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Module 2 · L6 of 22 ~55 min ⚡ +95 XP available

Area of Sectors, Annuli, and Composite Shapes

Three new area formulas, all built on the circle. Master the sector, the ring, and the triangle with an included angle.

Today's hook, 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?
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Recall, your gut answer first

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

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?

Without calculating write your initial thinking.

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The formulas you need to own
+5 XP to read

Three area formulas drive this lesson. The sector area and annulus area both build on the circle. The sine area rule handles triangles when the perpendicular height is unknown.

A sector is a fraction $\theta/360$ of the full circle, so its area is that same fraction of $\pi r^2$. An annulus (ring) has area = outer circle minus inner circle. The sine area rule $A = \tfrac{1}{2}ab\sin C$ works when two sides and the included angle are given.

Area of a sector and of an annulus.
Arc length and sector area: both use $\theta/360$. Arc → multiply by $2\pi r$; Area → multiply by $\pi r^2$.
Sector area
$A = \frac{\theta}{360} \times \pi r^2$. Fraction $\theta/360$ of the full circle area $\pi r^2$.
Annulus area
$A = \pi(R^2 - r^2)$. Outer circle area minus inner circle area.
Sine area rule
$A = \frac{1}{2}ab\sin C$. Use when two sides and the included angle are known.
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What you'll master
Know

Key facts

  • The sector area formula $A = (\theta/360) \times \pi r^2$
  • The annulus area formula $A = \pi(R^2 - r^2)$
  • The sine area rule $A = \tfrac{1}{2}ab\sin C$
Understand

Concepts

  • Why sector area is a fraction of $\pi r^2$, same fraction as arc length uses on circumference
  • Why annulus area = outer area − inner area
  • Why $A = \tfrac{1}{2}ab\sin C$ works when no perpendicular height is given
Can do

Skills

  • Calculate sector, annulus, and triangle areas using the correct formula
  • Identify when each formula applies
  • Solve composite area problems combining these formulas with L02 shapes
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Key terms
SectorA region bounded by two radii and an arc, a "pizza slice" of a circle.
AnnulusThe region between two concentric circles, a "ring" shape.
Concentric circlesTwo or more circles with the same centre but different radii.
Included angleThe angle formed between two known sides of a triangle, the angle sitting between them.
Exact formLeaving an answer as a multiple of $\pi$ (e.g. $18\pi$ cm²), used when a question says "leave in terms of $\pi$".
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Area of a Sector

A sector is a fraction of a full circle.

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Area of a Sector
core concept

A sector is a fraction of a full circle. The fraction is $\theta/360$. So the sector area is that same fraction of $\pi r^2$.

$$A = \frac{\theta}{360} \times \pi r^2$$
Arc length $\ell$
Fraction $\theta/360$ × circumference $2\pi r$ → linear units
Sector area $A$
Fraction $\theta/360$ × circle area $\pi r^2$ → square units
Comparing the two formulas: The fraction $\theta/360$ is identical in both. The only difference is whether you multiply by $2\pi r$ (circumference) or $\pi r^2$ (area). Keep this comparison in mind, you will never mix up the two formulas.
Unit check: Arc length → linear units (cm, m). Sector area → square units (cm², m²). If your "arc length" answer has squared units, you used the wrong formula. Always check units.
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Area of an Annulus

We just saw that sector area is a fraction of a full circle: A = (θ/360) × πr².

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Area of an Annulus
core concept
Beyond the syllabus. Annuli are not named in Applications of measurement (MST-11-05). A ring is just one composite shape, the outer circle minus the inner one, so this card is worth one careful read rather than hours of drill. What is examinable is the composite-area method itself: split the figure into shapes whose area you already know, then add or subtract. Learn it here and you can handle any ring, path or border the exam puts in front of you.

We just saw that sector area is a fraction of a full circle: A = (θ/360) × πr². That raises a question: what if the shape is a ring between two concentric circles, neither a sector nor a full circle? This card answers it → annulus area = outer circle area minus inner circle area = πR² − πr² = π(R² − r²), where R is the outer radius and r the inner radius.

