Quantitative Aptitude Ques 1525
Question: In a $ \Delta ABC, $ $ AD, $ $ BE $ and $ CF $ are three medians. The perimeter of is always $ \Delta ABC $ is always
Options:
A) Equal to $ (\overline{AD}+\overline{BE}+\overline{CF}) $
B) Greater than $ (\overline{AD}+\overline{BE}+\overline{CF}) $
C) Less than $ (\overline{AD}+\overline{BE}+\overline{CF}) $
D) None of the above
Show Answer
Answer:
Correct Answer: B
Solution:
- (b)
Let sides AB, BC and CA be denoted by a, b and c, respectively and median AD, BE and CF be denoted by mb, mc and ma, respectively.
We know that, $ 3(a^{2}+b^{2}+c^{2}) $ $ =4(ma^{2}+mb^{2}+mc^{2}) $ On analysing, $ ma+mb+mc<a+b+c $
$ \therefore $ Perimeter of $ \Delta ABC $ is always greater than $ (\overline{AD}+\overline{BE}+\overline{CF}) $
BETA
