# Triangle Centres and Square Corners +-- {.image} [[TriangleCentresandSquareCorners.png:pic]] > The corners of the red square are at the centres of the equilateral triangles. What's the total shaded area? =-- ## Solution by [[Properties of Squares]] and [[Lengths in Equilateral Triangles]] +-- {.image} [[TriangleCentresandSquareCornersLabelled.png:pic]] =-- 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 $3$.