Two equal circles with a radius of $1256 \mathrm{~m}$ intersect such that the distance between their centers is $50 \mathrm{~m}$. Calculate the area common to both circles.
The area common to the two circles is the sum of the areas of the two equal circular segments lying on the common chord; thus

$$
2 t=2\left(r^{2} \pi \frac{\alpha}{360^{\circ}}-\frac{r^{2} \sin \alpha}{2}\right)
$$

or

$$
2 t=r^{2}\left(\pi \frac{\alpha}{180^{\circ}}-\sin \alpha\right)
$$

where $\alpha$ can be determined from the following equation:

$$
\cos \frac{\alpha}{2}=\frac{c}{2 r}
$$

Substituting the given values:

$$
\frac{\alpha}{2}=42^{\circ} 53^{\prime} 45^{\prime \prime}, 2 t=582.3 \mathrm{~m}^{2}
$$

The problem was also solved by: Bauss O., Dénes A., Freibauer E., Friedmann B., Goldziher K., Hofbauer E., Kántor B., Kántor N., Kertész L., Krátky Gy., Petrogalli G., Probst E., Spitzer Ö., Szabó K., Weisz J.