This is a really interesting question! One thing that we can notice right off the bat is that each of the circles has the same amount of area swept out of it - namely, the amount swept out by one of the interior angles of the hexagon. Let’s call that interior angle θ. We know that the amount of area swept out in the circle is proportional to the angle swept out - mathematically
θ/360 = a/A
Where “a” is the area swept out by θ, and A is the area of the whole circle, which, given a radius of r, is πr^2. Substituting this in, we have
θ/360 = a/(πr^2)
Solving for “a”:
a = π(r^2)θ/360
So, we have the formula for the area of one of those sectors; all we need to do now is find θ and multiply our result by 6, since we have 6 circles. We can preempt this but just multiplying both sides of the formula by 6:
6a = 6π(r^2)θ/360
Which simplifies to
6a = π(r^2)θ/60
Now, how do we find θ? Let’s look first at the exterior angles of a hexagon. Imagine if you were taking a walk around a hexagon. At each corner, you turn some angle and keep walking. You make 6 turns in all, and in the end, you find yourself right back at the same place you started; you turned 360 degrees in total. On a regular hexagon, you’d turn by the same angle at each corner, which means that each of the six turns is 360/6 = 60 degrees. Since each interior and exterior angle pair up to make 180 degrees (a straight line), we can simply subtract that exterior angle from 180 to find θ, obtaining an angle of 180 - 60 = 120 degrees.
Finally, we substitute θ into our earlier formula to find that
6a = π(r^2)120/60
Or
6a = 2πr^2
So, the area of all six sectors is 2πr^2, or the area of two circles with radii r.
You multiply them just as you would multiply any normal polynomials, except there is a special rule. Since we know that i = √-1, the rule is that i² = -1. So, let's multiply:
