Question

In a $△ABC,P$ and $Q$ are points on sides $AB$ and $AC$ respectively, such that $PQ\parallel BC$. If

$AP=2.4cm,AQ=2cm,QC=3cm$ and $BC=6cm$, find $AB$ and $PQ.$

Answer 1

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

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

$\frac{2}{3}=\frac{2.4}{PB}$

$PB=3.6cm$

$AB=AP+PB=2.4+3.6=6cm$

As $PQ\parallel BC$

$\angle AQP=\angle ACB$

$\angle APQ=\angle ABC$

So by $AAA$ $△AQP\sim △ACB$

Hence $\frac{AQ}{AC}=\frac{QP}{CB}$

$\frac{2}{5}=\frac{QP}{6}$

$QP=2.4cm$

So $AB=6cm$ and $QP=2.4cm$