Perimeter and Arc Length
Trace the boundary. Every edge counts — straight or curved. The arc is just a fraction of the full circumference, and the fraction is determined by the angle.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
A pizza slice (sector) has a curved crust and two straight edges going in to the centre. If you wanted to know the total length of pastry needed to frame the slice, which edges would you measure? Is the crust the only part that matters?
Without calculating — write your initial thinking.
Three formulas drive every question in this lesson. Lock these in — arc length is a fraction of the circumference, and sector perimeter always includes the two straight radii.
The circumference $C = 2\pi r$ is the full boundary of a circle. An arc is a fraction of that boundary: the fraction is $\theta/360$. The perimeter of a sector includes the arc and the two straight radius edges.
Key facts
- The circumference formula $C = 2\pi r$ and its equivalents
- The arc length formula $\ell = (\theta/360) \times 2\pi r$
- What a composite perimeter problem requires
Concepts
- Why an arc is a fraction of the full circumference — and why that fraction is $\theta/360$
- Why the perimeter of a sector includes two radii, not just the arc
- Why you must account for every boundary edge, including straight edges
Skills
- Calculate arc length for any sector
- Find the perimeter of a sector (arc + two radii)
- Find the perimeter of composite shapes involving straight sides and arcs
Perimeter is the total length of the boundary. For any polygon — add every side. For composite shapes — trace the outer boundary and add every edge you cross.
Composite Perimeter Strategy:
- Trace the outer boundary of the shape with your finger
- Every edge your finger crosses is part of the perimeter
- Any internal construction lines are not part of the perimeter
- Add every boundary edge — straight and curved sections separately, then combine
An arc is a curved portion of a circle's circumference. A sector with central angle $\theta$ contains an arc that is exactly $\theta/360$ of the full circumference.
A sector has three edges — two straight radii and one curved arc. The perimeter includes all three.
Running Track Example: A rectangle with a semicircle on each short end. The outer perimeter consists of:
- Two long straight sides of the rectangle
- Two semicircular arcs — which together make one full circle
- The short sides of the rectangle are NOT included — they are replaced by the semicircles (internal)
What to write in your book
- Circumference: $C = 2\pi r$ (or $\pi d$). The full boundary of a circle.
- Arc length: $\ell = \frac{\theta}{360} \times 2\pi r$ — arc is fraction $\theta/360$ of the circumference.
- Sector perimeter: $P = 2r + \ell$. Three edges — two radii plus the arc. Write this formula first.
- Composite perimeters: trace the outer boundary — do not count interior edges shared between two shapes.
- Two semicircles of same radius = one full circle ($2\pi r$).
Did you get this? True or false: the arc length of a semicircle (180°) with radius $r$ is $\pi r$.
Worked examples · 3 in a row, reveal as you go
Find the arc length of a sector with radius 9 cm and central angle 80°. Give your answer correct to 2 decimal places.
Find the perimeter of a sector with radius 12 m and central angle 135°. Give your answer correct to 2 decimal places.
A running track consists of a rectangle 60 m long and 20 m wide, with a semicircle attached to each short end. Find the perimeter of the outside edge of the track correct to 2 decimal places.
What to write in your book
- Sector perimeter key rule: $P = 2r + \ell$ — always write this before substituting.
- Composite perimeter: trace boundary physically. Every edge your finger crosses is included; interior shared edges are not.
- Two semicircles of equal radius = one full circle. Use $C = 2\pi r$ directly.
- Keep answers as multiples of $\pi$ until the final step to avoid rounding error.
Quick check: A sector has radius 8 m and arc length 12 m. Its perimeter is:
Common errors · the 3 traps that cost marks
What to write in your book
- Always write $P = 2r + \ell$ before substituting for any sector perimeter question.
- Trace the outer boundary before writing numbers — do not rely on a diagram.
- Arc length has linear units; sector area has square units. Mismatched units signal the wrong formula.
Fill the gap: The arc length of a sector with $r = 6$ cm and $\theta = 90°$ is $\ell = \frac{90}{360} \times 2\pi \times 6 = $ cm (exact).
Quick-fire practice · 8 calculations
Find the arc length: $r = 6$ cm, $\theta = 90°$.
