We are asked to find the area of the minor segment of a circle. The central angle of the segment is $60^\circ$ and the radius of the circle is 22 cm. The area of the minor segment is the area of the sector minus the area of the triangle.

1. Problem Description

We are asked to find the area of the minor segment of a circle. The central angle of the segment is $60^\circ$ and the radius of the circle is 22 cm. The area of the minor segment is the area of the sector minus the area of the triangle.

2. Solution Steps

First, we find the area of the sector. The area of a sector with central angle $\theta$ (in degrees) and radius $r$ is given by:

$A_{sector} = \frac{\theta}{360} \pi r^2$

In this case, $\theta = 60^\circ$ and $r = 22$ cm. Thus,

$A_{sector} = \frac{60}{360} \pi (22)^2 = \frac{1}{6} \pi (484) = \frac{484 \pi}{6} = \frac{242 \pi}{3}$

Next, we find the area of the triangle. Since the central angle is $60^\circ$ and the two sides are equal to the radius, the triangle is an isosceles triangle. Since the angles opposite the equal sides are equal, the two other angles must be $\frac{180^\circ - 60^\circ}{2} = \frac{120^\circ}{2} = 60^\circ$. Thus, the triangle is equilateral. The area of an equilateral triangle with side $s$ is $\frac{\sqrt{3}}{4} s^2$. Since the side length is the radius, $s = 22$.

$A_{triangle} = \frac{\sqrt{3}}{4} (22)^2 = \frac{\sqrt{3}}{4} (484) = 121 \sqrt{3}$

The area of the minor segment is the area of the sector minus the area of the triangle:

$A_{segment} = A_{sector} - A_{triangle} = \frac{242\pi}{3} - 121\sqrt{3} \approx \frac{242(3.14159)}{3} - 121(1.73205) \approx 253.63 - 209.58 = 44.05$

$A_{segment} = \frac{242\pi}{3} - 121\sqrt{3} = 121 (\frac{2\pi}{3} - \sqrt{3})$ cm$^2$.

Using $\pi \approx 3.14159$ and $\sqrt{3} \approx 1.73205$,

$A_{segment} \approx 121 (\frac{2(3.14159)}{3} - 1.73205) = 121(2.0944 - 1.73205) = 121(0.36235) = 43.84435$.

3. Final Answer

$\frac{242\pi}{3} - 121\sqrt{3}$ cm$^2$ or approximately 43.84 cm$^2$.