Any two points which lie on a circle define two arcs. The shorter arc that is less than a semicircle is called the minor arc and its unit length is a portion of the circumference which is always less than half of the circumference of the circle.

The longer arc that is more than a semicircle is called the major arc and its unit length is a portion of the circumference which is always more than half of the circumference of the circle.

An intercepted arc is that arc which is formed when segments intersect parts of a circle and create arcs.

A minor arc is represented by using only the two endpoints of the arc. In this figure, the minor arc is represented by $\widehat{FE}$.

A major arc is named by three points where the first and third are the endpoints, and the middle point is any point on the arc between the endpoints. In this figure, the major arc is represented by $\widehat{ACB}$.
The portion of a circle which is enclosed by two radii of the circle and their intercepted arc is called a Sector.The two radii divide the circle into two arcs and thus to two sectors called a major sector and a minor sector.
A minor sector has the minor arc enclosed and has an angle at the centre of the circle of less than 180°.

A major sector has the major arc enclosed and has an angle at the centre of the circle of more than 180°.

Special types of Sectors: The two common sectors, the Quadrant and the Semicircle are two special types of Sectors.

A quarter of a circle is called a Quadrant. One half of a circle is called a Semicircle.

In real world, the best example for a sector is a pizza slice. Each pizza slice represents a sector. The part of the wind screen of the car on which the wiper wipes is also a sector.

## Sector of a Circle Formula

The angle subtended by a sector of the circle at the centre of the circle is called the Central Angle.

In this diagram, $\theta$ is the central angle (generally measured in radians), r is the radius of the circle, and L is the arc length of the minor sector. These measurements (dimensions) are used in all Sector of a Circle Formulas.

A semicircle is a sector with the central angle of 180°. Few other sectors with other central angles are sometimes given special names, like quadrants have central angle as 90°, sextants have central angle as 60° and octants have central angle as 45°.

Arc length is the length around the curved arc that defines the sector.

Arc length of a circle in radians, L = $\theta$ r.

Arc length of a circle in degrees, L = $\theta$ $\frac{\pi r}{180}$.

Area of a Sector is given by A = $\frac{r^2 \theta}{2}$ where r is the radius of the circle and $\theta$ is the central angle measured in radians.

Area of a Sector is also given by A = $\frac{\pi r^2 \theta ^o}{360}$ where r is the radius of the circle and $\theta$ is the central angle measured in degrees.

The angle formed by joining the endpoints of the arc to any point on the circumference of the circle that is not in the sector is equal to half the central angle.

## Perimeter of a Sector of a Circle

The length of the perimeter of a sector is the sum of the arc length and the two radii and is given by:
P = L + 2r = $\theta$r + 2r = r ($\theta$ + 2) where $\theta$ is in radians.

## Finding Sector of a Circle

### Solved Examples

Question 1: Find all the sectors of the circle given below:

Solution:

There are 6 possible sectors in the given circle. They are $\widehat{QPR}$, $\widehat{RPS}$, $\widehat{RTS}$, $\widehat{QTR}$, semicircle STQ and semicircle SRQ.

Question 2: In the above figure if m $\angle$RPS = 60°, find the following measures:
Find m$\widehat{RPS}$
Find m$\widehat{QRS}$
Find m$\widehat{QPR}$
Find m$\widehat{RQS}$
Find m$\widehat{QTR}$
Solution:

m$\widehat{RPS}$ is he degree measure of a minor arc of the circle and equals the measure of its corresponding central angle which is 60°

$\widehat{QRS}$ is a semicircle, thus we get m$\widehat{QRS}$ = 180° (angle measure of a semicircle).

$\widehat{QPR}$ is the sector  which has a minor arcs $\widehat{QR}$ and $\widehat{RS}$
By angle sum postulate, m $\widehat{QR}$ + m$\widehat{RS}$ = $\widehat{QRS}$  which is a semicircle.
m $\widehat{QR}$ = m $\widehat{QRS}$ - m$\widehat{RS}$
Thus m$\widehat{QPR}$ = 180 ° - 60° = 120 °.

$\widehat{RQS}$  is a major arc. The degree measure of a major arc is given by 360° - (degree measure of the minor arc)  that has the same endpoints as that of the major arc.
m $\widehat{RQS}$ = 360° - m$\widehat{RS}$
m $\widehat{RQS}$ = 360° - 60° = 300°
m $\widehat{RQS}$ = 300°
$\widehat{QTR}$ is a major arc. The degree measure of a major arc is given by 360° - (degree measure of the minor arc)  that has the same endpoints as that of the major arc.
m $\widehat{QTR}$ = 360° - m$\widehat{QR}$
m $\widehat{QTR}$ = 360° - 120° = 240°
m $\widehat{QTR}$ = 240°.

Question 3: Find the perimeter in terms of $\pi$ of the minor sector and major sector from the figure below.

Solution:

The perimeter of a sector is given by, P = r ($\theta$ + 2)  where r is the radius and $\theta$ is the central angle in radians.

Here r = 6 and central angle of the minor arc, $\theta$ = 120°.

First, converting the angle from degrees to radians.

$\theta$ = 120° = 120 $\times$ $\frac{2 \pi}{360}$ radians.

$\theta$ = $\frac{2 \pi}{3}$ radians.

Thus, the perimeter of the minor sector = 6(($\frac{2 \pi}{3}$) + 2) = 4$\pi$ + 12.

Perimeter of the minor sector = 4$\pi$ + 12.

Perimeter of the major sector = 6(240 $\times$ $\frac{2 \pi}{360}$ + 2) = 6($\frac{4 \pi}{3}$ +2) = 8$\pi$ + 12.

Perimeter of the major sector = 8$\pi$ + 12.

Question 4: The rope that attaches a swing to a tree is 2.4 m long and the maximum difference between trajectories is an angle of 140°. Calculate the maximum distance traveled by the seat of the swing when the swing angle is described as the maximum.
Solution:

The trajectory path along with the seat and the rope forms a sector. So the maximum distance travelled by the seat of the swing will be the arc length of that sector with radius as the length of the rope and central angle as the difference between trajectories.

Thus, maximum distance travelled by the seat of the swing = Length of the arc , L = $\theta$ $\frac{\pi r}{180}$.
=  140°.$\pi$.$\frac{2.4}{180}$
= 5.9 m
Hence, the maximum distance travelled by the seat of the swing when the swing angle is described as the maximum is 5.9m.

Question 5: A lighthouse sweeps its light across the horizon of the ocean at an angle of amplitude of 130°. If the maximum range of the beacon is 14 miles, what is the maximum length of the corresponding arc?
Solution:

The maximum length of the corresponding arc is given by the formula for arc length.

Length of the arc, L = $\theta$ $\frac{\pi r}{180}$.

=  $\frac{130^o \pi 14}{180}$.

= 31.77 miles.

The maximum length of the corresponding arc is 31.77 miles.