Question

P and Q are points on the sides CA and CB respectively of ΔABC right angled at C. AQ²  + BP² equals

(a) BC² + PQ² 

(b) AB² + PC² 

(c) AB² + PQ² 

(d) BC² + AC²

Answer 1

(c) AB² + PQ²

In rt. ∠d ΔACQ, 

⇒AC²+ CQ² = AQ²           ...(i) 

In rt. ∠d PCB, 

PC² + CB² = PB²          ...(ii) 

Adding eqn (i) and (ii) 

AC² + CQ² + PC² + CB² = AQ² + PB² 

⇒ (AC² + CB²) + (CQ² + PC²) = AQ² + PB² 

AB² + PQ² =  AQ² + PB² 

⇒ (rt∠d ΔABC)(rt ∠d ΔPQC)     (Pythagoras’ Theorem)