Question

In an equilateral triangle PQR, prove that PS² = 3(QS)².

Answer 1

Given: PS is the altitude of ΔPQR.

In ΔPSQ and ΔPSR,

∠PSQ ≅ ∠PSR   ......[Each angle is equal to 90°]

PS ≅ SP   ......[Common side]

PQ ≅ PR  ......[Sides of an equilateral triangle]

By R.H.S. criterion of congruence,

ΔPSQ ≅ ΔPSR

∴ QS ≅ SR  ......[C.S.C.T.]

Now, QS + SR = QR

QS + QS = QR  .......[∵ SR = QS]

2QS = QR

QS = `(QR)/2`  ......(i)

In right-angled triangle PQS, by Pythagoras theorem,

PS² + QS² = PQ²

PS² + QS² = QR²  ......[∵ PQ = QR]

PS² = QR² – QS²

ps² = (2QS)² – QS²  ......[∵ QR = 2QS]

PS² = 4QS² – QS²

PS² = 3QS² 

Hence proved.