Question

$ABCD$ is a parallelogram $P$ is a point on $AB$ such that $AP:PB=3:2$. $Q$ is a point on $CD$ such that $CQ:QD=7:3$. If $PQ$ meets $AC$ at $R$, then $AR:AC$ is

• A. $1:2$
• B. $6:13$
• C. $5:8$
• D. $4:7$

Answer 1

The correct option is B $6:13$

Since $ABCD$ is a parallelogram, $AB\parallel CD$ and $AD\parallel BC$.
Since $AB\parallel CD$ and $AC$ is Transvaal so $\angle PAQ=\angle QCR$
Since $ABCD$ is a parallelogram, then $AB=CD$ and $AD=BC$.

[Opposite sides of parallelogram are equal]
Let $AB=CD=X$
So $AP=\frac{3X}{5}$ and $PB=\frac{2X}{5}$ ...(As $AP:PB=3:2$)

$CQ=\frac{7X}{10}$ and $QD=\frac{3X}{10}$ ...( As $CQ:QD=7:3$)

$\Rightarrow \Delta CQR\sim \Delta APR$

$\therefore \frac{CR}{AR}=\frac{CQ}{AP}=\frac{\frac{7X}{10}}{\frac{3X}{5}}=\frac{7}{6}$

$\Rightarrow \frac{CR}{AR}+1=\frac{7}{6}+1=\frac{7+6}{6}=\frac{13}{6}$

$\Rightarrow 13AR=6AC$

$\Rightarrow \frac{AR}{AC}=\frac{6}{13}$

Hence, option B.