Circumferences and Areas of Circular Shapes
Circles appear in wheels, pipes, pizzas and running tracks. This NSW Year 9 Core maths subtopic revises circumference and area from Stage 4 (NESA outcomes MA4‑LEN‑C‑01 and MA4‑ARE‑C‑01) and applies them to semicircles, quadrants, sectors, rings and shapes built from them.
You will use \(C = 2\pi r\) and \(A = \pi r^2\) from a radius or a diameter, remember that a semicircle's perimeter includes its straight edge, and find arc lengths and sector areas from the angle at the centre, using \(\frac{\theta}{360}\). You will also find the area of a ring-shaped path, the waste when a circle is cut from a square, the radius of a tank from its circumference, and the perimeter and area of a running track made of a rectangle and two semicircles. It is part of Year 9 Core Surface Area and Volume.
Theory
The circumference of a circle is \(C = 2\pi r\) and its area is \(A = \pi r^2\). This Year 9 maths page revises both, then uses them for semicircles, quadrants and sectors, perimeters with straight edges, rings, and working back from a circumference to the radius.
The radius \(r\) runs from the centre of a circle to its edge. The diameter \(d\) goes right across through the centre, so \(d = 2r\).
The circumference is the distance all the way round the circle. The area is the space inside it.
\(\pi\) (pi) is the number of times the diameter fits around the circumference, about \(3.14159\). Use the \(\pi\) key on your calculator.
A semicircle is half a circle. A quadrant is a quarter. A sector is a slice cut by two radii, with an angle \(\theta\) at the centre. Its curved edge is an arc.
A sector with angle \(\theta\) is \(\dfrac{\theta}{360}\) of the circle:
| Shape | Area | Curved edge | Perimeter |
|---|---|---|---|
| circle | \(\pi r^2\) | \(2\pi r\) | \(2\pi r\) |
| semicircle | \(\dfrac{1}{2}\pi r^2\) | \(\pi r\) | \(\pi r + 2r\) |
| quadrant | \(\dfrac{1}{4}\pi r^2\) | \(\dfrac{1}{2}\pi r\) | \(\dfrac{1}{2}\pi r + 2r\) |
| sector of \(\theta\) | \(\dfrac{\theta}{360}\pi r^2\) | \(\dfrac{\theta}{360} \times 2\pi r\) | arc \(+ \, 2r\) |
How to solve a circle problem
- Find the radius. If you are given a diameter, halve it.
- Decide what fraction of a circle the shape is: whole, half, quarter, or \(\dfrac{\theta}{360}\).
- Use \(2\pi r\) for an edge or \(\pi r^2\) for an area, then take that fraction.
- Add any straight edges for a perimeter.
- Round once, at the end, to the accuracy asked for.
(i) its circumference
(ii) its area, each correct to two decimal places.
(i) With the diameter, use \(C = \pi d\).
| \(C\) | \(=\) | \(\pi \times 22\) |
| \(\) | \(=\) | \(69.115\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{69.12} \text{ cm}\) |
(ii) The area formula needs the radius, so halve the diameter first: \(r = 11\) cm.
| \(A\) | \(=\) | \(\pi r^2 = \pi \times 11^2\) |
| \(\) | \(=\) | \(121\pi = 380.132\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{380.13} \text{ cm}^2\) |
Using \(22\) as the radius would give an area four times too big.
(i) its area
(ii) its perimeter, each correct to two decimal places.
(i) The radius is \(7\) cm. Take half of a circle's area.
| \(A\) | \(=\) | \(\dfrac{1}{2} \times \pi \times 7^2\) |
| \(\) | \(=\) | \(76.969\ldots \approx \mathbf{76.97} \text{ cm}^2\) |
(ii) The perimeter is the curved edge plus the straight diameter.
| \(P\) | \(=\) | \(\dfrac{1}{2} \times \pi \times 14 + 14\) |
| \(\) | \(=\) | \(7\pi + 14\) |
| \(\) | \(=\) | \(35.991\ldots \approx \mathbf{35.99} \text{ cm}\) |
(i) its arc length
(ii) its area
(iii) its perimeter, each to two decimal places.
The sector is \(\dfrac{45}{360} = \dfrac{1}{8}\) of the whole circle.
| \(\text{arc}\) | \(=\) | \(\dfrac{1}{8} \times 2\pi \times 10 = 7.853\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.85} \text{ cm}\) |
| \(A\) | \(=\) | \(\dfrac{1}{8} \times \pi \times 10^2 = 39.269\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{39.27} \text{ cm}^2\) |
| \(P\) | \(=\) | \(7.853\ldots + 10 + 10\) |
| \(\) | \(\approx\) | \(\mathbf{27.85} \text{ cm}\) |
Keep the full arc value in the calculator until the end, then round once.
(i) Find the area of the ring between circles of radius \(5\) cm and \(8\) cm.
(ii) A circle has a circumference of \(50\) cm. Find its radius. Give both answers to two decimal places.
(i) The ring is the big circle minus the small one.
| \(A\) | \(=\) | \(\pi \times 8^2 - \pi \times 5^2\) |
| \(\) | \(=\) | \(64\pi - 25\pi = 39\pi\) |
| \(\) | \(=\) | \(122.522\ldots \approx \mathbf{122.52} \text{ cm}^2\) |
(ii) Substitute into \(C = 2\pi r\), then divide by \(2\pi\).
| \(50\) | \(=\) | \(2\pi r\) |
| \(r\) | \(=\) | \(\dfrac{50}{2\pi} = 7.957\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.96} \text{ cm}\) |
Common pitfalls
Frequently asked questions
What is the formula for the circumference of a circle?
\(C = 2\pi r\), or \(C = \pi d\). A circle with a diameter of \(22\) cm has a circumference of about \(69.12\) cm.
How do you find the area of a circle from the diameter?
Halve the diameter to get the radius, square it, then multiply by \(\pi\). For \(d = 22\) cm, \(r = 11\) cm and \(A = 121\pi \approx 380.13 \text{ cm}^2\).
How do you find the perimeter of a semicircle?
Add half the circumference to the diameter: \(P = \pi r + 2r\). For \(d = 14\) cm, \(P = 7\pi + 14 \approx 35.99\) cm.
How do you find the area of a sector?
Multiply the area of the whole circle by \(\dfrac{\theta}{360}\). A \(45^\circ\) sector of radius \(10\) cm is \(\dfrac{1}{8} \times 100\pi \approx 39.27 \text{ cm}^2\).
How do you find the radius from the circumference?
Divide the circumference by \(2\pi\): \(r = \dfrac{C}{2\pi}\). A circumference of \(50\) cm gives \(r \approx 7.96\) cm.