How calculate the shaded area in this picture?

Solution 1:

Let $(APD)$ be the area of the figure $APD$.

And let $x,y,z$ be $(KEPM),(PAD),(MPD)$ respectively.

First, we have $$(\text{square}\ ABCD)=a^2=x+4y+4z.\tag1$$

Second, we have $$(\text{sector}\ BDA)=\frac{\pi a^2}{4}=x+2y+3z.\tag2$$

Third, note that $KA=KD=a$ and that $(\triangle KAD)=\frac{\sqrt 3}{4}a^2$ since $\triangle KAD$ is a equilateral triangle.

So, since we have $$\begin{align}(K(E)AD(M))&=(\text{sector}\ AKD)+(\text{sector}\ DKA)-(\triangle KAD)\\&=\frac{\pi}{6}a^2+\frac{\pi}{6}a^2-\frac{\sqrt 3}{4}a^2\\&=\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2,\end{align}$$ we have $$\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2=x+y+2z.\tag3$$

Solving $(1),(2),(3)$ gives us $$(KEPM)=x=\left(1+\frac{\pi}{3}-\sqrt 3\right)a^2.$$

Solution 2:

Here is a geometric solution.

Let's find the angle $EDK$. Since $KA=KD=a=AD$ hence the triangle $AKD$ is equiliteral and the angle $KDA$ is $\pi/3$. Therefore angle $KDC$ is $\pi/6$. In the same way the angle $EDA$ is also $\pi/6$ and we get that the angle $EDK$ is $\pi/6$.

Now the shaded area $$(EPMK) = ({\rm square\,} EPMK) + 4\times ({\rm segment\,}EKE).$$

The edge of the square $EPMK$ (from the triangle $EDK$) is $$EK=2a\sin\frac{\pi}{12}=a\frac{\sqrt{3}-1}{\sqrt{2}},$$ hence the area $$({\rm square\,} EPMK) =(2-\sqrt{3})a^2.$$

And the area $$({\rm segment\,}EKE) = \frac{1}{2}\left(\frac{\pi}{6}-\sin\frac{\pi}{6}\right)a^2=\frac{1}{4}\left(\frac{\pi}{3}-1 \right)a^2.$$

So $$(EPMK) = (2-\sqrt{3})a^2 + \left(\frac{\pi}{3}-1 \right)a^2=\left(\frac{\pi}{3}+1-\sqrt{3}\right)a^2.$$