Question

ΔPQR is a right- angled triangle in which angle Q = 90 ° and S is a mid-point of side QR . Show that PR²= 4PS² - 3PQ²

Answer 1

Step-by-step explanation:

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