PQR is a triangle right angled at P and M is a point on QR such that PM \perp QR. Show that PM^2

For Competitive Exams

Question 2 PQR is a triangle right angled at P and M is a point on QR such that PM $\perp$ QR. Show that $PM^2$ = QM . MR.

Solution:

Given, ΔPQR is right angled at P is a point on QR such that PM $\perp$ QR

We have to prove, $PM^2$ = QM × MR

In ΔPQM, by Pythagoras theorem

$PQ^2 = PM^2 + QM^2$

Or, $PM^2 = PQ^2 – QM^2$ ……………………………..(i)

In ΔPMR, by Pythagoras theorem

$PR^2 = PM^2 + MR^2$

Or, $PM^2 = PR^2 – MR^2$ ………………………………………..(ii)

Adding equation, (i) and (ii), we get

$2PM^2 = (PQ^2 + PM^2) – (QM^2 + MR^2)$

= $QR^2 – QM^2 – MR^2 [∴ QR^2 = PQ^2 + PR^2]$

= $(QM + MR)^2 – QM^2 – MR^2$

= 2QM × MR

$PM^2$ = QM × MR

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