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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

In the given figure, seg PS is the median of ∆PQR and PT ⊥ QR. Prove that, PR2 = PS2 + QR × ST + QR(QR2)2 - Geometry Mathematics 2

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प्रश्न

In the given figure, seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

PR2 = PS2 + QR × ST + `("QR"/2)^2`

बेरीज

उत्तर

Seg PS is the median of ∆PQR   ...(Given)

∴ QS = SR = `1/2`QR   ...(1) [S is the midpoint of side QR]

In ∆PTS, ∠PTS = 90°   ...(Given)

∴ by Pythagoras theorem,

PS2 = PT2 + TS2   ...(2)

In ∆PTR, ∠PTR = 90°   ...(Given)

∴ by Pythagoras theorem,

PR2 = PT2 + TR2

∴ PR2 = PT2 + (TS + SR)2   ...(T-S-R)

∴ PR2 = PT2 + TS2 + 2ST2.SR + SR2   ...[(a + b)2 = a2 + 2ab + b2]

∴ PR2 = (PT2 + TS2) + 2ST.SR + SR2

∴ PR2 = PS2 + 2ST.`(("QR")/2) + (("QR")/2)^2`   ...[From (1) and (2)]

∴ PR2 = PS2 + QR × ST + `(("QR")/2)^2`

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पाठ 2: Pythagoras Theorem - Practice Set 2.2 [पृष्ठ ४३]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
पाठ 2 Pythagoras Theorem
Practice Set 2.2 | Q 3. (1) | पृष्ठ ४३

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