A Weird Area

let $A$ = area of the shaded region enclosed by the curve

$A_H$ = area of the hexagon

$A_R$ = area of the red region

$A_W$ = area of the white region inside the hexagon

From the figure, we can see that $\boxed{A = A_H - A_W + A_R}$ .

Solving for the area of the hexagon

$A_H =$ $6(area$ $of$ $one$ $equilateral$ $triangle)$

$A_H = 6(0.5)(2^2)(\frac{\sqrt{3}}{2}) = 6\sqrt{3}$

Solving for the area of the white region inside the hexagon

$A_W = 3(\frac{120}{360})(π)(1^2) = π$

Solving for the area of the red region

$A_R = 3(\frac{240}{360})(π)(1^2) = 2π$

Solving for the area of the shaded region enclosed by the curve

$A = A_H - A_W + A_R$

$A = 6\sqrt{3} - π + 2π =$ $\boxed{6\sqrt{3} + π}$

Red each covers 240 degree sector of circle r=1.
White each eats awy 120 degree sector of circle r=1.
So surplus each 120 degrees over Hexagon area.
There are 3 such sectors to give 3 120==360 degrees, that is a circle area $\pi*1^2 =\pi$ .
Total colored area =Hexagon area+surplus area=6 equilateral triangles sides = 2 +surplus area.
Total colored area= $6*\dfrac{\sqrt3} 4*2^2+\pi=\Large \color{#D61F06}{6\sqrt3+\pi}.$