Question

In the adjoining figure, ∠DFE = 90° , FG ⊥ ED. If GD = 8, FG = 12, find 

i. EG 

ii. FD, and 

iii. EF

Answer 1

i. In ∆DEF, ∠DFE = 90° and FG ⊥ ED [Given] 

∴ FG = GD × EG [Theorem of geometric mean] 

∴ 122 = 8 × EG . 

∴ EG = 144/8

∴ EG = 18 units 

ii. In ∆FGD, ∠FGD = 90° [Given] 

∴ FD² = FG² + GD² [Pythagoras theorem] 

= 122 + 82 = 144 + 64 

= 208 

∴ FD = \({\sqrt {208}}\) [Taking square root of both sides] 

∴ FD = 4\({\sqrt {13 }}\) units

iii. In ∆EGF, ∠EGF = 90° [Given] 

∴ EF² = EG² + FG² [Pythagoras theorem]

= 182 + 122 = 324 + 144 

= 468 

∴ EF = \({\sqrt{468}}\) [Taking square root of both sides] 

∴ EF = 6\({\sqrt{13 }}\) units