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 ?
Answer & explanation
Correct answer: option 1
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