We can make this into an equilateral triangle by joining O and other end of the chord. Since the angle between side of an equilateral triangle is $60^{\circ}$, we can fit $6$ triangle side by side (as a hexagon), so there will be $6$ shaded regions and $6$ triangle within the circle.
Let us assume the area of shaded region be $A$.
Then, Area of circle = $6\cdot A + 6\cdot$Area of equilateral triangles
$\implies6\cdot A =$ Area of circle – $6 \cdot$Area of equilateral triangles
$\implies 6\cdot A = \pi\cdot (4)^2 - 6\cdot \dfrac{\sqrt{3}}{4}\cdot (4)^2$
$\implies6\cdot A = \pi\cdot 16 - 24\cdot \dfrac{\sqrt{3}}{4}$
$\implies A = \dfrac{\pi\cdot 16}{6} - \dfrac{24\cdot \sqrt{3}}{6}$
$\implies A = \dfrac{8\pi}{3} - 4\sqrt{3}$
Or we can directly calculate the area of that sector by taking the proportion of area of circle as, $\dfrac{60}{360}\cdot \pi \cdot 4^2$
Aread of sector$= \dfrac{8\pi}{3}$
Area of equilatera triangle $=\dfrac{\sqrt{3}}{4}4^2 = 4\sqrt{3}$
Area of shaded region $=\dfrac{8\pi}{3} - 4\sqrt{3}$
