Question

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Answer 1

$\because$ DE ∥ BC

$$\begin{gathered} \therefore \frac{AD}{DB}=\frac{AE}{EC}(\mathrm{Basic}\mathrm{proportionality}\mathrm{theorem}) \\ \frac{x}{3x+4}=\frac{x+3}{3x+19} \\ \Rightarrow x(3x+19)=(x+3)(3x+4) \\ \Rightarrow 3x^2+19x=3x^2+4x+9x+12 \\ \Rightarrow 19x-13x=12 \\ \Rightarrow 6x=12 \\ \Rightarrow x=2 \end{gathered}$$