InterviewSolution
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InterviewSolution
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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. |
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Answer» 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 |
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