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.

Given: ∆PQR is an equilateral triangle.

QS = 1/3 QR

To prove: 9PS² = 7PQ²

Answer 1

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