Who Ate My Pi?

Let $AB=BC=AC=2\sqrt[4]{12}=2r$ . $r=\sqrt[4]{12}$ . We note that the top and down segments (in white) of the small circle are identical. The area of one of this small segment is

$\begin{aligned} A_1 & = \text{Area of }60^\circ \text{ sector} - \text{Area of equilateral triangle} \\ & = \frac {\pi r^2}6 - \frac 12r^2 \sin 60^\circ \\ & = \frac {\pi r^2}6 - \frac {\sqrt 3 r^2}4 \end{aligned}$

Since the big segment $AB$ (in white) of the big circle has a radius of $2r$ , its area is \(A_2 = 2^2A_1 = 4A_1). And the area of the blue area is given by:

\(\begin{align} A_1 & = \text{Area of the }{\color{blue}\text{blue }} \text{circle} - 2A_1 - \color{blue}A_2 & \small \color{blue} \text{Since }A_2 = 4A_1 \\ & = \pi r^2 - 6A_1 \\ & = \pi r^2 - 6 \left(\frac {\pi r^2}6 - \frac {\sqrt 3 r^2}4\right) \\ & = \frac {3 \sqrt 3 r^2}2 \\ & = \frac {3\sqrt 3}2 \left(\sqrt[4]{12}\right)^2 \\ & = \frac {3\sqrt 3}2 \left(2\sqrt 3\right) \\ & = \boxed 9 \end{align} \)