20. Michael was walking in Marrakesh when he saw a tiling formed by tessellating the square tile as shown.
The tile has four lines of symmetry and the length of each side is $8 \mathrm{~cm}$. The length of $X Y$ is $2 \mathrm{~cm}$. The point $Z$ is such that $X Z$ is a straight line and $Y Z$ is parallel to the sides of the square.

What is the area of the central grey octagon?
A $6 \mathrm{~cm}^{2}$
B $7 \mathrm{~cm}^{2}$
C $8 \mathrm{~cm}^{2}$
D $9 \mathrm{~cm}^{2}$
E $10 \mathrm{~cm}^{2}$
Solution
$\mathbf{E}$
Let $T, U, V$ and $W$ be the points shown in the diagram.
Because the tile has four lines of symmetry and $Y Z$ is parallel to the sides of the square, so also is $X U$, and $X U=Y Z$. Therefore $X U Z Y$ is a rectangle with $U Z=X Y=2 \mathrm{~cm}$.
Because $X Z$ is a straight line, by the symmetry of the tile, it follows that $U Y$ is also a straight line. Therefore $X Z$ and $U Y$ are the diagonals of the rectangle $X U Z Y$ and they divide the rectangle into four triangles with equal areas.
Also, by the symmetry of the tile, $U V W Z$ is a square, $X T$ is a straight line and $X U=V T$. Therefore, because $U V=U Z=2 \mathrm{~cm}$ and $X T=8 \mathrm{~cm}$, it follows that $X U=3 \mathrm{~cm}$.
The shaded octagon is made up of the $2 \mathrm{~cm} \times 2 \mathrm{~cm}$ square $U V W Z$ and four congruent triangles. Each of these four triangles has the same area as one quarter of the rectangle $X U Z Y$. Therefore the area of the shaded octagon equals the area of $U V W Z$ plus the area of the $2 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangle $X U Z Y$.
Hence the area of the octagon is $(2 \times 2+2 \times 3) \mathrm{cm}^{2}$, that is, $10 \mathrm{~cm}^{2}$.