Question

which one greater in triangles sum of medians or perimeter​

Answer 1

Answer:

We know that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.

∴ (AB+BC+AC)>(AD+BE+CF).

Hence, the perimeter of a triangle is greater than the sum of its three medians.

Answer 2

In the triangle ABC, D,E and F are the midpoints of sides BC,CA and AB respectively.

We know that the sum of two sides of a triangle is greater than twice the median bisecting the third side,

Hence in triangle ABD, AD is a median

⇒AB+AC>2(AD)

Similarly, we get

⇒BC+AC>2(CF)

⇒BC+AB>2(BE)

On adding the above inequations, we get

(AB+AC)+(BC+AC)+(BC+AB)>2AD+2CF+2BE

2(AB+BC+AC)>2(AD+BE+CF)

∴AB+BC+AC>AD+BE+CF

Then perimeter of a triangle is greater than sum of its three medians

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