There are a series of two different triangle in the picture. ΔABE ≅ ΔEBF ≅ ΔBFC (the orange shapes are those 3 triangles). The other triangles, ΔEFD and ΔCFD are 30-60-90 triangles and ΔEFD ≅ ΔCFD.
To calculate one of those, use the equation for the area of a triangle (Atriangle)
Atriangle = (1/2)bh where b is the base and h is the height.
For the orange section, we can use the triangle of the upper left triangle, ΔABE. The base (b) can be thought of as side AB and the height (h) can be thought of as side AE but both are the same because ABEF is a square. To find the side length, divide the diagonal by √2. So b = 12/√2 and h = 12/√2 so Atriangle = (1/2)(12/√2)(12/√2) = 144/(2•√4) = 144/(2•2) = 144/4 = 36 square units.
Since ΔABE ≅ ΔEBF ≅ ΔBFC then the area of the orange section is 3•36 = 108 square units.
On a 30-60-90 triangle, if we know the short leg, then the longer leg is the short side times √3. The short side of ΔEFD is side EF, which we already determined was 12/√2. That means the longer leg is (12/√2)•√3.
So the area of ΔEFD = (1/2)(12/√2)(12/√2)(√3) = 144√3/4 = 36√3. Since there are 2 of these triangles, the area of the green part is 2•36√3 = 72√3.
So the entire area is 108 + 72√3 square units.
