$PQRS$ is a square. $T$ and $U$ are the mid-points of $PS$ and $QR$ respectively. Find the area of $\Delta OTS$,if $PQ = 8 \, cm$,where $O$ is the point of intersection of $TU$ and $QS$.

$(8 CM^2)$ Given that $PQRS$ is a square with side length $PQ = 8 \, cm$.
Since $T$ and $U$ are the mid-points of $PS$ and $QR$ respectively,$TU$ is parallel to $PQ$ and $SR$.
$ST = \frac{1}{2} PS = \frac{1}{2} \times 8 = 4 \, cm$.
Since $TU$ is parallel to $PQ$ and $SR$,and $T, U$ are mid-points,$TU = PQ = 8 \, cm$.
In $\Delta OTS$ and $\Delta QPS$,since $TU \parallel PQ$,by the property of similar triangles,$\Delta OTS \sim \Delta QPS$.
Since $T$ is the mid-point of $PS$,the ratio of similarity is $\frac{ST}{SP} = \frac{1}{2}$.
Thus,$OT = \frac{1}{2} PQ = \frac{1}{2} \times 8 = 4 \, cm$.
Since $PQRS$ is a square,$\angle TSP = 90^\circ$,thus $\angle OTS = 90^\circ$.
Area of $\Delta OTS = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times ST \times OT$.
Area of $\Delta OTS = \frac{1}{2} \times 4 \times 4 = 8 \, cm^2$.