SSC CPO 2023 PYQ — Question 103

The correct answer is option 2 i.e. $\sqrt{63}$.

Given that

In an equilateral ∆PQR

QR = 3QS. If PQ = 9 cm

PQ = QR = RP = 9 cm

3QS = 9 cm

QS = 9/3 = 3 cm

In ∆PQS, By using the cosine rule,

∠Q = ∠P = ∠R = 60º

⇒ cos 60º = (PQ² + QS² - PS²)/(2 × PQ × QS)

⇒ (1/2) × 2 × 9 × 3 = 81 + 9 - PS²

⇒ PS² = 90 - 27 = 63

⇒ PS = √63