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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 $\frac{AD}{DB}\ =\ \frac{3}{4}$ and $AC\ =\ 15\ cm$, find $AE$.
"
Given:
In a $Δ\ ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\ ||\ BC$.
$\frac{AD}{DB}\ =\ \frac{3}{4}$ and $AC\ =\ 15\ cm$.
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,
$\frac{AD}{DB} + 1 = \frac{AE}{EC} + 1$
$\frac{3}{4} + 1 =\frac{AE+EC}{EC}$ ($\frac{AD}{DB}\ =\ \frac{3}{4}$)
$\frac{3+4}{4} =\frac{AC}{EC}$ ($AE+EC=AC$)
$\frac{7}{4} =\frac{15}{EC}$
$EC= \frac{15\times4}{7}$
$EC=\frac{60}{7} cm$
$AE=AC-EC$
$AE=15-\frac{60}{7} cm$
$AE=\frac{15\times7-60}{7}$
$AE=\frac{105-60}{7}$
$AE=\frac{45}{7} cm$
The measure of $AC$ is $\frac{45}{7} cm$.
Updated on: 10-Oct-2022
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