In the diagram, $P Q R S$ is a square with side length 8 . Points $T$ and $U$ are on $P S$ and $Q R$ respectively with $Q U=T S=1$. The length of $T U$ is closest to
(A) 8.5
(B) 9.9
(C) 10
(D) 10.6
(E) 11.3

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-057.jpg?height=398&width=420&top_left_y=663&top_left_x=1292)
We draw a line through $T$ to point $W$ on $Q R$ so that $T W$ is perpendicular to $Q R$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-112.jpg?height=407&width=439&top_left_y=260&top_left_x=905)

Since $T W R S$ has three right angles (at $W, R$ and $S$ ), then it must be a rectangle.

Therefore, $W R=T S=1$ and $T W=S R=8$.

Since $Q U=1$, then $U W=Q R-Q U-W R=8-1-1=6$.

Now, $\triangle T W U$ is right-angled at $W$.

By the Pythagorean Theorem, we have $T U^{2}=T W^{2}+U W^{2}$.

Thus, $T U^{2}=8^{2}+6^{2}=64+36=100$.

Since $T U>0$, then $T U=\sqrt{100}=10$.

ANSWER: (C)