Question

In a ABC, medians AD, BE and CF intersect each other at point G.

Prove that 3(AB + BC + CA) > 2(AD + BE + CF).​

Answer 1

Answer:

Given that in ∆ABC , medians AD, BE and CF intersects at a point G.

AG = 6 cm, BE = 9 cm and GF = 4.5 cm

As we know that if the medians in a triangle intersect at a point then the point is the centroid of the triangle and the centroid divides the median in 2:1 ratio.

Here, G is the centroid of the ∆ABC.

And AG:GD = 2:1,

BG:GE = 2:1 ,

CG:GF = 2:1

Now,

GD=3 cm

Again,

⇒ CG=4.5×2=9

CG=9 cm

BE = BG+GE………………(1)

⇒BG=2GE

Putting in equation (1)

9 = 2GE+GE

⇒ 3GE=9

⇒ GE = 3

GE = 3 cm

Hence, GD=3 cm, CG=9 cm and GE = 3 cm