Area of Circles and Sectors
Area of Circles and Sectors in NSW Year 10 Core maths revises \(A = \pi r^2\) and the area of a sector, which is the fraction \(\frac{\theta}{360}\) of the whole circle (NESA outcome MA4‑ARE‑C‑01, revising Stage 4). It then uses them in Stage 5 composite-area problems (MA5‑ARE‑C‑01), with answers exact in terms of \(\pi\) or rounded.
You will find the areas of sectors, major sectors, semicircles and rings, work backwards from an area to a radius or a sector angle, and find shaded regions where circles are cut from rectangles and squares. You will also find the lawn watered by a sprinkler, compare the value of two pizza sizes, cost mulch for a garden bed around a pond and price the glass for an arched window. It is part of Year 10 Core Measurement.
Theory
The area of a circle is \(\pi r^{2}\), and a sector is the fraction \(\dfrac{\theta}{360}\) of that circle. This Year 10 maths page revises the earlier formulas and extends them: exact answers in terms of \(\pi\), major sectors, rings and composite shapes, and working backwards to a radius or an angle.
The radius \(r\) of a circle runs 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.
The area of a circle is \(A = \pi r^{2}\). The number \(\pi \approx 3.14159\) is the same for every circle.
A sector is a slice of a circle between two radii, like a slice of pizza. The angle \(\theta\) at the centre tells you what fraction of the circle it is: \(\dfrac{\theta}{360}\). A semicircle is half a circle (\(180^\circ\)) and a quadrant is a quarter (\(90^\circ\)). A major sector has an angle bigger than \(180^\circ\).
An exact answer leaves \(\pi\) in, such as \(18\pi\text{ cm}^2\). An approximate answer uses the \(\pi\) key and rounds.
This page includes harder composite shapes and problems that work backwards from an area.
| Shape | Fraction of the circle | Area |
|---|---|---|
| Circle | \(1\) | \(\pi r^{2}\) |
| Semicircle | \(\dfrac{180}{360} = \dfrac{1}{2}\) | \(\dfrac{1}{2}\pi r^{2}\) |
| Quadrant | \(\dfrac{90}{360} = \dfrac{1}{4}\) | \(\dfrac{1}{4}\pi r^{2}\) |
| Sector, angle \(\theta\) | \(\dfrac{\theta}{360}\) | \(\dfrac{\theta}{360} \times \pi r^{2}\) |
| Ring (annulus) | big circle minus small circle | \(\pi R^{2} - \pi r^{2}\) |
How to find the area of a circle or sector
- Find the radius. Halve a diameter; use \(r = \dfrac{C}{2\pi}\) if you know the circumference.
- Find the angle. For the larger part of a cut circle, use \(360^\circ\) minus the cut-out angle.
- Substitute into \(\dfrac{\theta}{360} \times \pi r^{2}\). Simplify the fraction first for an exact answer in \(\pi\).
- Round only at the end if a decimal answer is asked for, and give square units.
Composite shapes and problems
- Add the parts (a rectangle with a semicircle on top) or subtract the holes (a square minus a quadrant, a big circle minus small circles).
- Two semicircles with the same diameter make one full circle.
- A ring is the big circle minus the small circle. A part of a ring is that fraction of the ring.
- Cost or value: multiply the area by the price per square unit, or divide the price by the area to compare value for money. For example, \(12.6\text{ m}^2\) of turf at $15 per square metre costs $189; a $20 pizza with an area of \(800\text{ cm}^2\) costs \(2000 \div 800 = 2.5\) cents per \(\text{cm}^2\).
- Equal areas: set the two area expressions equal and solve for the unknown radius or angle.
(i) A round table has a diameter of \(1.8\) m. Find its area, correct to two decimal places.
(ii) Find the exact area of a sector with radius \(9\) cm and angle \(80^\circ\).
(i) Halve the diameter: \(r = 0.9\) m.
| \(A\) | \(=\) | \(\pi \times 0.9^{2}\) |
| \(\) | \(=\) | \(2.5446\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{2.54}\text{ m}^2\) |
(ii) For an exact answer, leave \(\pi\) in.
| \(A\) | \(=\) | \(\dfrac{80}{360} \times \pi \times 9^{2}\) |
| \(\) | \(=\) | \(\dfrac{2}{9} \times 81\pi\) |
| \(\) | \(=\) | \(\mathbf{18\pi}\text{ cm}^2\) |
The shaded part is a major sector. Its angle is what is left of \(360^\circ\).
| \(\theta\) | \(=\) | \(360^\circ - 120^\circ = 240^\circ\) |
| \(A\) | \(=\) | \(\dfrac{240}{360} \times \pi \times 6^{2}\) |
| \(\) | \(=\) | \(\dfrac{2}{3} \times 36\pi\) |
| \(\) | \(=\) | \(\mathbf{24\pi}\text{ cm}^2\) |
| \(\) | \(\approx\) | \(\mathbf{75.4}\text{ cm}^2\) |
The two semicircles have diameter \(60\) m, so together they make one full circle of radius \(30\) m.
| \(\text{rectangle}\) | \(=\) | \(80 \times 60 = 4800\text{ m}^2\) |
| \(\text{circle}\) | \(=\) | \(\pi \times 30^{2} = 900\pi\) |
| \(A\) | \(=\) | \(4800 + 900\pi\) |
| \(\) | \(=\) | \(7627.4333\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7627.4}\text{ m}^2\) |
(i) A sector of radius \(12\) cm has an area of \(30\pi\text{ cm}^2\). Find its angle \(\theta\).
(ii) A circle has an area of \(200\text{ cm}^2\). Find its radius, correct to two decimal places.
(i) Compare the sector with the whole circle.
| \(\text{whole circle}\) | \(=\) | \(\pi \times 12^{2} = 144\pi\) |
| \(\text{fraction}\) | \(=\) | \(\dfrac{30\pi}{144\pi} = \dfrac{5}{24}\) |
| \(\theta\) | \(=\) | \(\dfrac{5}{24} \times 360^\circ = \mathbf{75^\circ}\) |
(ii) Undo the formula: divide by \(\pi\), then take the square root.
| \(\pi r^{2}\) | \(=\) | \(200\) |
| \(r^{2}\) | \(=\) | \(\dfrac{200}{\pi}\) |
| \(r\) | \(=\) | \(\sqrt{\dfrac{200}{\pi}}\) |
| \(\) | \(=\) | \(7.9788\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.98}\text{ cm}\) |
Check: \(\pi \times 7.98^{2} \approx 200\).
Common pitfalls
Frequently asked questions
How do you find the area of a circle?
Square the radius and multiply by \(\pi\). If you are given the diameter, halve it first. A circle of radius \(5\) cm has an area of \(25\pi \approx 78.5\text{ cm}^2\).
How do you find the area of a sector?
Work out what fraction of the circle the sector is, \(\dfrac{\theta}{360}\), then multiply by \(\pi r^{2}\).
What does give your answer in terms of pi mean?
It means leave \(\pi\) as a symbol instead of using a decimal, for example \(18\pi\text{ cm}^2\). This is the exact answer.
How do you find the area of a semicircle?
Find the area of the whole circle with \(\pi r^{2}\) and halve it. Use the radius, which is half the straight edge of the semicircle.
How do you find the radius from the area of a circle?
Divide the area by \(\pi\), then take the square root: \(r = \sqrt{\dfrac{A}{\pi}}\).