Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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 $\Delta DEF \: \:and \: \: \Delta ABC$

Answers (1)

best_answer

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

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

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

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

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

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

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

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

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

Posted by

seema garhwal

View full answer

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10 board exams from February 15 to March 18; download date sheet 2025 PDF

anam-khan

CBSE Date Sheet 2025 Live: CBSE Class 10, 12 board exam timetable soon at cbse.gov.in; practical dates

anam-khan

CBSE Date Sheet 2025 Live: CBSE to announce Class 10, 12 board exam dates soon at cbse.gov.in; sample paper

Latest Question