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Suppose the side length $a$ of the square is 10mm. A circle is tangent to all four sides of the square. And two quarter-circles with the same radius of 10mm have centers on the opposite vertices.

enter image description here

It may be easier to view it in the picture on the right.

What’s the area of shaded region? Using some trigonometrical calculation, I got a complex formula

$$S=\left[\frac{1}{2}(\pi -\arccos(-\frac{\sqrt{2}}{4}))+\sqrt{2}\sin(\arccos(\frac{5\sqrt{2}}{8}))-2\arccos(\frac{5\sqrt{2}}{8})\right]a^2$$ which gives 29.276$mm^2$. The way is far from beautiful. Don’t know if there are any simpler ways to do that? Is there any principles that I am not aware of?

Thank you.

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    $\begingroup$ Please, tell us more. What have you tried? What would you like to try, if you knew how? What's stopping you? $\endgroup$
    – Arthur
    Jun 17 '19 at 14:31
  • $\begingroup$ It would be nice if you could provide some more info about your question, such as how is the ellipse-like shape constructed. $\endgroup$ Jun 17 '19 at 14:31
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    $\begingroup$ I believe that it is a circle tangent to all 4 sides and two quarter-circles with radius $a$ that have centers on opposite vertices. $\endgroup$
    – ETS1331
    Jun 17 '19 at 14:34
  • $\begingroup$ @AjayMishra typo. It's mm² $\endgroup$
    – ZYX
    Jun 17 '19 at 15:12
  • $\begingroup$ @ETS1331 thank u. I quote yours. $\endgroup$
    – ZYX
    Jun 17 '19 at 15:13
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The area can be expressed in various forms of inverse trigonometric functions depending on methods. Below is maybe the most concise (assuming unit square.)

$$\frac14\left(\sqrt{7}-\cos^{-1}\frac{393}{4096}\right)$$

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$\hskip 1.9 in$enter image description here

Method 1: Via geometry

Here $GC = \frac{a}{\sqrt{2}}, GH = \frac{a}{2}, CH = a$

By using cosine rule, angle $\alpha$ and $\beta$ can be found. Now area of sector $CHI$ is $A = \cfrac{\alpha a^2 }{2}$, area of $ \Delta CHG=A_1 = \cfrac{ \sin \alpha a^2}{2 \sqrt{2} } \Rightarrow $ Area of figure $CHGI$ would be $2A_1$ .Now, area of figure $GHI = A_3 = A - 2A_1$.

From simple geometry $\angle HGI = \gamma = 2 \pi - 2 \beta$, now area of sector $GHI = A_4 = \cfrac{\gamma a^2 }{8} $ which implies area of shaded portion on the right is $A_4 - A_3$, hence total area is $2(A_4 - A_3)$

Method 2: Via calculus with some coordinate geometry

You can setup the system on cartesian coordinates with origin$(0,0)$ at the center of smaller circle $C_1$ , $- x$-axis along segment $CG$, and considering only one bigger circle $C_2$ , its centre would be at $ ( - a/ \sqrt{2} , 0 )$.

Equation of the circles are:

$$ \begin{align} C_1 & : x^2 + y^2 = \cfrac{a^2}{4} \\ C_2 & : \bigg(x + \cfrac{a}{\sqrt{2}} \bigg)^2 + y^2 = a^2 \end{align} $$

On solving $x = \cfrac{a}{4 \sqrt{2}}$, which is the $x$-coordinate of point of intersection. I hope now you can take from here. Can you?

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  • $\begingroup$ Only thing you have to calculate is $\alpha$ and $\beta$ $\endgroup$
    – u_sre
    Jun 17 '19 at 16:27
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I'm quite sure your solution is incorrect. I believe the correct value of the area is given by $$2\left(\overbrace{(2r)^2}^{\text{area of square}}-\overbrace{\frac{\pi(2r)^2}{4}}^{\text{area of larger quarter circle}}-\overbrace{(r^2-\frac{\pi r^2}{4})}^{\text{area between smaller circle}\\\,\,\,\,\,\text{ and lower left corner}}-2c\right)$$ where $c$ is the area between the bottom edge of the square and where the intersection of the two circles form an 'X' shape. The value of $c$ can be calculated by the following integrals $$c=\int_r^{(3+\sqrt{7})r/4}\left(r-\sqrt{r^2-(x-r)^2}\right)\mathrm{d}x+\int_{(3+\sqrt{7})r/4}^{2r}\left(2r-\sqrt{4r^2-(x-2r)^2}\right)\mathrm{d}x$$ where the two integrands are the equations of the bottom half of each circle in question. The final value for the area highlighted in blue simplifies to $$r^2\left(\sqrt{7}+\frac{3\pi}{2}+4\arctan{\left(\frac{2+\sqrt{7}}{3}\right)}-16\arctan{\left(\frac{4+\sqrt{7}}{9}\right)}\right)$$

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