Question

Q1) In the adjoining figure, ∠DFE = 90°, FG ⊥ ED. If GD = 8, FG = 12, find i. EG ii. FD, and iii. EF​

Answer 1

Step-by-step explanation:

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 = 208−−−√208 [Taking square root of both sides] ∴ FD = 413−−√13 units

iii. In ∆EGF, ∠EGF = 90° [Given] ∴ EF2 = EG2 + FG2 [Pythagoras theorem] = 182 + 122 = 324 + 144 = 468 ∴ EF = 468−−−√468 [Taking square root of both sides] ∴ EF = 613−−√13 units