Find the value of $ x $ for which $ \mathrm{DE} \| \mathrm{AB} $ in figure below.
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To do:
We have to find the value of \( x \) for which \( \mathrm{DE} \| \mathrm{AB} \).
Solution:
$D E \| AB$
This implies, by Basic Proportionality Theorem,
$\frac{C D}{A D} =\frac{C E}{B E}$
$\frac{x+3}{3 x+19} =\frac{x}{3 x+4}$
$(x+3)(3 x+4) =x(3 x+19)$
$3 x^{2}+4 x+9 x+12=3 x^{2}+19 x$
$19 x-13 x =12 $
$6 x =12$
$x=\frac{12}{6}$
$x=2$
Therefore, the value of \( x \) for which \( \mathrm{DE} \| \mathrm{AB} \) is $2$.
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