Square semicircle equilateral triangle

Let $A$ and $E$ be the midpoints of the vertical sides of the square. The $AE$ divides that semicircle and equilateral triangle, hence the blue area, into halves. Label the rest of figure as shown. We note that $AB=1$ because it is the radius of the semicircle with a diameter of 2. Also that the height of the equilateral triangle of side length 2, $DE=\sqrt 3$ . Since $AE=2$ , $AD= AE-DE = 2 - \sqrt 3$ .

We note that the area of the wedge $[DBC]$ is half of the blue area $A_{\color{#3D99F6}\text{blue}}$ . And we have:

$\begin{aligned} [DBC] & = {\color{#3D99F6}[ABC]} - \color{#D61F06} [ABD] & \small \color{#3D99F6} [ABC] = \text{area of sector }ABC \\ & = {\color{#3D99F6} \frac {\angle BAC} 2} - \color{#D61F06} \frac {AB\times AD \times \sin \angle BAC}2 & \small \color{#D61F06} [ABD] = \text{area of }\triangle ABD \\ & = \frac {\color{#3D99F6}\theta}2 - \frac {2-\sqrt 3}2 \sin \color{#3D99F6} \theta & \small \color{#3D99F6} \text{where }\angle BAC = \theta \\ \implies A_{\color{#3D99F6}\text{blue}} & = \theta - (2-\sqrt 3) \sin \theta \end{aligned}$

To find $\theta$ , applying sine rule in $\triangle ABD$ :

$\begin{aligned} \frac {\sin \angle ABD}{AD} & = \frac {\sin \color{#3D99F6} \angle ADB}{AB} & \small \color{#3D99F6} \text{Note that }\angle BDC = \frac \pi 6 \implies \angle ADB = \frac {5\pi}6 \\ \frac {\sin \left(\frac \pi 6-\theta \right)}{2-\sqrt 3} & = \frac {\sin \frac {5\pi}6}1 \\ \sin \left(\frac \pi 6-\theta \right) & = \frac {2-\sqrt 3}2 \end{aligned}$

$\implies \theta = \dfrac \pi 6 - \sin^{-1} \left(\dfrac {2-\sqrt 3}2\right) \approx 0.389220118$ radians

$\implies A_{\color{#3D99F6}\text{blue}} = \theta - (2-\sqrt 3) \sin \theta \approx \boxed{0.2875}$