Find the area.

The general area of a regular n-sided polygon with side length S is given as $|Area|=\frac{nS^2}{4}Cot(\frac{\pi}{n})\\$ The Area of the Yellow Pentagon is given by $\frac{5S^2}{4}Cot(\frac{\pi}{5})=500\\$

Each Blue Triangle is isosceles and has a base length of S. The Internal Angle of the Pentagon is given by $\frac{(n-2)\pi}{n} = \frac{3\pi}{5}$ so the base angle of the triangle is $\pi-\frac{3\pi}{5}=\frac{2\pi}{5}\\$

The Area of each Blue Triangle is given by $(\frac{S}{2})(\frac{S}{2})Tan(\frac{2\pi}{5})=\frac{S^2}{4}Tan(\frac{2\pi}{5})\\$

There are 5 Blue Triangles so the area of the Blue Region is given by $|Blue|=\frac{5S^2}{4}Tan(\frac{2\pi}{5})\\$

Dividing the two equations we have $\frac{|Blue|}{500}=\frac{\frac{5S^2}{4}Tan(\frac{2\pi}{5})}{\frac{5S^2}{4}Cot(\frac{\pi}{5})}\\$

Rearranging yields $|Blue|=500Tan(\frac{\pi}{5})Tan(\frac{2\pi}{5})\\$

There doesn't seem to be a nice identity simplifying $Tan(x)Tan(2x)$ so the answer is simply calculated as $\boxed{1118}\\$

It does, however, lead to a nice general solution for the ratio of the areas of the "Points" to the "Central area" of any regular polygon as \frac{|Points|}{|Centre|} = Tan(\frac{\pi}{n})Tan{\frac{2\pi}{n})\\