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 \[(\overline{AD}+\overline{BE}+\overline{CF})\] |
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