Question

In the figure, given, $P$ is a point on $AB$ such that $AP:PB=4:3$. $PQ$ is parallel to $AC$.
(i) Calculate the ratio $PQ:AC$, giving reason for your answer
(ii) In triangle $ARC,\angle ARC={90}^o$ and in triangle $PQS,\angle PSQ={90}^o$. Given $QS=6cm$, calculate the length of $AR$.

Answer 1

(i) In $△BPQ$ and $△BAC$

$\angle BPQ=\angle BAC[\because PQ\parallel AC]$

$\angle B=\angle B\left[common\right]$

$\therefore$ $△BPQ\sim △BAC$ (By AA similarity)

$\frac{PQ}{AC}=\frac{BP}{BA}\left[BySSST\right]⟶\left(2\right)$

Also, $\frac{AP}{BP}=\frac{4}{3}\Rightarrow \frac{AP}{BP}+1=\frac{4}{3}+1$

$\Rightarrow \frac{AP+PB}{PB}=\frac{7}{3}\Rightarrow \frac{AB}{PB}=\frac{7}{3}\Rightarrow \frac{PB}{AB}=\frac{3}{7}⟶\left(2\right)$

from (1) and (2), $\frac{PQ}{AC}=\frac{3}{7}$

(ii) In $△RAC$ and $△PSQ$

$\angle ARC=\angle PAQ\left[{90}^0\right]$

$\angle RAC=\angle QPS(PQ\parallel AC)$

$\therefore △RAC\sim △QPS$ [By AA Similarity]

$\therefore \frac{AR}{QS}=\frac{AC}{PQ}\Rightarrow \frac{AR}{6}=\frac{7}{3}$

$\Rightarrow AR=\frac{7\times 6}{3}=14cm$