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
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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
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