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

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Sum

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

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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.

  Is there an error in this question or solution?
2023-2024 (March) Model set 1 by shaalaa.com

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