Question

In triangle pqr right angled at q pr²=_×ps²-_pq²​

Answer 1

Answer:

Given in right triangle PQR, QS =  SR

By Pythagoras theorem, we have

PR2 = PQ2 + QR2 → (1)

In right triangle PQS, we have

PS2 = PQ2 + QS2

      = PQ2 + (QR/2)2    [Since QS =  SR = 1/2 (QR)]

      = PQ2 + (QR2/4)

4PS2 =  4PQ2 + QR2

∴ QR2 = 4PS2 −  4PQ2 → (2)

Put (2) in (1), we get

PR2 = PQ2 + (4PS2 −  4PQ2)

∴ PR2 = 4PS2 −  3PQ2

Step-by-step explanation:

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