Notice that, within the diagram, we have four identical right triangles (shaded below)

So, our strategy will be to assign a "nice" value to the sides of square ABCD, and then find the areas of the 4 shaded right triangles.

Let's say that each side of square ABCD has length 2.
So, the area of a square ABCD = (2)(2) = 4


Now notice that we have 2 SIMILAR triangles hiding within our diagram.



Let's pull these two triangles out of the diagram to get the following:

ASIDE: Since we already let each side of square ABCD have length 2, we know that DC = 2 and FC = 1
Also, once we apply the Pythagorean theorem to this right triangle, we find that side DF has length √5


To find the values of x and y, we'll use the fact that, if we have two similar triangles, then the ratios of their corresponding sides will always be equal.
For example, we can write √5/2 = 2/x
Cross multiply to get: (√5)(x) = (2)(2)
Solve: x = 4/√5

Likewise, to find the value of y, we can write: √5/2 = 1/y
Solve to get: y = 2/√5

We now have:


Area of triangle = (base)(height)/2
So, the area of our shaded BLUE triangle = (4/√5)(2/√5)/2 = 4/5

This means the total area of all 4 shaded regions = (4)(4/5) = 16/5

So, the area of the shaded square = 4 - 16/5 = 4/5

So the fraction of square ABCD that is shaded = (4/5)/4 = 1/5

Answer: C

Cheers,
Brent