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In Fig. 1, $DE||BC, AD=1\ cm$ and $BD=2\ cm$. What is the ratio of the $ar( \vartriangle ABC)$ to the $ar( \vartriangle ADE)$?
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Given: $\vartriangle ABC$ where, $DE∥BC, AD=1\ cm, BD=2\ cm$

To do: To find the ratio of the $ar( \vartriangle ABC)$ to the
 $ar( \vartriangle ADE)$.

Solution:
In $\vartriangle ABC and \vartriangle ADE$,

$\angle A=\angle A                             ( common)$

$\angle ABC=\angle ADE                  ( \because DE∥BC)$

$\angle ACB=\angle AED                  ( \because DE∥BC)$

$\therefore \vartriangle ABC~\vartriangle ADE           (AAA rule)$

$\Rightarrow\frac{ar( \vartriangle ABC)}{ar( \vartriangle ADE)}=(\frac{AB}{AD})^{2}$


$\Rightarrow \frac{ar( \vartriangle ABC)}{ar( \vartriangle ADE)}=(\frac{AD+BD}{AD})^{2}$

$\Rightarrow \frac{ar( \vartriangle ABC)}{ar( \vartriangle ADE)}=(\frac{AD+BD}{AD})^{2}$

$\Rightarrow \frac{ar( \vartriangle ABC)}{ar( \vartriangle ADE)}=(\frac{1+2}{1})^{2}$

$\Rightarrow \frac{ar( \vartriangle ABC)}{ar( \vartriangle ADE)}=(\frac{3}{1})^{2}$


$\Rightarrow \frac{ar( \vartriangle ABC)}{ar( \vartriangle ADE)}=\frac{9}{1}$

$\therefore$, The ratio of the areas of triangles $\vartriangle ABC$ and $ \vartriangle ADE$ is $9:1$

 
 

Updated on: 10-Oct-2022

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