#### Solution

There are a couple of good solutions here. First, let’s notice that because the points on the sides are midpoints, like $F$ is a midpoint of $CD$, and the segments $DH$ and $FB$ are at the same angle – so are parallel, points on the newly created segments are also their midpoints – like $J$ is a midpoint of $CK$. That means the triangle $\triangle CFJ$ is similar to $\triangle CDK$, and the similarity ratio is $\frac12$. Because of that we can say that the segment $FJ$ is a half of the segment $DK$ and so we can rotate the triangle $\triangle CFJ$ up, like in the picture below. It is immediately visible that the newly created quadrangle $DKJM$ is a square, and it is a square identical to the square in the centre. Thus, we can divide the whole big square into $5$ regions, each of which has the same area as the one we are looking for. The wanted area is therefore a fifth of the total area of the square.

We could also be more specific and say that the area of $\triangle CFJ$ is a quarter of the area of the $\triangle CDK$ and is obviously equal to the area of $\triangle DKE$, which is just identical to it. However, the triangles $\triangle CDK$ and $\triangle DKE$ together make up $\triangle CDE$, whose area is a fourth of the area of the large square (it is $\frac12 \frac{a}2 \times a$, where $a$ is a square’s side). Therefore, if $x$ is the area of $\triangle CFJ$, $5x = \frac14$, so $x = \frac1{20}$ and the area of $\triangle CDK$ is $4 \times \frac1{20} = \frac15$. The whole square is divided into $4$ such triangles and the square in the middle, so its area fraction has to be $1 – 4 \times \frac15 = \frac15$. $\\$

Anyway, the large square’s area is $100$, so the area of the square in the centre is $100 \div 5 = 20$.