Question

In triangle \( PQR \), the lengths of \( PT \) and \( TR \) are in the ratio \( 3:2 \). ST is parallel to QR. Two semicircles are drawn with \( PS \) and \( PQ \) as diameters, as shown in the figure. Which one of the following statements is true about the shaded area \( PQS \)? (Note: The figure shown is representative.) 

• A. The shaded area is \( \frac{16}{9} \) times the area of the semicircle with the diameter \( PS \).
• B. The shaded area is equal to the area of the semicircle with the diameter \( PS \).
• C. The shaded area is \( \frac{14}{9} \) times the area of the semicircle with the diameter \( PS \).
• D. The shaded area is \( \frac{14}{25} \) times the area of the semicircle with the diameter \( PQ \).

Hint:

When dealing with similar triangles and geometric areas, convert the ratios into lengths and then into areas using known formulas. Be cautious with semicircle areas: use \( A = \frac{1}{2} \pi r^2 \).

Answer 1

The Correct Option is A

Given \( PT : TR = 3 : 2 \), the total length \( PR = PT + TR = 3x + 2x = 5x \). 
Since ST is parallel to QR, the triangle \( PST \sim PQR \) (by AA similarity). So the side ratios are the same: \[ \frac{PS}{PQ} = \frac{PT}{PR} = \frac{3}{5} \Rightarrow \frac{PQ}{PS} = \frac{5}{3} \] Let the diameter of the semicircle on \( PS \) be \( d \), so its area is: \[ A_{PS} = \frac{1}{2} \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{8} \] Then, \( PQ = \frac{5}{3}d \), so the area of the semicircle with diameter \( PQ \) is: \[ A_{PQ} = \frac{1}{2} \pi \left( \frac{5d}{6} \right)^2 = \frac{25 \pi d^2}{72} \] Shaded area = \( A_{PQ} - A_{PS} \): \[ = \frac{25\pi d^2}{72} - \frac{\pi d^2}{8} = \pi d^2 \left( \frac{25}{72} - \frac{1}{8} \right) = \pi d^2 \left( \frac{25 - 9}{72} \right) = \frac{16\pi d^2}{72} = \frac{2\pi d^2}{9} \] Compare this to \( A_{PS} = \frac{\pi d^2}{8} \): \[ \frac{{Shaded area}}{A_{PS}} = \frac{2\pi d^2}{9} \cdot \frac{8}{\pi d^2} = \frac{16}{9} \]