How do I compute the area of a cross which contains several disks which may overlap?

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 $AD=r-x$