1.

In the adjoining figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4 PM2 – 3 PR2 .

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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