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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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