Question

In the figure, line $PQ$ is parallel to side $BC$. Find the value of '$x$'. If $AP=3$, $PB=6$, $AQ=5$, $QC=x$.

Answer 1

Since $PQ\parallel BC$,

Thus using Basic Proportionality Theorem

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

$\Rightarrow \frac{5}{x}=\frac{3}{6}$

$\Rightarrow 3x=30$, cross multiply

$\therefore x=10$