Mathematics Standard • Year 11 • Module 2 • Lesson 6
Area of Sectors, Annuli, and Composite Shapes
Build fluency with the three core area formulas: sector A = (θ/360) × πr², annulus A = π(R² − r²), and the sine area rule A = ½ab sin C.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each formula. Use θ for central angle, r for radius (and R for outer radius where applicable).
Sector area: A = __________________ Annulus area: A = __________________
Sine area rule (triangle): A = __________________
Q1.2 A sector and an arc both use the fraction θ/360. Write what comes after the fraction in each case.
Arc length ℓ = (θ/360) × __________ Sector area A = (θ/360) × __________
Q1.3 In the sine area rule A = ½ab sin C, what does the letter C represent? Tick one.
2. Worked example — sector area
Follow each line of working. Every step has a reason on the right.
Problem. Find the area of a sector with radius 10 cm and central angle 144°. Give your answer in exact form and to 2 decimal places.
Step 1 — Write the formula and substitute.
A = (144/360) × π × 10²
Reason: sector area formula. θ = 144°, r = 10 cm.
Step 2 — Simplify the fraction before multiplying.
144/360 = 2/5 so A = (2/5) × 100π
Reason: cleaner working, fewer calculator slips.
Step 3 — Evaluate.
A = 40π cm² (exact) = 125.6637... ≈ 125.66 cm²
Reason: keep π exact for the intermediate step; only convert to decimal once.
Conclusion. The sector area is 40π cm² ≈ 125.66 cm².
3. Faded example — fill in the missing steps
An annulus has outer radius R = 9 cm and inner radius r = 5 cm. Find the area in exact form and to 2 decimal places. Fill in each blank line. 4 marks
Step 1 — Identify and label.
R = ________ cm (outer), r = ________ cm (inner)
Step 2 — Apply the factored formula: A = π(R² − r²) = π( ________ − ________ ) = π( ________ )
Step 3 — Evaluate the exact form: A = ________ π cm²
Step 4 — Convert to decimal (2 d.p.): A ≈ ________________ cm²
Conclusion. Annulus area = ________ π cm² ≈ ________ cm².
4. Graduated practice — sector, annulus and sine area rule
Show your working in the space below each part. Round decimals to 2 d.p. unless otherwise told.
Foundation — single-step substitutions (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Find a sector area with r = 6 cm and θ = 90°. Leave in exact form. | |
| 4.2 1 | Find a sector area with r = 4 m and θ = 180°. Leave in exact form. | |
| 4.3 1 | Find an annulus area with R = 7 cm and r = 3 cm. Leave in exact form. | |
| 4.4 1 | Find the area of a triangle with sides 8 cm, 5 cm and included angle 90°. (sin 90° = 1.) |
Standard — typical HSC difficulty (6 questions)
Show at least one line of substitution. Label your final answer with square units.
4.5 Find the area of a sector with r = 5 cm and θ = 72°. Give exact form and 2 d.p. 2 marks
4.6 Find the area of a sector with r = 12 m and θ = 150° to 2 d.p. 2 marks
4.7 Find the area of an annulus with R = 10 cm and r = 6 cm to 2 d.p. 2 marks
4.8 A circular path surrounds a garden of radius 4 m. The path extends 1.5 m beyond the garden's edge. Find the path's area to 2 d.p. 2 marks
4.9 Find the area of a triangle with sides 7 cm, 9 cm and included angle 40° to 2 d.p. 2 marks
4.10 Find the area of an isosceles triangle with two equal sides of 15 m and included angle 100° to 2 d.p. 2 marks
Extension — reverse and composite (2 questions)
4.11 A sector has area 30π cm² and radius 6 cm. Find the central angle θ in degrees. 3 marks
4.12 A square of side 8 cm has a quarter-circle sector of radius 8 cm removed from one corner. Find the remaining area in exact form and to 2 d.p. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Three formulas
Sector area: A = (θ/360) × πr². Annulus area: A = π(R² − r²). Sine area rule: A = ½ab sin C.
Q1.2 — Same fraction, different second factor
Arc length: (θ/360) × 2πr (circumference). Sector area: (θ/360) × πr² (full circle area).
Q1.3 — Meaning of C
The angle between sides a and b — the included angle. Using any other angle gives a wrong area.
Q3 — Faded example (annulus R = 9, r = 5)
Step 1: R = 9 cm, r = 5 cm.
Step 2: A = π(9² − 5²) = π(81 − 25) = π(56).
Step 3: Exact form = 56π cm².
Step 4: 56π ≈ 175.93 cm².
Q4.1 — Sector, r = 6, θ = 90°
A = (90/360) × π × 36 = ¼ × 36π = 9π cm² (≈ 28.27 cm²).
Q4.2 — Sector, r = 4, θ = 180° (semicircle)
A = (180/360) × π × 16 = ½ × 16π = 8π m² (≈ 25.13 m²).
Q4.3 — Annulus R = 7, r = 3
A = π(49 − 9) = 40π cm² (≈ 125.66 cm²).
Q4.4 — Right-angled triangle, sides 8 and 5
A = ½ × 8 × 5 × sin 90° = ½ × 40 × 1 = 20 cm².
Q4.5 — Sector, r = 5, θ = 72°
A = (72/360) × π × 25 = ⅕ × 25π = 5π cm² ≈ 15.71 cm².
Q4.6 — Sector, r = 12, θ = 150°
A = (150/360) × π × 144 = (5/12) × 144π = 60π ≈ 188.50 m².
Q4.7 — Annulus R = 10, r = 6
A = π(100 − 36) = 64π ≈ 201.06 cm².
Q4.8 — Circular path around garden
Garden radius r = 4 m. Outer radius R = 4 + 1.5 = 5.5 m.
A = π(5.5² − 4²) = π(30.25 − 16) = 14.25π ≈ 44.77 m².
Q4.9 — Triangle, sides 7, 9, angle 40°
A = ½ × 7 × 9 × sin 40° = 31.5 × 0.64279... ≈ 20.25 cm².
Q4.10 — Isosceles triangle, sides 15, 15, angle 100°
A = ½ × 15 × 15 × sin 100° = 112.5 × 0.98481... ≈ 110.79 m².
Q4.11 — Reverse sector (find angle)
30π = (θ/360) × π × 36. Divide both sides by π: 30 = (θ/360) × 36.
So θ/360 = 30/36 = 5/6, giving θ = 5/6 × 360 = 300°.
Q4.12 — Square minus quarter-circle
Square area = 8² = 64 cm². Sector (θ = 90°, r = 8): A = ¼ × π × 64 = 16π cm².
Remaining = 64 − 16π cm² (exact) ≈ 64 − 50.27 = 13.73 cm².
Common error: writing 64 − 16π as a single decimal early loses an exact-form mark.