, 2019

Begin by constructing $AV$ perpendicular to $TX$ and $UB$ perpendicular to $YW$.

The four segments $TX, UB, AV$, and $YW$ divide $PQRS$ into 9 identical squares.

Label the intersections of the perpendicular pairs of these four segments as points $C, D, E$, and $F$.

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$A$ lies on $PS$ such that $PA = AY = YS$. $B$ lies on $SR$ such that $SX = XB = BR$. $C$ is the intersection of segments $UB$ and $YW$. $D$ is the intersection of segments $TX$ and $YW$. $E$ is the intersection of segments $TX$ and $AV$. $F$ is the intersection of $UB$ and $AV$.
Segment $UY$ is a diagonal of square $PUCY$ and so $UY$ passes through $E$, the centre of square $PUCY$.
Segment $UE$ is a diagonal of square $TUFE$.

Segment $TW$ is a diagonal of square $TQWD$ and $TF$ is a diagonal of square $TUFE$.

The diagonals in any square divide the square into 4 identical triangles.

For example, the diagonals $TF$ and $UE$ divide the square $TUFE$ into 4 identical triangles, 3 of which are shaded.

Similarly, we can show that diagonals $FW$ and $VC$ divide square $FVWC$ into 4 identical triangles, 3 of which are shaded.

We may construct the missing diagonals in each of the 9 squares.

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These diagonals divide square $PQRS$ into $9\times4=36$ identical triangles.

Since 10 of these triangles are shaded, then $\frac{10}{36}=\frac{5}{18}$ of square $PQRS$ is shaded.