find the area

By cosine law,

$(\sqrt{300})^2=x^2+x^2-2(x)(x)(\cos 120)$

$x^2=100$

Area of the star is equal to the area of the square minus the area of the four congruent triangles.

$A=\sqrt{300} - 4 \times \dfrac{1}{2} \times x^2 \times \sin 120 = \sqrt{300} - 2 \times 100 \times \sin 120 \approx \boxed{127}$