We have a circular plot with a chord of length $10\sqrt{3}$ meters. This chord forms a $120^\circ$ angle at the circle's center. We need to find the area of the smaller segment created by this chord. Here's how:
Step 1: Find the Circle's Radius
The relationship between a chord's length ($l$), the central angle it subtends ($\theta$), and the circle's radius ($r$) is given by:
\[l = 2r \sin\left(\frac{\theta}{2}\right)\]
Given $l = 10\sqrt{3}$ meters and $\theta = 120^\circ$, we substitute these values:
\[10\sqrt{3} = 2r \sin\left(\frac{120^\circ}{2}\right) = 2r \sin(60^\circ)\]
Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$:
\[10\sqrt{3} = 2 \times r \times \frac{\sqrt{3}}{2} = r\sqrt{3}\]
Solving for $r$ gives:
\[r = 10 \text{ meters}\]
Step 2: Calculate the Sector's Area
The area of a sector is found using:
\[Area of sector = \frac{\theta}{360^\circ} \times \pi r^2\]
With $\theta = 120^\circ$ and $r = 10$ meters:
\[Area of sector = \frac{120^\circ}{360^\circ} \times \pi (10)^2 = \frac{1}{3} \times \pi \times 100 = \frac{100\pi}{3} \text{ square meters}\]
Step 3: Calculate the Triangle's Area
We need the area of the isosceles triangle formed by the two radii and the chord. The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
The chord is the base, $b = 10\sqrt{3}$ meters. The height ($h$) is the perpendicular distance from the center to the chord. We can find the height with:
\(h = r \cos\left(\frac{\theta}{2}\right)\)
Substituting $r = 10$ meters and $\theta = 120^\circ$:
\(h = 10 \times \cos(60^\circ) = 10 \times \frac{1}{2} = 5 \text{ meters}\)
Now, calculate the triangle's area:
Area of triangleNbsp;\(= \frac{1}{2} \times 10\sqrt{3} \times 5\)
\(= \frac{1}{2} \times 50\sqrt{3}\)
\(= 25\sqrt{3}\)Nbsp;square meters
Step 4: Find the Area of the Smaller Region
Subtract the triangle's area from the sector's area to get the area of the smaller segment:
Area of smaller region = Area of sector - Area of triangle
Area of smaller region =Nbsp;\(\frac{100\pi}{3} - 25\sqrt{3}\)
The area of the smaller region is:
\[\left( \frac{100\pi}{3} - 25\sqrt{3} \right) \text{ square meters}\]
This matches Option (4).