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Question
In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2.
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Solution
Given: M is the midpoint of QR. ∠PRQ = 90°.
To prove: PQ2 = 4PM2 – 3PR2
Proof:
In ∆PRM,
∠PRM = 90° ...(Given)
By Pythagoras theorem,
∴ PM2 = PR2 + RM2
∴ RM2 = PM2 − PR2 ...(1)
In ∆PRQ,
∠PRQ = 90° ...(Given)
By Pythagoras theorem,
∴ PQ2 = PR2 + RQ2
∴ PQ2 = PR2 + (RM + MQ)2 ...[M is the midpoint of QR]
∴ PQ2 = PR2 + (RM + RM)2
∴ PQ2 = PR2 + (2RM)2
∴ PQ2 = PR2 + 4RM2
∴ PQ2 = PR2 + 4(PM2 − PR2) ...(from 1)
∴ PQ2 = PR2 + 4PM2 − 4PR2
∴ PQ2 = 4PM2 − 3PR2
Hence, PQ2 = 4PM2 – 3PR2.
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