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

Asked by Pragya Singh | 1 year ago |  75

##### Solution :-

Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR

We have to prove, PM2 = QM × MR

In ΔPQM, by Pythagoras theorem

PQ2 = PM2 + QM2

Or, PM2 = PQ2 – QM2 ……(i)

In ΔPMR, by Pythagoras theorem

PR2 = PM2 + MR2

Or, PM2 = PR2 – MR2 …….(ii)

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

2PM2 = (PQ2 + PM2) – (QM2 + MR2)

= QR2 – QM2 – MR2      [∴ QR2 = PQ2 + PR2]

= (QM + MR)2 – QM2 – MR2

= 2QM × MR

PM2 = QM × MR

Answered by Sudhanshu | 1 year ago

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