flower of life geometry question Attached is the question - with so many circles i haven't figured out a way to calculate what's asked. Would really appreciate some help. Thanks]1
 A: Area above horizontal line of radius length r is only the outer area left out near periphery marked yellow on one of six minor segments.
$$ A_{segment}=\pi r^2/6- r^2 \sqrt{3}/4= r^2(\pi/6-\sqrt {3}/4) $$
Area of  an equilateral triangle is known $ =(r^2 \sqrt{3}/4$ )
The smaller radius segments are a third in length and 1/9 in area. There are three such small areas/patches, so remaining area is  $A_s-3\cdot \dfrac19 A_s = \dfrac23 A_s$
There are six such areas in a regular hexagon totalling to
$$6 \cdot \frac23 A_s = 4 A_s = \frac23 r^2 (\pi-3 \sqrt {3}/2)$$
where we plugged in from above value for $A_{segment}.$ 

Fraction of total area = $\dfrac{4 A_s}{\pi r^2} = \dfrac23-\dfrac{\sqrt3}{\pi}= 11.5338 \,$%
A: There are six regions around the exterior that are not part of any of the smaller circles.  You are asked what percentage of the large circle they represent.  The inner area is divided into lenses and triangles with curved sides, which we can call deltas.  You should compute the area of a lens and a delta, count how many of each there are, and add up the areas to get the total area of the small circles.
