Mathematics Advanced • Year 11 • Module 2 • Lesson 2

Arc Length & Area of Sectors

Build procedural fluency in the radian formulas l = rθ and A = ½ r² θ, including the convert-first rule.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the two radian formulas:

Arc length: l = ____________

Sector area: A = ____________

Q1.2 Which unit must θ be in for both formulas above to hold without any extra factors?

________________________________

Q1.3 Write the rearranged versions of each formula (for solving for θ):

From l = rθ: θ = ____________

From A = ½ r² θ: θ = ____________

Stuck? Revisit lesson § The two moves and § Arc length and sector area.

2. Worked example — sector area with a degree-given angle

Follow each line of algebra. Every step has a reason on the right.

Problem. A sector has radius 10 cm and central angle 60°. Find its area in exact form.

Step 1 — Convert the angle to radians first.

60° × π/180 = 60π/180 = π/3 rad.

Reason: the formula A = ½ r² θ requires radians (Trap 01 in the lesson).

Step 2 — Write the area formula and substitute.

A = ½ r² θ = ½ × 10² × π/3

Reason: substitute r = 10 and θ = π/3 directly.

Step 3 — Compute step by step, keeping π exact.

= ½ × 100 × π/3 = 50 × π/3 = 50π/3.

Reason: never collapse π to 3.14 in HSC exact-value working.

Step 4 — Add the unit.

A = 50π/3 cm².

Conclusion. A = 50π/3 cm² (≈ 52.36 cm²).

3. Faded example — arc length with mixed units

A sector has radius 8 cm and central angle π/4 rad. Find the arc length. Fill in each blank. 4 marks

Step 1 — Check the angle's units.

θ = π/4 is already in ____________; no conversion needed.

Step 2 — Write and substitute into the arc-length formula.

l = rθ = ________ × ________

Step 3 — Simplify the product.

= ________π/________ = ________π.

Step 4 — Add the unit.

l = ________π cm.

Conclusion. Arc length = ________________ cm (≈ ________ cm to 2 d.p.).

Stuck? Revisit lesson § Worked Example 1 — Arc Length.

4. Graduated practice — arc length and area

Find the unknown quantity. Show one line of substitution and one of simplification. Keep answers in exact form.

Foundation — substitute and simplify (4 questions)

Q	Given	Working (1 line)	Answer (exact)
4.1 1	r = 6, θ = π/3. Find l.
4.2 1	r = 5, θ = π. Find l.
4.3 1	r = 4, θ = π/2. Find A.
4.4 1	r = 10, θ = 3π/4. Find A.

Standard — typical HSC difficulty (6 questions)

Convert any degree-given angles to radians first. Show your working.

4.5 A sector has r = 12 cm and θ = 45°. Find both l and A (exact form). 2 marks

4.6 A sector has r = 9 cm and θ = 2π/3 rad. Find both l and A. 2 marks

4.7 A sector has arc length l = 8π cm and r = 4 cm. Find θ (in radians, exact form), and interpret in one sentence. 2 marks

4.8 A sector has area A = 50π cm² and θ = π/2 rad. Find r. 2 marks

4.9 A sector has area A = 27π cm² and r = 6 cm. Find θ (radians, exact form). 2 marks

4.10 A sector has arc length l = 15 cm and θ = 5π/6 rad. Find r (exact form). 2 marks

Extension — combine concepts (2 questions)

4.11 A sector has perimeter (two radii + arc) equal to 40 cm and radius 12 cm. Find the central angle θ in radians (exact form). 3 marks

4.12 A sector has area 18π cm² and arc length 6π cm. Find both r and θ (exact form). Show your method — you'll need both formulas simultaneously. 3 marks

Stuck on 4.12? Divide A = ½ r² θ by l = rθ to eliminate θ and find r first.

5. Self-check the easy 3

Tick the first three once you've verified your method works.

For 4.1 I got l = 2π cm. I substituted θ in radians (not degrees) before multiplying.

For 4.3 I got A = 4π cm². I included the factor of ½ (didn't fall into Trap 03).

For 4.4 I got A = 75π/2 cm². I kept π exact and didn't switch to a decimal.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Two radian formulas

Arc length: l = rθ. Sector area: A = ½ r² θ. (Both with θ in radians.)

Q1.2 — Required angle unit

Radians. Using degrees directly is Trap 01 in the lesson; the answers will be wrong by a factor of π/180.

Q1.3 — Rearranged formulas

θ = l / r (from arc-length formula). θ = 2A / r² (from area formula, after multiplying both sides by 2 / r²).

Q3 — Faded example: l for r = 8, θ = π/4

Step 1: θ is already in radians; no conversion needed.
Step 2: l = rθ = 8 × π/4.
Step 3: = 8π/4 = 2π.
Step 4: l = 2π cm.
Conclusion: arc length = 2π cm (≈ 6.28 cm to 2 d.p.).

Q4.1 — r = 6, θ = π/3, find l

l = 6 × π/3 = 6π/3 = 2π cm.

Q4.2 — r = 5, θ = π, find l

l = 5 × π = 5π cm. (This is a semicircular arc; equals half the circumference 2πr/2 = πr.)

Q4.3 — r = 4, θ = π/2, find A

A = ½ × 4² × π/2 = ½ × 16 × π/2 = 8 × π/2 = 4π cm². (A quarter-circle of radius 4; check: πr²/4 = 16π/4 = 4π. ✓)

Q4.4 — r = 10, θ = 3π/4, find A

A = ½ × 10² × 3π/4 = ½ × 100 × 3π/4 = 50 × 3π/4 = 75π/2 cm² (≈ 117.81 cm²).

Q4.5 — r = 12, θ = 45°

Convert: 45° = π/4 rad. l = 12 × π/4 = 3π cm. A = ½ × 144 × π/4 = 72 × π/4 = 18π cm².

Q4.6 — r = 9, θ = 2π/3

l = 9 × 2π/3 = 18π/3 = 6π cm. A = ½ × 81 × 2π/3 = 81 × π/3 = 27π cm².

Q4.7 — l = 8π, r = 4, find θ

θ = l/r = 8π/4 = 2π rad. Interpretation: 2π rad is a full revolution, so the "sector" is actually the entire circle (the arc wraps all the way round).

Q4.8 — A = 50π, θ = π/2, find r

50π = ½ r² × π/2 = π r² / 4. Multiply both sides by 4/π: r² = 200. r = √200 = 10√2 cm.

Q4.9 — A = 27π, r = 6, find θ

θ = 2A / r² = 2 × 27π / 36 = 54π/36 = 3π/2 rad. (Equivalent to 270°.)

Q4.10 — l = 15, θ = 5π/6, find r

r = l / θ = 15 / (5π/6) = 15 × 6/(5π) = 90/(5π) = 18/π cm (≈ 5.73 cm).

Q4.11 — Perimeter 40 cm, r = 12 cm

Perimeter = 2r + l = 40 ⇒ l = 40 − 2(12) = 16 cm. Then θ = l/r = 16/12 = 4/3 rad (≈ 76.4°).

Q4.12 — A = 18π cm², l = 6π cm

Divide A by l to cancel θ: A/l = (½ r² θ) / (rθ) = r/2. So r = 2 A/l = 2(18π)/(6π) = 36π/(6π) = 6 cm.
Then θ = l/r = 6π/6 = π rad. (Check area: ½ × 36 × π = 18π ✓.) The sector is a semicircle of radius 6.