Intersection Area of Venn Diagram

Graph of the circles

In the Quadrant $\text{I}$ , the arcs cross at $0.5$ of the x-axis. Using this information, we can get the angle of the circles' segments that make up the overlapped area.

Let's use the Sine Rule:

$\begin{aligned} \dfrac{1}{\sin90°} = & \dfrac{0.5}{\sin(90-x)°} \\ 1 = & \dfrac{0.5}{\sin(90-x)°} \\ x = & 60 \end{aligned}$

We can now get the area of the segment.

$\frac{120°π}{360}-\frac{\sin120°}{2} = \frac{2π}{6} - \frac{\sqrt{3}}{4} \approx 0.614185$

Since there are two identical segments creating the area, we double the result above. The area of each circle is $π$ , so we just divide the two to get our answer.

$\frac{\frac{2π}{3}-\frac{\sqrt{3}}{2}}{π} \approx 0.391002 \space \small{\text{or}} \space \boxed{39.1\%}$

If the image doesn't appear, here it is .

WLOG $r=1$ . Then, we can find the area of the shaded region by $2\cdot [\text{Area of the }120^\circ\text{ sector}]-[\text{The two equilateral }\triangle].$ The area of the sectors, since we let $r=1$ , is $2\left(\frac 13 \pi\right)=\frac 23 \pi.$ The area of an equilateral triangle is given by $\frac{\sqrt{3}}4 s^2,$ where $s$ is the length of one side, so the area of the two equilateral triangles is $2\left(\frac{\sqrt 3}{4}\right)=\frac{\sqrt{3}}{2}.$ Thus, we have the answer as $\frac{\frac 23\pi-\frac{\sqrt{3}}2}{\pi}\approx 39.1\%.$