The problem asks us to find the area of the minor segment of a circle. We are given that the central angle of the segment is $60^\circ$ and the radius of the circle is $22$ cm.

1. Problem Description

The problem asks us to find the area of the minor segment of a circle. We are given that the central angle of the segment is $60^\circ$ and the radius of the circle is $22$ cm.

2. Solution Steps

The area of the minor segment can be found by subtracting the area of the triangle formed by the radii and the chord from the area of the sector formed by the central angle.

First, we find the area of the sector. The area of a sector is given by the formula:

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

where $\theta$ is the central angle in degrees and $r$ is the radius. In our case, $\theta = 60^\circ$ and $r = 22$ cm.

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

Next, we find the area of the triangle. Since the central angle is $60^\circ$ and the other two sides are equal to the radius, the triangle is an isosceles triangle. Since one angle is $60^\circ$, the other two angles must also be $60^\circ$, making it an equilateral triangle.

The area of an equilateral triangle with side length $s$ is given by:

$A_{triangle} = \frac{\sqrt{3}}{4} s^2$

In our case, $s = 22$ cm, so:

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

Now, we subtract the area of the triangle from the area of the sector to find the area of the minor segment:

$A_{segment} = A_{sector} - A_{triangle} = \frac{242\pi}{3} - 121\sqrt{3} = 121(\frac{2\pi}{3} - \sqrt{3})$

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

$A_{segment} \approx 121(\frac{2(3.14159)}{3} - 1.73205) = 121(\frac{6.28318}{3} - 1.73205) = 121(2.09439 - 1.73205) = 121(0.36234) \approx 43.84314$

3. Final Answer

The area of the minor segment is $\frac{242\pi}{3} - 121\sqrt{3}$ cm$^2$, which is approximately $43.84$ cm$^2$.

Final Answer: The final answer is $\frac{242\pi}{3} - 121\sqrt{3}$