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In the Given Figure, ∠Dfe = 90°, Fg ⊥ Ed, If Gd = 8, Fg = 12, Find (1) Eg (2) Fd and (3) Ef - Geometry

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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=\[6\sqrt{13}\]

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Solution In the Given Figure, ∠Dfe = 90°, Fg ⊥ Ed, If Gd = 8, Fg = 12, Find (1) Eg (2) Fd and (3) Ef Concept: Pythagoras Theorem.
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