Circumferences and Areas of Circular Shapes
A circle's circumference is \(C = 2\pi r\) and its area is \(A = \pi r^2\), and a sector takes the matching fraction of each. This NSW Year 9 Advanced maths subtopic revises these Stage 4 skills (NESA outcomes MA4‑LEN‑C‑01 and MA4‑ARE‑C‑01) with harder shapes.
You will find circumferences, arc lengths, sector areas and perimeters, the area of a pizza slice and of a circular path, and a radius from a circumference. You will find perimeters and areas of shapes made from semicircles, rings and sectors, find a radius from a sector's perimeter, show a Pythagoras result with semicircles, and find the area a goat can graze around a shed. It is part of Year 9 Advanced Surface Area and Volume.
Theory
A circle's circumference is \(2\pi r\) and its area is \(\pi r^2\). A sector is a fraction \(\dfrac{\theta}{360}\) of the circle, so its arc and area are that fraction too. This Year 9 Advanced maths page revises the earlier circle work and extends it to sectors, rings, exact answers and working backwards.
The radius \(r\) runs from the centre to the circle; the diameter \(d = 2r\) goes right across. The circumference is the distance around the circle, and \(\pi\) is the ratio of circumference to diameter, about \(3.14159\).
A sector is a slice bounded by two radii and an arc. A semicircle is a sector of \(180^\circ\); a quadrant is \(90^\circ\). A sector bigger than a semicircle is a major (reflex) sector, with angle \(360^\circ\) minus the small angle.
An annulus is a ring between two circles with the same centre. A segment is the region between a chord and its arc: sector minus triangle.
An exact answer keeps \(\pi\) as a symbol, such as \(12\pi~\text{cm}^2\). A rounded answer uses the calculator's \(\pi\) and rounds at the end.
For a sector with angle \(\theta\):
Annulus (ring) with outer radius \(R\) and inner radius \(r\):
How to solve a circle problem
- Find the radius: halve a diameter if needed.
- Find the fraction: \(\dfrac{\theta}{360}\) for a sector. For a slice of something cut into \(n\) equal pieces, \(\theta = \dfrac{360}{n}\). For a major sector use \(360 - \theta\).
- Apply the formula: fraction \(\times 2\pi r\) for a length, fraction \(\times \pi r^2\) for an area.
- Add straight edges for a perimeter: radii, sides of squares, or the straight ends of a ring.
- Give the answer exactly (in terms of \(\pi\)) or rounded, as asked.
Composite and backwards problems
- Shaded regions: whole shape minus the white parts (circles cut from a circle, sectors cut from a square, a segment cut from a circle).
- Combining arcs: two semicircles of the same size make one full circle.
- Radius from \(C\) or \(A\): divide by \(2\pi\), or divide by \(\pi\) and take the square root.
- Radius in two places: if a sector's perimeter is given, write \(2r + \dfrac{\theta}{360} \times 2\pi r\), take out \(r\) as a common factor and divide.
- Grazing or reach problems: split the region into sectors, one for each corner the rope bends around, with the rope length shortened each time.
The sector is \(\dfrac{135}{360} = \dfrac{3}{8}\) of the circle.
| \(\ell\) | \(=\) | \(\dfrac{3}{8} \times 2\pi \times 14 = 10.5\pi \approx \mathbf{32.99}~\text{cm}\) |
| \(P\) | \(=\) | \(14 + 14 + 10.5\pi \approx \mathbf{60.99}~\text{cm}\) |
| \(A\) | \(=\) | \(\dfrac{3}{8} \times \pi \times 14^2 = 73.5\pi \approx \mathbf{230.91}~\text{cm}^2\) |
The perimeter needs both straight radii as well as the arc.
Substitute into \(A = \pi r^2\) and undo the operations: divide by \(\pi\), then take the square root.
| \(\pi r^2\) | \(=\) | \(5\) |
| \(r^2\) | \(=\) | \(\dfrac{5}{\pi} = 1.5915\ldots\) |
| \(r\) | \(=\) | \(\sqrt{1.5915\ldots} = 1.2615\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.26}~\text{m}\) |
Use the unrounded radius for the circumference.
| \(C\) | \(=\) | \(2\pi \times 1.2615\ldots = 7.9266\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.93}~\text{m}\) |
The fraction of a circle is \(\dfrac{120}{360} = \dfrac{1}{3}\). Area: big sector minus small sector.
| \(A\) | \(=\) | \(\dfrac{1}{3}\pi(5^2 - 3^2) = \dfrac{16\pi}{3}\) |
| \(\) | \(\approx\) | \(\mathbf{16.76}~\text{m}^2\) |
Perimeter: two arcs plus two straight ends of \(5 - 3 = 2\) m.
| \(P\) | \(=\) | \(\dfrac{1}{3}(2\pi \times 5) + \dfrac{1}{3}(2\pi \times 3) + 2 + 2\) |
| \(\) | \(=\) | \(\dfrac{16\pi}{3} + 4 \approx \mathbf{20.76}~\text{m}\) |
(i) Find the length of the inside edge, correct to \(2\) decimal places.
(ii) How much further is one lap around the outside edge of the lane?
(i) The two semicircles make one full circle.
| \(P\) | \(=\) | \(2 \times 85 + 2\pi \times 32\) |
| \(\) | \(=\) | \(170 + 64\pi \approx \mathbf{371.06}~\text{m}\) |
(ii) The straights are the same; only the radius grows, to \(33.2\) m.
| \(\text{extra}\) | \(=\) | \(2\pi \times 33.2 - 2\pi \times 32\) |
| \(\) | \(=\) | \(2\pi \times 1.2 \approx \mathbf{7.54}~\text{m}\) |
The extra distance depends only on the lane width, not on the radius. That is why races in outside lanes start further ahead.
Common pitfalls
Frequently asked questions
What is the formula for the circumference of a circle?
\(C = 2\pi r = \pi d\), where \(r\) is the radius and \(d\) is the diameter.
How do you find the area of a sector?
Multiply the area of the whole circle, \(\pi r^2\), by the fraction \(\dfrac{\theta}{360}\), where \(\theta\) is the angle at the centre in degrees.
How do you find the arc length of a sector?
Multiply the full circumference, \(2\pi r\), by \(\dfrac{\theta}{360}\). For the perimeter of the sector, add the two radii.
How do you find the radius from the area of a circle?
Divide the area by \(\pi\), then take the square root: \(r = \sqrt{\dfrac{A}{\pi}}\). An area of \(50~\text{cm}^2\) gives \(r \approx 3.99\) cm.
What does exact answer in terms of pi mean?
Leave \(\pi\) as a symbol instead of using a decimal, for example \(36\pi~\text{cm}^2\). It is the most precise form of the answer.
How do you find the area of a ring?
Subtract the area of the inner circle from the area of the outer circle: \(\pi R^2 - \pi r^2 = \pi(R^2 - r^2)\).