Question

In the adjoining figure, M is the midpoint of QR. ∠PRQ = 90°. 

Prove that, PQ² = 4 PM² – 3 PR² .

Answer 1

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