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.

Label clearly: $R$ = outer (larger) radius, $r$ = inner (smaller) radius. Always identify and write both before substituting. The subtraction is always $R^2 - r^2$, subtracting in the wrong order gives a negative area.
Rounding tip: Compute $R^2 - r^2$ first (exact), then multiply by $\pi$ at the final step. Avoid computing $\pi R^2$ and $\pi r^2$ separately, this introduces two rounding errors instead of one.
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The Sine Area Rule and Composite Areas

We just saw annulus area using π(R² − r²) to subtract the inner circle from the outer.

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The Sine Area Rule and Composite Areas
core concept

We just saw annulus area using π(R² − r²) to subtract the inner circle from the outer. That raises a question: some triangles in composite shapes give you two sides and an included angle but no perpendicular height, how do you find the area then? This card answers it → use the sine area rule: A = ½ab sin C, valid whenever two sides and the included angle are known.

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.

When $C = 90°$: $\sin 90° = 1$, so the formula becomes $A = \tfrac{1}{2}ab$, consistent with a right-angled triangle. The sine rule reduces to the standard formula as a special case.

The included angle check: Before applying $A = \tfrac{1}{2}ab\sin C$, confirm that $C$ is at the vertex where sides $a$ and $b$ meet. If the angle is at a different vertex, the formula gives the wrong answer.

Composite Areas, Combining All Formulas:

The strategy from earlier lessons still applies: identify components, decide add or subtract, calculate each, combine.

Rectangle − sector
Rectangle area − sector area (corner cut)
Triangle + sector
Triangle area + sector area (attached)
Logo problem
Triangle area − 3 × sector area (corner sectors removed)
Negative area = wrong decision: If you get a negative area, you have added when you should have subtracted. Never write a negative area as a final answer, use it as a signal to re-examine your add/subtract choice.

Sector area = (θ/360) × πr². The sector is a fraction of the full circle’s area, with the fraction equal to θ/360. Use exact values throughout and round only in the final answer.

Pause, copy the sine area rule A = ½ab sin C and its usage condition (two sides and the included angle known; no perpendicular height available) into your book.

Simplify the fraction before multiplying, cleaner working, fewer calculator errors.
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PROBLEM 2 · AREA OF AN ANNULUS

An annulus has an outer radius of 9 cm and an inner radius of 5 cm. Find its area correct to 2 decimal places.

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$R = 9\text{ cm}$ (outer), $r = 5\text{ cm}$ (inner)
Always write both radii explicitly before substituting. $R$ is always larger.
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PROBLEM 3 · COMPOSITE AREA, LOGO

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. (b) Find the total area of the three sectors. (c) Find the logo area.

a
$A_\triangle = \tfrac{1}{2} \times 8 \times 8 \times \sin 60° = 32\sin 60° = 32 \times 0.86602\ldots = 27.71\text{ cm}^2$
Equilateral triangle: $a = b = 8$, included angle $C = 60°$. The sine area rule handles this exactly.
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Quick check: An annulus has outer radius 8 m and inner radius 3 m. Its area (to 2 d.p.) is:

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Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Trap 01
Arc length formula instead of sector area
Using $(\theta/360) \times 2\pi r$ when asked for area gives a length answer. Check units: area must have square units. If you get cm when the question asks for cm², you used the arc formula.
Trap 02
Wrong subtraction order in annulus
$\pi(r^2 - R^2)$ gives a negative area. Always write $R$ = outer and $r$ = inner before substituting, then subtract $R^2 - r^2$ (larger minus smaller).
Trap 03
Non-included angle in sine area rule
Using an angle that is not between the two known sides gives a wrong area even though the formula looks correct. The included angle $C$ must be at the vertex where sides $a$ and $b$ meet.

Fill the gap: A sector with $r = 6$ cm and $\theta = 120°$ has area $A = \frac{120}{360} \times \pi \times 6^2 = \frac{1}{3} \times$ $= 12\pi$ cm².

