1. **Problem statement:** Find the area of the segment AYB of a circle with radius $21$ cm and central angle $\angle AOB = 120^\circ$. 2. **Formula and explanation:** The area of a segment is the area of the sector minus the area of the triangle formed by the two radii and the chord. - Area of sector $AOB = \frac{\theta}{360^\circ} \times \pi r^2$ where $\theta$ is the central angle in degrees. - Area of triangle $AOB = \frac{1}{2} r^2 \sin(\theta)$ where $\theta$ is in radians or degrees (using sine of the angle). 3. **Calculate the area of the sector:** $$\text{Area of sector} = \frac{120}{360} \times \pi \times 21^2 = \frac{1}{3} \times \pi \times 441 = 147\pi$$ 4. **Calculate the area of the triangle:** $$\text{Area of triangle} = \frac{1}{2} \times 21^2 \times \sin 120^\circ = \frac{1}{2} \times 441 \times \sin 120^\circ$$ Since $\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $$\text{Area of triangle} = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} = \frac{441 \sqrt{3}}{4}$$ 5. **Calculate the area of the segment:** $$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle} = 147\pi - \frac{441 \sqrt{3}}{4}$$ 6. **Final answer:** The area of segment AYB is $$\boxed{147\pi - \frac{441 \sqrt{3}}{4} \text{ cm}^2}$$