A square is cut along a diagonal and reassembled to form the parallelogram $P Q R S$ as shown in the diagram. If $P R=90 \mathrm{~mm}$, what is the area of the original square, in $\mathrm{mm}^{2}$ ?
(A) 324
(B) 1620
(C) 1800
(D) 2025
(E) 2700

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-058.jpg?height=313&width=485&top_left_y=1356&top_left_x=1251)
Suppose that the original square had side length $x \mathrm{~mm}$.

We extend $P Q$ and draw a line through $R$ perpendicular to $P Q$, meeting $P Q$ extended at $T$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-114.jpg?height=306&width=485&top_left_y=1403&top_left_x=880)

$S R T Q$ is a square, since it has three right angles at $S, Q, T$ (which makes it a rectangle) and since $S R=S Q$ (which makes the rectangle a square).

Now $R T=S Q=x \mathrm{~mm}$ and $P T=P Q+Q T=2 x \mathrm{~mm}$.

By the Pythagorean Theorem, $P R^{2}=P T^{2}+R T^{2}$ and so $90^{2}=x^{2}+(2 x)^{2}$.

Therefore, $5 x^{2}=8100$ or $x^{2}=1620$.

The area of the original square is $x^{2} \mathrm{~mm}^{2}$, which equals $1620 \mathrm{~mm}^{2}$.

ANswer: (B)