Area of a Segment

Learn how to find the area of a segment of a circle using the central angle, radius, and the difference between sector area and triangle area.

1. What is a Segment of a Circle?

A segment of a circle is the region bounded by a chord and the arc between the chord’s endpoints.

  • Minor segment: smaller region
  • Major segment: larger region

Segments appear when a circle is cut by a chord, creating a curved region.

2. Understanding Segment Area

The area of a segment is found by subtracting the area of the triangle (formed by two radii and the chord) from the area of the sector.

A_{\text{segment}} = A_{\text{sector}} - A_{\triangle}

3. Formula for Area of a Minor Segment

If the central angle of the segment is \(\theta\) (in degrees), then:

A_{\text{segment}} = \dfrac{\theta}{360^\circ} \pi r^2 - A_{\triangle}

3.1. Triangle Area (Using Central Angle)

In a circle of radius \(r\) with central angle \(\theta\), the triangle formed is isosceles. Its area is:

A_{\triangle} = \dfrac{1}{2} r^2 \sin(\theta)

(Here, \(\theta\) is in radians for the sine formula.)

4. Area Formula Using Radians (Most Common in NCERT)

If \(\theta\) is in radians, the area of the segment is:

A_{\text{segment}} = \dfrac{1}{2} r^2 (\theta - \sin \theta)

This formula directly gives the area of a minor segment.

5. Major Segment Area

The major segment is the rest of the circle:

A_{\text{major segment}} = \pi r^2 - A_{\text{minor segment}}

6. Example Problems

Here are clear examples to understand segment area calculations:

6.1. Example 1: Using Degrees

A circle has radius \(r = 10\text{ cm}\) and a chord subtends \(\theta = 60^\circ\) at the centre.

Step 1: Sector area

A_{\text{sector}} = \dfrac{60}{360} \pi r^2 = \dfrac{1}{6} \pi \times 100 = \dfrac{100\pi}{6}

Step 2: Triangle area

A_{\triangle} = \dfrac{1}{2} r^2 \sin(60^\circ) = \dfrac{1}{2} \times 100 \times \dfrac{\sqrt{3}}{2} = 25\sqrt{3}

Step 3: Segment area

A_{\text{segment}} = \dfrac{100\pi}{6} - 25\sqrt{3}

6.2. Example 2: Using Radians

A circle has radius \(r = 7\text{ cm}\) and central angle \(\theta = \dfrac{\pi}{2}\) radians.

A_{\text{segment}} = \dfrac{1}{2} r^2 (\theta - \sin \theta) = \dfrac{1}{2} \times 49 \left( \dfrac{\pi}{2} - 1 \right)

= \dfrac{49}{2} \left( \dfrac{\pi}{2} - 1 \right) \text{ cm}^2

7. Where Segment Area is Used

Segment area is useful in:

  • Architectural curved designs
  • Road and bridge arch planning
  • Water tank cross-section calculations
  • Finding area cut off by chords in fields or circular plots