In figure, DEF is a right -angled triangle with E =90^ . FE is pr… - Vidyadip

Question

In figure, DEF is a right -angled triangle with $\angle E =90^{\circ}$. $FE$ is produced to $G$ and $GH$ is drawn perpendicular to $DE =8 cm , DH =8 cm , DH =6 cm$ and $HF =4 cm$, find $\frac{\operatorname{Ar} \triangle DEF }{\operatorname{Ar} \triangle GHF }$

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Answer

$\operatorname{In} \triangle DEF \text { and } \triangle GHF _{,}$
$\angle DEF =\angle GHF \left(90^{\circ} \text { each) }\right.$
$\angle DEF =\angle GHF \quad \ldots \text { (common) }$
$\triangle DEF =\triangle GHF \quad \ldots .( AA \text { corollary) }$
$\therefore \frac{ Ar \triangle DEF }{ Ar \triangle GHF }=\frac{ EF ^2}{ HF ^2}\ldots(1)$
[The ration of areas of two similar triangle is equal to the ratio of square of their corresponding sides.]
In right $\triangle D E F$, (By Pythagoras theorem)
$D E^2+E F^2=D F^2$
$E F^2=10^2-8^2$
$E F^2=36$
$E F=6$
From ( 1),
$\frac{\operatorname{Ar} \triangle DEF }{\operatorname{Ar} \triangle GHF }=\left(\frac{6}{4}\right)^2=\frac{9}{4}$
i.e. $9: 4$

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