From square $A B C D$, isosceles triangles were cut out as shown in the figure, leaving rectangle $P Q R S$. Knowing that the total area of what was cut out measures $200 \mathrm{~cm}^{2}$, what is the length of $P R$, in $\mathrm{cm}$?
(a) $\sqrt{200}$
(b) 200
(c) $\sqrt{800}$
(d) 25
(e) 88

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-010.jpg?height=417&width=414&top_left_y=531&top_left_x=1415)
The correct option is (b).

First, we notice that the triangles $\triangle A P S$ and $\triangle C Q R$ are congruent, as they have three equal angles (one of them being a right angle) and also one of their sides $(P S=Q R)$. Similarly, the triangles $\triangle B P Q$ and $\triangle D R S$ are also congruent. Let $A P=x$ and $B P=y$. Then the area of the triangle $\triangle A P S$ is $\frac{1}{2} x^{2}$ and the area of the triangle $\triangle B P Q$ is $\frac{1}{2} y^{2}$, and the area cut out is $2\left(\frac{1}{2} x^{2}+\frac{1}{2} y^{2}\right)=x^{2}+y^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-047.jpg?height=414&width=420&top_left_y=2312&top_left_x=1412)

Thus, we establish that $x^{2}+y^{2}=200$. Now we notice that $P R$ is the hypotenuse of the right triangle $\triangle P S R$. To calculate $P R$, we just need to know the lengths of the legs $P S$ and $R S$. But $P S$ is the hypotenuse of the right triangle $\triangle A P S$ and from the Pythagorean Theorem it follows that $(P S)^{2}=(A S)^{2}+(A P)^{2}=x^{2}+x^{2}=2 x^{2}$. Similarly, we obtain $(R S)^{2}=2 y^{2}$. Therefore,

$$
(P R)^{2}=(P S)^{2}+(R S)^{2}=2 x^{2}+2 y^{2}=2\left(x^{2}+y^{2}\right)=2 \times 200=400
$$

that is, $P R=\sqrt{400}=20 \mathrm{~m}$.