16) Determine the area of the polygon obtained as the union of two squares, both with side length 1, having the same center and rotated by $45^{\circ}$ relative to each other.
(A) $4-2 \sqrt{2}$
(B) $\frac{9}{8}$
(C) $2-\frac{\sqrt{3}}{2}$
(D) $\frac{5}{4}$
(E) the above answers are incorrect.

![](https://cdn.mathpix.com/cropped/2024_04_17_0927328f2a5ca041b809g-2.jpg?height=269&width=298&top_left_y=1100&top_left_x=1074)
16) The answer is (A). The segments $A B, B C, C D$ and $E D$ have the same length, which we denote as $x$. The side $B D$ is the hypotenuse of the right-angled and isosceles triangle $B C D$, so its length is $x \sqrt{2}$. The length of the side $A E$, which we know is 1, is equal to the sum of the lengths of $A B, B D$ and $D E$, that is, we have: $1=x+x \sqrt{2}+x$. Solving this, we get $x=\frac{1}{2+\sqrt{2}}$. The area of the triangle $B D C$ is:

$$
\begin{gathered}
\frac{1}{2} x \cdot x=\frac{1}{2}\left(\frac{1}{2+\sqrt{2}}\right)^{2}=\frac{1}{2}\left(\frac{2-\sqrt{2}}{(2+\sqrt{2})(2-\sqrt{2})}\right)^{2}=\frac{1}{2}\left(\frac{2-\sqrt{2}}{2}\right)^{2}= \\
\frac{1}{2}\left(1-\frac{\sqrt{2}}{2}\right)^{2}=\frac{1}{2}\left(1+\frac{1}{2}-\sqrt{2}\right)=\frac{3}{4}-\frac{1}{2} \sqrt{2}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_d4fc67ca786af90b2fcdg-4.jpg?height=471&width=429&top_left_y=1015&top_left_x=1493)

The area of the figure is obtained by adding the area of a square with side length 1 to the areas of 4 triangles with the same area as $B C D$, so the required area is: $1+4\left(\frac{3}{4}-\frac{1}{2} \sqrt{2}\right)=4-2 \sqrt{2}$.