In an equilateral triangle ABC, P is the centroid of this triangle. Side of ΔABC is $16\sqrt{3}$ cm. What is the distance of point P from side BC ?

8 cm 12 cm 9 cm 10 cm

8 cm

Height = AD = \(\frac{\sqrt {3}}{2}\) × Side = Height = \(\frac{\sqrt {3}}{2}\) × 16√3 = AD = 24 cm = AP : PD = 2 : 1 Let, AP = 2m PD = m, AD = 3m = 3m = 24 = m = 8