## 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}