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

As shown in figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, then (1) EG = ? (2) FD = ? (3) EF = ?

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प्रश्न

As shown in figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, then (1) EG = ? (2) FD = ? (3) EF = ?

बेरीज
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उत्तर

(1) In ∆DEF,

∠DFE = 90° and seg FG ⊥ hypotenuse ED  ...[Given]

∴ FG2 = EG × GD   ...[By theorem of geometric mean]

∴ (12)2 = EG × 8   ...[Given]

∴ 144 = EG × 8

∴ EG = `144/8`

∴ EG = 18 units

(2)  In ∆DGF,

∠DGF = 90°   ...[ ⸪ FG ⊥ ED]

∴ FD2 = FG2 + GD2   ...[Pythagoras theorem]

∴ FD2 = (12)2 + (8)2   ...[Given]

∴ FD2 = 144 + 64

∴ FD2 = 208

∴ FD = `sqrt(16 xx 13)`   ...[Taking square root of both sides]

∴ FD = `4sqrt(13)` units

(3) In EGF,

 ∠EGF = 90°   ...[⸪ FG ⊥ ED]

∴ EF2 = EG2 + FG2    ...[Pythagoras theorem]

∴ EF2 = (18)2 + (12)2    ...[From (i) and given]

∴ EF2 = 324 + 144

∴ EF2 = 468

∴ EF = `sqrt(36 xx 13)`   ...[Taking square root of both sides]

∴ EF = `6sqrt(13)` units

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Theorem of Geometric Mean
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Pythagoras Theorem - Q.3 (B)
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