#### Question

In the given figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF

#### Solution

We know that,

In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Here, seg GF ⊥ seg ED

\[\therefore {GF}^2 = EG \times GD\]

\[ \Rightarrow {12}^2 = EG \times 8\]

\[ \Rightarrow 144 = EG \times 8\]

\[ \Rightarrow EG = \frac{144}{8}\]

\[ \Rightarrow EG = 18\]

Hence, EG = 18.

Now,

According to Pythagoras theorem, in ∆DGF

\[{DG}^2 + {GF}^2 = {FD}^2 \]

\[ \Rightarrow 8^2 + {12}^2 = {FD}^2 \]

\[ \Rightarrow 64 + 144 = {FD}^2 \]

\[ \Rightarrow {FD}^2 = 208\]

\[ \Rightarrow FD = 4\sqrt{13}\]

In ∆EGF

\[{EG}^2 + {GF}^2 = {EF}^2 \]

\[ \Rightarrow {18}^2 + {12}^2 = {EF}^2 \]

\[ \Rightarrow 324 + 144 = {EF}^2 \]

\[ \Rightarrow {EF}^2 = 468\]

\[ \Rightarrow EF = 6\sqrt{13}\]

Hence, FD =\[4\sqrt{13}\] and EF=\[6\sqrt{13}\]