Question

The perpendicular from P on the side QR of PQR intersects QR at S such that QS =3RS.Prove that PQ^2=QR^2+2PR^2​

Answer 1

Answer:

2PQ² = QR² + 2PR²

Step-by-step explanation:

Correct Question :

2PQ² = QR² + 2PR²

As PS is ⊥ QR

Hence using Pythagorus theorem

PQ² = PS² + QS²

=> 2PQ² = 2PS² + 2QS²

QS  = 3RS

=> 2PQ² = 2PS² + 2(3RS)²

=> 2PQ² = 2PS² + 18RS²

=> 2PQ² = 2PS² + 2RS² + 16RS²

=> 2PQ² = 2(PS² + RS²) + 16RS²

=> 2PQ² = 2PR² + (4RS)²

Now QS = 3RS

=> QR = QS + RS

=> QR  = 3RS + 4S

=> QR = 4RS

Putting this in above

2PQ² = 2PR² + QR²

2PQ² = QR² + 2PR²