Question

In the given figure, triangle PQR is right-angled at Q. S is the mid-point of side QR. Prove that QR² = 4(PS² – PQ²).

Answer 1

Given: In triangle PQR, ∠PQR = 90° and S is the mid-point of QR.

To prove: QR² = 4(PS² – PQ²)

in right-angled ΔPQS, by Pythagoras theorem,

PQ² + QS² = PS²

⇒ QS² = PS² – PQ²   .......(i)

Since S is the mid-point of side QR,

∴ QS = `(QR)/2`

Substituting the value of QS in equation (i),

`((QR)/2)^2 = PS^2 - PQ^2`

`(QR^2)/4 = PS^2 - PQ^2`

QR² = 4(PS² – PQ)²

Hence proved.