Question

In the figure (not drawn to scale) given below, P is a point on AB such that AP: PB = 4:3. PQ is parallel to AC and QD is parallel to CP. In $\Delta ARC,\angle ARC={90}^{\circ }$, and in $\Delta PQS,\angle PSQ={90}^{\circ }$. The length of QS is 6 cms. What is ratio AP: PD?

• A. 10 : 3
• B. 2 : 1
• C. 7 : 3
• D. 8 : 3

Answer 1

The correct option is C 7 : 3

From figure,


Given that
$PQ||AC,\therefore \frac{CQ}{QB}=\frac{AP}{PB}=\frac{4}{3}Again,QD||CP,\therefore \frac{PD}{BD}=\frac{CQ}{QB}=\frac{4}{3}As\frac{PD}{BD}=\frac{4}{3}\Rightarrow \frac{PD}{DB+PD}=\frac{4}{3+4}\Rightarrow \frac{PD}{PB}=\frac{4}{7}\Rightarrow PD=\frac{4}{7}PB\therefore \frac{AP}{PD}=\frac{AP}{\frac{4}{7}PB}=\frac{7}{4}\times \frac{AP}{PB}=\frac{7}{4}\times \frac{4}{3}=7:3$