MCQ
In the given figure value of $x$ for which $\ce{DE \| BC}$ is
- A$3$
- ✓$2$
- C$4$
- D$1$
✓
Answer
Correct option: B.
$2$
In $\triangle \ce{ABC , DE \| BC}$
$\therefore \frac{AD}{DB}=\frac{AE}{EC}$
$\Rightarrow \frac{x+3}{3 x+19}=\frac{x}{3 x+4}$
$\Rightarrow(x+3)(3 x+4)=x(3 x+19)$
$\Rightarrow 3 x^2+4 x+9 x+12=3 x^2+19 x$
$\Rightarrow 3 x^2+13 x+12=3 x^2+19 x$
$\Rightarrow 12=3 x^2+19 x-3 x^2-13 x$
$\Rightarrow 12=6 x$
$\Rightarrow x=\frac{12}{6}=2$
$\therefore x=2$
$\therefore \frac{AD}{DB}=\frac{AE}{EC}$
$\Rightarrow \frac{x+3}{3 x+19}=\frac{x}{3 x+4}$
$\Rightarrow(x+3)(3 x+4)=x(3 x+19)$
$\Rightarrow 3 x^2+4 x+9 x+12=3 x^2+19 x$
$\Rightarrow 3 x^2+13 x+12=3 x^2+19 x$
$\Rightarrow 12=3 x^2+19 x-3 x^2-13 x$
$\Rightarrow 12=6 x$
$\Rightarrow x=\frac{12}{6}=2$
$\therefore x=2$
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