Question

In $\Delta ABC,\overline{PQ}||\overline{BC}$ and $AP=3x-19,PB=x-5,AQ=x-3,QC=3cm$. Find $x$.

Answer 1

Given:- $ABC$ is a $△$ in which $PQ\parallel BC$ and $AP=3x-19,PB=x-5,AQ=x-3,QC=3cm$

To find:- $x=?$

Solution:-

From basic proportionality theorem,

$\frac{AP}{PB}=\frac{AQ}{QC}$

$\frac{3x-19}{x-5}=\frac{x-3}{3}$

$3(3x-19)=(x-3)(x-5)$

$9x-57=x^2-3x-5x+15$

$x^2-8x+15-9x+57=0$

$x^2-17x+72=0$

$x^2-9x-8x+72=0$

$(x-9)(x-8)=0$

$x=8$ or $x=9$

Hence the valuen of $x$ is $8$ or $9$.