G is the centroid of the equilateral ΔABC. If AB = 10 cm then length of AG is ?

According to question,

Given :
AB = BC = CA = 10 cm
G = Centroid
AG = 2 units
GD = 1 unit
AD = 3 units = Height
As we know that the height of the equilateral triangle is
$$\eqalign{ & = \frac{{\sqrt 3 }}{2} \times 10 = 5\sqrt 3 \cr & \therefore 3\,{\text{units}} = 5\sqrt 3 \cr & \,\,\,\,\,\,1\,{\text{unit}} = \frac{{5\sqrt 3 }}{3} \cr & \,\,\,\,\,\,2\,{\text{units}} = \frac{{5\sqrt 3 }}{3} \times 2 \cr & \,\,\,\,\,\,2\,{\text{units}} = \frac{{10\sqrt 3 }}{3} \cr & \therefore {\text{AG}} = \frac{{10\sqrt 3 }}{3}\,cm \cr} $$