Quantitative Aptitude Ques 232
Question: BE, CF are the two medians of $ \Delta ABC $ and G is their point of intersection. EF cuts AG at O. The ratio of AO and OG is equal to
Options:
A) 1 : 3
B) 2 : 3
C) 3 : 1
D) 1 : 2
Show Answer
Answer:
Correct Answer: C
Solution:
- F and E are mid-points of AB and AC, respectively So, $ EF\parallel BC $ Also in $ \Delta ADB, $ $ FO\parallel BD $
$ \Rightarrow $ $ \frac{AF}{FB}=\frac{AO}{OD} $
$ \Rightarrow $ $ \frac{AF}{FB}+1=\frac{AO}{OD}+1 $
$ \Rightarrow $ $ \frac{AF+FB}{FB}=\frac{OA+OD}{OD} $
$ \Rightarrow $ $ \frac{AB}{FB}=\frac{AD}{OD} $
$ \Rightarrow $ $ \frac{2FB}{FB}=\frac{AD}{OD} $
So, O is mid-point of AD.
$ \therefore $ $ OA=OD $
$ \Rightarrow $ $ OA=OG+DG $
$ \Rightarrow $ $ OA=OG+\frac{AG}{2} $ $ ( \because \frac{AG}{GD}=\frac{2}{1} ) $
$ \Rightarrow $ $ OA=OG+\frac{AO+OG}{2} $
$ \Rightarrow $ $ 2OA=3OG+AO $
$ \Rightarrow $ $ \frac{AO}{AG}=\frac{3}{1}=3:1 $
BETA
