imc

We notice that the square can be split into $4$ congruent smaller squares, with the altitude of the equilateral triangle being the side of this smaller square. Note that area of shaded part in a quarter square is the total area of quarter square minus a white triangle (which has already been split in half). When we split an equilateral triangle in half, we get two $30^{\circ }-60^{\circ }-90^{\circ }$ triangles. Therefore, the altitude, which is also the side length of one of the smaller squares, is $\sqrt{3}$. We can then compute the area of the two triangles as $2\cdot \frac{1\cdot \sqrt{3}}{2}=\sqrt{3}$. The area of the each small squares is the square of the side length, i.e. ${\left(\sqrt{3}\right)}^2=3$. Therefore, the area of the shaded region in each of the four squares is $3-\sqrt{3}$. Since there are $4$ of these squares, we multiply this by $4$ to get $4\left(3-\sqrt{3}\right)=\boxed{12-4\sqrt{3}}$ as our answer.