Question

Please solve this
It's urgent!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Given in right triangle PQR, QS =  SR

By Pythagoras theorem, we have

PR^2 = PQ^2 + QR^2 → (1)

In right triangle PQS, we have

PS^2 = PQ^2 + QS^2

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

      = PQ^2 + (QR^2/4)

4PS^2 =  4PQ^2 + QR^2

∴ QR^2 = 4PS^2 −  4PQ^2 → (2)

Put (2) in (1), we get

PR^2 = PQ^2 + (4PS^2 −  4PQ^2)

∴ PR^2 = 4PS^2 −  3PQ^2

proved