Congruent rectangles $A B C D$ and $E F G H$ are placed such that their congruent sides are parallel. Points $I, J, K, L, M$, and $N$ are the intersections of the extended sides as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_f2ed6898bbb1d7496803g-4.jpg?height=514&width=874&top_left_y=2070&top_left_x=594)

The area of rectangle $B N H M$ is $12 \mathrm{~cm}^{2}$, the area of rectangle $M B C K$ is $63 \mathrm{~cm}^{2}$, and the area of rectangle $M L G H$ is $28 \mathrm{~cm}^{2}$.

Determine the area of rectangle $I F J D$.

(E. Semerádová)

Hint. What are the areas of the other rectangles?
The area of rectangle $H N C K$ is $63-12=51\left(\mathrm{~cm}^{2}\right)$, the area of rectangle $B L G N$ is $28-12=16\left(\mathrm{~cm}^{2}\right)$. The ratio of the areas of rectangles $N G J C$ and $H N C K$ is the same as the ratio of the lengths of segments $N G$ and $H N$, and that is the same as the ratio of the areas of rectangles $B L G N$ and $M B N H$. Therefore,

$$
S_{N G J C}: 51=16: 12=4: 3,
$$

and thus the area of rectangle $N G J C$ is $51 \cdot 4: 3=68\left(\mathrm{~cm}^{2}\right)$.

Since rectangles $A B C D$ and $E F G H$ are congruent and shifted, segments like $I E$ and $C J$ are congruent, and the same applies to other pairs. Therefore, rectangles like $I E M A$ and $N G J C$ are congruent, particularly they have the same area. The same applies to other pairs:

![](https://cdn.mathpix.com/cropped/2024_04_17_f2ed6898bbb1d7496803g-5.jpg?height=508&width=868&top_left_y=977&top_left_x=594)

The area of rectangle $I F J D$ is $12+2 \cdot 51+2 \cdot 16+4 \cdot 68=418\left(\mathrm{~cm}^{2}\right)$.