The problem is 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.

1. Problem Description

The problem is 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.

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.

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

$A_{sector} = \frac{\theta}{360} \pi r^2$, where $\theta$ is the central angle in degrees and $r$ is the radius of the circle.

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

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

Next, calculate the area of the triangle. Since the two sides of the triangle are radii, the triangle is an isosceles triangle. The area of a triangle can be calculated using the formula $A = \frac{1}{2}ab\sin C$, where $a$ and $b$ are the lengths of two sides and $C$ is the angle between them.

In this case, $a = 22$ cm, $b = 22$ cm, and $C = 60^{\circ}$. Therefore,

$A_{triangle} = \frac{1}{2}(22)(22)\sin(60^{\circ}) = \frac{1}{2}(484) \frac{\sqrt{3}}{2} = 242 \frac{\sqrt{3}}{2} = 121\sqrt{3}$ cm$^2$.

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

$A_{segment} = A_{sector} - A_{triangle} = \frac{242\pi}{3} - 121\sqrt{3} = \frac{242\pi - 363\sqrt{3}}{3}$

$A_{segment} \approx \frac{242(3.1416) - 363(1.732)}{3} = \frac{760.2672 - 628.916}{3} = \frac{131.3512}{3} \approx 43.78$ cm$^2$

We have $A_{segment} = \frac{242\pi}{3} - 121\sqrt{3} = 121(\frac{2\pi}{3} - \sqrt{3})$. Using $\pi = 3.1416$ and $\sqrt{3} = 1.732$, we have

$A_{segment} = 121 (\frac{2(3.1416)}{3} - 1.732) = 121 (\frac{6.2832}{3} - 1.732) = 121(2.0944 - 1.732) = 121(0.3624) = 43.8504$ cm$^2$.

3. Final Answer

The area of the minor segment is approximately 43.85 cm$^2$.

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

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

$A_{segment} \approx 43.85$ cm$^2$