The diagram below shows a shaded region and an equilateral triangle of

I solved like this:

lets cut the triangle in 4 equilateral triangles on the midpoints of sides.
Area of the little triangle = \(\frac{\sqrt{3}}{4}\) (since side equals 1).
Let's calculate shaded are within, let's say, leftmost little triangle. \(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{6}\)*2
the latter being the sixth of the circle, multiplied by 2, because 2 such sectors fall into leftmost little triangle.
Then we have 2 more such vertex triangles, so 3 in total and 1 in the middle. Multiply the above expression by 3 for vertex little triangles and for the middle: \(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{2}\), since there are 3 sixths sectors falling in the middle little triangle.

So finally we have to sum these and add the outside circle sectors, which is π\(r^2\)*\(\frac{5}{6}\)*3.

So: (\(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{6}\)*2)*3+\(\frac{\sqrt{3}}{4}\)-\(\frac{πr^2}{2}\)+π\(r^2\)*\(\frac{5}{6}\)*3=\(\sqrt{3}\)+\(πr^2\)

insert r=0.5, \(\sqrt{3}\)+\(0.25π\)

What am I missing?

Edit: the question is asking the difference between triangle area and the one calculacted above. so \(\sqrt{3}\)+\(0.25π\) - \(\sqrt{3}\)=\(0.25π\), which is B.