Question
Find the value of x for which $DE || AB$ in:

Answer


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

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