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In a $Δ\ ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\ ||\ BC$.
If $\frac{AD}{DB}\ =\ \frac{2}{3}$ and $AC\ =\ 18\ cm$, find $AE$.
"
Given:
In a $Δ\ ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\ ||\ BC$.
$\frac{AD}{DB}\ =\ \frac{2}{3}$ and $AC\ =\ 18\ cm$.
To do:
We have to find the measure of $AE$.
Solution:
$DE\ ||\ BC$ (given)
Therefore,
By Basic proportionality theorem,
$\frac{AD}{DB}\ =\ \frac{AE}{EC}$
Adding $1$ on both sides,
$\frac{AD}{DB} + 1 = \frac{AE}{EC} + 1$
$\frac{2}{3} + 1 =\frac{AE+EC}{EC}$ ($\frac{AD}{DB}\ =\ \frac{2}{3}$)
$\frac{2+3}{3} =\frac{AC}{EC}$ ($AE+EC=AC$)
$\frac{5}{3} =\frac{18}{EC}$
$EC= \frac{18\times3}{5}$
$EC=\frac{54}{5} cm$
$AE=AC-EC$
$AE=18-\frac{54}{5} cm$
$AE=\frac{18\times5-54}{5}$
$AE=\frac{90-54}{5}$
$AE=\frac{36}{5} cm$
$AE=7.2 cm$
The measure of $AC$ is $7.2 cm$.
Updated on: 10-Oct-2022
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