Question

pqr is a triangle right angle at P and M is point on QR such that pm perpendicular to QR so that p m square equals to given in to mi

Answer 1

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

Answer 2

Given PM perpendicular to QR

To prove :PMsquare =QM. MR

Proof: in ΔPQM and ΔPMR

ΔPQM.~ΔPMR(theorem 6.7)

PM/QM=MR/PN (by bpt)

PM2=QM.QR

Hence Proved