GATE CS 2025 Set 1 Q.7: Quantitative Aptitude

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.

Triangle PQR with vertex P at the top and base QR at the bottom. A horizontal line segment ST connects point S on side PQ to point T on side PR, parallel to QR at half the height of triangle PQR.

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{2}{5}$

$\frac{1}{2}$

Official answer and solution

Official answer: A
From: Official GATE 2025 answer key

Solution

Use the similarity of $\triangle PST$ and $\triangle PQR$ and the fact that the ratio of areas of similar triangles equals the square of the ratio of their altitudes.

Let $H$ be the altitude (shortest distance) from $P$ to line $QR$ . Since $ST \parallel QR$ and the distance between $ST$ and $QR$ is $\frac{H}{2}$ , the altitude of $\triangle PST$ from $P$ to $ST$ is:

h = H - \frac{H}{2} = \frac{H}{2}

Because $ST \parallel QR$ , $\triangle PST \sim \triangle PQR$ with linear scale factor $\frac{h}{H} = \frac{1}{2}$ . Therefore:

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

Since $\text{Area}(\text{trapezium } SQRT) = \text{Area}(\triangle PQR) - \text{Area}(\triangle PST) = 4 - 1 = 3$ parts, the required ratio is:

\frac{\text{Area}(\triangle PST)}{\text{Area}(\text{trapezium } SQRT)} = \frac{1}{3}

Final answer

Option (A)

Common trap

Taking the ratio of $\triangle PST$ to $\triangle PQR$ ( $1/4$ ) instead of $\triangle PST$ to trapezium $SQRT$ ( $1/3$ ).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question: 7 of 65
Section: General Aptitude

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