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If triangle AOB is an equilaterlal triangle (O is the and A is a point on the x axis ) then the centriod of the triangle will be
ASSUMING O IS THE ORIGIN: A IS DISTANCE a UNITS FROM THE ORIGIN.coordinate of A is (a,0) coordinate of b is (a/2,31/2a/2)...easily attainable....u knw that ob makes an angle of pi/3 with x axis and that the distance of ob is a units...... NOW, THE X COORDINATE OF CENTROID IS a/2. from symetry. or now u knw the points u can find centroid usin- g=((x1+x2+x3)/3,(y1+y2+y3)/3)
ASSUMING O IS THE ORIGIN:
A IS DISTANCE a UNITS FROM THE ORIGIN.coordinate of A is (a,0)
coordinate of b is (a/2,31/2a/2)...easily attainable....u knw that ob makes an angle of pi/3 with x axis and that the distance of ob is a units......
NOW, THE X COORDINATE OF CENTROID IS a/2. from symetry.
or now u knw the points u can find centroid usin- g=((x1+x2+x3)/3,(y1+y2+y3)/3)
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