Question

The perpendicular PS on the base QR of PQR intersects QR in S such that QS = 3SR. Prove that : 2 PQ2 = 2 PR2 + QR2.​

Answer 1

We form our diagram from given information , As :

Here

QS = 3 RS ( Given )

and

QR = QS + RS = 3 RS + RS = 4 RS , So

RS = QR4 ---- ( A )

Now we apply Pythagoras theorem in triangle PQS and get

PQ2 = PS2 + QS2

PS2 = PQ2 - QS2 ---- ( 1 )

And apply Pythagoras theorem in triangle PRS and get

PR2 = PS2 + RS2

PS2 = PR2 - RS2 , Substitute value from equation 1 we get

⇒PQ2 - QS2 = PR2 - RS2

⇒PQ2 = PR2 - RS2 + QS2

⇒PQ2 = PR2 - RS2 + ( QR - RS )2

⇒PQ2 = PR2 - RS2 + QR2 + RS2 - 2 QR ×RS

⇒PQ2 = PR2 + QR2 - 2 QR ×RS

Now substitute value from equation A and get

⇒PQ2 = PR2 + QR2 - 2 QR ×QR4

⇒PQ2 = PR2 + QR2 - QR22

⇒PQ2 = 2 PR2 + 2 QR2 − QR22⇒PQ2 = 2 PR2 + QR2 2⇒2 PQ2 =2 PR2 + QR2