## Area of Circles Filling a Sector

A circle radius 10cm is cut into six equal sectors. Inside each sector, circles of decreasing radius are drawn to fill the sector as shown.

The radius of the largest circle is the solution to
$r+\frac{r}{sin \pi /6}=3r=10 \rightarrow r= \frac{10}{3} cm$
.
The smaller circles extend out to
$10-2r=10- 2 \frac{10}{3}=\frac{10}{3} cm$
from the centre.

Repeating the above process for the next largest circle gives us a radius for this circle of
$\frac{1}{3} \frac{10}{3} = \frac{10}{9}$
.
In general the radius of the kth largest circle will be
$\frac{10}{3^k}$
and the area of the circle is
$\pi (\frac{10}{3^k})^2 = \frac{100 \pi }{9^k}$
.
The areas of the circles form a geometric sequence with first term
$a = \frac{100 \pi }{9}$
and common ratio
$r = \frac{1}{9}$
.
The area of all the circles is
$A= \frac{a}{1-r}=\frac{\frac{100 \pi }{9}}{1-1/9}=\frac{800 \pi}{9}$
.