Question

In the adjacent figure given, if $PQ\parallel AC,PC=x-3,BC=2x,AQ=x-2$ and $AB=2x+3$, then find ${}^{\prime }{x}^{\prime }$.

Answer 1

Here, $AQ=x-2$

$AB=2x+3$

$\therefore BQ=2x+3-x+2$

$\therefore BQ=x+5$

Also, $BC=2x$

$PC=x-3$

$\therefore BP=2x-x+3=x+3$

Since $PQ||AC$

$\therefore \frac{AQ}{BQ}=\frac{CP}{BP}$

$\therefore AQ\times BP=CP\times BQ$

$\therefore (x-2)(x+3)=(x-3)(x+5)$

$\therefore x^2+x-6=x^2+2x-15$

$\therefore 15-6=x$

$\therefore x=9$

Value of $x=9$.