The Deathly Hallows

The length from the centre of the circle $O$ to one of the vertices is $\frac{2}{3}L$ . Connect $O$ to all three vertices we have 3 equal triangles, and each of them has the area of $\frac{1}{2}(\frac{2}{3}L)(\frac{2}{3} L)\sin 60^\circ = \frac{\sqrt{3}}{9} L^2$ , so the area of the triangle is $\frac{3\sqrt{3}}{9}L^2 = \frac{\sqrt{27}}{9}L^2$ .

From here you can already found the answer, but for the sake of a complete solution, I'll do the calculation for the circle as well.

The radius of the circle is then $\frac{1}{3} L$ , so the area is $\pi \frac{1}{9}L^2$ . Hence the area of the shaded region is $\frac{\sqrt{27}-\pi}{9}L^2$

By pythagorean theorem, we have

$(2x)^2=x^2+L^2$ $\implies$ $4x^2=x^2+L^2$ $\implies$ $x^2=\dfrac{L^2}{3}$ $\implies$ $x=\dfrac{L}{\sqrt{3}}$ or $2x=\dfrac{2L}{\sqrt{3}}$

$\tan 30 =\dfrac{r}{x}$ $\implies$ $r=x \tan 30 = x\left(\dfrac{\sqrt{3}}{3}\right)=\dfrac{\sqrt{3}x}{3}=\dfrac{L}{3}$

The area of the shaded region is equal to the area of the triangle minus the area of the circle. We have

$A[triangle]=\dfrac{1}{2}(2x)(L)=\dfrac{1}{2}\left(\dfrac{2L}{\sqrt{3}}\right)(L)=\dfrac{L^2}{\sqrt{3}}$

$A[circle]=\pi r^2=\pi \left(\dfrac{L}{3}\right)^2=\dfrac{\pi L^2}{9}$

$A[shaded]=\dfrac{L^2}{\sqrt{3}}-\dfrac{\pi L^2}{9}=\dfrac{9L^2-\pi L^2\sqrt{3}}{9\sqrt{3}}$

Multiplying the fraction by $\dfrac{\sqrt{3}}{\sqrt{3}}$ , we get

$A[shaded]=\dfrac{9\sqrt{3}L^2-\pi L^2(3)}{27}$

Simplifying further, we have

$A[shaded]=\dfrac{9\sqrt{3}L^2-3\pi L^2}{27}=\dfrac{3\sqrt{3}L^2-\pi L^2}{9}=\dfrac{\sqrt{27}L^2-\pi L^2}{9}=\boxed{\dfrac{L^2(\sqrt{27}-\pi)}{9}}$

Tell me where I made a mistake,

we know that de area of the triangle is $A_1=\frac{\sqrt{3}}{4}L^2$ , we know that de radius of the circle is $r=\frac{\sqrt{3}}{6}L$

So the area of the circle is $A_2=\frac{\pi}{12}L^2$

Hence the shaded region must be $A_1-A_2=\frac{(\sqrt{27}-\pi)L^2}{12}$