PQR is a right angled
triangle, right angled at Q. If QRST is a
square on side QR and PRUV is a square on
PR, then PS
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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 )
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 )
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