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In the adjoining figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4 PM2 – 3 PR2 . |
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Answer» Proof: In ∆PQR, ∠PRQ = 90° [Given] PQ2 = PR2 + QR2 (i) [Pythagoras theorem] RM = QR [M is the midpoint of QR] ∴ 2RM = QR (ii) ∴ PQ2 = PR2 + (2RM)2 [From (i) and (ii)] ∴ PQ2 = PR2 + 4RM2 (iii) Now, in ∆PRM, ∠PRM = 90° [Given] ∴ PM2 = PR2 + RM2 [Pythagoras theorem] ∴ RM2 = PM2 – PR2 (iv) ∴ PQ2 = PR2 + 4 (PM2 – PR2 ) [From (iii) and (iv)] ∴ PQ2 = PR2 + 4 PM2 – 4 PR2 ∴ PQ2 = 4 PM2 – 3 PR2 |
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