Question
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides AB and AC respectively such that DE || BC.
If $\frac{\text{AD}}{\text{DB}}=\frac{3}{4}$ and AC = 15cm, find AE.

Answer

Given $\frac{\text{AD}}{\text{DB}}=\frac{3}{4}$ and AC = 15cm
$\text{AB}=\text{AD}+\text{DB}$
$=3+4$
$=7\text{cm}$
We know that,
$\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
$\Rightarrow\frac{3}{7}=\frac{\text{AE}}{15}$
$\Rightarrow\text{AE}=\frac{3\times15}{7}$
$\Rightarrow\text{AE}=\frac{45}{7}$
$\Rightarrow\text{AE}=6.428$
$\Rightarrow\text{AE}=6.43\text{cm}$

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