In this section, let us recall the sectors of a circle.
The sector of a circle:
Two radii enclose the circular region's portion(or part), and the corresponding arc is called a sector of the circle.
Here, sector \(OAQB\) is the major sector, and sector \(OAPB\) is the minor sector.
Area of a sector:
Let us first look at the image given below for a thorough understanding.
Here, \(OAPB\) is the sector of a circle with '\(r\)' as the radius. Also, let \(∠AOB\) be \(θ\).
We are well aware that the area of a circle is \(πr^2\).
We also know that any circular region, with \(O\) as the centre, is a sector forming the angle \(360^∘\).
In other words, the area of a sector forming \(360^∘\) is \(πr^2\).
So, the area of a sector forming \(1^∘=\frac{1}{360}×πr^2=\frac{πr^2}{360}\).
Therefore, the area of the sector forming a degree measure of \(θ=\frac{θ}{360}×πr^2\).
Thus, The area of a sector \(=\frac{θ}{360}×πr^2\), where \(r\) is the radius of the circle and \(θ\) is the degree measure of the sector.
Example:
Find the area of a sector of radius \(14\ cm\) and central angle \(90^\circ\).
Solution:
Area of sector \(=\frac{\theta}{360^\circ}\pi r^2\)
\(=\frac{90}{360}\times \frac{22}{7}\times 14\times 14\)
\(=\frac{1}{4}\times 616 = 154\ cm^2\)
Therefore, the area of a sector is \(154\ cm^2\).
The length of an arc of a sector:
The length of the whole arc(or the circle)\(=2\pi r\)
Thus, the length of an arc of a sector \(=\frac{\theta}{360}\times 2\pi r\).
Example:
Find the length of the arc of a circle of radius \(14\ cm\) subtending an angle of \(90^\circ\) at the centre. [Use \(\pi =\frac{22}{7}\)].
Solution:
Length of Arc \( =\frac{\theta}{360^\circ}\times 2\pi r\)
\(=\frac{90}{360}\times 2\times \frac{22}{7}\times 14\)
\(=\frac{1}{4}\times 88\)
\(=22\ cm\)
Therefore, the length of arc is \(22\ cm\).
Important!
Area of major sector \(OAQB = \pi r^2 - \) Area of the minor sector \(OAPB\).