1. **State the problem:** We need to find the area of a segment of a circle with radius $12$ cm and a central angle of $60^\degree$. 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 with central angle $\theta$ (in degrees) and radius $r$ is: $$\text{Area}_{\text{sector}} = \frac{\theta}{360} \times \pi r^2$$ - Area of the triangle formed by two radii and the chord when $\theta$ is known: $$\text{Area}_{\text{triangle}} = \frac{1}{2} r^2 \sin(\theta)$$ 3. **Calculate the sector area:** $$\text{Area}_{\text{sector}} = \frac{60}{360} \times \pi \times 12^2 = \frac{1}{6} \times \pi \times 144 = 24\pi$$ 4. **Calculate the triangle area:** $$\text{Area}_{\text{triangle}} = \frac{1}{2} \times 12^2 \times \sin(60^\degree) = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2} = 72 \times \frac{\sqrt{3}}{2} = 36\sqrt{3}$$ 5. **Calculate the segment area:** $$\text{Area}_{\text{segment}} = \text{Area}_{\text{sector}} - \text{Area}_{\text{triangle}} = 24\pi - 36\sqrt{3}$$ **Final answer:** The area of the segment is $$24\pi - 36\sqrt{3} \text{ cm}^2$$