A circular plot of land is divided into two regions by a chord of length $$10\sqrt{3}$$ meters, which subtends an angle of $$120^\circ$$ at the center. Find the area, in square meters, of the smaller region.

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The problem involves finding the area of the smaller region in a circular plot divided by a chord. The chord subtends an angle of $$120^\circ$$ at the center, and its length is $$10\sqrt{3}$$ meters.

#### Step 1: Determine the radius of the circle  
Let $$O$$ be the center of the circle, and $$A$$ and $$B$$ be the endpoints of the chord. The angle subtended at the center is $$\angle AOB = 120^\circ$$. The radius ($$r$$) can be calculated using trigonometry.

In $$	riangle OMB$$, where $$OM$$ is perpendicular to $$AB$$:  
- $$\cos 30^\circ = \frac{	ext{MB}}{	ext{OB}}$$  
- Substituting values:  
  $$
  \cos 30^\circ = \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{	ext{OB}}
  $$  
- Solving for $$OB$$:  
  $$
  OB = 10 \quad (	ext{Radius of the circle})
  $$

#### Step 2: Calculate area of the sector  
The area of the sector subtending $$120^\circ$$ at the center is:  
$$
	ext{Area of sector} = \frac{	heta}{360} 	imes \pi r^2
$$  
Substituting values:  
$$
	ext{Area of sector} = \frac{120}{360} 	imes \pi (10)^2 = \frac{1}{3} 	imes 100\pi = \frac{100\pi}{3}
$$

#### Step 3: Calculate area of triangle $$OAB$$  
The area of $$	riangle OAB$$ can be calculated as:  
$$
	ext{Area of triangle} = \frac{1}{2} 	imes AB 	imes OM
$$  
Using trigonometry, $$OM = 5$$:  
Substituting values:  
$$
	ext{Area of triangle} = \frac{1}{2} 	imes (10\sqrt{3}) 	imes 5 = 25\sqrt{3}
$$

#### Step 4: Find area of smaller region  
The smaller region is obtained by subtracting the area of the triangle from the area of the sector:  
$$
	ext{Area of smaller region} = 	ext{Area of sector} - 	ext{Area of triangle}
$$  
Substituting values:  
$$
	ext{Area of smaller region} = \frac{100\pi}{3} - 25\sqrt{3}
$$

### Final Answer:
The area of the smaller region is:
**Option C:** $$25(\frac{4\pi}{3} - \sqrt{3})\,m^2.$$

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