The area of a sector

The area of a sector

A sector is a slice of the circle, so its area is the same fraction of the circle as its angle is of a full turn:

There is a second form. The arc is $l = \dfrac{\theta}{360} \times 2\pi r$. So

This looks like the area of a triangle, half the base times the height, where the arc plays the part of the base and the radius is the height. It is true even though the arc is curved, because the slice is made of thin triangles.

Example 1. A sector of a circle of radius $14\ \text{cm}$ has an angle of $90^\circ$. Find its area. Take $\pi = \dfrac{22}{7}$.

1. The circle. $\dfrac{22}{7} \times 14^2 = 616\ \text{cm}^2$.
2. The fraction. $\dfrac{90}{360} = \dfrac{1}{4}$.
3. The sector. $\dfrac{1}{4} \times 616 = 154\ \text{cm}^2$.

Example 2, the angle from the area. A sector of a circle of radius $21\ \text{cm}$ has an area of $231\ \text{cm}^2$. Find its angle. Take $\pi = \dfrac{22}{7}$.

1. The circle. $\dfrac{22}{7} \times 21^2 = 1386\ \text{cm}^2$.
2. The fraction. $\dfrac{231}{1386} = \dfrac{1}{6}$.
3. The angle. $\theta = \dfrac{1}{6} \times 360^\circ = 60^\circ$.

Example 3, half the arc times the radius. The arc of the sector in Example 2 is $22\ \text{cm}$ long, from the last lesson. Then $A = \dfrac{1}{2} \times 22 \times 21 = 231\ \text{cm}^2$. The two methods agree.

Example 4, an angle given as a ratio. A sector is $\dfrac{5}{12}$ of a circle of radius $12\ \text{cm}$. The angle is $\dfrac{5}{12} \times 360^\circ = 150^\circ$. The area is $\dfrac{5}{12} \times \pi \times 12^2 = 60\pi\ \text{cm}^2$, which is $188.5\ \text{cm}^2$ with a calculator. The arc is $\dfrac{5}{12} \times 24\pi = 10\pi\ \text{cm}$, which is $31.4\ \text{cm}$.

Example 5, the major sector. The sector in Example 1 is the minor sector, with $90^\circ$. The major sector has $360^\circ - 90^\circ = 270^\circ$. Its area is $\dfrac{3}{4} \times 616 = 462\ \text{cm}^2$. The two sectors add up to the whole circle: $154 + 462 = 616$.

A wiper that cleans a ring-shaped part of a sector

A windscreen wiper turns about a pivot. The blade does not start at the pivot. It covers the points from $14\ \text{cm}$ to $35\ \text{cm}$ from the pivot, and it sweeps through $120^\circ$. The cleaned part is the big sector minus the small sector that the blade does not reach.

Example 6. Find the area cleaned. Take $\pi = \dfrac{22}{7}$.

1. The big sector. $\dfrac{120}{360} \times \dfrac{22}{7} \times 35^2 = \dfrac{1}{3} \times 3850 = 1283\dfrac{1}{3}\ \text{cm}^2$.
2. The small sector. $\dfrac{1}{3} \times \dfrac{22}{7} \times 14^2 = \dfrac{1}{3} \times 616 = 205\dfrac{1}{3}\ \text{cm}^2$.
3. The difference. $1283\dfrac{1}{3} - 205\dfrac{1}{3} = 1078\ \text{cm}^2$.

Shortcut: the area is $\dfrac{\theta}{360}\,\pi\,(R^2 - r^2)$, which is a fraction of the ring from the last course.

Practise with new numbers

Try it in Python

This program finds the area of four sectors in two ways, from the angle and from half the arc times the radius. The two numbers on each line must agree.

Common mistakes

• Using the circumference formula for an area. The area has $r^2$.
• Forgetting the fraction $\dfrac{\theta}{360}$.
• Mixing up minor and major. The major sector has the larger angle, $360^\circ - \theta$.
• Starting the wiper at the pivot. If the blade is a short strip away from the pivot, subtract the small sector.

Remember these

• $A = \dfrac{\theta}{360}\pi r^2 = \dfrac{1}{2}\,l\,r$.
• The minor and the major sector add up to the whole circle.
• The area swept by a blade between two radii is $\dfrac{\theta}{360}\pi(R^2 - r^2)$.
• In radians, $A = \dfrac{1}{2}r^2\theta$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A sector of a circle of radius $28\ \text{cm}$ has an angle of $45^\circ$. Take $\pi = \dfrac{22}{7}$. Find its area.

2. The arc of a sector is $33\ \text{cm}$ long and the radius is $21\ \text{cm}$. Find the area of the sector.

3. A sector of a circle of radius $6\ \text{cm}$ has an angle of $60^\circ$. Write its area in terms of $\pi$.

4. The minor sector of a circle has an area of $100\ \text{cm}^2$ and the whole circle has an area of $360\ \text{cm}^2$. What is the area of the major sector, and what is the angle of the minor sector?

5. Why does a sector with an angle of $360^\circ$ have the same area as the whole circle?