VSA2 marksYear: 2020MediumCBSE Class 10
trianglessimilaritypythagorasproof
Question
In Figure-5, △PQR is right-angled at P. M is a point on QR such that PM is perpendicular to QR. Show that PQ2=QM×QR.

Solution
1
△PQM∼△RQP [By AA similarity]
+1 mark2
∴RQPQ=PQQM⇒PQ2=QM×QR
+1 mark