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

We are given that the lines QR and ST are parallel to each other, and the shortest distance between these two lines is half the shortest distance between the point P and the line QR.

Let the height of the trapezium SQRT be \( h_1 \), which is the shortest distance between the lines QR and ST, and let the height of the triangle PST be \( h_2 \), which is the shortest distance between the point P and the line QR.

According to the problem, \( h_2 = 2h_1 \) because the height of the triangle is twice the height of the trapezium.

The area of a triangle is given by: \[ \text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height}. \] Similarly, the area of a trapezium is given by: \[ \text{Area of Trapezium} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}. \] Since the areas of the triangle and trapezium are proportional to their respective heights and the common base, the ratio of the areas of the triangle PST and the trapezium SQRT depends on the heights. The ratio of the areas is: \[ \frac{\text{Area of Triangle PST}}{\text{Area of Trapezium SQRT}} = \frac{h_2}{h_1} = \frac{2h_1}{h_1} = 2. \] Thus, the ratio of the areas is \( \frac{1}{3} \).

Therefore, the correct answer is \( \boxed{A} \).