Question

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 the 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

 

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{2}{5} \)
• D. \( \frac{1}{2} \)

Hint:

When dealing with geometric shapes, ratios of areas depend on the ratio of the heights when the bases are parallel.

Answer 1

The Correct Option is A

To solve this problem, we need to find the ratio of the area of the triangle PST to the area of the trapezium SQRT, given that the lines QR and ST are parallel and the shortest distance between them is half the shortest distance between point P and line QR.

1. Let \( d \) be the shortest distance between line QR and point P. Hence, the shortest distance between the parallel lines QR and ST is \( \frac{d}{2} \).

2. Since ST is parallel to QR, triangle PST and trapezium SQRT are similar in shape. The heights from P to line QR and from S to line QR are in the ratio of the shortest distances: \( \frac{1}{2} \).

3. Since ST is parallel to QR, triangle PST and triangle PQR are similar by AAA similarity criterion.

4. The ratio of their areas equals the square of the ratio of their corresponding heights (or sides). Therefore, the ratio of the area of triangle PST to triangle PQR is:

   \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\)

5. Now observe that trapezium SQRT is obtained by subtracting the area of triangle PST from triangle PQR. So its area is \( \frac{3}{4} \times \) (the area of triangle PQR).

6. Finally, the ratio of the area of triangle PST to the area of trapezium SQRT is:

   \(\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}\)

Hence, the correct answer is \(\frac{1}{3}\).