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' \[\frac{Area\,\,of\,\Delta APQ}{Area\,\,of\,\Delta ABC}\] \[={{\left( \frac{AP}{AB} \right)}^{2}}={{\left( \frac{1}{3} \right)}^{2}}=\frac{1}{9}\,\,\,\,\,\,\,\,[\because \,\,\,\,3AP=AB]\] \[=\,\,\,\,\mathbf{1:9}\]You need to login to perform this action.
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