Finding the area of a square that has a circle inside itself I tried to solve the following problem:

I think the image is self-descriptive. I tried to draw a vertical line from the top-end of $\theta$ angle to the horizontal line, then tried to use the similarity of triangles to solve the problem, but with no luck. I also tried to find the area of the sector, but with no much success too.
 A: The radius of the circle is half the length of a side of the square. The arc length formula might be useful
$$s=r\theta$$
Since we are given an arc length. We just need the corresponding angle in radians. That is where the other two side lengths.
$$tan(\theta)=3/2$$
$$r=12/tan^{-1}(3/2)$$
$$A=l^2=(2r)^2$$
This answer will be in square meters so you will need to do a unit conversion to express the answer in terms of square centimeters.
A: We have:
$\tan \theta = \dfrac{3}{2}, \tag{1}$
whence
$\theta = \arctan(\dfrac{3}{2}): \tag{2}$
now
$12m = r\theta$
$= r \arctan(\dfrac{3}{2}), \tag{3}$
where $r$ is the radius; thus
$r = \dfrac{12m}{\arctan(\frac{3}{2})}; \tag{4}$
the side of the square is $2r$; the area is thus
$4r^2 = (2r)^2$
$= \dfrac{576m^2}{\arctan^2(\frac{3}{2})}; \tag{5}$
since $1m = 100cm$, $1m^2 = 10,000cm^2$, so the area becomes
$\dfrac{5,760,000cm^2}{\arctan^2(\frac{3}{2})}. \tag{7}$
A: in right triangle, we have $$\tan \theta=\frac{3}{2}$$ $$\implies \color{red}{\theta=\tan^{-1}\left(\frac{3}{2}\right)}$$ Let, $R$ be the radius of circle then the length of given arc $$=\text{(aperture angle)}\times \text{(radius of circle)}=12$$ $$\implies \tan^{-1}\left(\frac{3}{2}\right)\times R=12$$ $$R=\frac{12}{\tan^{-1}\left(\frac{3}{2}\right)}$$ Now, the length of each side of square $ABCD$ $$=2R=\frac{24}{\tan^{-1}\left(\frac{3}{2}\right)}$$ Hence, the area of square $ABCD$ (in square cm.) $$=\text{(side of square)}^2=\left(\frac{2\times 1200}{\tan^{-1}\left(\frac{3}{2}\right)}\right)^2$$$$=\color{blue}{\frac{5760000}{\left(\tan^{-1}\left(\frac{3}{2}\right)\right)^2}\approx 5963452.099\ cm^2}$$
