What is the area of the marked region?



The triangle \(ABC\) has \(BC = \dfrac{2}{3}\), \(AC = 1\), and \(\widehat{BCA} = 60^\circ\).
Hence its area equals \[\dfrac{\dfrac{2}{3} \cdot \sin(60^\circ)}{2} = \dfrac{\sqrt{3}}{3}\,.\]