该图展示了一个婴儿活动区，其四周由 PQ、QR 和 PS 处的隔板围成，其中 PQ = 18 m，PS = 40 m，角 SPQ = 角 QRS = $90^\circ$，且锐角 PSR = $q$ 可以变化。

1. Set up a geometric construction to find the length of QR.

Draw a perpendicular from $P$ to $SR$, meeting $SR$ at $X$. Draw a perpendicular from $Q$ to $PX$, meeting $PX$ at $Y$.

2. Find the length of PX using triangle PSX.

In right-angled $\triangle PSX$, $\angle PSX = \theta$.
$PX = PS \sin \theta = 40 \sin \theta$

3. Find the length of PY using triangle PQY.

Since $PQ \perp PS$ and $PX \perp SR$, the angle between $PQ$ and $PX$ is equal to the angle between $PS$ and $SR$, so $\angle QPY = \theta$.
In right-angled $\triangle PQY$,
$PY = PQ \cos \theta = 18 \cos \theta$

4. Express QR in terms of theta and find the total length L.

Since $QR \perp SR$ and $PX \perp SR$, $QR$ is parallel to $PX$. With $QY \perp PX$, $YXRQ$ forms a rectangle.
$QR = YX = PX - PY = 40 \sin \theta - 18 \cos \theta$
The total length of the partitions is $L = PQ + PS + QR$.
$L = 18 + 40 + (40 \sin \theta - 18 \cos \theta)$
$L = 58 + 40 \sin \theta - 18 \cos \theta$ (Shown)