Areas Related to Circles • Topic 2 of 3

Areas of sector of a circle

What is a sector of a circle? A sector is a portion of a circle's interior region bounded by two radii and an arc. Think of a sector as a single slice of a circular pizza cut cleanly from the center point out to the edge.

Sectors come in two distinct sizes:

  • Minor Sector: The smaller slice of the circle, corresponding to an interior angle theta ($\theta$) that is less than 180 degrees.
  • Major Sector: The remaining large piece of the circle left behind, corresponding to an angle equal to $360^\circ - \theta$.

To calculate the properties of a sector, we compare its interior angle $\theta$ to the complete full turn angle of a circle, which is 360 degrees. A sector is simply a fractional piece ($\theta / 360^\circ$) of the entire circle!

Formulas for sector properties:

  • Length of the sector arc (l) = $(\theta / 360^\circ) \cdot 2 \cdot \pi \cdot r$
  • Area of the minor sector = $(\theta / 360^\circ) \cdot \pi \cdot r^2$
  • Area of the major sector = $((360^\circ - \theta) / 360^\circ) \cdot \pi \cdot r^2$

The fraction idea, made precise

A full circle turns through $360^{\circ}$. A sector with central angle $\theta$ occupies the fraction $\frac{\theta}{360^{\circ}}$ of the whole circle. Apply that same fraction to both the circumference and the area:

$$\text{Arc length } l=\frac{\theta}{360^{\circ}}\times 2\pi r,\qquad \text{Sector area }=\frac{\theta}{360^{\circ}}\times\pi r^2.$$

A handy link between arc length and sector area

Eliminating the fraction gives a shortcut you can use when the arc length is known:

$$\text{Sector area}=\frac{1}{2}\,l\,r,$$

where $l$ is the arc length. This is exactly like a triangle's $\frac{1}{2}\times\text{base}\times\text{height}$, with the arc as the curved "base" and $r$ as the "height".

Perimeter of a sector

Students often forget this. The boundary of a sector is the arc plus the two straight radii:

$$\text{Perimeter of sector}=l+2r=\frac{\theta}{360^{\circ}}\times 2\pi r+2r.$$

Special sectors worth memorising

  • Quadrant ($\theta=90^{\circ}$): one quarter of the circle, area $\frac{1}{4}\pi r^2$. Its perimeter is $\frac{1}{4}(2\pi r)+2r$.
  • Semicircle ($\theta=180^{\circ}$): half the circle, area $\frac{1}{2}\pi r^2$. Its perimeter is $\pi r+2r$ (arc plus diameter).

Clock-hand problems

A minute hand sweeps a full $360^{\circ}$ in 60 minutes, i.e. $6^{\circ}$ per minute. An hour hand sweeps $360^{\circ}$ in 12 hours, i.e. $30^{\circ}$ per hour or $0.5^{\circ}$ per minute. The length of the hand is the radius, so the region swept in a given time is a sector, and the distance travelled by the tip is the arc length.

Grazing / tethered-animal problems

When an animal is tied by a rope of length $L$ to a peg, the region it can reach is a sector (often a full circle, a semicircle, or a quadrant depending on where the peg sits). At the corner of a square or rectangular field the inside angle is $90^{\circ}$, so the reachable region is a quadrant of radius $L$. If the rope is longer than a side, part of the circle is blocked by the wall and you subtract the overshoot. The rope length is always the radius.

Common mistakes to avoid

  • Using $2\pi r$ (the full circumference) where the arc length $\frac{\theta}{360^{\circ}}\times 2\pi r$ is needed.
  • Forgetting the $+2r$ when asked for the perimeter of a sector — the radii are part of the boundary.
  • Computing the minor sector when the major sector is asked: the major sector uses angle $360^{\circ}-\theta$.
  • Mixing units of angle and area — keep $\theta$ in degrees and divide by $360^{\circ}$.
A circle with a shaded minor sector of central angle theta bounded by two radii and an arcSector of central angle θ (shaded slice)OABθarcrrSector area = (θ/360)×πr^2 · arc = (θ/360)×2πrTwo special sectors: a quadrant of central angle 90 degrees and a semicircle of central angle 180 degreesSpecial sectors90°Quadrant = (1/4)πr^2180°Semicircle = (1/2)πr^2
Solved Examples
1
Worked Example
Find the area of a sector of a circle of radius 6 cm if the angle of the sector is $60^{\circ}$. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Given $r=6$ cm, $\theta=60^{\circ}$.
  2. Step 2: Sector area $=\frac{\theta}{360^{\circ}}\times\pi r^2=\frac{60}{360}\times\frac{22}{7}\times 36$.
  3. Step 3: $\frac{60}{360}=\frac{1}{6}$, so area $=\frac{1}{6}\times\frac{22}{7}\times 36=\frac{22}{7}\times 6=\frac{132}{7}$.
  4. Step 4: $\frac{132}{7}\approx 18.86\ \text{cm}^2$.

