In the image, $ABCD$ is a square of side $4$. $\triangle DFC$ is equilateral, then the shaded area is?
My try was extending $DF$ and $CF$, meeting $AB$ and then i tried using the $30-60-90$ triangle relations, but i couldn't find anything besides the sides of those triangles. Any hints?
Since $\measuredangle ECF=15^{\circ}$ and $\measuredangle F=60^{\circ}$, we obtain: $$S_{\Delta ECF}=\frac{4^2\sin15^{\circ}\sin60^{\circ}}{2\sin75^{\circ}}=\frac{8\cdot\frac{\sqrt3-1}{2\sqrt2}\cdot\frac{\sqrt3}{2}}{\frac{\sqrt3+1}{2\sqrt2}}=8\sqrt3-12.$$