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

1.	In the adjoining figure, ∠DFE = 90°, FG ⊥ ED. If GD = 8, FG = 12, find i. EG ii. FD, and iii. EF
Answer» i. In ∆DEF, ∠DFE = 90° and FG ⊥ ED [Given] ∴ FG² = GD × EG [Theorem of geometric mean] ∴ 122 = 8 × EG. ∴ EG = \(\frac{144}{8}\) ∴ 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.

In the adjoining figure, ∠DFE = 90°, FG ⊥ ED. If GD = 8, FG = 12, find i. EG ii. FD, and iii. EF

Answer»

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

∴ FG² = GD × EG [Theorem of geometric mean]

∴ 122 = 8 × EG.

∴ EG = \(\frac{144}{8}\)

∴ 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.