In the diagram, hexagon $P Q R S T U$ has interior right angles at $P, Q, S, T$, and $U$ and an exterior right angle at $R$. Also, $P U=U T, P Q=S T=10$, and $Q S=8$. The perimeter of $P Q R S T U$ is closest to
(A) 48
(B) 56
(C) 63
(D) 71
(E) 72

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-009.jpg?height=347&width=385&top_left_y=2060&top_left_x=1312)
Extend $P Q$ and $T S$ to meet at point $X$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-017.jpg?height=390&width=393&top_left_y=827&top_left_x=926)

Since quadrilateral $Q R S X$ has three right angles (at $Q, R$ and $S$ ), it must have a fourth right angle at $X$.

Thus, $Q R S X$ is a rectangle, which means that $X S=Q R$ and $Q X=R S$.

The perimeter of $P Q R S T U$ is

$$
\begin{aligned}
P Q+Q R+R S+S T+T U+U P & =P Q+X S+Q X+S T+T U+U P \\
& =(P Q+Q X)+(X S+S T)+T U+U P \\
& =P X+X T+T U+U P
\end{aligned}
$$

which is the perimeter of quadrilateral of $P X T U$.

But quadrilateral $P X T U$ has four right angles, and so is a rectangle.

Also, $P U=U T$, so $P X T U$ is a square, and so the perimeter of $P X T U$ equals

$$
4 \times P X=4 \times(P Q+Q X)=4 \times(10+Q X)=40+4 \times Q X
$$

Finally, $Q X=P X-P Q=P X-10=X T-10=X T-S T=X S$, which means that $\triangle Q X S$ is isosceles as well as being right-angled at $X$.

By the Pythagorean Theorem, $Q X^{2}+X S^{2}=Q S^{2}$ and so $2 \times Q X^{2}=8^{2}$ or $Q X^{2}=32$.

Since $Q X>0$, then $Q X=\sqrt{32}=\sqrt{16 \times 2}=\sqrt{16} \times \sqrt{2}=4 \sqrt{2}$.

Thus, the perimeter of $P Q R S T U$ is $40+4 \times 4 \sqrt{2}=40+16 \sqrt{2} \approx 62.6$.

(We could have left this as $40+4 \sqrt{32} \approx 62.6$.)

Of the given choices, this is closest to 63 .