Question

2. In fig M is the midpoint of QR, ZPRQ=90° Prove that PQ+= 4 PM? – 3 PR²
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Answer 1

CORRRCT QUESTION

In fig M is the midpoint of QR,∠PRQ=90° Prove that PQ² = 4 PM²– 3 PR²

ANSWER

GIVEN

M is the midpoint of QR,

ZPRQ=90°

PROOF

In ∆PQR, ∠PRQ = 90° [Given]

PQ2 = PR2 + QR2 (i) [Pythagoras theorem]

RM = QR [M is the midpoint of QR]

∴ 2RM = QR (ii)

∴ PQ2 = PR2 + (2RM)2 [From (i) and (ii)]

∴ PQ2 = PR2 + 4RM2 (iii) Now, in ∆PRM, ∠PRM = 90° [Given]

∴ PM2 = PR2 + RM2 [Pythagoras theorem]

∴ RM2 = PM2 – PR2 (iv)

∴ PQ2 = PR2 + 4 (PM2 – PR2 ) [From (iii) and (iv)]

∴ PQ2 = PR2 + 4 PM2 – 4 PR2

∴ PQ2 = 4 PM2 – 3 PR2