Xor Gate?

Relevant wiki: Length and Area - Composite Figures

Notice that lines $OA,OP,OB,AP$ and $BP$ are all radii of the unit circle, meaning they have a length of $1$ .

Since all the lines have the same length, we can say that $\triangle OAP$ and $\triangle OBP$ are equilateral triangles.

Now, notice that the area of the shaded region is given as below:

Therefore, we need to calculate each component first.

Area of unit circle $=\pi(1)^2=\pi$

Area of sector $OAPB = \dfrac{120}{360}\times \pi(1)^2= \dfrac{\pi}{3}$

Area of $\triangle OAP = \dfrac{1}{2}(1)^2(\sin60^{\circ}) = \dfrac{1}{2}\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\sqrt{3}}{4}$

Area of shaded region

$=2\Big($ Area of unit circle $-2 \times$ Area of sector $OAPB+2\times$ Area of $\triangle OAP\Big)$

$=2\left(\pi - 2 \times \dfrac{\pi}{3} + 2 \times \dfrac{\sqrt{3}}{4}\right)\\ =2\left(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}\right)\\ =\dfrac{2\pi}{3}+\sqrt{3}\\ =\dfrac{2\pi+3\sqrt{3}}{3}\\ =\boxed{\dfrac{4\pi+6\sqrt{3}}{6}}$

Relevant wiki: Length and Area - Composite Figures

$B \ and \ C$ are the centers of our circles. Triangles $ABC \ and\ BCD$ are equilateral triangles with each angle=60.

We know the mentioned sides that are equal to the radius. With the application of Pythagoras, we get $AD= \sqrt{3}r$ and angle $ABD=120$ $\implies$ the area of the intersection of the two circles(the one highlighted in yellow) is $r^2\times(\frac{2\pi}{3}-\frac{\sqrt{3}}{2})$ = $2[\frac{r^2}{2}(\frac{2\pi}{3}-sin(\frac{2\pi}{3}))]$ .The area of the two circles is $2\pi r^2$ .

Thus, the area of one white part or one green part in our problem is $\pi r^2- r^2\times(\frac{2\pi}{3}-\frac{\sqrt{3}}{2})= \frac{2\pi+3\sqrt{3}}{6}$ . Keeping in mind that $r=1$ . Multiply this area by $2$ , we obtain $\boxed{\Large\frac{2\pi+3\sqrt{3}}{3}}$ = $\boxed{\Large\frac{4\pi+6\sqrt{3}}{6}}$

Relevant wiki: Circles - Area

Consider figure $A$ , the distance between the centers of the circle is $1$ .

Consider figure $B$ , the area of one shaded part is equal to the area of a circle minus the area of two segments of a circle. We need the value of $\angle \theta$ ,

$\cos \theta = \dfrac{\frac{1}{2}}{1}=\dfrac{1}{2}$

$\theta = \cos^{-1}\dfrac{1}{2} = 60$ .

It follows that $2 \theta = 120$ .

We need the value of $a$ , by pythagorean theorem

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

It follows that $2a=\sqrt{3}$

Consider figure $C$ . The area of one segment is $\dfrac{120}{360} \pi - \dfrac{1}{2}\left(\dfrac{1}{2}\right)(\sqrt{3})=\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}$ . So the area of two segments is $2\left(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}\right)=\dfrac{2 \pi}{3}-\dfrac{\sqrt{3}}{2}$

So the area of one shaded part is $\pi -\left(\dfrac{2 \pi}{3}-\dfrac{\sqrt{3}}{2}\right)=\pi - \dfrac{2 \pi}{3} + \dfrac{\sqrt{3}}{2}=\dfrac{2\pi + 3\sqrt{3}}{6}$ .

Finally, the area of two shaded part is $2\left(\dfrac{2\pi + 3\sqrt{3}}{6}\right) = \boxed{\dfrac{4 \pi + 6\sqrt{3}}{6}}$

The white area in the center, $A_w$ , is two circular segments, each with angle $120^\circ$ and radius $1$ .

$A_w=2\times\frac{R^2}{2}(\frac{\alpha\pi}{180}-\sin\alpha)=2\times\frac{1}{2}(\frac{120\pi}{180}-\sin120^\circ)=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$

Green area on the left is area of circle minus the white area: $\pi-(\frac{2\pi}{3}-\frac{\sqrt{3}}{2})=\frac{\pi}{3}+\frac{\sqrt{3}}{2}$

Total green area is double that. $2\times(\frac{\pi}{3}+\frac{\sqrt{3}}{2})=\frac{2\pi}{3}+\sqrt{3}=\frac{4\pi+6\sqrt{3}}{6}$

OP can you post a solution? I apologize but I'm not sure about the answer. In my opinion the answer should be $\frac{4 \pi + 6 \sqrt{3}}{6}$ , and I would like to see where I am wrong. Thanks.