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]

∴ FG^{2} = 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]

∴ FD^{2} = FG^{2} + GD^{2} ......[Pythagoras theorem]

∴ FD^{2} = (12)^{2} + (8)^{2} ......[Given]

∴ FD^{2} = 144 + 64

∴ FD^{2} = 208

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

∴ FD = `4sqrt(13)` units

(iii) In EGF,

∠EGF = 90° ......[⸪ FG ⊥ ED]

∴ EF^{2} = EG^{2} + FG^{2} ......[Pythagoras theorem]

∴ EF^{2} = (18)^{2} + (12)^{2} ......[From (i) and given]

∴ EF^{2} = 324 + 144

∴ EF^{2} = 468

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

∴ EF = `6sqrt(13)` units