What is the area for the shaded region HIML. Each side of the square is $100$, taking $\pi = 3.142$.
Quarter circle ALMC center at D, r = 100.
Circle HKJI, r = 50.
H, I , J, K are the mid-points of AB, BC, CD and DA respectively.
I know how to find the area for a sector or arc, but can't figure out to solve this one. A solution/suggestions will be appreciated.
Assume unit square for simplicity. Note that the shaded area is enclosed between the circles $x^2+y^2=1$ and $(x-\frac12)^2 + (y-\frac12)^2 =\frac14$. In polar coordinates, they become $r=1$ and $r=\frac12( \sin \theta +\cos\theta+\sqrt{\sin2\theta})$. Then, the area is within the angles $(\tan^{-1}\frac12, \frac\pi2- \tan^{-1}\frac12)$ and can be integrated as
\begin{align} &\int_{\tan^{-1}\frac12}^{\frac\pi2-\tan^{-1}\frac12} \frac12 \left[ r^2_1(\theta) - r^2_2(\theta) \right]d\theta \\ =&\frac12\int_{\tan^{-1}\frac12}^{\frac\pi2-\tan^{-1}\frac12} \left[\frac14 (\sin \theta +\cos\theta+\sqrt{\sin2\theta})^2-1\right] d\theta \\= &\frac14-\frac{3\pi}{16}+ \tan^{-1}\frac12 \end{align}
Edit: Geometric solution
Let [.] denote areas. Then, the shaded area is
$$A= [DHBI] - [HBI] - [DLM]\tag1$$
where $[DHBI] = \frac12$, $ [HBI]= \frac14 ( 1- \pi(\frac12)^2)= \frac14- \frac\pi{16}$, and $$ [DLM] = \frac 12\angle LDM \cdot 1^2 = \frac12(\frac\pi2-2\angle CDI )= \frac\pi4- \tan^{-1} \frac12$$
Plug the three individual results into (1) to arrive at
$$A= \frac14-\frac{3\pi}{16}+ \tan^{-1}\frac12$$