Question

In the figure, ∆PQR is a right angled triangle with angle Q=90°. If QRST is a square on side QR and PRMN is a square on side PR of the triangle, prove that PS=QM

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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

also by this method

In ∆ PQR , we apply Pythagoras theorem and get

PR2 = PQ2 + QR2                         --------------------- ( 1 )

Now we apply Pythagoras in ∆ PQS , we get

PS2 = PQ2 + QS2

⇒PQ2 = PS2 - QS2

⇒PQ2 = PS2 - (QR2)2                        ( Given QS = SR  = QR2  )

⇒4PQ2 = 4PS2 - QR2 

⇒QR2 = 4PS2 - 4PQ2                        -------------- ( 2 )

Now we substitute value from equation 2 in equation 1 and get

⇒PR2 = PQ2 + 4PS2 - 4PQ2 

⇒PR2 = 4PS2 - 3PQ2                                ( Hence proved )