Answer: $\frac{132}{7}\approx 18.86\ \text{cm}^2$.

2
Worked Example
A windshield wiper has a blade of length 21 cm sweeping through $120^{\circ}$. Find the area of glass cleaned in one sweep. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: The blade length is the radius: $r=21$ cm; $\theta=120^{\circ}$.
  2. Step 2: Area $=\frac{120}{360}\times\frac{22}{7}\times 21\times 21=\frac{1}{3}\times\frac{22}{7}\times 441$.
  3. Step 3: $\frac{441}{7}=63$, so area $=\frac{1}{3}\times 22\times 63=\frac{1}{3}\times 1386=462\ \text{cm}^2$.

Answer: $462\ \text{cm}^2$.

3
Worked Example
The minute hand of a clock is 14 cm long. Find the area swept by it in 15 minutes. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: In 60 minutes the hand turns $360^{\circ}$, so per minute it turns $6^{\circ}$.
  2. Step 2: In 15 minutes, $\theta=15\times 6=90^{\circ}$; radius $r=14$ cm.
  3. Step 3: Area $=\frac{90}{360}\times\frac{22}{7}\times 14\times 14=\frac{1}{4}\times\frac{22}{7}\times 196$.
  4. Step 4: $=\frac{1}{4}\times 22\times 28=\frac{1}{4}\times 616=154\ \text{cm}^2$.

Answer: $154\ \text{cm}^2$.

4
Worked Example
Find the length of the arc of a sector of angle $60^{\circ}$ in a circle of radius 21 cm. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Arc length $l=\frac{\theta}{360^{\circ}}\times 2\pi r=\frac{60}{360}\times 2\times\frac{22}{7}\times 21$.
  2. Step 2: $\frac{60}{360}=\frac{1}{6}$ and $2\times\frac{22}{7}\times 21=132$.
  3. Step 3: $l=\frac{1}{6}\times 132=22$ cm.

Answer: Arc length $=22$ cm.

5
Worked Example
Find the perimeter of a quadrant of a circle of radius 7 cm. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: A quadrant has $\theta=90^{\circ}$. Arc length $=\frac{90}{360}\times 2\pi r=\frac{1}{4}\times 2\times\frac{22}{7}\times 7$.
  2. Step 2: $=\frac{1}{4}\times 44=11$ cm.
  3. Step 3: Perimeter $=$ arc $+2r=11+2\times 7=11+14=25$ cm.

Answer: Perimeter $=25$ cm.

6
Worked Example
A sector has area $\frac{1}{6}$ of a circle of radius 12 cm. Find the central angle of the sector.
Solution
  1. Step 1: Sector area $=\frac{\theta}{360^{\circ}}\times\pi r^2$ and this equals $\frac{1}{6}\pi r^2$.
  2. Step 2: So $\frac{\theta}{360^{\circ}}=\frac{1}{6}$.
  3. Step 3: $\theta=\frac{360^{\circ}}{6}=60^{\circ}$.

Answer: $\theta=60^{\circ}$.

7
Worked Example
A sector of a circle of radius 21 cm has a central angle of $120^{\circ}$. Find its area and arc length. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Area $=\frac{120}{360}\times\frac{22}{7}\times 21\times 21=\frac{1}{3}\times\frac{22}{7}\times 441$.
  2. Step 2: $\frac{441}{7}=63$, so area $=\frac{1}{3}\times 22\times 63=462\ \text{cm}^2$.
  3. Step 3: Arc $=\frac{120}{360}\times 2\times\frac{22}{7}\times 21=\frac{1}{3}\times 132=44$ cm.

