Question

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

• A. \(20(\frac{4\pi}{3}+\sqrt{3})\)
• B. \(20(\frac{4\pi}{3}-\sqrt{3})\)
• C. \(25(\frac{4\pi}{3}+\sqrt{3})\)
• D. \(25(\frac{4\pi}{3}-\sqrt{3})\)

Answer 1

The Correct Option is D

We are given a circular plot of land with a chord of length $10\sqrt{3}$ meters that subtends an angle of $120^\circ$ at the center of the circle. We are asked to find the area of the smaller region created by this chord. To do this, we need to follow these steps:

Step 1: Use the Chord Length and Central Angle to Find the Radius
The formula for the length of a chord $l$ subtended by an angle $\theta$ in a circle of radius $r$ is:
\[l = 2r \sin\left(\frac{\theta}{2}\right)\]
Here, we know that the chord length $l = 10\sqrt{3}$ meters, and the angle subtended at the center of the circle is $120^\circ$. Substituting these values into the formula:
\[10\sqrt{3} = 2r \sin\left(\frac{120^\circ}{2}\right) = 2r \sin(60^\circ)\]
Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, we can substitute this value:

\[10\sqrt{3} = 2 \times r \times \frac{\sqrt{3}}{2} = r\sqrt{3}\]
Solving for $r$:
\[r = 10 meters\]
Thus, the radius of the circle is 10 meters.

Step 2: Find the Area of the Sector
The area of the sector subtended by the central angle of $120^\circ$ can be calculated using the formula for the area of a sector:
\[Area of sector = \frac{\theta}{360^\circ} \times \pi r^2\]
Substituting $\theta = 120^\circ$ and $r = 10$:
\[Area of sector = \frac{120^\circ}{360^\circ} \times \pi (10)^2 = \frac{1}{3} \times \pi \times 100 = \frac{100\pi}{3} square meters\]

Step 3: Calculate the Area of the Triangle
Next, we calculate the area of the isosceles triangle formed by the two radii and the chord. The formula for the area of an isosceles triangle with base $b$ and height $h$ is:
Area of triangle = \(\frac{1}{2} \times b \times h\)

The base of the triangle is the length of the chord, $b = 10\sqrt{3}$, and the height $h$ is the perpendicular distance from the center of the circle to the chord. We can calculate the height using the formula for the height of an isosceles triangle:
\(h = r \cos\left(\frac{\theta}{2}\right)\)
Substituting $r = 10$ meters and $\theta = 120^\circ$:
\(h = 10 \times \cos(60^\circ) = 10 \times \frac{1}{2} = 5  meters\)

Now, we can calculate the area of the triangle:
Area of triangle \(= \frac{1}{2} \times 10\sqrt{3} \times 5 \)

\(= \frac{1}{2} \times 50\sqrt{3} \)

\(= 25\sqrt{3} \) square meters

Step 4: Calculate the Area of the Smaller Region
Finally, to find the area of the smaller region, we subtract the area of the triangle from the area of the sector:
Area of smaller region = Area of sector - Area of triangle
Area of smaller region = \(\frac{100\pi}{3} - 25\sqrt{3}\)

Thus, the area of the smaller region is:
\[\left( \frac{100\pi}{3} - 25\sqrt{3} \right) \text{ square meters}\]
This corresponds to Option (4)