4 semis in a square

Let the square be a unit square with side length and area of 1 and the radius of the semicircles be $r$ , We note that the centers of two adjacent semicircles and the square vertex between them form a right triangle with side lengths of $1:\sqrt 3: 2$ .

Therefore, we have $r+\sqrt 3 r = 1$ , $\implies r = \dfrac 1{1+\sqrt 3}$ and the ratio of the area of the 4 semicircles to the area of the square is: $\dfrac {2A_\bigcirc}{A_\square} = \dfrac {2\pi}{(1+\sqrt 3)^2} \approx \boxed{0.842}$ , since $A_\square = 1$ .

In right $\triangle{ABC}, \:\ BC = 2r$ and $AC = r \implies AB = \sqrt{3} \implies BD = r$

$\implies A_{square} = (\sqrt{3} + 1)^2 r^2$ and $A_{circles} = 2\pi r^2 \implies \dfrac{A_{circles}}{A_{square}} = \dfrac{2\pi}{(\sqrt{3} + 1)^2} \approx \boxed{0.84178721}$

If circle radius is taken as unity, we can show by construction ( joining centers of two tangent adjoining circles) that side of square=2/{(√3)-1}.

This gives the area ratio of four semi circles to square as (π/2)[{(√3)-1}^2]=0.8417.

Answer=0.8417

The centers of two tangent semicircles and their point of tangency are collinear because the two radii that join the centers to the point of tangency are both perpendicular to the same tangent at the same point. Therefore, the image to the left is a right triangle. If the radius of the circle is $r$ , the length of the longer leg is $(\sqrt{3})r$ .

The length of the entire square's side is $(\sqrt{3})r+r = (\sqrt{3}+1)r$ . The desired proportion is $\frac{4(\frac{1}{2}\pi r^2)}{((\sqrt{3}+1)r)^2} \approx \boxed{0.8418}$ .