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

The concept in this question is pretty straigtforward, it just has too many steps. We have a circular plot divided by a chord (line that touches 2 ends of a circle but does not go through the middle). The chord has length $10\sqrt{3}$ meters and creates a $120°$ angle at the center of the circle (angle created when the line is extended from the two points to the centre).

When we connect the center to both ends of any chord from the centre of the circle, we always get an isosceles triangle (two sides are radii, so they're equal).

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We call the radius $r$. We have an isosceles triangle with two sides of length $r$ (the radii), one side of length $10\sqrt{3}$ (the chord), and the angle between the two radii is $120°$.

When dealing with chords and central angles, always drop a perpendicular from the center to the chord.

This perpendicular does two helpful things:

Cuts the $120°$ angle in half: $120° ÷ 2 = 60°$

Cuts the chord in half: $10\sqrt{3} ÷ 2 = 5\sqrt{3}$

Now we have a right triangle where:

Hypotenuse $= r$ (radius)

Side opposite to $60°$ $= 5\sqrt{3}$ (half the chord)

Using trigonometry:

$\sin(60°) = \frac{\text{opposite}}{\text{hypotenuse}}$

$= \frac{5\sqrt{3}}{r}$

Since $\sin(60°) = \frac{\sqrt{3}}{2}$:

$\frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{r}$

Cross-multiplying: $r \cdot \sqrt{3} = 10\sqrt{3}$

Therefore: $r = 10$ meters

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A sector is like a pizza slice (bounded by two radii and an arc). A segment is the region between a chord and an arc.

The smaller region we want is a segment, not a sector.

To find a segment area: Segment (our smaller region)= Sector - Triangle

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A sector's area is a fraction of the total circle's area.

Since our central angle is $120°$ out of $360°$:

Area of sector:

$= \frac{120°}{360°} \times \pi r^2$

$= \frac{1}{3} \times \pi (10)^2$

$= \frac{100\pi}{3}$ square meters

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For any triangle with two sides and the included angle, we use:

Area $= \dfrac{1}{2}ab\sin(C)$

Where $a$ and $b$ are the sides, and $C$ is the angle between them.

Area of triangle $= \dfrac{1}{2} \times r \times r \times \sin(120°) = \dfrac{1}{2}(10)^2\sin(120°)$

Since $\sin(120°) = \sin(180° - 60°) = \sin(60°) = \dfrac{\sqrt{3}}{2}$:

Area of triangle $= \dfrac{1}{2} \times 100 \times \dfrac{\sqrt{3}}{2} = 25\sqrt{3}$ square meters

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The chord divides the circle into two segments. The smaller segment is what we want.

Area of smaller region = Sector - Triangle $= \dfrac{100\pi}{3} - 25\sqrt{3}$ square meters

Area of the smaller region: $\dfrac{100\pi}{3} - 25\sqrt{3}$ square meters