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. Note that the drawing in the image has a radius of 19cm, but the problem states it is 22cm, so we must use 22cm.

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. Note that the drawing in the image has a radius of 19cm, but the problem states it is 22cm, so we must use 22cm.

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.

Area of sector = $(\theta/360) \times \pi r^2$, where $\theta$ is the central angle in degrees and $r$ is the radius.

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

Area of sector = $(60/360) \times \pi (22)^2 = (1/6) \times \pi (484) = (484/6) \pi = (242/3) \pi$

Since the central angle is $60^\circ$ and the other two sides of the triangle are radii of the circle (and hence equal), the triangle is an isosceles triangle. Also, since the base angles of an isosceles triangle are equal, the remaining two angles in the triangle are equal. Let each of these angles be $x$. Then $60 + x + x = 180$, so $2x = 120$, and $x = 60$. This means that the triangle is an equilateral triangle, and all sides are of length 22 cm.

Area of an equilateral triangle = $(\sqrt{3}/4) \times side^2$.

Area of the triangle = $(\sqrt{3}/4) \times (22)^2 = (\sqrt{3}/4) \times 484 = 121 \sqrt{3}$.

Area of minor segment = Area of sector - Area of triangle

Area of minor segment = $(242/3) \pi - 121 \sqrt{3} = 121 (\frac{2\pi}{3} - \sqrt{3})$

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

Area of minor segment = $121 (\frac{2 \times 3.14159}{3} - 1.73205) = 121 (\frac{6.28318}{3} - 1.73205) = 121 (2.09439 - 1.73205) = 121 (0.36234) = 43.84314$

Alternatively:

Area of sector = $(60/360) \pi (22)^2 = (1/6) \pi (484) = (242/3)\pi$

Area of triangle = $(1/2) r^2 \sin(\theta) = (1/2) (22)(22) \sin(60) = (1/2) (484) (\sqrt{3}/2) = (484/4) \sqrt{3} = 121 \sqrt{3}$

Area of segment = Area of sector - Area of triangle = $(242/3)\pi - 121 \sqrt{3} \approx (242/3)(3.14159) - 121(1.732) \approx 253.307 - 209.572 = 43.735$

3. Final Answer

The area of the minor segment is approximately $43.74 cm^2$.

$121(\frac{2\pi}{3}-\sqrt{3}) \text{ cm}^2$ or approximately $43.73 \text{ cm}^2$