Area of A Spinner

let $y$ be the area of the red region (1 part). One part is in the form of a sector of a circle. So

$y=\dfrac{240}{360} \pi (1^2)=\dfrac{2}{3} \pi$ , since there are three parts, $3y=3\left(\dfrac{2}{3}\right) \pi=2 \pi$

let $x$ be the area of the white region (1 part). One part is in the form of a sector of a circle. So

$x=\dfrac{120}{360} \pi (1^2)=\dfrac{1}{3}\pi$ , since there are three parts, $3x=3\left(\dfrac{1}{3}\right)=\pi$

let $H$ be the area of the regular hexagon. Since a regular hexagon is composed of $6$ equilateral triangles, we have

$H=6\left(\dfrac{1}{2}\right)(2^2)\left(\dfrac{\sqrt{3}}{2}\right)=(3)(4)\left(\dfrac{\sqrt{3}}{2}\right)=6\sqrt{3}$

Let $A$ be the area of the shaded part (red and green). then

$A=H-x+y=6\sqrt{3}-\pi+2 \pi=$ $\color{#3D99F6}\large \boxed{6\sqrt{3}+\pi}$

The area of the regular hexagon is: $\dfrac{3}{2}a^2*\sqrt{3}=\dfrac{3}{2}*4*\sqrt{3}=6\sqrt{3}$ The shaded region (which looks like a fidget spinner :) ) is equal to the area of the hexagon + 3 sector with an area of $t$ . Since three sector with an area of $t$ define a circle with radii $\dfrac{2}{2}=1$ , the area of the shaded region is: $6\sqrt{3}+3t=6\sqrt{3}+r^2*\pi=6\sqrt{3}+1\pi=\boxed{6\sqrt{3}+\pi}$