Can you find the green area?

Problem:

Solution:

Fun problem for the morning.

First, our goal is the area on the side:

The equilateral triangle has area is $\frac{1}{2}a^2 \sin \frac{\pi}{3} = \frac{\sqrt{3}}{4}a^2$

The sector have area $\frac{\frac{pi}{6}}{2\pi}\pi a^2 = \frac{1}{12}\pi a^2$

Therefore the remaining area on the side is $a^2 - \frac{\sqrt{3}}{4}a^2 - 2(\frac{1}{12}\pi a^2) = a^2(1 - \frac{\sqrt{3}}{4} - \frac{\pi}{6})$

Next, we want the area of the petal.

The remaining area on the corner is $a^2 - \frac{1}{4}\pi a^2 = a^2(1 - \frac{\pi}{4})$

Therefore the petal area is $a^2(1 - \frac{\pi}{4}) - 2(a^2(1 - \frac{\sqrt{3}}{4} - \frac{\pi}{6})) = a^2(\frac{\sqrt{3}}{2} + \frac{\pi}{12} - 1)$

Finally, the area we wanted is $a^2 - 2a^2(1 - \frac{\pi}{4}) - 2a^2(\frac{\sqrt{3}}{2} + \frac{\pi}{12} - 1) = a^2(1 - \sqrt{3} + \frac{\pi}{3})$