Question

In a ∆PQR, PS⊥QR, prove that PQ2+RS2= QS2+PR2

Answer 1

Answer:

→ PQ2 = 2 PR2 + 2 QR2 - QR22 PQ2 = 2 PR2 + QR2 2 2 PQ2 =2 PR2 + QR2

Step-by-step explanation:

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 equation 1 we get و Substitute value from

→ 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 XRS

*Now substitute value from equation A and get

→ PQ2 = PR2 + QR2 - 2 QR XQR4

→PQ2 = PR2 + QR2 - QR22

→ PQ2 = 2 PR2 + 2 QR2 - QR22 PQ2 = 2 PR2 + QR2 2 2 PQ2 =2 PR2 + QR2