Question

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

Answer 1

Compute the required value:

Given: $DE\parallel AB$

Using basic proportionality theorem we have,

$\frac{CD}{AD}=\frac{CE}{BE}$

$$\begin{gathered} \Rightarrow \frac{x+3}{3x+19}=\frac{x}{3x+4} \\ \Rightarrow \left(x+3\right)\left(3x+4\right)=x\left(3x+19\right) \end{gathered}$$

$\Rightarrow 3x^2+4x+9x+12=3x^2+19x$

Combining the like terms,

$$\begin{gathered} \Rightarrow 19x-13x=12 \\ \Rightarrow 6x=12 \\ \Rightarrow x=2 \end{gathered}$$

Hence, the value of $x$ is $2$.