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Q9 of 45 Page 38

In the figure 2.22, M is themidpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2

In ∆PRQ, ∠PRQ = 900


PQ2 = PR2 + QR2 – – – 1


In ∆PRM, ∠PRM = 900


PM2 = PR2 + MR2


⇒ PM2 = PR2 + 2 [ M is midpoint]


⇒ 4(PM2 – PR2) = QR2 – – – 2


1 And 2 implies


PQ2 = PR2 + 4(PM2 – PR2)


⇒ PQ2 = 4PM2 – 3PR2


PROVED.


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Questions · 45
2. Pythagoras Theorem
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