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.).