What is the altitude of an equilateral triangle whose side is 12√3 cm?

$$\triangle$$ ABC is equilateral triangle with AB = AC = BC = $$12\sqrt{3}$$ cm

AD is altitude of the triangle and also, in an equilateral triangle altitude and median coincides. Thus, AD is the median

=> CD = $$\frac{12\sqrt{3}}{2} = 6\sqrt{3}$$ cm

In $$\triangle$$ ACD, => $$(AD)^2 = (AC)^2 - (CD)^2$$

=> $$(AD)^2 = (12\sqrt{3})^2 - (6\sqrt{3})^2$$

=> $$(AD)^2 = 432 - 108 = 324$$

=> $$AD = \sqrt{324} = 18$$ cm

=> Ans - (D)