6. Let $A B C D$ be a square. Points $E, F, G$ and $H$ divide the sides $\overline{D A}, \overline{A B}, \overline{B C}$ and $\overline{C D}$ in the ratio 4 : 3, respectively. Points $I, J, K$ and $L$ are the midpoints of the sides of the square $E F G H$. The area of the shaded square $I J K L$ inscribed in the square $E F G H$ is $200 \mathrm{dm}^{2}$. Calculate the area of the square $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_06_03_0e9513bf766e1ec9edb5g-2.jpg?height=594&width=583&top_left_y=637&top_left_x=728)
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_b51751bd0eca2a2c4a9cg-3.jpg?height=480&width=483&top_left_y=1328&top_left_x=821)

From the area of square $I J K L$, we can calculate the length of its side. Let the length of the side of square IJKL be $y$. From the condition of the problem, we have:

$y^{2}=200$, so $y=10 \sqrt{2} \text{ cm}$.

2 POINTS

Let the length of the side of square EFGH be $a$. Let $x=\frac{a}{2}$. Then we have:

$y=x \sqrt{2}$, so $10 \sqrt{2}=x \sqrt{2}$, which means $x=10 \text{ cm}$.

2 POINTS

The length of the side of square EFGH is $a=20 \text{ cm}$.

1 POINT

The ratio of the legs of the right triangle $A F E$ is $3: 4$. Let $|E A|=3 k$ and $|A F|=4 k$.

By the Pythagorean theorem, we have $(3 k)^{2}+(4 k)^{2}=20^{2}$,

so $25 k^{2}=400$, which means $k^{2}=16, k=4$.

The length of the side of square $A B C D$ is $|A F|+|F B|=|A F|+|E A|=16+12=28 \text{ cm}$.

##