Question
In $\triangle A B C, D$ and $E$ are points on the sides $A B$ and $A C$ respectively such that $D E \| B C$ If $\frac{ AD }{ DB }=\frac{3}{4}$ and $AC =15 cm$ find $AE$
✓
Answer
Let $A E$ be $x$
$\therefore EC =15- x$
In $\triangle ABC$ we have $DE \| BC$
By Basic proportionality theorem, we have
$
\begin{aligned}
& \frac{ AD }{ DB }=\frac{ AE }{ EC } \\
& \frac{3}{4}=\frac{x}{15-x} \\
& 4 x =3(15- x ) \\
& 4 x =45-3 x \\
& 7 x =45 \\
& \Rightarrow x =\frac{45}{7} \\
& =6.43
\end{aligned}
$
The value of $x=6.43$
$\therefore EC =15- x$
In $\triangle ABC$ we have $DE \| BC$
By Basic proportionality theorem, we have
$
\begin{aligned}
& \frac{ AD }{ DB }=\frac{ AE }{ EC } \\
& \frac{3}{4}=\frac{x}{15-x} \\
& 4 x =3(15- x ) \\
& 4 x =45-3 x \\
& 7 x =45 \\
& \Rightarrow x =\frac{45}{7} \\
& =6.43
\end{aligned}
$
The value of $x=6.43$
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