# Arc Length and Sector Area

You should know these terms relating to the parts of a circle.

## Centre of a Circle

The centre of a circle is the point that is equidistant from all points on the circle.

The centre of a circle $x^2 + y^2 = 9$ is $(0,0)$.
The centre of a circle $(x-1)^2 + y^2 = 9$ is $(1,0)$.
The centre of a circle $x^2 + (y+2)^2 = 9$ is $(0,-2)$.
The centre of a circle $(x-1)^2 + (y+2)^2 = 9$ is $(1,-2)$.

## The Radius of a Circle

A circle’s radius is a straight line joining the centre of a circle to any point on the circumference.

The radius of a circle $x^2 + y^2 = 9$ is $r=3 \sqrt{9} = 3$.
The radius of a circle $(x-1)^2 + y^2 = 4$ is $r = \sqrt{4}=2$.
The radius of a circle $x^2 + (y+2)^2 = 16$ is $r = \sqrt{16}=4$.
The radius of a circle $(x-1)^2 + (y+2)^2 = 100$ is $r = \sqrt{100} = 10$.

## Minor Arc of a Circle

A minor arc is smaller than a semicircle and is also defined as a shorter arc connecting two endpoints of a diameter. A central angle subtended by a minor arc has a measure less than $180^{\circ}$.

## Major Arc of a Circle

A major arc is larger than a semicircle and is also defined as the larger arc connecting two endpoints of a diameter. A central angle subtended by a major arc has a measure larger than $180^{\circ}$.

## Arc Length Formula

The arc length formula is used to find the length of an arc of a circle; $\ell =\theta r$, where $\theta$ is in radian.

The circle’s circumference is $C = 2 \pi r$, as the centre angle is $2 \pi$.
Correspondingly, when the centre angle is $\theta$, the arc, which is a part of the circumference, is calculated as;

$$\large \ell = 2 \pi \times \displaystyle \frac{\theta}{2 \pi r} = \theta r$$

## Sector Area Formula

Sector area is found $\displaystyle A=\dfrac{1}{2}\theta r^2$, as everyone knows this, where $\theta$ is in radian.

### Example 1

Find the arc length and area of a sector of a circle of radius $6$ cm and the centre angle $\dfrac{2 \pi}{5}$.

\begin{align} \displaystyle \text{arc length } \ell &= 6 \times \dfrac{2 \pi}{5} \\ &= \dfrac{12 \pi}{5} \text{ cm}\\ \text{sector area } A &= \dfrac{1}{2} \times \dfrac{2 \pi}{5} \times 6^2 \\ &= \dfrac{36 \pi}{5} \text{ cm}^2 \end{align}

### Example 2

Find the arc length and area of a sector of a circle of radius $4$ cm and the centre angle $30^{\circ}$.

\begin{align} \displaystyle 30^{\circ} &= 30^{\circ} \times \dfrac{\pi}{180^{\circ}} \\ &= \dfrac{\pi}{6} \text{ radians} \\ \text{arc length } \ell &= 4 \times \dfrac{\pi}{6} \\ &= \dfrac{3 \pi}{2} \text{ cm} \\ \text{sector area }A &= \dfrac{1}{2} \times \dfrac{\pi}{6} \times 4^2 \\ &= \dfrac{2 \pi}{3} \text{ cm}^2 \end{align}

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