A4 Square $A B C D$ has side length $10 \mathrm{~cm}$. Points $W, X, Y$, and $Z$ are the midpoints of segments $A B, B C, C D$ and $W X$, respectively. Determine the area of quadrilateral $X C Y Z$ (in $\mathrm{cm}^{2}$ ).
Answer: 25

Solution 1:
We break up the quadrilateral $X C Y Z$ into two triangles; triangle $Y Z X$ and triangle $Y X C$. As $Z$ is on the midpoint of line $W X$, the areas of triangle $Y Z X$ and $Y Z W$ are equal; they are both half of the area of triangle $W Y X$, which is a right isosceles triangle with bases of length $5 \sqrt{2} \mathrm{~cm}$. Thus, the area of triangle $Y Z X$ is
$$
\frac{1}{2} \times \frac{1}{2}(5 \sqrt{2})^{2}=\frac{25}{2} \mathrm{~cm}^{2}
$$

The area of triangle $Y X C$ is also $\frac{25}{2} \mathrm{~cm}^{2}$, since it is a right isosceles triangle with bases of length $5 \mathrm{~cm}$. Thus, the total area of quadrilateral $Y Z X C$ is $25 \mathrm{~cm}^{2}$.
Solution 2:
As $Z$ is the midpoint of $W X$, the areas of the triangles having equal bases and equal heights are equal: $[Y W Z]=[Y X Z]$. Clearly $\triangle W B X \cong \triangle Y C X$, so $[W B X]=[Y C X]$. Hence
$$
[X C Y Z]=\frac{1}{2}[W B C Y]=\frac{1}{4}[A B C D]=25 \mathrm{~cm}^{2} .
$$

Solution 3:
We place the points on a coordinate grid, with $B$ being the origin, line $B C$ being the $x$ axis, and line $A B$ being the $y$-axis. We see that $X=(5,0), W=(0,5), C=(10,0)$, and $Y=(10,5)$. Thus, $Z=\left(\frac{5}{2}, \frac{5}{2}\right)$. By the Shoelace Formula, the area of quadrilateral $X C Y Z$ is equal to
$$
\left|\frac{1}{2}\left(\left(5 \cdot \frac{5}{2}+\frac{5}{2} \cdot 5+10 \cdot 0+10 \cdot 0\right)-\left(0 \cdot \frac{5}{2}+\frac{5}{2} \cdot 10+5 \cdot 10+0 \cdot 5\right)\right)\right|=\left|\frac{1}{2}(25-75)\right|=25
$$