question_answer

In a\[\Delta ABC,\]\[AD,\] \[BE\] and \[CF\] are three medians. The perimeter of  is always\[\Delta ABC\] is always                      [SSC (CGL) 2014]

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

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, \[3\,\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]\[=4\,\,(m{{a}^{2}}+m{{b}^{2}}+m{{c}^{2}})\]

On analysing, \[ma+mb+mc<a+b+c\]

\[\therefore \]Perimeter of \[\Delta ABC\]is always greater than \[(\overline{AD}+\overline{BE}+\overline{CF})\]