CEMC Cayley, 2022

Extend $PQ$ and $TS$ to meet at point $X$.

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 $\text{QX gt;0}$, 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.