In the image, three overlapping squares are depicted. Determine their areas, given that:

$\triangleright$ the side of the largest square is $1 \mathrm{~mm}$ longer than the side of the middle square and $2 \mathrm{~mm}$ longer than the side of the smallest one,

$\triangleright$ the common part of the largest and middle square is a square with an area of $100 \mathrm{~mm}^{2}$,

$\triangleright$ the common part of the middle and smallest square is a square with an area of $64 \mathrm{~mm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f31b3216c3ca95aeff1dg-1.jpg?height=440&width=443&top_left_y=1565&top_left_x=841)

(It does not matter to measure, the image is inaccurate.)
We solve the example by writing the known data into the sketch, i.e., the areas of the common parts: $100 \mathrm{~mm}^{2}, 64 \mathrm{~mm}^{2}$.

If the area of the square is $100 \mathrm{~mm}^{2}$, then its side measures $10 \mathrm{~mm}$. If the area of the square is $64 \mathrm{~mm}^{2}$, then its side measures $8 \mathrm{~mm}$. Therefore, we further write the sides $8 \mathrm{~mm}$ and $10 \mathrm{~mm}$ into the diagram. From this, it follows that the side of the middle square measures $8+10=18 \mathrm{~mm}$. The middle square, in terms of size, is also the square located in the center of the diagram; otherwise, it would not meet the conditions of the problem. The largest square, therefore, has a side length of $1+18=19 \mathrm{~mm}$, and its area is then $S=19 \cdot 19=361 \mathrm{~mm}^{2}$. The middle square has an area of $S=18 \cdot 18=324 \mathrm{~mm}^{2}$. The smallest square has a side length of $18-1=17 \mathrm{~mm}$, and its area is then $S=17 \cdot 17=289 \mathrm{~mm}^{2}$. The areas of the three given squares are: $361 \mathrm{~mm}^{2}, 324 \mathrm{~mm}^{2}$, and $289 \mathrm{~mm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f31b3216c3ca95aeff1dg-2.jpg?height=663&width=665&top_left_y=385&top_left_x=730)

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