# How do I compute the area of a cross which contains several disks which may overlap? I have found this in a Facebook group :) The only thing we are given is the marked distance and we need to find the shaded area. I have tried several options but I am always missing some data. Then I tried the following: I considered two different conditions:

1. The two side circles do not overlap; they are tangent to each other. In this case, the given distance corresponds to 3 diameters, so each is 3.33 units and the requested area is 100 sq. units.

2. The two side circles overlap by 100%. Again the area is 100 sq. units.

It is therefore obvious that the area is independent from the amount of overlapping, i.e. from the angle of the specific line segment to the horizontal line.

Can you provide a geometrical solution? Many thanks!!

• which group (I ask so that I could join)? – Invisible Nov 24 at 18:19

Let's call $$r$$ the radius of a circle, and $$2x$$ the overlap region between two circles. Then the area is $$A=(2r)^2+4(2r)(4r-2x)=4r^2+32r^2-16rx=36r^2-16rx$$ Now let's take a look at one of the circles on the sides. Draw the perpendicular from the intersection point to the axis connecting circles. This has a length $$h$$. Using Pythagoras, $$h^2=r^2-(r-x)^2=2rx-x^2$$ Now looking at the given line, just take half, and we apply Pythagoras again: $$h^2+(3r-x)^2=5^2$$ Using $$h^2$$ from above, $$5^2=2rx-x^2+9r^2-6rx+x^2=9r^2-4rx$$ Then area is exactly four times this, so $$A=4\cdot5^2=100$$ In the above picture $$DE=x$$, $$CD=h$$, $$AC=r$$. Then $$AD=r-x$$