# As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then EG = ? - Geometry

Sum

As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then EG = ?

#### Solution

(i) 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

(ii)  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

(iii) 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

Concept: Theorem of Geometric Mean
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