Area of Sector of a Circle

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