Area of square as a function of $\hat{a}$? Suppose $ABCD $   is square ,and $AM=DN=QB=PC$   so $$A'B'C'D'$$
 is a square too.
Can someone help me to find area of $\bf{smaller -square}$(or $\color{red} {\Box A'B'C'D'}
$)  as a function of angle $\hat{a}$ .It seems to be easy ,but I get stuck on this problem.Thanks in advance.

 A: 
Let $\alpha=\angle NAD$, $a=|AB|$. Then
\begin{align}
S_{A'B'C'D'}&=S_{ABCD}-4S_{ADD'}
\\
&=
a^2-4\,\tfrac12 a\,a\,\cos\alpha\sin\alpha
\\
&=a^2(1-\sin2\alpha)
.
\end{align}
A: Label the points $A, B, C, D, E, F, G, H, I, J, K, L$ as in this diagram:

Now, let $AD = x$ and let $\angle DAF = \alpha$. 
We wish to express the area of square $IJKL$ in terms of $x$ and $\alpha$. 

First, let $DF = AE = y$. The tangent of $\alpha$ can then be written
$$\tan \alpha = \frac{y}{x} \,\,\implies\,\, y = x\tan\alpha$$
Triangle $ADF$ is right, so we can apply the Pythagorean theorem to find that $$AF = \sqrt{x^2+y^2}$$
To find $IL$, we will find $AI$ and $LF$ and subtract both from $AF$. 

To find $AI$, note that triangles $AIE$ and $ADF$ are similar. This means that 
$$\frac{AE}{AI} = \frac{AF}{AD}$$
$$\frac{y}{AI}=\frac{\sqrt{x^2+y^2}}{x}$$
$$AI = \frac{xy}{\sqrt{x^2+y^2}}$$

To find $LF$, note that triangles $DLF$ and $ADF$ are similar. This means that 
$$\frac{DF}{LF} = \frac{AF}{DF}$$
$$\frac{y}{LF} = \frac{\sqrt{x^2+y^2}}{y}$$
$$LF = \frac{y^2}{\sqrt{x^2+y^2}}$$

Finally we can find $IL$:
$$IL = AF - AI - LF$$
$$IL = \sqrt{x^2+y^2} - \frac{y^2+xy}{\sqrt{x^2+y^2}}$$
This means that our final answer for the area of square $IJKL$ is:
$$(IL)^2 = \boxed{\left(\sqrt{x^2+y^2} - \frac{y^2+xy}{\sqrt{x^2+y^2}}\right)^2\,}$$
where $x = AF$ and $y = x\tan \alpha$, with $\alpha = \angle DAF$.
