# Given that the radius of the circumcircle (large circle) is r, evaluate the ratio of the area of A regions to B regions - (sumA_i)/(sumB_i)?

Nov 9, 2016

Radius of the large circumcircle$= r$

Area of this large circle $= \pi {r}^{2}$

The length of each side of the square at the center is equal to the radius of the large circumcircle.

So area of the square ${r}^{2}$

Radius of each of four small circle$= \frac{r}{2}$

So its area $= \pi {\left(\frac{r}{2}\right)}^{2} = \frac{1}{4} \pi {r}^{2}$

From the figure it is obvious that

$\sum {A}_{i} = \text{area of large circle"-"area of square"-2xx"area of small circle}$

$= \pi {r}^{2} - {r}^{2} - 2 \times \frac{1}{4} \pi {r}^{2} = \left(\frac{\pi}{2} - 1\right) {r}^{2}$

Each of the petal of flower at the center is composed of two similar segments of small circle and 8 such segments form the complete flower.

Area of each such segment
$= \frac{1}{4} \left(\frac{1}{4} \pi {r}^{2}\right) - \frac{1}{2} {\left(\frac{r}{2}\right)}^{2}$

$= \left(\frac{1}{16} \pi - \frac{1}{8}\right) {r}^{2}$

So $\sum {B}_{i} = 8 \times \left(\frac{1}{16} \pi - \frac{1}{8}\right) {r}^{2}$

$= \left(\frac{\pi}{2} - 1\right) {r}^{2}$

$\frac{\sum {A}_{i}}{\sum {B}_{i}} = \frac{\left(\frac{\pi}{2} - 1\right) {r}^{2}}{\left(\frac{\pi}{2} - 1\right) {r}^{2}} = \frac{1}{1}$