Three new area formulas — all built on the circle. Master the sector, the ring, and the triangle with an included angle.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A circular garden has a circular pond in the centre. You want to turf the garden but not the pond. You know both radii. How would you find the area to turf — and why can't you just measure it directly?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Come back to this at the end of the lesson.
Wrong: Sector area = (theta/360) x pi x r^2 without the 1/2 factor.
Right: The correct formula is A = (theta/360) * pi * r² for degrees, or A = 0.5 * r² * theta for radians.
Core Content
A sector is a fraction of a full circle. The fraction is $\theta/360$. So the sector area is that same fraction of $\pi r^2$.
$$A = \frac{\theta}{360} \times \pi r^2$$| Formula | What follows $\theta/360$ |
|---|---|
| Arc length $\ell$ | Circumference: $2\pi r$ |
| Sector area $A$ | Full circle area: $\pi r^2$ |
An annulus is the ring between two concentric circles. Its area = outer circle area − inner circle area.
$$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$$Both forms are correct. The factored form $\pi(R^2 - r^2)$ is more efficient on a calculator — compute the bracket first, then multiply by $\pi$ once.
The basic formula $A = \tfrac{1}{2}bh$ needs the perpendicular height. When two sides and the included angle are given instead, use the sine area rule.
$$A = \frac{1}{2}ab\sin C$$where $a$ and $b$ are two known side lengths and $C$ is the included angle — the angle sitting between those two sides.
$\sin 90° = 1$, so the formula becomes $A = \tfrac{1}{2}ab$ — consistent with a right-angled triangle where the two sides are the base and height. The sine rule reduces to the standard formula as a special case.
The strategy from L02 still applies: identify components, decide add or subtract, calculate each, combine.
| Common combination | Method |
|---|---|
| Rectangle with sector cut from corner | Rectangle area − sector area |
| Triangle with sector attached | Triangle area + sector area |
| Circle with hole (= annulus) | $\pi(R^2 - r^2)$ directly |
| Logo with three corner sectors removed | Triangle area − 3 × sector area |
Worked Examples
Find the area of a sector with radius 10 cm and central angle 144°. Give your answer correct to 2 decimal places.
An annulus has an outer radius of 9 cm and an inner radius of 5 cm. Find its area correct to 2 decimal places.
Find the area of a triangle with two sides of 8 cm and 11 cm and an included angle of 65°. Give your answer correct to 2 decimal places.
A logo is made from an equilateral triangle with side length 8 cm. A sector of radius 3 cm and central angle 60° is drawn at each vertex.
(a) Find the triangle area using the sine area rule. (b) Find the total area of the three sectors. (c) Find the logo area (triangle minus sectors).
Label which formula you are using before substituting.
Section A — Sector Area
1 Find the area of a sector with $r = 5$ cm and $\theta = 72°$. Give in exact form and to 2 decimal places.
2 Find the area of a sector with $r = 12$ m and $\theta = 150°$. Answer to 2 decimal places.
3 A sector has area $30\pi$ cm² and radius 6 cm. Find the central angle $\theta$.
Section B — Annulus Area
4 Find the area of an annulus with $R = 10$ cm and $r = 6$ cm. Answer to 2 decimal places.
5 A circular path surrounds a garden of radius 4 m. The path extends 1.5 m beyond the garden edge. Find the area of the path to 2 decimal places.
Section C — Sine Area Rule
6 Find the area of a triangle with sides 7 cm and 9 cm and included angle 40°. Answer to 2 decimal places.
7 Find the area of a triangle with sides 15 m and 15 m and included angle 100°. Answer to 2 decimal places.
Section D — Composite Areas
8 A square of side 8 cm has a sector of radius 8 cm and angle 90° removed from one corner. Find the remaining area to 2 decimal places.
9 A shape consists of a rectangle 10 cm × 6 cm, with a semicircle of diameter 6 cm added to one short end and a triangle (base 6 cm, two sides 6 cm each, included angle 50°) removed from the other short end. Find the total area to 2 decimal places.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Multiple Choice
1 A sector has radius 6 cm and central angle 120°. Its area is:
? Regarding this topic, 1 A sector has radius 6 cm and central angle 120°. Its area is:
2 An annulus has outer radius 8 m and inner radius 3 m. Its area, correct to 2 decimal places, is:
? Regarding this topic, 2 An annulus has outer radius 8 m and inner radius 3 m. Its area, correct to 2 decimal places, is:
3 A triangle has two sides of 10 cm and 14 cm with an included angle of 30°. Its area is:
? Regarding this topic, 3 A triangle has two sides of 10 cm and 14 cm with an included angle of 30°. Its area is:
Short Answer
SA 4 2 marks Find the area of a sector with radius 9 m and central angle 200°. Give your answer correct to 2 decimal places.
SA 5 3 marks A circular table has diameter 1.6 m. A circular lazy Susan with diameter 0.8 m sits in the centre. Find the area of the table not covered by the lazy Susan, correct to 2 decimal places.
SA 6 4 marks A logo is made from an equilateral triangle with side length 8 cm. A sector of radius 3 cm and central angle 60° is drawn at each vertex.
(a) Find the area of the equilateral triangle using the sine area rule, correct to 2 decimal places. (2 marks)
(b) Find the total area of the three sectors. (1 mark)
(c) Find the area of the logo (triangle minus sectors), correct to 2 decimal places. (1 mark)
Scale the platforms using your knowledge of sectors, annuli and composite shapes. Pool: lessons 1–6.
Tick when you have finished the lesson and checked your answers.