Question

In the adjoining figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = 1/3 QR. Prove that: 9 PS² = 7 PQ².

Answer 1

Given: ∆PQR is an equilateral triangle. 

QS = 1/3 QR 

To prove: 9PS² = 7PQ²

Proof: 

∆PQR is an equilateral triangle [Given] 

∴ ∠P = ∠Q = ∠R = 60° (i) [Angles of an equilateral triangle] 

PQ = QR = PR (ii) [Sides of an equilateral triangle]

In ∆PTS, ∠PTS = 90° [Given] 

PS² = PT² + ST² (iii) [Pythagoras theorem] 

In ∆PTQ, 

∠PTQ = 90° [Given] 

∠PQT = 60° [From (i)] 

∴ ∠QPT = 30° [Remaining angle of a triangle]

∴ ∆PTQ is a 30° – 60° – 90° triangle 

∴ PT = √3/2 PQ (iv) [Side opposite to 60°] 

QT = 1/2 PQ (v) [Side opposite to 30°] 

QS + ST = QT [Q – S – T] 

∴ 1/3 QR + ST = 1/2 PQ [Given and from (v)] 

∴ PQ + ST = 1/2 PQ [From (ii)]

∴ ST = PQ/2 – PQ/2 

∴ ST = (3PQ - 2PQ)/6 

∴ ST = PQ/6 (vi) 

PS² = |(√3/2PQ)² + (PQ/6)² [From (iii), (iv) and (vi)]

∴ PS² = (3PQ²/4) + (PQ²/36) 

∴ PS² = (27PQ²/36) + (PQ²/36)

∴ PS² = (28PQ²)/36 

∴ PS² = 7/3PQ² 

∴PS² = 7/3 PQ²