Skip to content
M
hscscience Maths Std · Y11
0/100daily goal
0
0
0 due
0
L1 · 0 XP
KJ
Your weak spots
Insights load after your first practice round.
Module 2 · L5 of 22 ~45 min ⚡ +95 XP available

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.

Today's hook — 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?
0/5QUESTS
Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

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.

auto-saved
02
The formulas you need to own
+5 XP to read

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.

CIRCUMFERENCE ARC LENGTH SECTOR PERIM r C = 2πr or πd θ r r ℓ=(θ/360)×2πr fraction of C r r P = 2r + ℓ 2 radii + arc
$P_{\text{sector}} = 2r + \dfrac{\theta}{360} \times 2\pi r$
Circumference
$C = 2\pi r = \pi d$. The full boundary of a circle. Starting point for all arc calculations.
Arc length
$\ell = \frac{\theta}{360} \times 2\pi r$. The arc is a fraction $\theta/360$ of the full circumference.
Sector perimeter
$P = 2r + \ell$. Three edges: arc + two radii. The most common error is forgetting the two straight sides.
03
What you'll master
Know

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
Understand

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
Can do

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
04
Key terms
PerimeterThe total length of the boundary of a shape — the distance around the outside.
CircumferenceThe perimeter of a circle — $C = 2\pi r$ or $C = \pi d$.
ArcA portion of the circumference — a curved section of the boundary.
SectorA "pizza slice" region of a circle — bounded by two radii and an arc.
Subtended angleThe central angle $\theta$ formed by two radii — determines what fraction of the circle you have.
05
Perimeter of Polygons and Composite Shapes
core concept

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.

Rectangle: $P = 2\ell + 2w$  |  Square: $P = 4s$  |  Triangle: $P = a+b+c$  |  Circle: $C = 2\pi r$

Composite Perimeter Strategy:

  1. Trace the outer boundary of the shape with your finger
  2. Every edge your finger crosses is part of the perimeter
  3. Any internal construction lines are not part of the perimeter
  4. Add every boundary edge — straight and curved sections separately, then combine
The trap: In a rectangle-plus-semicircle, the diameter of the semicircle is an internal edge — it is not on the outer boundary. Only three sides of the rectangle plus the curved arc form the perimeter.
06
Arc Length
core concept

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.

$$\ell = \frac{\theta}{360} \times 2\pi r$$
$\theta = 360°$
Full circle → $2\pi r$ (the full circumference)
$\theta = 180°$
Semicircle → $\pi r$
$\theta = 90°$
Quarter circle → $\tfrac{1}{2}\pi r$
Sense check: When $\theta = 360°$, the arc formula gives $\frac{360}{360} \times 2\pi r = 2\pi r$ — the full circumference. The formula always works, even for the complete circle.
07
Perimeter of a Sector and Composite Arcs
core concept

A sector has three edges — two straight radii and one curved arc. The perimeter includes all three.

$$P = 2r + \ell = 2r + \frac{\theta}{360} \times 2\pi r$$
Most common error: Students calculate the arc length and write it as the "perimeter of the sector." The arc is only one of three edges. Write $P = 2r + \ell$ at the top of your working to commit to accounting for all three boundaries.

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)
Efficiency trick: Two semicircles of the same radius = one full circle. Instead of calculating two separate semicircle arcs, find the circumference of one full circle: $2\pi r$. This saves a step and reduces rounding error.
Distinguish arc length from sector area: Arc length → uses $2\pi r$ → answer in linear units (cm, m). Sector area → uses $\pi r^2$ → answer in square units (cm², m²). If your "arc length" answer has square units, you used the wrong formula.
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 $\pi r$.

PROBLEM 1 · ARC LENGTH OF A SECTOR

Find the arc length of a sector with radius 9 cm and central angle 80°. Give your answer correct to 2 decimal places.

1
$\ell = \dfrac{\theta}{360} \times 2\pi r$   |   $\theta = 80°$, $r = 9$ cm
Identify and write the arc length formula. This is an arc length question.
PROBLEM 2 · PERIMETER OF A SECTOR

Find the perimeter of a sector with radius 12 m and central angle 135°. Give your answer correct to 2 decimal places.

1
$P = 2r + \ell$   (two radii + arc)
Write the plan first — commits you to all three boundary edges.
PROBLEM 3 · COMPOSITE PERIMETER — RUNNING TRACK

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.

1
Boundary: 2 long sides + 2 semicircular arcs (short sides are interior — not included)
The semicircles replace the short ends. The two short sides of the rectangle are internal edges — they are not on the outer boundary.
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:

Trap 01
Only calculating the arc for a sector perimeter
A sector has three edges: arc + two radii. Write $P = 2r + \ell$ before calculating anything. The arc alone is not the perimeter.
Trap 02
Including internal edges in a composite perimeter
Trace the boundary physically before writing numbers. Any edge shared between two joined shapes is interior — do not count it.
Trap 03
Confusing arc length formula with sector area formula
Both start with $\theta/360$. Arc length: multiply by $2\pi r$ → linear units. Sector area: multiply by $\pi r^2$ → square units. Check your answer's units to catch this error.
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).

1

Find the arc length: $r = 6$ cm, $\theta = 90°$.

2

Find the arc length: $r = 15$ m, $\theta = 120°$.

3

Find the arc length to 2 d.p.: $r = 8$ cm, $\theta = 45°$.

4

Find the perimeter of a sector: $r = 10$ cm, $\theta = 90°$.

5

Find the perimeter to 2 d.p.: $r = 7$ m, $\theta = 150°$.

6

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.

7

A sector has radius 6 m and angle 240°. Find its perimeter to 2 d.p.

8

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?

10
Revisit your thinking

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?

auto-saved
01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 32 marks

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)

auto-saved
ApplyBand 43 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)

auto-saved
AnalyseBand 54 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)

auto-saved
📖 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].

01
Boss battle — Perimeter & Arc Length!
earn bronze · silver · gold

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 arena
02
Science Jump · platform challenge

Climb platforms by answering perimeter and arc length questions. Pool: lessons 1–5.

Mark lesson as complete

Tick when you've finished the practice and review.