Mathematics Standard • Year 11 • Module 2 • Lesson 5
Perimeter and Arc Length
Apply perimeter and arc length to realistic scenarios — fences, garden edges, running tracks, fairground rides, sector windows.
Problem 1 — Fencing a rectangular yard
A backyard is rectangular, 18.5 m long and 12 m wide. A gate (1.2 m wide) goes in along one short end, replacing that section of fence.
Set up: What are we solving for?
(i) Calculate the full perimeter of the yard (without the gate adjustment). 1 mark
(ii) Subtract the gate width to find the actual length of fencing needed. 2 marks
(iii) Fencing panels are 1.8 m long. How many full panels are required (round UP, justify direction)? 2 marks
Stuck on (iii)? Divide fencing length by panel length, then round up because a partial panel must be a whole panel.Problem 2 — School running track (composite perimeter)
A school's running track has a rectangular infield 100 m long and 50 m wide, with a semicircle at each short end (diameter = 50 m).
Set up: What are we solving for?
(i) Identify which edges form the OUTER boundary. 1 mark
(ii) Calculate the total outer perimeter (one lap distance) to 2 d.p. Use the trick that two equal semicircles = one full circle. 3 marks
(iii) A student runs 5 laps for warmup. Find the total distance to the nearest metre. 2 marks
Stuck on (i)? The short ends of the rectangle are replaced by the semicircles — they are NOT on the outer boundary.Problem 3 — Pizza-slice frame (sector perimeter)
A novelty pizza-slice picture frame is shaped like a sector with radius 28 cm and central angle 45°. Decorative tape is fitted around the outside.
Set up: What are we solving for?
(i) Calculate the arc length (curved crust) to 2 d.p. 2 marks
(ii) Calculate the total perimeter (arc + two radii) to 2 d.p. 2 marks
(iii) Decorative tape costs $0.85 per cm and is sold by the cm (any whole-cm length). Find the cost (round UP to the next whole cm of tape needed). 2 marks
Stuck? Revisit lesson § Perimeter of a Sector — P = 2r + ℓ. The MOST common error is forgetting the two radii.Problem 4 — Garden path with curved end
A garden path is bounded by three straight sides (8 m, 5 m, and 8 m) and a curved end on the fourth side. The curved end is a 60° arc of a circle with radius 5 m, with its centre at the midpoint of the 5 m side.
Set up: What are we solving for?
(i) Sketch and identify the four boundary edges (three straight + one arc). 1 mark
(ii) Calculate the arc length to 2 d.p. 2 marks
(iii) Calculate the total perimeter (three straight sides + arc) to 2 d.p. 2 marks
Stuck on (ii)? The arc formula is ℓ = (60/360) × 2π × 5 = (1/6) × 10π. Keep π exact.Problem 5 — Ferris wheel ride distance (arc length real-world)
A Sydney Easter Show Ferris wheel has a radius of 18 m. A passenger sits in a carriage as the wheel rotates by 270° (three-quarters of a full turn) before being briefly stopped.
Set up: What are we solving for?
(i) What fraction of a full circle is 270°? 1 mark
(ii) Calculate the arc length traced by the passenger (distance travelled along the curve) to 2 d.p. 3 marks
(iii) The wheel takes 4 minutes for a complete revolution (360°). How long (in seconds) does the 270° trip take? 2 marks
Stuck on (iii)? 270° is 3/4 of a full turn, so the time is 3/4 of 4 minutes — then convert to seconds.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Fencing a yard
Set up. We are finding the perimeter, subtracting the gate width, then converting to whole panels.
(i) Full perimeter = 2(18.5) + 2(12) = 37 + 24 = 61 m.
(ii) Fencing needed = 61 − 1.2 = 59.8 m.
(iii) Panels = 59.8 / 1.8 = 33.222... → round UP to 34 panels. A partial panel can't be installed as fence; you must buy a whole panel even if you only use part of the last one.
Problem 2 — Running track
Set up. We are finding one lap (outer perimeter) then 5 laps.
(i) Outer boundary: 2 long straight sides (100 m each) + 2 semicircular arcs. The two short ends of the rectangle are interior.
(ii) Two equal semicircles (r = 25) = one full circle: C = 2π × 25 = 50π = 157.08 m. Two long sides = 200 m. Total = 200 + 50π = 200 + 157.08 = 357.08 m (to 2 d.p.).
(iii) 5 laps = 5 × 357.08 = 1785.4 ≈ 1785 m (to nearest m).
Problem 3 — Pizza-slice frame
Set up. We are finding the arc length, the total sector perimeter, then the tape cost.
(i) ℓ = (45/360) × 2π × 28 = (1/8) × 56π = 7π = 21.99 cm (to 2 d.p.).
(ii) P = 2r + ℓ = 2(28) + 21.99 = 56 + 21.99 = 77.99 cm.
(iii) Tape needed = 78 cm (round 77.99 up). Cost = 78 × $0.85 = $66.30.
Problem 4 — Garden path
Set up. We are finding the perimeter of a four-edged shape (three straight, one arc).
(i) Boundary = 8 m + 5 m + 8 m straight sides + one arc of a circle with r = 5, θ = 60°.
(ii) ℓ = (60/360) × 2π × 5 = (1/6) × 10π = 10π/6 = 5π/3 = 5.24 m (to 2 d.p.).
(iii) P = 8 + 5 + 8 + 5.24 = 26.24 m (to 2 d.p.).
Problem 5 — Ferris wheel
Set up. We are finding the arc length for 270°, then the time it takes given a 4-minute full revolution.
(i) 270/360 = 3/4.
(ii) ℓ = (3/4) × 2π × 18 = (3/4) × 36π = 27π = 84.82 m (to 2 d.p.).
(iii) Time = (3/4) × 4 min = 3 min = 180 seconds.