question_answer

In a triangle ABC, a line PQ is drawn 66, parallel to BC. Points P, Q being on AB and AC respectively. If AB = 3AP, then what is the ratio of the area of triangle APQ to the area of triangle ABC?

A)  \[1:3\]                          

B)  \[1:5\]

C)  \[1:7\]                          

D)  \[1:9\]

Correct Answer: D

Solution :

According to Thale's theorem \[PQ||BC\] then, \[\Delta \,APQ\tilde{\ }\Delta \,ABC\] Now'     \[\frac{Area\,\,of\,\Delta APQ}{Area\,\,of\,\Delta ABC}\] \[={{\left( \frac{AP}{AB} \right)}^{2}}={{\left( \frac{1}{3} \right)}^{2}}=\frac{1}{9}\,\,\,\,\,\,\,\,[\because \,\,\,\,3AP=AB]\] \[=\,\,\,\,\mathbf{1:9}\]