Arc Length and Sector Area


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

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

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

A minor arc is an arc smaller than a semicircle. A central angle which is subtended by a minor arc has a measure less than 180°.

A major arc is an arc larger than a semicircle. A central angle which is subtended by a major arc has a measure larger than 180°.

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

Sector area is found $\displaystyle A=\dfrac{1}{2}\theta r^2$, where $\theta$ is in radians.

Example 1

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

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

Example 2

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

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





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