I just drew some circles, and discovered I had no idea how to calculate the area of the marked area of these circumferences.

First, draw three circles, sharing intersection points as centers. (don't really know how to express this).
Then, at each three outer intersections points of these three circles, draw a circle with the same radius.
Calculate the area of the outer circles' zone not colliding with any of the other circles.

The radius of the circles is $5$ $cm$, but using $R$ would be just as valid, or prefferable.

enter image description here

  • $\begingroup$ What have you tried? A good way to find the area of weird regions is to break it down into bits that you know how to solve... $\endgroup$ – Frank Oct 5 '16 at 20:50
  • $\begingroup$ By symmetry all the points form at grid of equilateral triangles. Therefore we have a lot of $60^\circ$ angles in play. Thus one of the shaded regions is half a circle minus two times a circular segment corresponding to $60^\circ$. The formula for one shaded region will be $$\left(\frac{\sqrt3}2+\frac\pi6\right)R^2$$ and you have three of them. $\endgroup$ – String Oct 5 '16 at 21:00
  • $\begingroup$ @Frank That was what I wanted to do, but the only thing I have is area of a circle : $A = \pi r^2$, and a segment of an area : $Z=\frac{A*\alpha}{2\pi}$ $\endgroup$ – mazunki Oct 5 '16 at 21:08
  • $\begingroup$ @String I saw the 60 degree symmetry, but I have no idea how to reach from that statement to the formula you come up with. $\endgroup$ – mazunki Oct 5 '16 at 21:11
  • $\begingroup$ @mazunki: You should use the formula for a circular segment. Then just plug in $60^\circ=\pi/3\operatorname{rad}$, multiply by two and subtract from half a circle. $\endgroup$ – String Oct 5 '16 at 21:22

The marked area is equal to the area of 3 semicircles minus the area of 6 small segments. enter image description here


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