A Circle and a Square sharing an Area

The area of the shade region $A$ is made up two parts rectangle $BFGE$ and circular segment $FG$ (in pale blue). To find the area of segment $A_{\text{segment}}$ , we minus the area of $\triangle AFG$ , $A_\triangle$ , from the area of sector $AFG$ , $A_{\text{sector}}$ . In equation, we have:

$\begin{aligned} A & = A_{BFGE} + A_{\text{segment}} & \small \blue{\text{Note that }\triangle AFG \text{ is equilateral.}} \\ & = BF \times BE + A_{\text{sector}} - A_\triangle & \small \blue{\implies \angle FAG = 60^\circ \text{ and }h = \frac {\sqrt 3}2} \\ & = \left(h-\frac 12\right) \times 1 + \frac {60^\circ}{360^\circ}\pi - \frac 12 (1)(1) \sin 60^\circ \\ & = \frac {\sqrt 3}2 - \frac 12 + \frac \pi 6 - \frac {\sqrt 3}4 \\ & = \frac {\sqrt 3}4 - \frac 12 + \frac \pi 6 \\ & \approx \boxed{0.457} \end{aligned}$

To get the area of the shaded part, we need to get the area of the rectangle $BFGE$ and the segment of $\angle GAF$ and add them.

Line segment $\overline{FG}$ has the same length as the side of the square which is $1$ . Since $\overline{AG}$ and $\overline{AF}$ are radii, they form an equilateral triangle together with $\overline{FG}$ which means $\angle GAF$ must be $60°$ . We can now then compute for the area of the segment.

$1^{2} \left(\frac{60π}{360} - \frac{\sin(60°)}{2} \right) \approx 0.090586$

In order to find the area of rectangle $BFGE$ , we must first find the length of $\overline{BF}$ .

The length between $K$ and $J$ is the same as the length of $\overline{FH}$ .

$\overline{FH} = \overline{KJ} = x$

Line segment $\overline{LJ}$ is the circle's diameter. Thus, $\overline{KJ} = x$ and $\overline{LK} = 2-x$ . We can use the Intersecting Chords Theorem to get the value of $x$ .

$\begin{aligned} \overline{LK} × \overline{KJ} & = \overline{FK} × \overline{KG} \\ (2-x) × x & = \frac{1}{2} × \frac{1}{2} \\ -x^{2}+2x & = \frac{1}{4} \\ -x^{2}+2x-\frac{1}{4} & = 0 \end{aligned}$

The roots of the quadratic equation above are $1+\frac{\sqrt{3}}{2}$ and $1-\frac{\sqrt{3}}{2}$ . Since the value we're looking for $x$ is less than $1$ , we take the latter root which is approximately $0.133975$ . The length of $\overline{BF}$ is $\frac{1}{2}$ , which is the length of $\overline{BH}$ , minus $x$ . The rectangle's area can now be calculated and it is $0.366025$ . We add up the segment's area and the rectangle's and we get the answer, rounded to the nearest ten-thousandths, $\boxed{0.4566}$ .

Let the green region be defined by: $\frac{1}{2} \le x \le \sqrt{1-y^2}, -\frac{1}{2} \le y \le \frac{1}{2}$ such that orgin of the $xy-$ plane lies at $A$ above. Integrating with respect to $y$ yields:

$\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-y^2} - \frac{1}{2} dy = \frac{y}{2} \sqrt{1-y^2} + \frac{1}{2} \arcsin(y) - \frac{y}{2}|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\frac{\pi}{6} - \frac{1}{2} + \frac{\sqrt{3}}{4}}.$