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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Construction : Draw a perpendicular from A and G and name it as M and N respectively. We know that medians of a triangle intersect at a ratio of 2:1. Thus, AG/GD = 2/1 or, AG + GD/GD = 2+1/1 or, AD/GD = 3/1 .......................... (i)

Now, in triangles AMD and GND, <AMN = <GND <ADM is common Thus triangle AMD ~ GND (AA)

So, we get AM/GN = AD/GD or, AM/GN = 3/1 [From (i)] Next ar(ABC) : ar(BGC) = 1/2 x BC x AM : 1/2 x BC x GN = AM : GN = 3 : 1 or, ar(GBC) = 1/3ar(ABC)

2)in the fig.consider tri.ABC AD is the median,so ar.ABD=arACG (1) now,consider tri.BGC GD is the median so, ar.BGD=CGD (2) subtracting 2 from 1,we get, ar.AGB=AGC llly,Ar. BGC=AGC i.e.Ar[AGB=AGC=BGC] and ar.[AGB+G=BGC+AGC=ABC] hence,the areas of all the triangles so formed is 1/3 Ar of ABC