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Revisit your thinking

Quick-fire practice · 9 calculations

Quick-fire practice · 9 calculations

1

Find the area of a sector: $r = 5$ cm, $\theta = 72°$. Give exact and decimal form.

2

Find the area of a sector to 2 d.p.: $r = 12$ m, $\theta = 150°$.

3

A sector has area $30\pi$ cm² and radius 6 cm. Find the central angle $\theta$.

4

Find the area of an annulus to 2 d.p.: $R = 10$ cm, $r = 6$ cm.

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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 d.p.

6

Find the area of a triangle to 2 d.p.: sides 7 cm and 9 cm, included angle 40°.

7

Find the area to 2 d.p.: sides 15 m and 15 m, included angle 100°.

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 d.p.

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 d.p.

Odd one out: Three of these statements about area formulas are correct. Which one is wrong?

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Revisit your thinking

Go back to your Think First answer. The annulus formula $A = \pi(R^2 - r^2)$ gives exactly the turf area you need, outer circle minus inner circle. You can't measure it directly because the shape is bounded by two curves, not straight edges.

What did you get right? What surprised you? How has your thinking about area changed?

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 32 marks

SA 1. Find the area of a sector with radius 9 m and central angle 200°. Give your answer correct to 2 decimal places. (2 marks)

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ApplyBand 43 marks

SA 2. 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. (3 marks)

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AnalyseBand 54 marks

SA 3. 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)

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📖 Comprehensive answers (click to reveal)

Drill 1: $A = \frac{1}{5} \times 25\pi = 5\pi = \mathbf{15.71\text{ cm}^2}$  ·  2: $A = \frac{5}{12} \times 144\pi = 60\pi = \mathbf{188.50\text{ m}^2}$  ·  3: $30\pi = \frac{\theta}{360} \times 36\pi$ → $\frac{\theta}{360} = \frac{30}{36} = \frac{5}{6}$ → $\theta = \mathbf{300°}$

Drill 4: $A = \pi(100-36) = 64\pi = \mathbf{201.06\text{ cm}^2}$  ·  5: $A = \pi(5.5^2 - 4^2) = \pi(30.25-16) = 14.25\pi = \mathbf{44.77\text{ m}^2}$  ·  6: $A = \frac{1}{2} \times 63 \times \sin 40° = \mathbf{20.25\text{ cm}^2}$  ·  7: $A = \frac{1}{2} \times 225 \times \sin 100° = \mathbf{110.79\text{ m}^2}$

Drill 8: Square: $64$ cm²; Sector: $\frac{1}{4}\pi \times 64 = 16\pi = 50.27$ cm²; Remaining $= 64 - 50.27 = \mathbf{13.73\text{ cm}^2}$  ·  9: Rectangle: $60$; Semicircle: $4.5\pi = 14.14$; Triangle: $18 \times \sin 50° = 13.79$; Total $= 60 + 14.14 - 13.79 = \mathbf{60.35\text{ cm}^2}$

SA 1 (2 marks): $A = \frac{5}{9} \times 81\pi = 45\pi = \mathbf{141.37\text{ m}^2}$ [2].

SA 2 (3 marks): $R = 0.8$ m, $r = 0.4$ m [1]; $A = \pi(0.64 - 0.16) = 0.48\pi = \mathbf{1.51\text{ m}^2}$ [2].

SA 3(a) (2 marks): $A = \frac{1}{2} \times 8 \times 8 \times \sin 60° = 32 \times \frac{\sqrt{3}}{2} = 27.7128\ldots = \mathbf{27.71\text{ cm}^2}$ [2]. (b) (1 mark): One sector: $\frac{1}{6} \times 9\pi = \frac{3\pi}{2}$; Three: $\frac{9\pi}{2} = \mathbf{14.14\text{ cm}^2}$ [1]. (c) (1 mark): $27.7128\ldots - 14.1371\ldots = \mathbf{13.58\text{ cm}^2}$ [1].