Answer: Area $=462\ \text{cm}^2$ and arc length $=44$ cm.

8
Worked Example
The arc length of a sector of a circle of radius 14 cm is 22 cm. Find the area of the sector using the relation $\text{area}=\frac{1}{2}lr$.
Solution
  1. Step 1: Given $l=22$ cm and $r=14$ cm.
  2. Step 2: Area $=\frac{1}{2}\,l\,r=\frac{1}{2}\times 22\times 14$.
  3. Step 3: $=11\times 14=154\ \text{cm}^2$.

Answer: $154\ \text{cm}^2$.

9
Worked Example
Find the area of the major sector of a circle of radius 7 cm in which the minor sector has a central angle of $90^{\circ}$. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Major-sector angle $=360^{\circ}-90^{\circ}=270^{\circ}$.
  2. Step 2: Area $=\frac{270}{360}\times\frac{22}{7}\times 7\times 7=\frac{3}{4}\times 22\times 7$.
  3. Step 3: $=\frac{3}{4}\times 154=115.5\ \text{cm}^2$.

Answer: $115.5\ \text{cm}^2$.

10
Worked Example
The hour hand of a clock is 6 cm long. Find the area swept by it between 8 a.m. and 11 a.m. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: From 8 a.m. to 11 a.m. is 3 hours. The hour hand turns $360^{\circ}$ in 12 hours, i.e. $30^{\circ}$ per hour.
  2. Step 2: $\theta=3\times 30^{\circ}=90^{\circ}$; radius $r=6$ cm.
  3. Step 3: Area $=\frac{90}{360}\times\frac{22}{7}\times 6\times 6=\frac{1}{4}\times\frac{22}{7}\times 36$.
  4. Step 4: $=\frac{1}{4}\times\frac{792}{7}=\frac{198}{7}\approx 28.29\ \text{cm}^2$.

Answer: $\frac{198}{7}\approx 28.29\ \text{cm}^2$.

11
Worked Example
The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.
Solution
  1. Step 1: Perimeter $=l+2r$, so $27.2=l+2\times 5.7=l+11.4$.
  2. Step 2: Arc length $l=27.2-11.4=15.8$ m.
  3. Step 3: Area $=\frac{1}{2}\,l\,r=\frac{1}{2}\times 15.8\times 5.7$.
  4. Step 4: $=\frac{1}{2}\times 90.06=45.03\ \text{m}^2$.

Answer: $45.03\ \text{m}^2$.

12
Worked Example
A horse is tied to a peg at one corner of a square grass field of side 15 m by a 5 m long rope. Find the area of the field the horse can graze. (Take $\pi=3.14$).
Solution
  1. Step 1: At a corner of the square the inside angle is $90^{\circ}$, so the horse grazes a quarter-circle (quadrant) of radius 5 m.
  2. Step 2: Grazed area $=\frac{90}{360}\times\pi r^2=\frac{1}{4}\times 3.14\times 5\times 5$.
  3. Step 3: $=\frac{1}{4}\times 3.14\times 25=\frac{1}{4}\times 78.5=19.625\ \text{m}^2$.

Answer: $19.625\ \text{m}^2$.

Key Points

  • A sector is a portion of a circle bounded by two radii lines and an outer connecting arc.
  • The area of a sector depends directly on its central interior angle theta ($\theta$).
  • Every sector is calculated as a fraction of a full circle's area: $\frac{\theta}{360^\circ} \cdot \pi r^2$.
  • The length of a sector's curved edge line is called its arc length, calculated as $\frac{\theta}{360^\circ} \cdot 2\pi r$.
  • A minor sector spans an angle less than 180°, while a major sector covers the remaining angle up to 360°.
  • Sector area also equals $\frac{1}{2}lr$, where $l$ is the arc length — useful when $l$ is known.
  • The perimeter of a sector is arc $+2r$; never forget to add the two radii.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The area of a sector of angle $\theta$ is:
Explanation: Fraction of the circle's area.
Q2.The length of an arc of angle $\theta$ is:
Explanation: Fraction of the circumference.
Q3.A quadrant of a circle is a sector of angle:
Explanation: A quarter $=90^\circ$.
Q4.A semicircle is a sector of angle:
Explanation: Half the circle.
Practice more MCQs → 📝 Assignment · 35 marks