Check the picture 1.
The regions contained in an odd number of these 5 circles are
the pink colored region, contained only in the circle with radius 2,
and the yellow regions, each contained in the intersection of 2 small circles and the third large circle.
So what we need to determine is the "area pink + area yellow".
Consider the second picture.
Let A2 be the area of the right isosceles triangle, and A1 be half of the yellow region.
the region A2 + A1 is called a sector, and it is 1/4 of the area of a unit circle.
![A1=A_{sector}-A2= \frac{1}{4} \pi r^2- \frac{1}{2}.1.1= \frac{1}{4} \pi 1^2- \frac{1}{2}= \frac{1}{4}\pi- \frac{1}{2}](https://tex.z-dn.net/?f=A1%3DA_%7Bsector%7D-A2%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cpi%20r%5E2-%20%5Cfrac%7B1%7D%7B2%7D.1.1%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Cpi%201%5E2-%20%5Cfrac%7B1%7D%7B2%7D%3D%20%20%5Cfrac%7B1%7D%7B4%7D%5Cpi-%20%5Cfrac%7B1%7D%7B2%7D)
Thus, the overall yellow region is
![8A1=8(\frac{1}{4}\pi- \frac{1}{2})=2\pi-4](https://tex.z-dn.net/?f=8A1%3D8%28%5Cfrac%7B1%7D%7B4%7D%5Cpi-%20%5Cfrac%7B1%7D%7B2%7D%29%3D2%5Cpi-4)
The purple area is:
Area of large circle - 4*(Area of 1 small circle) + yellow region.
(because subtracting all 4 small circles means that each of the 4 separate yellow regions have been subtracted twice)
![A_{purple}=\pi.2^2-4(\pi.1^2)+2\pi-4=4\pi-4\pi+2\pi-4=2\pi-4](https://tex.z-dn.net/?f=A_%7Bpurple%7D%3D%5Cpi.2%5E2-4%28%5Cpi.1%5E2%29%2B2%5Cpi-4%3D4%5Cpi-4%5Cpi%2B2%5Cpi-4%3D2%5Cpi-4)
Finally, the total area is :
![2\pi-4+2\pi-4=4\pi-8](https://tex.z-dn.net/?f=2%5Cpi-4%2B2%5Cpi-4%3D4%5Cpi-8)
(units squared)
Answer:
![6\pi-8](https://tex.z-dn.net/?f=6%5Cpi-8)
square units