Question

If in fig. (i) and (ii), PQ $\parallel$ BC, find QC in (i) and AQ in (ii).

Answer 1

In $\Delta ABC,PQ\parallel BC$
$\therefore$ By basic Proportionality Theorem, we get
$\frac{AP}{PB}=\frac{AQ}{QC}$
$\Rightarrow \frac{1.5}{3}=\frac{1.3}{QC}$
$\Rightarrow QC=\frac{1.3\times 3}{15}=\frac{13}{5}=2.6cm$
Hence, $QC=2.6cm$

In fig. (ii) given above
In $\Delta ABC,PQ\parallel BC$
$\therefore$ By Basic Proportionality Theorem, we get
$\frac{AP}{PB}=\frac{AQ}{QC}\Rightarrow \frac{3}{6}=\frac{AQ}{5.3}$
$\Rightarrow AQ=\frac{3\times 5.3}{6}=\frac{5.3}{2}=2.65cm$