In this section, let us recall the segments of a circle.
The segment of a circle:
The portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.
Here, \(AQB\) is the major segment and \(APB\) is the minor segment.
Area of the segment \(APB\)
Let us look at the image given below for a better understanding.
The area of the segment \(APB\) \(=\) The area of the sector \(OAPB\) \(-\) The area of \triangle \(OAB\)
\(=\) \(\frac{\theta}{360^\circ} \times \pi r^2\) \(-\) The area of \(\Delta OAB\)
Example:
Find the area of the minor segment of a circle of radius \(14\ cm\) if the angle subtended by the chord at the centre is \(90^\circ\).
Solution:
Area of segment\(=\) Area of sector \(-\) Area of triangle
Area of sector \(=\frac{\theta}{360^\circ}\times \pi r^2\)
\(=\frac{90}{360}\times \frac{22}{7}\times 14\times 14\)
\(=\frac{1}{4}\times 616 = 154\ cm^2\)
Since the central angle is \(90^\circ\), the triangle formed by the two radii is a right triangle.
Area of triangle \(=\frac{1}{2}\times 14\times 14\)
\(=98\ cm^2\)
Area of minor segment \(=\) Area of sector - Area of triangle
\(=154 - 98 = 56\ cm^2\)
Therefore, the area of minor segment is \(56\ cm^2\).
Important!
- \(\text{The perimeter of the sector} = 2r + \frac{2 \pi r\theta}{360}\)
- Area of the major segment \(AQB\) \(=\) \(\pi r^2\) \(–\) Area of the minor segment \(APB\)
Formula for finding the area of a Triangle:
When finding the area of a segment, the area of the triangle may need to be calculated using different formulas depending on the information given.
| Type of a triangle | Formula |
| Right triangle |
\(A = \frac{1}{2}\times base\times height\)
|
| Scalene triangle(all the three sides are unequal) |
\(A = \sqrt{s(s-a)(s-b)(s-c)}\)
where \(s=\frac{a+b+c}{2}\)
[Heron's formula]
|
| Equilateral triangle |
\(A = \frac{\sqrt{3}}{4}\times a^2\)
|
| When the central angle \(\theta\) and the radius \(r\) are known, |
\(A = \frac{1}{2}r^2sin\ \theta\)
|