Center of the circle: ABC

=> AB = BC = AC = 2R = 4

=> ABC is a equilateral triangle

The area for the shaded region: S

The area for a sector definition by A and 2 tangential points: S_{A}

S = S_{ABC }- 3S_{A}

S_{ABC} = \(\dfrac{1}{2}\).4.4.\(\dfrac{\sqrt{3}}{2}\) = 4\(\sqrt{3}\)

S_{A }= \(\dfrac{60}{360}\)S_{circles }= \(\dfrac{1}{6}\)\(\pi\)2^{2 }= \(\dfrac{2\pi}{3}\)

^{=> S = 4\(\sqrt{3}\)} - 3\(\dfrac{2\pi}{3}\) = 4\(\sqrt{3}\) - 2\(\pi\)

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