The problem asks to find the area of the minor segment of a circle with a central angle of $60^\circ$ and a radius of $22$ cm.

1. Problem Description

The problem asks to find the area of the minor segment of a circle with a central angle of $60^\circ$ and a radius of $22$ cm.

2. Solution Steps

The area of the minor segment is the difference between the area of the sector and the area of the triangle formed by the two radii and the chord.

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

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

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

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

Next, calculate the area of the triangle. Since the triangle is formed by two radii of equal length ($22$ cm) and the angle between them is $60^\circ$, the triangle is an isosceles triangle with one angle being $60^\circ$. Since the other two angles are equal, they must each be $(180^\circ - 60^\circ)/2 = 60^\circ$. Therefore, the triangle is an equilateral triangle with side length $22$ cm.

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

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

In this case, $s = 22$ cm, so

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

The area of the minor segment is the difference between the area of the sector and the area of the triangle:

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

$A_{segment} = \frac{242\pi}{3} - 121\sqrt{3} \approx \frac{242(3.14159)}{3} - 121(1.732) \approx 253.64 - 209.572 = 44.068$

We can approximate using $\pi \approx 3.14$ and $\sqrt{3} \approx 1.732$. Then

$A_{segment} \approx \frac{242(3.14)}{3} - 121(1.732) = \frac{759.88}{3} - 209.572 \approx 253.29 - 209.572 = 43.718$

3. Final Answer

$\frac{242\pi}{3} - 121\sqrt{3}$ cm$^2$ which is approximately $43.72$ cm$^2$.