Area of a star

First examine one tenth of the star, as follows:

We are given that $AC = 100$ and $AE = BF = 20$ . By symmetry of the star, $\angle DOF = \frac{1}{10}360° = 36°$ . Also, $\angle DAC$ is the sum of half the interior angle and exterior angle of a regular pentagon, so $\angle DAC = \frac{1}{2}108° + 72° = 126°$ . Then $\angle DAE = \angle DAC - \angle EAB = 126° - 90° = 36°$ and $\angle BCF = 180° - \angle DAC - \angle DOF = 180° - 126° - 36° = 18°$ .

By trigonometry on $\triangle BCF$ , $BC = \frac{20}{\tan 18°} = 20\sqrt{5 + 2\sqrt{5}}$ , and by trigonometry on $\triangle DAE$ , $DE = 20 \tan 36° = 20\sqrt{5 - 2\sqrt{5}}$ . Then $EF = AB = AC - BC = 100 - 20\sqrt{5 + 2\sqrt{5}}$ , and $DF = DE + EF = 20\sqrt{5 - 2\sqrt{5}} + 100 - 20\sqrt{5 + 2\sqrt{5}}$ .

The area of trapezoid $ACFD$ is $A_{ACFD} = \frac{1}{2}(AC + DF)AE$ $=$ $\frac{1}{2}(100 + 20\sqrt{5 - 2\sqrt{5}} + 100 - 20\sqrt{5 + 2\sqrt{5}})20$ $=$ $200(10 + \sqrt{5 - 2\sqrt{5}} - \sqrt{5 + 2\sqrt{5}})$ .

The star consists of $10$ trapezoids congruent to trapezoid $ACFD$ , so its area is $A_{star} = 10 \cdot 200(10 + \sqrt{5 - 2\sqrt{5}} - \sqrt{5 + 2\sqrt{5}}) \approx \boxed{15297.71798}$ .