question_answer

G is the centroid of the equilateral \[\Delta \,ABC\] If AB = 10 cm, then length of AG is [SSC CGL Tier II,2014]

A) \[\frac{5\sqrt{3}}{3}\,cm\]

B) \[\frac{10\sqrt{3}}{3}\,cm\]

C) \[5\sqrt{3}\]

D) \[10\sqrt{3}\]

Correct Answer: B

Solution :

[b] in equilateral \[\Delta \] Altitude = Median So, length of altitude             \[(AD)=\frac{\sqrt{3}}{2}A=\frac{\sqrt{3}}{2}\times 10=5\sqrt{3}\] Now, \[AG=\frac{2}{3}\times AD=\frac{2\times 5\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}\]