GATE CS 2025 Set 1 Q7 — General Aptitude

Since the line $ST$ is parallel to $QR$, the triangle $\triangle PST$ is similar to the triangle $\triangle PQR$ (i.e., $\triangle PST \sim \triangle PQR$).

Let the shortest distance (height) from the point $P$ to the line $QR$ be $H$.

According to the problem:

• The shortest distance between the parallel lines $QR$ and $ST$ is $h = \dfrac{H}{2}$.
• Therefore, the height of $\triangle PST$ from vertex $P$ to the base $ST$ is: $$H_1 = H - h = H - \frac{H}{2} = \frac{H}{2}$$

Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding heights:

Let the area of $\triangle PQR$ be $A$. Then:

• $\text{Area}(\triangle PST) = \dfrac{A}{4}$
• $\text{Area}(\text{trapezium } SQRT) = \text{Area}(\triangle PQR) - \text{Area}(\triangle PST) = A - \dfrac{A}{4} = \dfrac{3A}{4}$

The required ratio of the area of $\triangle PST$ to the area of the trapezium $SQRT$ is:

Hence, the correct option is A.