In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT? Note: The figure shown is representative.

Let the perpendicular distance (height) from vertex $P$ to the line $QR$ be $H$.

Given that the shortest distance between the parallel lines $ST$ and $QR$ is $\frac{1}{2}H$, the height of $riangle PST$ from vertex $P$ to line $ST$ is:

Since $ST \parallel QR$, $riangle PST$ is similar to $riangle PQR$ ($riangle PST \sim riangle PQR$) with a scale factor of linear dimensions equal to:

The ratio of their areas is the square of the ratio of their heights:

$$\frac{\text{Area}( riangle PST)}{\text{Area}( riangle PQR)} = k^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$

Let $\text{Area}( riangle PST) = A$. Then $\text{Area}( riangle PQR) = 4A$.

The area of the trapezium $SQRT$ is the difference between the areas of $riangle PQR$ and $riangle PST$:

$$\text{Area}(SQRT) = \text{Area}( riangle PQR) - \text{Area}( riangle PST) = 4A - A = 3A$$

Therefore, the ratio of the area of $riangle PST$ to the area of trapezium $SQRT$ is:

$$\frac{\text{Area}( riangle PST)}{\text{Area}(SQRT)} = \frac{A}{3A} = \frac{1}{3}$$

Hence, (A) is the correct option.