Question

PQR IS A TRIANGLE RIGHT ANGLED AT P AND M IS A POINT ON QR SUCH THAT PM PERPENDICULAR TO QR.SHOW THAT PM^2=QM.RM​

Answer 1

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Answer 2

Answer:

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

To prove: PM2 = QM × MR

Proof: In ΔPQM, we have

PQ2 = PM2 + QM2 [By Pythagoras theorem]

Or, PM2 = PQ2 - QM2 ...(i)

In ΔPMR, we have

PR2 = PM2 + MR2 [By Pythagoras theorem]

Or, PM2 = PR2 - MR2 ...(ii)

Adding (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