This note covers the circumference and area of a circle, and the length of an arc and the area of a sector, which are fractions of the whole circle. It also shows how to combine circles with other shapes and how to work backwards to find a radius or an angle. Circles appear in wheels, dials, tracks, pipes and many other designs.
Key ideas
The radius \(r\) is the distance from the centre to the edge. The diameter \(d\) goes right across through the centre, so \(d = 2r\). The circumference is the distance around the circle.
- Circumference: \(C = 2\pi r\) or \(C = \pi d\).
- Area: \(A = \pi r^2\). Square the radius first, then multiply by \(\pi\).
- \(\pi\) is the number 3.14159... It is the ratio of any circle's circumference to its diameter. Use the \(\pi\) button on your calculator, not 3.14, unless the question tells you otherwise.
A sector is a slice of a circle between two radii. An arc is a part of the circumference. If the angle at the centre is \(\theta\) degrees, then the sector is the fraction \(\dfrac{\theta}{360}\) of the whole circle.
- Arc length \(= \dfrac{\theta}{360} \times 2\pi r\).
- Sector area \(= \dfrac{\theta}{360} \times \pi r^2\).
- The perimeter of a sector is the arc length plus the two radii. Do not forget the radii.
Exact answers. If a question asks for an answer "in terms of \(\pi\)", leave \(\pi\) as a symbol, for example \(6\pi\) cm. Otherwise use your calculator value of \(\pi\) and round to three significant figures.
Working backwards. You may be given a circumference, an area, an arc length or a sector area and asked for \(r\) or \(\theta\). Write the formula, substitute what you know, and rearrange. For an area, you will need to take a square root at the end.
Compound shapes. A semicircle has half the circumference of a full circle (plus the diameter if you need the full outline) and half the area. Many problems join semicircles or quarter circles to rectangles. Treat each part separately, then add or subtract.
Turning wheels give a practical use of circumference. In one full turn, a wheel moves forward by a distance equal to its circumference, so the distance after \(n\) turns is \(n \times 2\pi r\).
A quarter circle has angle \(90^\circ\) and a half circle has angle \(180^\circ\), so these fractions appear often. If a sector angle is a reflex angle, such as \(240^\circ\), the same formulas still work because the fraction is \(\dfrac{240}{360} = \dfrac{2}{3}\). Always check that your answer is sensible: an arc must be shorter than the full circumference, and a sector must have less area than the full circle.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(C = 2\pi r = \pi d\) | Circumference of a circle. |
| \(A = \pi r^2\) | Area of a circle. |
| \(\text{Arc length} = \dfrac{\theta}{360} \times 2\pi r\) | \(\theta\) is the angle at the centre in degrees. |
| \(\text{Sector area} = \dfrac{\theta}{360} \times \pi r^2\) | Same fraction of the area of the full circle. |
| Perimeter of a sector | Arc length \(+ 2r\). |
| Semicircle | Area \(\tfrac{1}{2}\pi r^2\). Curved edge \(\pi r\). |
Worked examples
Example 1 A bicycle wheel has radius 35 cm. How far does the bicycle travel when the wheel makes 100 complete turns? Give your answer in metres.
Step 1: circumference \(= 2\pi \times 35 = 70\pi \approx 219.91\) cm.
Step 2: distance for 100 turns \(= 100 \times 219.91 = 21\,991\) cm.
Step 3: convert to metres. \(21\,991 \div 100 = 219.91\) m.
Answer: about 220 m (3 s.f.)
Example 2 A sector of a circle has radius 15 cm and angle \(72^\circ\). Find its arc length, its area and its perimeter.
Step 1: the fraction of the circle is \(\dfrac{72}{360} = \dfrac{1}{5}\).
Step 2: arc length \(= \tfrac{1}{5} \times 2\pi \times 15 = \tfrac{1}{5} \times 30\pi = 6\pi \approx 18.85\) cm.
Step 3: area \(= \tfrac{1}{5} \times \pi \times 15^2 = \tfrac{1}{5} \times 225\pi = 45\pi \approx 141.37\) cm\(^2\).
Step 4: perimeter \(= 18.85 + 15 + 15 = 48.85\) cm.
Answer: arc length 18.8 cm; area 141 cm\(^2\); perimeter 48.8 cm (all to 3 s.f.).
Example 3 A running track is a rectangle 100 m long and 60 m wide with a semicircle added at each of the two short ends (the semicircles have diameter 60 m). Find the length of one lap and the total area.
Step 1: the two semicircles together make one full circle of diameter 60 m. Its circumference is \(\pi \times 60 = 60\pi \approx 188.50\) m.
Step 2: the two straight sides are each 100 m. Lap length \(= 200 + 188.50 = 388.50\) m.
Step 3: the area of the rectangle is \(100 \times 60 = 6000\) m\(^2\).
Step 4: the two semicircles make a circle of radius 30 m. Area \(= \pi \times 30^2 = 900\pi \approx 2827.43\) m\(^2\).
Step 5: total area \(= 6000 + 2827.43 = 8827.43\) m\(^2\).
Answer: lap length about 388 m; area about 8830 m\(^2\) (3 s.f.).
Common errors
- Using the diameter in \(\pi r^2\). Halve the diameter first.
- Multiplying by 2 before squaring, giving \((2r)^2\) or \(2\pi r^2\) instead of \(\pi r^2\).
- Mixing up the circumference formula and the area formula. Remember that area is in square units and has \(r^2\).
- Forgetting to add the two radii to the arc length when asked for the perimeter of a sector.
- Rounding \(\pi\) to 3.14 early and losing accuracy in the final answer.
Exam tip: for sector questions, first write the fraction \(\dfrac{\theta}{360}\) in its simplest form. It reduces the number work and makes it easy to check that your answer is smaller than the whole circle's value.
Check yourself
- Find the area of a circle with diameter 10 cm.
Answer
\(r = 5\), so \(A = \pi \times 5^2 = 25\pi \approx 78.5\ \text{cm}^2\). - Find the arc length of a sector with radius 9 cm and angle \(120^\circ\).
Answer
\(\dfrac{120}{360} \times 2\pi \times 9 = \tfrac{1}{3} \times 18\pi = 6\pi \approx 18.8\ \text{cm}\). - A sector has radius 8 cm and area 40 cm\(^2\). Find its angle.
Answer
\(40 = \dfrac{\theta}{360} \times \pi \times 64\), so \(\theta = \dfrac{40 \times 360}{64\pi} \approx 71.6^\circ\). - A sector has angle \(150^\circ\) and arc length 20 cm. Find the radius.
Answer
\(20 = \dfrac{150}{360} \times 2\pi r\), so \(r = \dfrac{20 \times 360}{150 \times 2\pi} \approx 7.64\ \text{cm}\).