The centre of gravity of an isosceles triangle with base (p) and sides (q) from its base is
A. $$\frac{{\sqrt {4{{\text{p}}^2} - {{\text{q}}^2}} }}{6}$$
B. $$\frac{{4{{\text{p}}^2} - {{\text{q}}^2}}}{6}$$
C. $$\frac{{{{\text{p}}^2} - {{\text{q}}^2}}}{4}$$
D. $$\frac{{{{\text{p}}^2} + {{\text{q}}^2}}}{4}$$
Answer: Option A

Correct answer is:
[√(4q² - p²)]/6
@Bashir Ahmad
We know that COG of triangle is h/3 from its base
Using Pythagorus theorm
h = [√(4q² - p²)]/2
=> COG = h/3 = 1/3 * ( [√(4q² - p²)]/2 )
= [√(4q² - p²)]/6
How