Problem 4. In the plane, a square with a side length of 1024 cm is given. A square with half the side length is drawn inside the given square, and this process continues until the side lengths, expressed in centimeters, are natural numbers (see the diagram on the right).

![](https://cdn.mathpix.com/cropped/2024_06_05_3a3538b5fdc92afac337g-07.jpg?height=228&width=543&top_left_y=549&top_left_x=937)
a) Determine the perimeter of the figure obtained in the manner described above.

b) Calculate the area of the square that is located in the center of the resulting sequence of squares.
Solution. a) If we add up the lengths of the right vertical segments, we will get the length of the side of the original square. Therefore, the perimeter of the resulting figure consists of two lengths of the side of the original square and twice the sum of the lengths of all the horizontal sides of the squares that are drawn. The lengths of the sides of all the drawn squares are: $1024 \mathrm{~cm}, 512 \mathrm{~cm}, 256 \mathrm{~cm}, 128 \mathrm{~cm}, 64 \mathrm{~cm}, 32 \mathrm{~cm}, 16 \mathrm{~cm}$, $8 \mathrm{~cm}, 4 \mathrm{~cm}, 2 \mathrm{~cm}$ and $1 \mathrm{~cm}$, so the perimeter of the resulting figure is

$$
\begin{aligned}
L & =2 \cdot 1024+2(1024+512+256+128+64+32+16+8+4+2+1) \\
& =2048+2 \cdot 2047=6142 \mathrm{~cm} .
\end{aligned}
$$

b) In the sequence, there are 11 squares drawn, and the length of the side of the middle square is equal to $32 \mathrm{~cm}$. According to this, the area of the middle square is $P=32 \cdot 32=1024 \mathrm{~cm}^{2}$.

## VII section