Filled With Circles 2

$2\times\angle CYO+\angle CZY=90^\circ$

$\angle CYO=\frac{1}{2}\times (90^\circ-arctan(\frac{1}{4}))=37.98188^\circ$

$r=2\times tan(\angle CYO)=1.56155$

$b=8-2r, \frac{a}{b}=\frac{2}{8}, a=\frac{8-2r}{4}=1.21922$

Area of trapezoid $ABYX$ is $A_t=\frac{1}{2}(4+2a)2R=10.0539$

Area of the circle centered at O is $A_c=\pi r^2=7.6606$

The ratio of the areas $\frac{A_c}{A_t}=0.761948$

This ratio remains the same for all the circles below as well, each within its own trapezoid, so the total area of circles is this ratio times the total area of $\triangle XYZ$

$A=0.761948\times\frac{1}{2}\times 4\times 8=12.19117$

Using the method of finding the radius in "Filled with Circles" calculate the areas of the 1st 2 circles.

Then find the ratio Area(B)/Area(A) and use these figures to calculate the sum of an infinite geometric series.

Sum = Area(A)/(1- r)