Question

In a ΔPQR, angle Q = 90° and S is the mid-point of QR. Prove that

PR2

= PS2

+ 3RS2​

Answer 1

Answer:

As per Pythagorean theorem

PT2=PQ2+QT2....(in triangle QPT)

⇒QT2=PT2−PQ2

And PR2=PQ2+QR2 ( in the triangle APR and given T is the mide point QR then QR=2QT) put value 

⇒PR2=PQ2+(2QT)2

⇒PR2=PQ2+4QT2

⇒PR2=PQ2+4(PT2−PQ2)

=PQ2+4PT2−4PQ2

⇒PR2=4PT2−3PQ2.