1MA1 · G17–G18Geometry & MeasuresFoundation & HigherGrades 4–7

Circles, Arcs & Sectors

The distance around a circle is always a little over three times its diameter — exactly π\pi times. From that one constant come the formulas for circumference and area, and from those come arcs and sectors, which are simply fractions of a whole circle.

45 min Mensuration

The big picture

An arc is part of the circumference, and a sector is part of the area — like a slice of pizza. Both are the same fraction of the whole circle: the angle at the centre divided by 360∘360^{\circ}. That means one idea handles every arc and sector question.

Edexcel often asks for answers "in terms of π\pi", which means leaving π\pi as a symbol, like 7π7\pi cm. That keeps answers exact and avoids rounding — and is often quicker, because you never need to multiply by 3.14159.

What you'll learn

  • Calculate the circumference and area of a circle
  • Calculate arc lengths and sector areas
  • Find the perimeter of a sector and of compound shapes with circular parts
  • Give answers in terms of π\pi and work backwards to find a radius or angle
1

Circumference and area

Circumference C=πd=2πrC = \pi d = 2\pi r. Area A=πr2A = \pi r^{2}.

Check whether you are given the radius or the diameter; the area formula needs the radius.

In terms of π\pi: a circle of radius 5 cm has area 25π25\pi cm² and circumference 10π10\pi cm.

C=2πrC = 2\pi r
Circumference.
A=πr2A = \pi r^{2}
Area.
Example 1A circular pond has diameter 12 m. Find its area to 1 decimal place.
1Radius =6= 6 m.
2A=π×36=113.097…A = \pi \times 36 = 113.097\dots
Answer113.1 m²
Your turnSame method, new numbers — try it before moving on.

Find the area of a circle with radius 7 cm, to 1 decimal place.

To 1 decimal place.

Show worked solution
Solution
  1. 1A=π×72=49πA = \pi \times 7^2 = 49\pi.
  2. 2=153.938…= 153.938\ldots

Answer: 153.9153.9 cm²

Tip — Square the radius before multiplying by π\pi. πr2\pi r^{2} is π×r2\pi \times r^{2}, not (πr)2(\pi r)^{2}.

2

Arcs and sectors

A sector with angle θ\theta at the centre is θ360\dfrac{\theta}{360} of the whole circle.

So arc length =θ360×2πr= \dfrac{\theta}{360} \times 2\pi r and sector area =θ360×πr2= \dfrac{\theta}{360} \times \pi r^{2}.

The perimeter of a sector is the arc length plus two radii.

l=θ360×2πrl = \dfrac{\theta}{360} \times 2\pi r
Arc length.
A=θ360×πr2A = \dfrac{\theta}{360} \times \pi r^{2}
Sector area.
Example 1A sector has radius 9 cm and angle 140∘140^{\circ}. Find, in terms of π\pi, its arc length and area, and find its perimeter to 1 d.p.
1Arc =140360×18π=7π= \dfrac{140}{360} \times 18\pi = 7\pi cm.
2Area =140360×81π=31.5π= \dfrac{140}{360} \times 81\pi = 31.5\pi cm².
3Perimeter =7π+18=39.99…= 7\pi + 18 = 39.99\dots
AnswerArc 7π7\pi cm, area 31.5π31.5\pi cm², perimeter 40.0 cm
Your turnSame method, new numbers — try it before moving on.

A sector has radius 6 cm and angle 100∘100^\circ. Find its arc length to 1 decimal place.

To 1 decimal place.

Show worked solution
Solution
  1. 1Arc =100360×2π×6= \frac{100}{360} \times 2\pi \times 6.
  2. 2=10.47…= 10.47\ldots cm.

Answer: 10.510.5 cm

Why this works

A semicircle is just a sector with θ=180∘\theta = 180^{\circ}, and a quadrant has θ=90∘\theta = 90^{\circ}. The fraction method covers them without separate formulas.

3

Compound shapes

Shapes combining rectangles and semicircles, or circles with parts removed, are handled by adding and subtracting areas.

For perimeters, include only the outer edges — not the lines where shapes join.

