17. We will start from a square with a perimeter equal to 2. Under these conditions, we get $h_{1}=\frac{1}{4}, r_{1}=h_{1} \sqrt{2}=\frac{\sqrt{2}}{4}$. Further calculations are conveniently recorded in the following table:

| $p$ | $h_{p}\left(h_{p+1}=\frac{h_{p}+r_{p}}{2}\right)$ | $r_{p}\left(r_{p+1}=\sqrt{\left.\overline{h_{p+1} r_{p}}\right)}\right.$ |
| :---: | :---: | :---: |
| 1 | 0.2500000 | 0.353554 |
| 2 | 0.3017767 | 0.365406 |
| 3 | 0.3112086 | 0.3203641 |
| 4 | 0.3172863 | 0.3188214 |
| 5 | 0.318038 | 0.3184372 |
| 6 | 0.3182455 | 0.3183413 |
| 7 | 0.3182934 | 0.3183173 |
| 8 | 0.318353 | 0.3183111 |
| 9 | 0.3183098 | 0.3183102 |
| 10 |  | 0.3183098 |
|  |  |  |

Thus, $\frac{1}{\pi}=0.3183098 \ldots$, from which we get: $\pi=3.1415926$.

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-097.jpg?height=691&width=671&top_left_y=246&top_left_x=356)

Fig. 44
17*. From the formula for the circumference: $C=2 \pi r$, we get: $\pi=$ $=\frac{C}{2 r}$ or $\frac{1}{\pi}=\frac{2 r}{C}$. Consider a square with a perimeter of 2 and sequentially find the values of $h_{p}$ and $r_{p}$ using the formulas from the previous problem. By repeatedly performing the calculations, achieve that the values of $h_{p}$ and $r_{p}$ coincide in the first five decimal places. This will find the value of $\frac{1}{\pi}=\frac{2 r}{\bar{c}} \approx \frac{2 r}{2}=r$ with the corresponding accuracy. (This method of calculating $\pi$ is called the method of isoperimeters.)