What is the area of the marked region?



The triangle \(ABC\) with \(AB = \dfrac{1}{3}\), \(BC = \dfrac{2}{3}\), and \(\widehat{ABC} = 180^\circ - 60^\circ = 120^\circ\). Hence its area equals \[\dfrac{\dfrac{1}{3} \cdot \dfrac{2}{3} \cdot \sin(120^\circ)}{2} = \dfrac{\sqrt{3}}{18}\,.\]