Quantitative Aptitude Ques 2326
Question: AB is a straight line, C and D are points the same side of AB such that CA perpendicular to AB and DB is perpendicular to AB. Let AD and BC meet at E. What is $ \frac{AE}{AD}+\frac{BE}{BC} $ equal to?
Options:
A) 2
B) 1.5
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
- Since, AB is a straight line and C and D are points such that $ AC\bot AB $ and $ BD\bot AB. $
$ \therefore $ $ AC\parallel BD $ So, ABCD forms trapezium. Now, by property of trapezium diagonals intersect each other in the ratio of lengths of parallel sides.
$ \therefore $ $ \frac{AE}{ED}=\frac{BE}{CE} $
$ \frac{AE}{AD-AE}=\frac{BE}{BC-BE} $
$ \frac{BC-BE}{BE}=\frac{AD-AE}{AE} $
$ \frac{BC}{BE}-1=\frac{AD}{AE}-1 $
$ \Rightarrow $ $ \frac{BC}{BE}=\frac{AD}{AE} $
$ \frac{AE}{AD}=\frac{BE}{BC} $
But the value of $ \frac{AE}{AD} $ or $ \frac{BE}{BC} $ cannot be determined.
So, we cannot find the value of $ \frac{AE}{AD}+\frac{BE}{BC}. $
BETA
