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04/06/2021 12:19 pm
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PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.
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04/06/2021 12:19 pm
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
This post was modified 4 years ago by Raavi Tiwari