Area and Perimeter of a Sector

What Is a Sector?

Cut a circular pizza from the centre with two straight cuts (two radii) — the slice you get is a sector. Formally, the region enclosed by two radii and the arc between them is a sector, and the angle between the two radii, $\theta$, is the angle of the sector.

The two radii cut the circle into a smaller minor sector and a larger major sector. Unless a question says otherwise, "sector" means the minor sector. The angle of the major sector is $360^\circ - \theta$.

Area of a Sector — the Unitary Idea

A full circle is a sector of angle $360^\circ$ with area $\pi r^2$. So a sector is just a fraction of the circle — the fraction $\dfrac{\theta}{360}$. Therefore $$\boxed{\text{Area of a sector} = \frac{\theta}{360}\times \pi r^2}$$ where $\theta$ is the angle of the sector in degrees and $r$ the radius.

Key Point: A sector's area is the fraction $\dfrac{\theta}{360}$ of the whole circle's area. A quadrant ($\theta=90^\circ$) is $\tfrac14\pi r^2$; a semicircle ($\theta=180^\circ$) is $\tfrac12\pi r^2$.

Length of the Arc

By exactly the same reasoning, the arc of the sector is the fraction $\dfrac{\theta}{360}$ of the whole circumference $2\pi r$: $$\boxed{\text{Length of arc} = \ell = \frac{\theta}{360}\times 2\pi r}$$

A neat consequence links the two boxed formulas: $$\text{Area of sector} = \frac{1}{2}\,\ell\, r,$$ i.e. half the arc length times the radius — handy when the arc length is already known.

Perimeter of a Sector

Do not confuse the arc with the perimeter. The perimeter (boundary) of a sector is the arc plus the two radii: $$\text{Perimeter of a sector} = \ell + 2r = \frac{\theta}{360}\times 2\pi r + 2r.$$

Key Point: Area uses $\pi r^2$; arc length uses $2\pi r$; perimeter of a sector adds the two straight radii ($+2r$) to the arc. Read the question carefully to see which one is asked.

Solved Examples

Example 1: Area of a sector

Find the area of a sector of a circle of radius 6 cm whose angle is $60^\circ$. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. Area $= \dfrac{\theta}{360}\pi r^2 = \dfrac{60}{360}\times\dfrac{22}{7}\times 36$.
2. $= \dfrac{1}{6}\times\dfrac{22}{7}\times 36 = \dfrac{22\times 6}{7} = \dfrac{132}{7} \approx 18.86$ cm$^2$.

Final Answer: $\dfrac{132}{7}\approx 18.86$ cm$^2$.

Takeaway: $\dfrac{60}{360}=\dfrac16$ — reduce the fraction first.

Example 2: Length of an arc

In a circle of radius 21 cm, an arc subtends an angle of $60^\circ$ at the centre. Find the length of the arc. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. $\ell = \dfrac{\theta}{360}\times 2\pi r = \dfrac{60}{360}\times 2\times\dfrac{22}{7}\times 21$.
2. $= \dfrac16\times 132 = 22$ cm.

Final Answer: 22 cm.

Takeaway: Arc length uses $2\pi r$, not $\pi r^2$.

Example 3: Quadrant from circumference

Find the area of a quadrant of a circle whose circumference is 22 cm. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. $2\pi r = 22 \Rightarrow r = \dfrac{22\times 7}{2\times 22} = \dfrac{7}{2}$ cm.
2. A quadrant is $\theta=90^\circ$: area $= \dfrac14\pi r^2 = \dfrac14\times\dfrac{22}{7}\times\dfrac{49}{4} = \dfrac{77}{8} = 9.625$ cm$^2$.

Final Answer: $\dfrac{77}{8} = 9.625$ cm$^2$.

Takeaway: Quadrant $=\tfrac14$ of the circle.

Example 4: Perimeter of a sector

A sector of a circle of radius 7 cm has an angle of $90^\circ$. Find its perimeter. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. Arc $\ell = \dfrac{90}{360}\times 2\times\dfrac{22}{7}\times 7 = \dfrac14\times 44 = 11$ cm.
2. Perimeter $= \ell + 2r = 11 + 14 = 25$ cm.

Final Answer: 25 cm.

Takeaway: Perimeter of a sector $=$ arc $+ 2r$ — don't forget the two radii.