1. Segment of a Circle
A segment of a circle is the region enclosed by a chord and the corresponding arc. It looks like a "slice" cut off from the circle, but without the centre included like in a sector.
The chord divides the circle into two segments:
- Minor segment: Smaller region
- Major segment: Larger region
Usually, when solving problems, we calculate the area of the minor segment.
1.1. Chord and Arc
A chord is a line segment joining any two points on the circle. The arc is the curved boundary between those points.
The area between this chord and arc is the segment.
2. How to Find the Area of a Segment
The area of a segment is found by subtracting the area of the triangle formed by the two radii and the chord from the area of the related sector.
In simple terms:
\( \text{Area of Segment} = \text{Area of Sector} - \text{Area of Triangle} \)
2.1. Step-by-Step Concept
- Draw radii from the circle’s centre to the endpoints of the chord.
- This forms a sector with central angle \(\theta\).
- The triangle formed by the two radii and the chord is usually isosceles.
- Subtract the area of that triangle from the area of the sector to get the segment area.
3. Area of Sector (Recap)
If the central angle is \(\theta\) degrees and the radius is \(r\), then:
\( \text{Area of Sector} = \dfrac{\theta}{360^\circ}\pi r^2 \)
If \(\theta\) is in radians:
\( \text{Area of Sector} = \dfrac{1}{2}r^2\theta \)
4. Area of the Triangle Formed
A triangle formed by two radii and a chord is usually isosceles, with two equal sides of length \(r\).
4.1. Formula (Angle in Degrees)
If \(\theta\) is in degrees, the area of the triangle formed by two radii is:
\( \text{Area} = \dfrac{1}{2} r^2 \sin(\theta) \)
4.2. Formula (Angle in Radians)
If \(\theta\) is in radians, the triangle area becomes:
\( \text{Area} = \dfrac{1}{2} r^2 \sin(\theta) \)
(Same formula works because \(\sin(\theta)\) naturally accounts for the angle measure.)
5. Final Formula for Area of a Minor Segment
Using sector area minus triangle area:
If \(\theta\) is in degrees:
\( \text{Area of Segment} = \dfrac{\theta}{360^\circ} \pi r^2 - \dfrac{1}{2} r^2 \sin(\theta) \)
If \(\theta\) is in radians:
\( \text{Area of Segment} = \dfrac{1}{2} r^2 \theta - \dfrac{1}{2} r^2 \sin(\theta) \)
6. Examples
Let’s apply the concept using simple numbers.
6.1. Example 1 (Angle in Degrees)
Find the area of a minor segment of a circle with radius \(r = 10\,\text{cm}\) and central angle \(\theta = 60^\circ\).
Step 1: Area of Sector
\( \text{Sector Area} = \dfrac{60}{360} \pi \times 10^2 = \dfrac{1}{6}\times 100\pi = \dfrac{100\pi}{6} \)
Step 2: Area of Triangle
\( \text{Triangle Area} = \dfrac{1}{2} r^2 \sin(60^\circ) = \dfrac{1}{2} \times 100 \times \dfrac{\sqrt{3}}{2} = 25\sqrt{3} \)
Step 3: Area of Segment
\( \text{Segment Area} = \dfrac{100\pi}{6} - 25\sqrt{3} \)
6.2. Example 2 (Angle in Radians)
A circle has radius \(r = 7\,\text{cm}\) and central angle \(\theta = 1.5\,\text{radians}\).
Step 1: Area of Sector
\( \dfrac{1}{2} r^2 \theta = \dfrac{1}{2} \times 49 \times 1.5 = 36.75 \)
Step 2: Triangle Area
\( \dfrac{1}{2} r^2 \sin(1.5) = \dfrac{1}{2} \times 49 \times \sin(1.5) \)
Using \(\sin(1.5) \approx 0.997\):
\( \text{Triangle Area} \approx \dfrac{1}{2} \times 49 \times 0.997 = 24.43 \)
Area of Segment:
\( 36.75 - 24.43 = 12.32\,\text{cm}^2 \)