, 2022

Let $P$ be the point with coordinates $(7,0)$ and let $Q$ be the point with coordinates $(0,5)$. [[IMAGE0]] Then $APDQ$ is a rectangle with width 7 and height 5, and so it has area $7
$\cdot$ 5 = 35$. Hexagon$ABCDEF is formed by removing two triangles from rectangle

APDQ$, namely$$\triangle$ BPC$and$$\triangle$ EQF$. Each of$$\triangle$ BPC$and$$\triangle$ EQF is right-angled, because each shares an angle with rectangle

APDQ$. Each of$$\triangle$ BPC$and$$\triangle$ EQF$has a base of length 3 and a height of 2. Thus, their combined area is$2 $\cdot \frac{1}{2} \cdot 3 \cdot$ 2 = 6. This means that the area of hexagon

ABCDEF$is$35 - 6 = 29$.
Since $\triangle PQS$ is right-angled at $P$, then by the Pythagorean Theorem, $$SQ^2 = SP^2 + PQ^2 = (x+3)^2 + x^2$$ Since $\triangle QRS$ is right-angled at $Q$, then by the Pythagorean Theorem, we obtain $$\begin{aligned} RS^2 & = SQ^2 + QR^2 \\ (x+8)^2 & = ((x+3)^2 + x^2) + 8^2 \\ x^2 + 16x + 64 & = x^2 + 6x + 9 + x^2 + 64 \\ 0 & = x^2 - 10x + 9 \\ 0 & = (x-1)(x-9)\end{aligned}$$ and so $x = 1$ or $x = 9$.

(We can check that if $x = 1$, $\triangle PQS$ has sides of lengths 4, 1 and $\sqrt{17}$ and $\triangle QRS$ has sides of lengths $\sqrt{17}$, 8 and 9, both of which are right-angled, and if $x = 9$, $\triangle PQS$ has sides of lengths 12, 9 and 15 and $\triangle QRS$ has sides of lengths 15, 8 and 17, both of which are right-angled.) In terms of $x$, the perimeter of $PQRS$ is $x + 8 + (x+8) + (x+3) = 3x + 19$. Thus, the possible perimeters of $PQRS$ are 22 (when $x = 1$) and 46 (when $x = 9$).