Question

1) In figure, DFE = 90°, FG 1 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