Example 1A running track is a rectangle 100 m by 60 m with a semicircle on each short end. Find the distance around the track.
1Two straights: 200200 m.
2Two semicircles make a full circle of diameter 60: 60π60\pi m.
3Total =200+60π=388.5= 200 + 60\pi = 388.5 m.
Answer200+60π≈388.5200 + 60\pi \approx 388.5 m
4

Working backwards

If the area or arc length is given, substitute into the formula and solve for the radius or the angle.

To find rr from an area, divide by π\pi then square root.

Example 1A sector has angle 120∘120^{\circ} and arc length 10π10\pi cm. Find its radius.
1120360×2πr=10π\dfrac{120}{360} \times 2\pi r = 10\pi.
22πr3=10π\dfrac{2\pi r}{3} = 10\pi, so r=15r = 15.
Answer15 cm
Your turnSame method, new numbers — try it before moving on.

A sector has angle 60∘60^\circ and arc length 4π4\pi cm. Find its radius, in cm.

Show worked solution
Solution
  1. 160360×2πr=4π\frac{60}{360} \times 2\pi r = 4\pi, so πr3=4π\frac{\pi r}{3} = 4\pi.
  2. 2r=12r = 12 cm.

Answer: 1212 cm

Tip — When the answer is "in terms of π\pi", divide both sides by π\pi early — the π\pi cancels and the algebra is simpler.

Think like an examiner

  • State whether you are using radius or diameter.
  • Leave answers in terms of π\pi when asked; otherwise round as instructed.
  • Include both radii in a sector perimeter.

Circles

C=2πr=πdC = 2\pi r = \pi d
Circumference.
A=πr2A = \pi r^{2}
Area.
θ360×2πr\tfrac{\theta}{360} \times 2\pi r
Arc length.
θ360×πr2\tfrac{\theta}{360} \times \pi r^{2}
Sector area.

Watch out for these

Using the diameter in πr2\pi r^{2}.
Halve the diameter first.
Giving the arc length as the sector perimeter.
Add two radii.
Using θ180\tfrac{\theta}{180} as the fraction.
Use θ360\tfrac{\theta}{360}.
Including internal joining lines in a compound perimeter.
Only the outer boundary counts.

Stretch yourself

A square of side 10 cm has a quarter circle of radius 10 cm drawn inside it, centred at one corner. Find the area of the region inside the square but outside the quarter circle, in terms of π\pi and to 1 d.p.

Hint — Subtract the quarter-circle area from the square area.

Reveal solution
SolutionSquare area =100= 100 cm². Quarter circle area =14×π×100=25π= \tfrac14 \times \pi \times 100 = 25\pi cm². Region =100−25π≈21.5= 100 - 25\pi \approx 21.5 cm². ■\blacksquare

Questions students ask

What does "in terms of π" mean?
Leave π\pi as a symbol in the answer, e.g. 16π16\pi cm², rather than converting to a decimal.
Is a semicircle’s perimeter half the circumference?
No — add the diameter too, the straight edge.
Which value of π should I use?
Use the π button on your calculator unless the question says otherwise.

Key takeaways

  • Circumference 2πr2\pi r; area πr2\pi r^{2}.
  • Arcs and sectors are θ360\tfrac{\theta}{360} of the circle.
  • Sector perimeter == arc +2r+ 2r.
  • Split compound shapes; count only outer edges for perimeter.
  • Work backwards by substituting and solving for rr or θ\theta.

Exam-style questions

Practise it the way it is examined

3 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 41 mark
  1. A circle has diameter 18 cm. Write down its circumference in terms of π\pi.[1]
    Show mark scheme

    Answer: 18π18\pi cm

Answer on paper first, then mark it.
Question 2Grade 63 marks
  1. A running track has two straight sections of 80 m joined by semicircular ends of diameter 50 m. Work out the length of one lap, to 1 decimal place.[3]
    Show mark scheme

    Answer: 317.1317.1 m

Answer on paper first, then mark it.
Question 3Grade 72 marks
  1. A sector of a circle of radius 12 cm has area 24π24\pi cm². Work out the angle of the sector.[2]
    Show mark scheme

    Answer: 60∘60^\circ

Answer on paper first, then mark it.

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