What is the area of the marked region?



The marked triangle \(ABC\) has \(BC = \dfrac{1}{3}\), \(AC = 1\), and \(\widehat{BCA} = 60^\circ\). Hence its area equals \[\dfrac{\dfrac{1}{3} \cdot \sin(60^\circ)}{2} = \dfrac{\sqrt{3}}{6}\,.\]