In figure 2.21, \(\angle DFE = 90^\circ\), FG ⊥ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF
Figure: Right-angled \(\triangle DEF\) with the right angle at F; G is the foot of the perpendicular from F on hypotenuse ED (E–G–D).
Answer: (1) EG = 18 (2) FD = \(4\sqrt{13}\) (3) EF = \(6\sqrt{13}\)
(1) By the theorem of geometric mean,
\[FG^2 = EG \times GD\] \[144 = EG \times 8 \Rightarrow EG = 18\](2) In right-angled \(\triangle FGD\),
\[FD^2 = FG^2 + GD^2 = 144 + 64 = 208\] \[FD = \sqrt{208} = 4\sqrt{13}\](3) In right-angled \(\triangle FGE\),
\[EF^2 = FG^2 + EG^2 = 144 + 324 = 468\] \[EF = \sqrt{468} = 6\sqrt{13}\]Log in or create a free account to answer this question.
