Question

In ΔPQR, PS is perpendicular to RQ extended. Prove that PR² = PQ² + QR² + 2QR.SQ.

Answer 1

Given △PQR, PS is perpendicular to RQ extended. 

△PSR is right angle triangle, right angle at S.

 Apply Pythagoras theorem in △PSR.

PR² = PS² + SR²

From the figure,

SR = SQ + QR

PR² = PS² + (SQ + QR)²   ...[Using an identiy, (a + b)² = a² + 2ab + b²]

PR² = PS² + SQ² + 2SQ.QR + QR²

Applying Pythagoras theorem in △PSQ.

PQ² = PS² + SQ²

PR² = PQ² + QR² + 2SQ.QR

Hence, proved.