488. Bretschneider gives an approximate construction for $\pi$ accurate to the fifth decimal place: $\frac{13}{50} \sqrt{146}$, if the radius of the circle is taken as unity; construct this approximation.

## Problems of Zelena.
488. Since $146=11^{2}+5^{2}$, the question reduces to constructing the expression

$$
\pi=\frac{13 \sqrt{11^{2}+5^{2}}}{50}
$$

We divide the diameter $A B$ (Fig. 113) into 10 equal parts and on its extension, we lay off $B C=\frac{1}{10} A B$ and $C D=\frac{2}{10} A B$. On the tangent at the end $A$ of the diameter, we lay off from $A$ the length of the radius $A M$. Then the hypotenuse $M C$ will represent $\frac{1}{5} \sqrt{11^{2}+5^{2}}$. We describe a circle with radius $M C$ from $M$, intersecting the tangent at $R$. From $M$, we draw a second tangent $M S$, construct the rectangle $A M F D$, and draw $R S$ parallel to the diagonal $A F$. The length of the tangent $M S=2 \pi$. Indeed, from the similarity of triangles $A M F$ and $R M S$, we have: $M S: M F=M R: M A$, but

$$
M F=\frac{13}{5}
$$

and

$$
M R=\frac{1}{5} \sqrt{11^{2}+5^{2}}, M A=1
$$

therefore,

$$
M S=\frac{13}{25} \sqrt{246}=12 \pi
$$

accurate to five decimal places.

Zhelen, E., abbot, a pedagogue-mathematician, author of several elementary courses.

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-210.jpg?height=406&width=828&top_left_y=1896&top_left_x=383)

Fig. 113