<s>[INST] The area not covered by six pointed star In a circle with radius $r$, two equi triangles overlapping each other in the form of a six pointed star touching the circumference is inscribed! What is the area that is not covered by the star?
Progress
By subtracting area of the star from area of circle , the area of the surface can be found! But how to calculate the area of the star?
 [/INST] If the six-pointed star is regular, then the answer is $r^2(\pi-\sqrt{3})$. If it is not, then the answer can be larger, up to a limit of $r^2\big(\pi-\frac{3}{4}\sqrt{3}\big)$.
Proof

The required area is the area of the circle ($\pi r^2$) minus the area of the star.
The area of the star is the area of a large equilateral triangle (A) plus the area of three small ones. Each small one has a side-length $\frac{1}{3}$  of the large triangle's and therefore an area $\frac{1}{9}$ of its area. So the area we want is $\pi r^2 - A(1+\frac{3}{9}) = \pi r^2 - \frac{4A}{3}$.
A is equal to 6 times the area of the right-angled triangle shown. Stand it on its short side. Its area is $B=\frac{\mathrm{base * height}}{2} = \frac{r^2\sin{30^\circ}\cos{30^\circ}}{2}=r^2(\frac{1}{2})\frac{\sqrt{3}}{2}\frac{1}{2} = r^2(\frac{\sqrt{3}}{8})$. So $A=6B=\frac{3r^2\sqrt{3}}{4}$
So the required area is $\pi r^2 - \big(\frac{3r^2\sqrt{3}}{4}\big)\frac{4}{3} = r^2(\pi-\sqrt{3})$
However, all you say is that the triangles overlap to make a star. You do not define a star. If we allow the overlap such that the points on the circle are unequally spaced, then we might get a shape like this:

At the limit, the area of the star is the area of a large triangle, A, giving required area $\pi r^2 - A = r^2\big(\pi-\frac{3}{4}\sqrt{3}\big)$
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