What is the area of the shaded region of the square?

Question

To find area of shaded portion in the below figure, the picture generate by following mathematica code.

Block[{cond = {x^2 + (y - 1/2)^2 < 1/4 && x > 0,
    (x - 1)^2 + (y - 1)^2 < 1 && x < 1 && y < 1,
    (x - 1)^2 + y^2 < 1 && y > 0 && x < 1}
   },
 RegionPlot[Evaluate@Append[cond, And @@ cond],
    {x, -#, #}, {y, -#, #}, PlotPoints -> 40,
    PlotRange -> {{-0.2, 1.2}, {-0.2, 1.2}},
    PlotStyle -> {None, None, None, Cyan},
    Axes -> 1, Frame -> 0] &@1.5
 ]

Answer 1

The leftmost point is $(\frac 12, 1-\frac 2{\sqrt 5})$ and the top point is $(\frac 25,\frac 45)$ from solving simultaneous equations. Now I would cut the region horizontally at $y=\frac 12$, integrate the area above that line, and double the result.

Answer 2

As the plots shows we can have the area as follows:

$$A=\int_{y=0.2}^{0.5}\int_{x=1-\sqrt{2y-y^2}}^{\sqrt{y-y^2}}+\int_{y=0.5}^{0.8}\int_{x=1-\sqrt{1-y^2}}^{\sqrt{y-y^2}}$$