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In a ∆ ABC, AD, BE and CF are three medians. The perimeter of ∆ ABC is always
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- equal to (AD + BE + CF)
- greater than (AD + BE + CF)
- less than (AD + BE + CF)
- None of these
Correct Option: B
On the basis of given in question , we draw a figure triangle ABC in which AD, BE and CF are three medians ,
We know that sum of any two sides of a triangle is greater than twice the mediun bisecting the third side.
Here, D, E and F are the midpoint of the sides BC, AC and AB.
∴ AB + AC > 2 AD
AB + BC > 2BE
BC + AC > 2 CF
Adding all three, we get
2(AB + BC + AC) > 2(AD + BE + CF)
⇒ AB + BC + AC > AD + BE + CF
