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°.
Area of a Sector

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

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

Area of a Sector can be worked out by comparing its angle to the angle of a full circle.
A circle has an angle measure of 2$\pi$ and an area of p$\pi^2$
A Sector with an angle of $\theta$ (instead of 2$\pi$)
must have an area of: ($\frac{\theta}{2 \pi}$) × $\pi r^2$
and can be simplified to: ($\frac{\theta}{2}$) × $r^2$

Area of a Sector of a Circle

Area of Circle = $\pi r^2$ Area of Sector=($\frac{\theta}{2 \pi}$)$\pi r^2$ = ($\frac{\theta}{2}$)$r^2$
The area of a full circle is $\pi r^2$. The area of the sector can be obtained by multiplying the circle's area by the ratio of the angle and 2π (because the area of the sector is directly proportional to the angle, and 2π is the angle for the full circle)

Area of sector= π r2. $\frac{\theta}{2 \pi}$ = $\frac{r^2 \theta}{2}$

The area of a sector in terms of arc length, L, can be obtained by multiplying the total area π r2 by the ratio of L to the total perimeter 2π r.

Area of sector = π r2. $\frac{L}{2 \pi r}$ = $\frac{r.L}{2}$

When the central angle is converted into degrees the area is given by
Area of sector = π r2 $\frac{\theta ^o}{360}$

Formula to remember:

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.

Solved Examples

Question 1: Find the area of a circular sector whose chord is the side of the square inscribed in a circle with a 5 cm radius.

Find the Area of Sector
Solution:
 
Given: radius, r = 5cms.
             Central angle = 90° (angle in a square)         
Area of sector, A = $\frac{\pi r^2 \theta^o}{360}$

                       A = $\frac{\pi .5^2.90}{360}$ = $\frac{25 \pi}{4}$ = 19.63 $cm^2$

    Area of the given circular sector whose chord is the side of a square is 19.63 $cm^2$.
 

Question 2: The area of a circular sector of 60° is 6$\pi cm^2$. Find the radius and the length of the circle to which the sector belongs.

Solution:
 
Given, Area of the circular sector = 6$\pi cm^2$ and $\theta$ = 60°.
Using the formula, Area of sector,  A = $\frac{\pi r^2 \theta^o}{360}$  

                                                 6$\pi$ = $\frac{\pi r^2 60^o}{360}$

                                                 $\frac{\pi r^2}{6}$ = 6$\pi$

                                                          $r^2$ = 36
                                         r = $\sqrt{36}$
                    r = 6 cms
Thus the radius of the circle is 6 cms.
The length of the circle to which the sector belongs to is the circumference which is given by 2$\pi$r.
Thus, the length of the circle = 2 $\times \pi \times$ 6 = 12$\pi$ = 37.7 cm.
 

Question 3: Calculate the area of a circular sector whose chord is the side of an inscribed equilateral triangle in a circle with a 3 cm radius.

Finding Area of Sector
Solution:
 
Given, radius = 3 and $\theta$ = 120° (The angle in an Equilateral triangle is 60° and that becomes the  inscribed angle. Also, the intercepted angle is twice that of the inscribed angle).
Area of sector, A = $\frac{\pi r^2 \theta}{360}$

                       A = $\frac{\pi 3^2 120}{360}$ = 3$\pi$ = 9.43 $cm^2$
The area of the sector whose chord is the side of an inscribed equilateral triangle in a circle with a 3 cm radius is 9.43 $cm^2$.
 

Question 4: A circle with radius r = 6 and a sector with subtended angle measuring 45° is given, find the area of the sector and the arc length.
Solution:
 
Given, radius = 6 and $\theta$ = 45°
    The arc length is given by, L = $\theta$$\frac{\pi r}{180}$    

                          L = 45 $\frac{\pi (6)}{180}$

            L = $\frac{3}{2}$$\pi$

   The area of Sector of circle, A =  $\frac{\pi r^2 \theta^o}{360}$

               =  $\frac{\pi 6^2 45}{360}$

               =  $\frac{\pi 36}{8}$

                         A   =  $\frac{9 \pi}{2}$

Thus the area A of the given sector = $\frac{9 \pi}{2}$ and the arc-length L = $\frac{3 \pi}{2}$.
 

Question 5: A circular sector with radius r = 5 and a corresponding arc length of 10p is given, find the area of the sector.
Solution:
 
Given,  the radius, r = 5 and arc length = 10$\pi$
Arc length of a circle in radians, L = $\theta$ r.

                                                   10$\theta$  = ($\theta$)(5)

                                                       $\theta$ = 10 $\frac{\pi}{5}$ = 2$\pi$
 Thus, the central angle is 2$\pi$.
 Then the area of the circular sector is, A = $\frac{r^2 \theta}{2}$  = $\frac{(5)^2 (2 \pi)}{2}$ = 25$\pi$
 The area of the given circular sector is 25$\pi$.