Question:
In $\Delta A B C, D E \| B C$ such that $\frac{A D}{D B}=\frac{3}{5}$.
If AC = 5.6 cm, then AE = ?
(a) 4.2 cm
(b) 3.1 cm
(c) 2.8 cm
(d) 2.1 cm
Solution:
(d) 2.1 cm
It is given that $\mathrm{DE} \| B C$.
Applying Thales' theorem, we get:
$\frac{A D}{D B}=\frac{A E}{E C}$
Let $A E$ be $x \mathrm{~cm}$.
Therefore, $E C=(5.6-x) \mathrm{cm}$
$\Rightarrow \frac{3}{5}=\frac{x}{5.6-x}$
$\Rightarrow 3(5.6-x)=5 x$
$\Rightarrow 16.8-3 x=5 x$
$\Rightarrow 8 x=16.8$
$\Rightarrow x=2.1 \mathrm{~cm}$
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