In a circle with radius $r=1 \mathrm{~m}$, construct a sector such that the angle corresponding to its arc is $60^{\circ}$. What is the area of the sector?
We obtain the sought area by subtracting the area of the equilateral triangle with side $r$ from the area of the circular sector corresponding to a $60^{\circ}$ arc.

Thus,

$$
t=\frac{r^{2} \pi}{6}-\frac{r^{2}}{4} \sqrt{3}
$$

In this case,

$$
t=\frac{\pi}{6}-\frac{\sqrt{3}}{4}=0.090 \mathrm{~m}^{2}
$$

(Rezső Paunz, Pécs.)