Mathematics Standard • Year 11 • Module 2 • Lesson 5
Perimeter and Arc Length
Build fluency in perimeter, circumference, arc length, and sector perimeter — trace the boundary, count every edge.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each formula:
Circumference: C = ____________ (or equivalently C = ____________ )
Arc length: ℓ = ____________ × 2πr Sector perimeter: P = ____________ + ℓ
Q1.2 What fraction of a full circle is each arc? (Write the fraction.)
90° arc = ______ 180° arc = ______ 60° arc = ______ 240° arc = ______
Q1.3 A sector has radius 5 cm and arc length 6 cm. Tick the perimeter:
☐ 6 cm ☐ 11 cm ☐ 16 cm ☐ 30 cm
2. Worked example — arc length
Follow each line — every step earns a method mark.
Problem. Find the arc length of a sector with radius 9 cm and central angle 80°, to 2 d.p.
Step 1 — Write the formula.
ℓ = (θ/360) × 2πr
Reason: arc length = fraction of circumference (θ/360 of the full circle).
Step 2 — Substitute.
ℓ = (80/360) × 2 × π × 9
Reason: θ = 80°, r = 9 cm.
Step 3 — Simplify the fraction first.
80/360 = 2/9. ℓ = (2/9) × 18π = 4π
Reason: cleaner working — exact form 4π carried until the very last step.
Step 4 — Evaluate and state with units.
ℓ = 4π = 12.5663...
ℓ = 12.57 cm (to 2 d.p.)
3. Faded example — perimeter of a sector
Find the perimeter of a sector with radius 12 m and central angle 135°. Fill in every blank. 4 marks
Step 1 — Plan: P = 2r + ℓ (two radii + arc)
Step 2 — Arc: ℓ = (135/360) × 2π × 12 = ______ × 24π = ______ π (exact)
Step 3 — Two radii: 2 × ______ = ______ m
Step 4 — Sum: P = ______ + ______ π = ______ + 28.27 = ______ m (to 2 d.p.)
Sense check: P must be greater than 2r (24 m) — the arc adds to that.
4. Graduated practice — arc length and perimeter
For every question: write the formula, substitute, then state the answer with the correct unit.
Foundation — arc length only (4 questions, 2 d.p.)
| Q | Problem | Answer (with unit) |
|---|---|---|
| 4.1 1 | Arc length: r = 6 cm, θ = 90°. | |
| 4.2 1 | Arc length: r = 15 m, θ = 120°. | |
| 4.3 1 | Arc length: r = 8 cm, θ = 45°. | |
| 4.4 1 | Circumference: full circle, r = 10 cm. |
Standard — perimeter problems (6 questions, 2 d.p.)
4.5 Perimeter of a sector: r = 10 cm, θ = 90°. 2 marks
4.6 Perimeter of a sector: r = 7 m, θ = 150°. 2 marks
4.7 Perimeter of a sector: r = 6 m, θ = 240°. 2 marks
4.8 Perimeter of a triangle with sides 7 cm, 24 cm, 25 cm. (Pythagoras hint: this is a right-angled triangle — but you only need to add the three sides.) 1 mark
4.9 A shape is a rectangle 8 cm × 5 cm with a semicircle of diameter 5 cm attached to one short end. Find the OUTSIDE perimeter to 2 d.p. (Hint: the short end where the semicircle attaches is interior — not on the boundary.) 3 marks
4.10 A quarter-circle (r = 4 cm) is removed from the corner of a square (side 4 cm). Find the outside perimeter of the resulting shape to 2 d.p. (Note: the two straight sides of the square that touched the corner are gone; instead the arc curves between them.) 3 marks
Extension — multi-step composite perimeters (2 questions)
4.11 A running track has a rectangular infield 60 m by 20 m, with a semicircle on each short end (diameter = 20 m). Find the perimeter of the outside edge to 2 d.p. (Hint: two equal semicircles combine into one full circle.) 3 marks
4.12 A shape is a square (side 10 cm) with a 60° sector (radius 10 cm) attached to one side. Find the outside perimeter to 2 d.p. (The shared side is interior — gone from the boundary; the two sector radii are new boundary edges.) 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Key formulas
C = 2πr (or πd). ℓ = (θ/360) × 2πr. P = 2r + ℓ.
Q1.2 — Fractions of a circle
90° = 1/4. 180° = 1/2. 60° = 1/6. 240° = 2/3.
Q1.3 — Sector perimeter (r = 5, ℓ = 6)
P = 2r + ℓ = 10 + 6 = 16 cm. Common error: ticking 6 cm — that's only the arc, not the perimeter.
Q3 — Faded sector perimeter
Step 2: ℓ = (135/360) × 24π = (3/8) × 24π = 9π m. Step 3: 2 × 12 = 24 m. Step 4: P = 24 + 9π = 24 + 28.27 = 52.27 m (to 2 d.p.).
Q4.1 — Arc, r = 6, θ = 90°
ℓ = (1/4) × 12π = 3π = 9.42 cm.
Q4.2 — Arc, r = 15, θ = 120°
ℓ = (1/3) × 30π = 10π = 31.42 m.
Q4.3 — Arc, r = 8, θ = 45°
ℓ = (1/8) × 16π = 2π = 6.28 cm.
Q4.4 — Full circumference, r = 10
C = 2π × 10 = 20π = 62.83 cm (to 2 d.p.).
Q4.5 — Sector perimeter, r = 10, θ = 90°
ℓ = (1/4) × 20π = 5π. P = 2 × 10 + 5π = 20 + 15.708 = 35.71 cm.
Q4.6 — Sector perimeter, r = 7, θ = 150°
ℓ = (5/12) × 14π = 35π/6 = 18.326... P = 14 + 18.33 = 32.33 m.
Q4.7 — Sector perimeter, r = 6, θ = 240°
ℓ = (240/360) × 12π = (2/3) × 12π = 8π = 25.133. P = 12 + 25.13 = 37.13 m.
Q4.8 — Triangle perimeter
P = 7 + 24 + 25 = 56 cm.
Q4.9 — Rectangle + semicircle
Boundary: 2 long sides (8 cm each) + 1 short side (5 cm) + semicircle arc.
Semicircle: r = 2.5, arc = π × 2.5 = 2.5π.
P = 16 + 5 + 2.5π = 21 + 7.854 = 28.85 cm (to 2 d.p.).
Q4.10 — Square minus quarter-circle
Boundary: 2 full sides of square (4 cm each) + nothing on the two sides that meet at the missing corner (the arc replaces that corner only — the rest of those two sides is still present along the full length where it does not meet the quarter-circle radius... wait, the quarter-circle has r = 4, equal to the side length, so the arc REPLACES the corner entirely — the two adjacent sides are entirely gone). Boundary: 2 full sides (4 cm each) + 1 quarter-arc (r = 4).
Arc = (1/4) × 2π × 4 = 2π = 6.283.
P = 2 × 4 + 2π = 8 + 6.28 = 14.28 cm.
Q4.11 — Running track perimeter
Two semicircles (diameter 20 → r = 10) combine into one full circle: C = 2π × 10 = 20π.
Two long sides: 2 × 60 = 120 m.
P = 120 + 20π = 120 + 62.83 = 182.83 m (to 2 d.p.).
Q4.12 — Square + sector composite
Boundary: 3 sides of square (4th side is interior — joined to sector) + 2 radii of sector + sector arc.
Arc = (60/360) × 2π × 10 = (1/6) × 20π = 10π/3.
3 sides + 2 radii = 30 + 20 = 50 cm.
P = 50 + 10π/3 = 50 + 10.472 = 60.47 cm (to 2 d.p.).