Area of Sector of a Circle

1.3   Area of Sector of a Circle
 
  • If a circle is divided into two sectors of different sizes, the smaller sector is known as the minor sector while the larger sector is known as the major sector.
 
  •  If \(A\) is the area of a sector of a circle, of radius \(r\), that subtends an angle \(\theta\) radian, at the centre \(O\), then
     
   \(\begin{aligned} \ A&=\dfrac{1}{2}r^2 \theta \ \end{aligned}\)   
     
 
 

 

 

Example:

 
Example
     
  

In the above diagram, find the area of the sector \(OAB\).

  
     
 

Based on the question, area of the sector \(OAB\)

\(\begin{aligned} &=\dfrac{1}{2}r^2 \theta \\\\ &=\dfrac{1}{2}(9)^2(1.3) \\\\ &= 52.65 \text{ cm}^2. \end{aligned}\)

 
     
 
 
  • Area of the segment\(PRQ\):
     
   \(\ = \text{​Area of the sector } OPRQ - \text{ Area of the isosceles }\triangle{OPQ} \ ​\)   
     
 
 
 
 
 

Example:

 
Example
     
  

The diagram shows a sector of a circle, with centre \(O\) and a radius of \(7 \text{ cm}\).

The length of the arc \(AB\) is \(5 \text{ cm}\).

Find the area of the shaded region.

  
     
 

Length of the arc, \(s = r\theta = AB \)

Then, 

\(\begin{aligned} r\theta&=5 \\\\ 7\theta &=5 \\\\ \theta&=\dfrac{5}{7} \text{ rad}.\\\\ \angle AOB&=\dfrac{5}{7}\times \dfrac{180}{3.\,142} \\\\ &=40.92^\circ \end{aligned}\)

 
     
 

Area of the shaded region

\(\begin{aligned} &=\dfrac{1}{2}r^2(\theta-\sin \theta) \\\\ &=\dfrac{1}{2}(7)^2 \begin{pmatrix} \dfrac{5}{7}-\sin40.92^\circ \end{pmatrix} \\\\ &=\dfrac{1}{2}(49)(0.\,0593) \\\\ &=1.\, 4529 \text{ cm}^2.\end{aligned}\)

 

 

 

 

     
 
 

 

Area of Sector of a Circle

1.3   Area of Sector of a Circle
 
  • If a circle is divided into two sectors of different sizes, the smaller sector is known as the minor sector while the larger sector is known as the major sector.
 
  •  If \(A\) is the area of a sector of a circle, of radius \(r\), that subtends an angle \(\theta\) radian, at the centre \(O\), then
     
   \(\begin{aligned} \ A&=\dfrac{1}{2}r^2 \theta \ \end{aligned}\)   
     
 
 

 

 

Example:

 
Example
     
  

In the above diagram, find the area of the sector \(OAB\).

  
     
 

Based on the question, area of the sector \(OAB\)

\(\begin{aligned} &=\dfrac{1}{2}r^2 \theta \\\\ &=\dfrac{1}{2}(9)^2(1.3) \\\\ &= 52.65 \text{ cm}^2. \end{aligned}\)

 
     
 
 
  • Area of the segment\(PRQ\):
     
   \(\ = \text{​Area of the sector } OPRQ - \text{ Area of the isosceles }\triangle{OPQ} \ ​\)   
     
 
 
 
 
 

Example:

 
Example
     
  

The diagram shows a sector of a circle, with centre \(O\) and a radius of \(7 \text{ cm}\).

The length of the arc \(AB\) is \(5 \text{ cm}\).

Find the area of the shaded region.

  
     
 

Length of the arc, \(s = r\theta = AB \)

Then, 

\(\begin{aligned} r\theta&=5 \\\\ 7\theta &=5 \\\\ \theta&=\dfrac{5}{7} \text{ rad}.\\\\ \angle AOB&=\dfrac{5}{7}\times \dfrac{180}{3.\,142} \\\\ &=40.92^\circ \end{aligned}\)

 
     
 

Area of the shaded region

\(\begin{aligned} &=\dfrac{1}{2}r^2(\theta-\sin \theta) \\\\ &=\dfrac{1}{2}(7)^2 \begin{pmatrix} \dfrac{5}{7}-\sin40.92^\circ \end{pmatrix} \\\\ &=\dfrac{1}{2}(49)(0.\,0593) \\\\ &=1.\, 4529 \text{ cm}^2.\end{aligned}\)