QT=2x$ 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 $\text{x 2 mm 2}$, which equals $\text{1620 mm 2}$.