Question : In a triangle ABC, AD, BE, and CF are three medians. The perimeter of ABC is:
Option 1: equal to $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 2: greater than $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 3: less than $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 4: none of these
Correct Answer: greater than $(\overline {AD}+\overline{BE}+\overline{CF})$
Solution :
We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side.
Here, D, E, and F are the midpoints of the sides BC, AC, and AB respectively.
∴ AB + AC > 2AD --(1)
AB + BC > 2BE --(2)
BC + AC > 2CF --(3)
Adding all three equations, we get
2(AB + BC + AC) > 2(AD + BE + CF)
⇒ AB + BC + AC > AD + BE + CF
⇒ AB + BC + AC > $(\overline {AD}+\overline{BE}+\overline{CF})$
Hence, the correct answer is greater than $(\overline {AD}+\overline{BE}+\overline{CF})$.
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