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Square $ABCD$ has side equal to $a$. Points $A$ and $D$ are centers of two Quarter-Circles (see image below), which intersect at point K. Find the area defined by side $CD$ and arcs $KC$ and $KD$.

enter image description here

Here's what I did: The darkened area can be found by Substracting area of figure defined by points $AKD$ from quarter-circle $CAD$. Area of quarter-circle $= \dfrac{a^2\pi}{4}$. Now onto the harder part:

The way I calculated the Area of $AKD$ is by noticing that it's half of an elipse (at least I'm pretty sure it is). With $R1 = \dfrac{a}{2}$ (by symmetry) and $R2=\dfrac{a\sqrt3}{2}$ (by Pythagoras). The area of $AKD$ will be half of an ellipse: $\dfrac{R1R2\pi}{2} = \dfrac{\ a^2\sqrt3}{8}\pi$

The area of darkened figure will be the difference between two areas: $\dfrac{a^2\pi}{4} - \dfrac{\ a^2\sqrt3}{8}\pi $.

But my answer, for some reason, is way off. What am I doing wrong? Does $AKD$ not represent a semi-ellipse?

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    $\begingroup$ It is not half of ellipse. Its boundary is not smooth. $\endgroup$ – JCAA Jul 5 at 12:20
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    $\begingroup$ AKD is not a half of the ellipse :) Ellipse has no sharp corners like point K. $\endgroup$ – Oldboy Jul 5 at 12:20
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    $\begingroup$ haha, well, yeah that makes sense.. $\endgroup$ – Ebrin Jul 5 at 12:22
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    $\begingroup$ And ellipse is not made of circular arcs :) $\endgroup$ – Oldboy Jul 5 at 12:42
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    $\begingroup$ @Ebrin Something that looks like an ellipse is not necessarily an ellipse. You can make a smooth, closed curve by concatenating circular arcs of different radii, but such curve is neither a circle nor an ellipse. Ellipse is not made of circular arcs! $\endgroup$ – Oldboy Jul 5 at 14:21
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enter image description here

Find area $S$ first:

$$S=\frac16 a^2\pi-P_{\triangle ADK}$$

Area of ADK is:

$$P_{ADK}=2S+P_{\triangle ADK}=2(\frac16 a^2\pi-P_{\triangle ADK})+P_{\triangle ADK}=\frac13 a^2\pi-P_{\triangle ADK}$$

$$P_{ADK}=\frac13 a^2\pi-\frac14a^2\sqrt3$$

Shaded area is simply:

$$P_{shaded}=P_{ADC}-P_{ADK}=\frac14 a^2\pi-(\frac13 a^2\pi-\frac14a^2\sqrt3)$$

$$P_{shaded}=\frac14a^2\sqrt3-\frac1{12}a^2\pi=\frac1{12}a^2(3\sqrt3-\pi)$$

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    $\begingroup$ Wow, I see now that the solution is.. fairly simple. I thank you for quick and simple response. $\endgroup$ – Ebrin Jul 5 at 12:47
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This solution may be weird, but I think it's kind of original. Let's define the function $$f(x)=\sqrt{a^2-(x-a)^2}.$$ It corresponds to the semi-circumference of radius $a$ with center in $x=a$. Now, we calculate the area of the quarter of circumference: $$A_{\text{quarter}}=\frac{\pi a^2}{4}.$$ Now, we integrate to get the part of that quarter "without the grey part" (white part): $$A_\text{white part}= 2\int_0^{a/2}f(x)dx=2\int_0^{a/2}\sqrt{a^2-(x-a)^2}dx=\frac{a^2(4\pi-3\sqrt{3})}{12}.$$ And finally we find the grey part area we wanted: $$A=A_{\text{quarter}}-A_\text{white part}=\frac{\pi a^2}{4}-\frac{a^2(4\pi-3\sqrt{3})}{12}=\boxed{\frac{a^2(3\sqrt{3}-\pi)}{12}}.$$

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    $\begingroup$ Thanks for showing the alternative way, however as of this moment i'm unfamiliar with calculus. $\endgroup$ – Ebrin Jul 5 at 12:50
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Alternatively, you can easily find the shaded area by Calculus.

Assume that the vertex A is at the origin $O$ then the equation of circle with center at $A$ (i.e. origin $O$) is $x^2+y^2=a^2$. Similarly, the equation of circle with center at $B$ (i.e. $(a,0)$) is $(x-a)^2+y^2=a^2$ or $x^2+y^2-2ax=0$. The quarter circles intersect each other at $(\frac a2, \frac{a\sqrt3}{2})$

The shaded area is equal to bounded area between quarter circles $$\int_{a/2}^a \left(\sqrt{2ax-x^2}-\sqrt{a^2-x^2}\right)\ dx$$ $$=\frac12\left((x-a)\sqrt{a^2-(x-a)^2}+a^2\sin^{-1}\left(\frac{x-a}{a}\right)-x\sqrt{a^2-x^2}-a^2\sin^{-1}\left(\frac{x}{a}\right)\right)_{a/2}^a$$ $$=\frac{a^2(3\sqrt3-\pi)}{12}$$

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