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