GATE 2025 EE GA Q.10: solution and answer

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Basic proportionality theorem (Thales): ST is parallel to QR, so it divides the sides PQ and PR in the same ratio. That gives $PS:SQ=PT:TR=3:2$.

2. Take $PS=3k$ and $SQ=2k$, so $PQ=PS+SQ=5k$.

3. A semicircle with diameter $d$ has area $\frac{1}{2}\pi {\left(\frac{d}{2}\right)}^2=\frac{\pi d^2}{8}$. So the semicircle on PS has area $\frac{\pi (3k{)}^2}{8}=\frac{9\pi k^2}{8}$, and the semicircle on PQ has area $\frac{\pi (5k{)}^2}{8}=\frac{25\pi k^2}{8}$.

4. Both semicircles lie on the same side of PQ and share the end point P. The circle on PS touches the circle on PQ from inside at P, so the small semicircle lies entirely inside the large one. The shaded region is the large semicircle with the small one removed: $$\text{Shaded}=\frac{25\pi k^2}{8}-\frac{9\pi k^2}{8}=\frac{16\pi k^2}{8}=2\pi k^2$$

5. Compare this with the semicircle on PS: $$\frac{\text{Shaded}}{\text{Semicircle on PS}}=\frac{16\pi k^2/8}{9\pi k^2/8}=\frac{16}{9}$$ This matches option (A).

6. Check the other options. Against the semicircle on PQ the ratio is $\frac{16}{25}$, not $\frac{14}{25}$, so (D) is false. Since $\frac{16}{9}\ne 1$ and $\frac{16}{9}\ne \frac{14}{9}$, (B) and (C) are false as well.