1. In the first square, a second square is inscribed whose vertices are the midpoints of the sides of the first square. Inside the second square, there is a third square whose sides are $2 \mathrm{~cm}$ away from the sides of the second square. Inside the third square, there is a fourth square whose sides are $3 \mathrm{~cm}$ away from the sides of the third square. What is the area of the first square if the area of the fourth square is $9 \mathrm{~cm}^{2}$?
1. 

![](https://cdn.mathpix.com/cropped/2024_05_30_3b064f6bc407e279e59eg-04.jpg?height=477&width=485&top_left_y=795&top_left_x=794)

From the area of the fourth square, which is $P_{4}=9 \mathrm{~cm}^{2}$, it follows that $a_{4}=3 \mathrm{~cm}$.

The sides of the third square are $3 \mathrm{~cm}$ away from the sides of the fourth square, so the length of the side of the third square is $a_{3}=a_{4}+3+3=3+3+3=9 \mathrm{~cm}$.

The sides of the second square are $2 \mathrm{~cm}$ away from the sides of the third square, so the length of the side of the second square is $a_{2}=a_{3}+2+2=9+2+2=13 \mathrm{~cm}$.

The area of the second square is $P_{2}=a_{2} \cdot a_{2}=13 \cdot 13=169 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_3b064f6bc407e279e59eg-04.jpg?height=466&width=468&top_left_y=1663&top_left_x=794)

The second square consists of four congruent triangles, and the first square consists of eight such triangles, so the area of the first square is twice the area of the second square, i.e.,

$P_{1}=2 \cdot P_{2}=2 \cdot 169=338 \mathrm{~cm}^{2}$.