Question

In the adjoining figure, PQ = QR and ∠PSQ = 90°.

Prove that : PR² = 2PQ.RS.

Answer 1

Given:

PQ = QR.

∠PSQ = 90° so PS ⟂ SQ.

To Prove: PR² = 2PQ.RS.

Proof [Step-wise]:

1. Place coordinates:

Let S = (0, 0), R = (r, 0). 

So, RS = r, Q = (q, 0) with 0 < q < r and because PS ⟂ SQ take P = (0, h).

2. By Pythagoras in right triangles PSQ and PRS.

PQ² = q² + h² and PR² = r² + h²

3. Also QR = r – q.

Given PQ = QR. 

Square both sides:

PQ² = QR²

⇒ q² + h² = (r – q)²

= r² – 2rq + q²

Hence, h^2 = r^2 – 2rq.

4. Substitute h² into PR²:

PR² = r² + h²

= r² + (r² – 2rq)

= 2r² – 2rq

= 2r(r – q)

5. But r = RS and r – q = RQ = QR = PQ by the given.

So, PR² = 2 × RS × PQ.

Therefore, PR² = 2PQ.RS, as required.