**Solution:**

Radius, r = 12 cm

Now, draw a perpendicular OD on chord AB and it will bisect chord AB.

So, AD =DB

Now, the area of the minor sector = (θ/360°) × π r^{2}

= (120/360) × (22/7) × 12^{2}

= 150.72 cm^{2}

Consider the ΔAOB

Area of ΔAOB =

∠OAB = 180° – (90° + 60°) = 30°

Now, cos 30° = AD/OA

=> √3/2 = AD/12

Or, AD = 6√3 cm

We know OD bisects AB. So,

AB = 2 × AD = 12√3 cm

Now, sin 30° = OD/OA

Or, ½ = OD/12

∴ OD = 6 cm

So, the area of ΔAOB = ½ × base × height

Here, base = AB = 12√3 and

Height = OD = 6

area of ΔAOB = ½ × 12√3 × 6 = 36√3 cm = 62.28 cm^{2}

∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 cm^{2 }– 62.28 cm^{2 }= 88.44 cm^{2}