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D, E and F are respectively the mid-points of sides AB, BC and CA of D ABC. Find the ratio of the areas of triangle DEF and triangle ABC.

Q5   D, E and F are respectively the mid-points of sides AB, BC and CA of \Delta ABC. Find the
       ratio of the areas of \Delta DEF \: \:and \: \: \Delta ABC

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 D, E and F are respectively the mid-points of sides AB, BC and CA of \Delta ABC.     ( Given )

DE=\frac{1}{2}AC       and          DE||AC

In \Delta BED \: \:and \: \: \Delta ABC,

\angle BED=\angle BCA                (corresponding angles )

\angle BDE=\angle BAC               (corresponding angles )

\Delta BED \: \:\sim \: \: \Delta ABC          (By AA)

\frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{DE^2}{AC^2}

\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{(\frac{1}{2}AC)^2}{AC^2}

\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{1}{4}

\Rightarrow ar(\triangle BED)=\frac{1}{4}\times ar(\triangle ABC)

Let {ar(\triangle ABC)  be x.

\Rightarrow ar(\triangle BED)=\frac{1}{4}\times x

Similarly,

\Rightarrow ar(\triangle CEF)=\frac{1}{4}\times x                       and          \Rightarrow ar(\triangle ADF)=\frac{1}{4}\times x

ar(\triangle ABC)=ar(\triangle ADF)+ar(\triangle BED)+ar(\triangle CEF)+ar(\triangle DEF)

\Rightarrow x=\frac{x}{4}+\frac{x}{4}+\frac{x}{4}+ar(\triangle DEF)

\Rightarrow x=\frac{3x}{4}+ar(\triangle DEF)

\Rightarrow x-\frac{3x}{4}=ar(\triangle DEF)

\Rightarrow \frac{4x-3x}{4}=ar(\triangle DEF)

\Rightarrow \frac{x}{4}=ar(\triangle DEF)

 

\frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{\frac{x}{4}}{x}

\Rightarrow \frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{1}{4}

 

 

 

 

 

 

 

 

 

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