question_answer

Points P and Q lie on side AB and AC of triangle ABC respectively such that segment PQ is parallel to side BC. If the ratio of areas of triangle APQ and triangle ABC is 9 : 16, then what is the ratio of AP : PB?

A)  3 : 4

B)  9 : 7

C)  9 : 16

D)  3 : 1

Correct Answer: D

Solution :

\[PQ||BC\] \[\therefore \]\[\angle APQ=\angle ABC\] \[\angle AQP=\angle ACB\] By AA-similarity. \[\Delta APQ-\Delta ABC\] \[\therefore \] \[\frac{\text{Area}\,\text{of}\,\Delta \Alpha PQ}{\text{Area}\,\text{of}\,\Delta \Alpha \Beta C}=\frac{A{{P}^{2}}}{A{{B}^{2}}}\] \[\Rightarrow \]\[\frac{A{{P}^{2}}}{A{{B}^{2}}}=\frac{9}{16}\Rightarrow \frac{AP}{AB}=\frac{3}{4}\] \[\Rightarrow \]\[\frac{AB}{AP}=\frac{4}{3}\Rightarrow \frac{AP+PB}{AP}=\frac{4}{3}\] \[\Rightarrow \]\[1+\frac{PB}{AP}=\frac{4}{3}\] \[\Rightarrow \]\[\frac{PB}{AP}=\frac{4}{3}-1=\frac{1}{3}\] \[\Rightarrow \]\[\frac{AP}{PB}=\frac{3}{1}\]