In the diagram, two pairs of identical isosceles triangles are cut off of square $A B C D$, leaving rectangle $P Q R S$. The total area cut off is $200 \mathrm{~m}^{2}$. The length of $P R$, in metres, is
(A) $\sqrt{200}$
(B) 20
(C) $\sqrt{800}$
(D) 25
(E) 15

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-112.jpg?height=406&width=436&top_left_y=1331&top_left_x=1278)
Let $A S=x$ and $S D=y$.

Since $\triangle S A P$ and $\triangle S D R$ are isosceles, then $A P=x$ and $D R=y$.

Since there are two pairs of identical triangles, then $B P=B Q=y$ and $C Q=C R=x$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-185.jpg?height=409&width=412&top_left_y=370&top_left_x=911)

$\triangle S D R$ is right-angled (since $A B C D$ is a square) and isosceles, so its area (and hence the area of $\triangle B P Q)$ is $\frac{1}{2} y^{2}$.

Similarly, the area of each of $\triangle S A P$ and $\triangle Q C R$ is $\frac{1}{2} x^{2}$.

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

Now, by the Pythagorean Theorem, used first in $\triangle P R S$, then in $\triangle S A P$ and $\triangle S D R$,

$$
\begin{aligned}
P R^{2} & =P S^{2}+S R^{2} \\
& =\left(S A^{2}+A P^{2}\right)+\left(S D^{2}+D R^{2}\right) \\
& =2 x^{2}+2 y^{2} \\
& =2(200) \\
& =400
\end{aligned}
$$

so $P R=20 \mathrm{~m}$.

ANswer: (B)