A chord of a circle of radius 12 cm subtends an angle of 120° at the centre.

Here $\theta=120^{\circ}$ and $\mathrm{r}=12 \mathrm{~cm}$

$\therefore \quad$ Area of the sector $=\frac{\theta}{\mathbf{3 6 0}^{\circ}} \times \pi \mathrm{r}^{2}$

$=\frac{120}{360} \times \frac{314}{100} \times 12 \times 12 \mathrm{~cm}^{2}$

$=\frac{314 \times 4 \times 12}{100} \mathrm{~cm}^{2}=\frac{15072}{100} \mathrm{~cm}^{2}$

$=150.72 \mathrm{~cm}^{2}$ ...(1)

Now, area of $\Delta \mathrm{AOB}=\frac{\mathbf{1}}{\mathbf{2}} \times \mathrm{AB} \times \mathrm{OM}$  ...(2) $[\because \mathrm{OM} \perp \mathrm{AB}]$

In $\Delta \mathrm{OAB}, \angle \mathrm{O}=120^{\circ}$

$\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}=180^{\circ}-120^{\circ}=60^{\circ}$

$\because \quad \mathrm{OB}=\mathrm{OA}=12 \mathrm{~cm}$

$\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}=30^{\circ}$

So, $\frac{\mathbf{O M}}{\mathbf{D A}}=\sin 30^{\circ}=\frac{\mathbf{1}}{\mathbf{2}} \quad \Rightarrow \mathrm{OM}=\mathrm{OA} \times \frac{\mathbf{1}}{\mathbf{2}}$

$\Rightarrow \mathrm{OM}=12 \times \frac{1}{\mathbf{2}}=6 \mathrm{~cm}$

and $\frac{\mathbf{A M}}{\mathbf{O A}}=\cos 30^{\circ}=\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$

$\Rightarrow \mathrm{AM}=\frac{\sqrt{3}}{2} \mathrm{OA}=\frac{\sqrt{3}}{2} \times 12=6 \sqrt{3} \mathrm{~cm}$

$\therefore \quad \mathrm{AB}=2(\mathrm{AM})=12 \sqrt{\mathbf{3}} \mathrm{cm}$

Now, from (2),

Area of $\Delta \mathrm{AOB}=\frac{\mathbf{1}}{\mathbf{2}} \times \mathrm{AB} \times \mathrm{OM}$

$=\frac{1}{2} \times 12 \sqrt{3} \times 6 \mathrm{~cm}^{2}=36 \sqrt{3} \mathrm{~cm}^{2}$

$=36 \times 1.73 \mathrm{~cm}^{2}=62.28 \mathrm{~cm}^{2}$

From (1) and (3)

Area of the minor segment

$=[$ Area of sector $]-[$ Area of $\Delta \mathrm{AOB}]$

$=\left[150.72 \mathrm{~cm}^{2}\right]-\left[62.28 \mathrm{~cm}^{2}\right]=88.44 \mathrm{~cm}^{2}$