Question

Find the value of x for which $DE\parallel AB$ in given figure.

Answer 1

Given, $DE\parallel AB$
$\therefore$ $\frac{CD}{AD}=\frac{CE}{BE}$ [by basic proportionally theoreum]
$\Rightarrow \frac{x+3}{3x+19}=\frac{x}{3x+4}$
$\Rightarrow (x+3)(3x+4)=x(3x+19)$
$\Rightarrow (x+3)(3x+4)=x(3x+19)$
$\Rightarrow 3x^2+4x+9x+12=3x^2+19x$
$\Rightarrow 19x-13x=12$
$\Rightarrow 6x=12$
$\therefore x=\frac{16}{2}=2$
Hence, the required value of x is 2.