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}) $

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