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. \[\Delta APQ-\Delta ABC\] \[\therefore \] \[\frac{\text{Area}\,\text{of}\,\Delta \Alpha PQ}{\text{Area}\,\text{of}\,\Delta \Alpha \Beta C}=\frac{A{{P}^{2}}}{A{{B}^{2}}}\] \[\Rightarrow \]\[\frac{A{{P}^{2}}}{A{{B}^{2}}}=\frac{9}{16}\Rightarrow \frac{AP}{AB}=\frac{3}{4}\] \[\Rightarrow \]\[\frac{AB}{AP}=\frac{4}{3}\Rightarrow \frac{AP+PB}{AP}=\frac{4}{3}\] \[\Rightarrow \]\[1+\frac{PB}{AP}=\frac{4}{3}\] \[\Rightarrow \]\[\frac{PB}{AP}=\frac{4}{3}-1=\frac{1}{3}\] \[\Rightarrow \]\[\frac{AP}{PB}=\frac{3}{1}\]You need to login to perform this action.
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