In a right - angled triangle PQR, ∠Q is right angled, and S is a point on the side PQ. Find the value of PS.

In a right - angled triangle PQR, ∠Q is right angled, and S is a point on the side PQ. Find the value of PS. Correct Answer √{RS² – PR² + (2 × PQ × PS)}

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In ΔPQR:

PR² = PQ² + QR²

QR² = PR² – PQ²           ----(1)

In ΔSQR:

RS² = QS² + QR²

QR² = RS² – QS²           ----(2)

From (1) and (2):

PR² – PQ² =  RS² – QS²

QS² = RS² + PQ² – PR²

(PQ – PS)² = RS² + PQ² – PR²

PQ² + PS² – (2 × PQ × PS) = RS² + PQ² – PR²

PS² = RS² – PR² + (2 × PQ × PS)

PS = √{RS² – PR² + (2 × PQ × PS)}