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.