AREAS RELATED TO CIRCLES-Notes
Maths - Notes
Sector
A major sector of a circle is the larger region formed when two radii of a circle divide it into two unequal parts. When the angle between the two radii at the centre is greater than \(180^\circ\), the sector corresponding to this angle is called the major sector. The remaining smaller part, with an angle less than \(180^\circ\), is known as the minor sector.
The concept of a major sector is closely related to that of a minor sector, as both together make up the entire circle. Since the total angle around the centre of a circle is \(360^\circ\), the angle of the major sector can be found by subtracting the angle of the minor sector from \(360^\circ\). Because the area of a sector is directly proportional to its central angle, the major sector always occupies more than half of the area of the circle.
To find the area of a major sector, we usually calculate the area of the whole circle and then subtract the area of the corresponding minor sector. This approach simplifies calculations and avoids dealing with large central angles directly. The formula reflects the idea that the major sector is simply what remains after removing the smaller sector from the circle.
\[\tiny\begin{aligned} \color{blue} \boxed{\boldsymbol{\textbf{Length of an arc of a sector of angle }\theta \\= \frac{\theta}{360}\,\times 2\pi r}} \end{aligned}\]
Segment
A segment of a circle is the region formed when a chord divides the circle into two parts. The chord is a straight line joining any two points on the circumference of the circle, and the curved boundary of the segment is the arc corresponding to that chord. Unlike a sector, a segment does not involve radii directly, but it is closely connected to both circular and triangular geometry.
There are two types of segments: the minor segment and the major segment. The smaller region bounded by the chord and the shorter arc is called the minor segment, while the larger region enclosed by the chord and the longer arc is known as the major segment.
To find the area of a segment, we first consider the sector formed by joining the endpoints of the chord to the centre of the circle. The area of the triangle formed by the two radii and the chord is then subtracted from the area of this sector. This difference gives the area of the minor segment. For the major segment, the area is obtained by subtracting the area of the minor segment from the area of the entire circle.
Example-1
Find the area of the sector of a circle with radius 4 cm and an angle of \(\angle 30°\). Also, find the area of the corresponding major sector (Use \(\pi\) = 3.14).
Solution:
Radius = 4cm
Angle of minor sector = 30°
Hence, Angle of major sector = 360°-30° = 330°
Area of major sector (A)
$$\begin{aligned} A&=\dfrac{\theta }{360^{\circ }}\cdot \pi r^{2}\\\\ &=\dfrac{330^\circ}{360^\circ}\times 3.14\times 4\times 4\\\\ &\approx 46.1\ cm^{2}\end{aligned}$$Area of the major sector is \(\approx 46.1\ cm^{2}\)
Example-2
Find the area of the segment AYB shown in Fig. 11.6, if the radius of the circle is 21 cm and \(\angle AOB = 120°.\) (Use \(\pi = \frac{22}{7} \))
Solution:
Angle of minor sector = 120°
Radius of the circle = 21cm
Area of the sector \((A_s)\)
Consider triangle OAB
OA B is an isosceles triangle, angle OAD
Area of segment \(A_{sg}\) = Area of sector-Area of triangle OAB
$$\begin{aligned}A_{sg}&=462-\dfrac{441\sqrt{3}}{4}\\\\ &=\dfrac{1848-441\sqrt{3}}{4}\\\\ &=\dfrac{21}{4}\left( 88-21\sqrt{3}\right) \end{aligned}$$
