, 2022

Extend $PQ$ and $TS$ to meet at point $X$. [[IMAGE0]] Since quadrilateral $QRSX$ has three right angles (at $Q$, $R$ and $S$), it must have a fourth right angle at $X$. Thus, $QRSX$ is a rectangle, which means that $XS = QR$ and $QX = RS$. The perimeter of $PQRSTU$ is $$\begin{align*} PQ + QR + RS + ST + TU + UP & = PQ + XS + QX + ST + TU + UP \\ & = (PQ + QX) + (XS + ST) + TU + UP \\ & = PX + XT + TU + UP\end{align*}$$ which is the perimeter of quadrilateral of $PXTU$. But quadrilateral $PXTU$ has four right angles, and so is a rectangle. Also, $PU = UT$, so $PXTU$ is a square, and so the perimeter of $PXTU$ equals $$4 \times PX = 4\times (PQ + QX) = 4 \times (10 + QX) = 40 + 4 \times QX$$ Finally, $QX = PX - PQ = PX - 10 = XT - 10 = XT - ST = XS$, which means that
$\triangle QXS$ is isosceles as well as being right-angled at $X$. By the Pythagorean Theorem, $QX^2 + XS^2 =
QS^2$and so$2 $\times$ QX^2 = 8^2$or$QX^2 = 32$. Since$QX>0$, then$QX = $\sqrt{32} = \sqrt{16 \times}$ 2} = $\sqrt{16} \times \sqrt{2} = 4\sqrt{2}\$.$

Thus, the perimeter of $PQRSTU$ 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.