Find the area of the segment of a circle of radius $12 \, cm$ whose corresponding sector has a central angle of $60^{\circ}$ (Use $\pi = 3.14$).

(N/A) Given that,radius of a circle $(r) = 12 \, cm$ and central angle of sector $(\theta) = 60^{\circ}$.
Area of sector $= \frac{\pi r^2 \theta}{360^{\circ}} = \frac{3.14 \times 12 \times 12 \times 60^{\circ}}{360^{\circ}} = 3.14 \times 2 \times 12 = 75.36 \, cm^2$.
Since the triangle formed by the two radii and the chord is an isosceles triangle with a vertex angle of $60^{\circ}$,it is an equilateral triangle.
Area of equilateral triangle $= \frac{\sqrt{3}}{4} \times (\text{side})^2 = \frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3} \, cm^2$.
Area of the segment $=$ Area of sector $-$ Area of triangle $= (75.36 - 36\sqrt{3}) \, cm^2$.