Question

In the given figure, ∠DFE $={90}^{\circ }$, $FG\perp ED$, If $GD=8$, $FG=12$, find (i) $EG$ (ii) $FD$ and (iii) $EF$

Answer 1

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.

i.e $G{F}^2=EG\times GD$

Here, seg $GF\perp ED$
∴ $G{F}^2=EG\times GD$

⇒${12}^2=EG\times 8$

⇒$144=EG\times 8$

⇒ $EG=\frac{144}{8}$

⇒ $EG=18$
Hence, $EG=18$.
Now,
According to Pythagoras theorem, in ∆DGF
$D{G}^2+G{F}^2=F{D}^2$ [Using Pythagoras theorem ]

⇒ $8^2+{12}^2=F{D}^2$

⇒ $64+144=F{D}^2$

⇒ $F{D}^2=208$

⇒ $FD=\sqrt{208}=14.42$
In ∆EGF, we have
$E{G}^2+G{F}^2=E{F}^2$ [Using Pythagoras theorem ]

⇒ ${18}^2+{12}^2=E{F}^2$

⇒ $324+144=E{F}^2$

⇒ $E{F}^2=468$

⇒ $EF=\sqrt{468}=21.63$ [Taking square root both sides]
Hence, $EG=18$, $FD=14.42EF=21.63$.