Find the arc length: $r = 15$ m, $\theta = 120°$.
Find the arc length to 2 d.p.: $r = 8$ cm, $\theta = 45°$.
Find the perimeter of a sector: $r = 10$ cm, $\theta = 90°$.
Find the perimeter to 2 d.p.: $r = 7$ m, $\theta = 150°$.
A shape is made from a rectangle 8 cm × 5 cm with a semicircle of diameter 5 cm attached to one short end. Find the perimeter of the outside edge to 2 d.p.
A sector has radius 6 m and angle 240°. Find its perimeter to 2 d.p.
A quarter-circle of radius 4 cm is removed from the corner of a square of side 4 cm. Find the perimeter of the resulting shape to 2 d.p.
Odd one out: Three of these statements about sector perimeter are correct. Which one is wrong?
Go back to your Think First answer. A pizza slice perimeter = curved crust (arc) + two straight edges (radii) = $P = 2r + \ell$. The crust is only one of three boundary edges.
What did you get right? What surprised you? How has your thinking about perimeter changed?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
SA 1. Find the arc length of a sector with radius 18 cm and central angle 40°. Give your answer correct to 2 decimal places. (2 marks)
SA 2. Find the perimeter of a sector with radius 14 cm and central angle 225°. Give your answer correct to 2 decimal places. (3 marks)
SA 3. A garden path is bounded by three straight sides (8 m, 5 m, and 8 m) and a curved arc on the fourth side. The arc is part of a circle with radius 5 m, centred at the midpoint of the 5 m side.
(a) Find the central angle of the arc. (1 mark)
(b) Find the arc length correct to 2 decimal places. (1 mark)
(c) Find the total perimeter correct to 2 decimal places. (2 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $\ell = \frac{1}{4} \times 12\pi = 3\pi = \mathbf{9.42\text{ cm}}$ · 2: $\ell = \frac{1}{3} \times 30\pi = 10\pi = \mathbf{31.42\text{ m}}$ · 3: $\ell = \frac{1}{8} \times 16\pi = 2\pi = \mathbf{6.28\text{ cm}}$ · 4: $\ell = \frac{1}{4} \times 20\pi = 5\pi$; $P = 20 + 5\pi = \mathbf{35.71\text{ cm}}$ · 5: $\ell = \frac{5}{12} \times 14\pi = \frac{35\pi}{6}$; $P = 14 + \frac{35\pi}{6} = \mathbf{32.33\text{ m}}$
Drill 6: Boundary: $2 \times 8 + 5 + \frac{1}{2} \times 2\pi \times 2.5 = 16 + 5 + 2.5\pi = \mathbf{28.85\text{ cm}}$ · 7: $\ell = \frac{240}{360} \times 12\pi = 8\pi$; $P = 12 + 8\pi = \mathbf{37.13\text{ m}}$ · 8: Three full sides + quarter arc: $P = 3 \times 4 + \frac{1}{4} \times 2\pi \times 4 = 12 + 2\pi = \mathbf{18.28\text{ cm}}$
SA 1 (2 marks): $\ell = \frac{40}{360} \times 2\pi \times 18 = \frac{1}{9} \times 36\pi = 4\pi = \mathbf{12.57\text{ cm}}$ [2].
SA 2 (3 marks): $\ell = \frac{225}{360} \times 2\pi \times 14 = \frac{5}{8} \times 28\pi = 17.5\pi = 54.98$ cm [1]; $P = 28 + 54.98 = \mathbf{82.98\text{ cm}}$ [2].
SA 3(a): The two radii (5 m each) and the chord (5 m) form an equilateral triangle. Central angle = 60° [1]. (b) $\ell = \frac{60}{360} \times 2\pi \times 5 = \frac{10\pi}{6} = \mathbf{5.24\text{ m}}$ [1]. (c) $P = 8 + 5 + 8 + 5.24 = \mathbf{26.24\text{ m}}$ [2].
Face the boss using your knowledge of perimeter and arc length calculations. Pool: lessons 1–5. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%).
⚔ Enter the arenaClimb platforms by answering perimeter and arc length questions. Pool: lessons 1–5.
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