Notes
triangle centres and square corners solution

Triangle Centres and Square Corners

Solution by Properties of Squares and Lengths in Equilateral Triangles

As the diagonal of the square is $6$, the side length is $3 \sqrt{2}$. This is also the side length of the equilateral triangles. Using the relationship between the lengths in an equilateral triangle, the length of $C D$ is $\frac{1}{3} \times \frac{\sqrt{3}}{2} \times 3 \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$. Triangle $C D F$ is an isosceles right-angled triangle, so $D F$ has the same length as $C D$. The area of triangle $C D F$ is then $\frac{1}{2} \times \frac{\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{3}}{\sqrt{2}} = \frac{3}{4}$. The shaded region comprises four of these triangles, so it has area $3$.