There is a point $P$ inside equilateral $\triangle ABC$ such that $PA=PB=PC=x.$ We know $[ABC]=\dfrac{\sqrt 3}{3}.$ Find $x.$

sqrt (3) / 3 = (1/2) s^2 * sqrt (3) / 2       where s is the side of the triangle

1/3 = (1/4)s^2

4/3  = s^2

We have triangle APB

angle PAB = angle PBA = 30

angle APB = 120

Law of Cosines

s^2 = 2x^2 - 2x^2 * cos (120)

s^2 = 2x^2 - 2x^2 *(-1/2)

s^2 = 3x^2

4/3 = 3x^2

4/9 = x^2

2/3 = x