A) \( \Large 3\sqrt{3}cm \) |
B) \( \Large 2\sqrt{3}cm \) |
C) \( \Large \sqrt{3}cm \) |
D) \( \Large \frac{\sqrt{3}}{2}cm \) |
B) \( \Large 2\sqrt{3}cm \) |
Here 'O' is the centroid of triangle and centroid of triangle divides line segment in the ratio 2 : 1 Here BO : OM = 2 : 1
Now, in \( \Large \triangle ABD \)
\( \Large \left(AD\right)^{2} = \sqrt{AB^{2}-BD^{2}} \)
= 36 - 9
=27
\( \Large AD = 3\sqrt{3} cm = BM \)
\( \Large \because \triangle ABC\ is\ an\ equilateral\ triangle \)
Now BM = BO + OM
\( \Large 3\sqrt{3} = 2x+x \)
\( \Large \therefore x = \sqrt{3} \)
\( \Large \therefore OB = 2\sqrt{3} cm \)