What is the formula for the circumference of a circle in Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, the circumference of a circle is calculated using the formula C = 2πr, where 'C' represents the circumference and 'r' is the radius of the circle. Alternatively, if the diameter 'd' is known, the formula can be expressed as C = πd.

How do you calculate the area of a circle in Mensuration?

To calculate the area of a circle in Mensuration, use the formula A = πr², where 'A' is the area and 'r' is the radius of the circle. This formula is essential for solving problems related to circles in the Cambridge IGCSE Mathematics curriculum.

What is the difference between the radius and diameter of a circle?

The radius of a circle is the distance from the center of the circle to any point on its circumference, while the diameter is the distance across the circle, passing through the center. The diameter is twice the length of the radius, expressed as d = 2r.

How can you find the length of an arc in a circle?

The length of an arc in a circle can be found using the formula L = θ/360° × 2πr, where 'L' is the arc length, 'θ' is the central angle in degrees, and 'r' is the radius of the circle. This concept is part of the Mensuration topic in Cambridge IGCSE Mathematics.

What is the significance of π (pi) in circle calculations?

In circle calculations, π (pi) is a mathematical constant approximately equal to 3.14159. It represents the ratio of the circumference of any circle to its diameter and is crucial in formulas for calculating the circumference and area of circles in Cambridge IGCSE Mathematics.

Why is "Using diameter instead of radius (or vice versa)" a common mistake in Circles?

Why it happens: Reading the question too quickly. How to avoid it: Note: C = d = 2 r. If the diagram gives diameter, halve it before using A = r^2.

Why is "Using radians in the $\theta/360$ formula" a common mistake in Circles?

Why it happens: Calculator left in radian mode. How to avoid it: 0580 always uses degrees. Set the calculator to DEG mode at the start of the paper. Source: 0580/42 — recurring

Why is "Calling a sector a segment" a common mistake in Circles?

Why it happens: Mixing up the two regions. How to avoid it: Sector = two RADII + arc. Segment = one CHORD + arc.

Why is "Using $\pi r^{2}$ for arc length" a common mistake in Circles?

Why it happens: Mixing up area and length formulae. How to avoid it: Length comes from circumference 2 r. Area comes from r^2.

Cambridge IGCSE Mathematics (0580)

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Short Notes - Circles

Detailed Study Notes

Detailed notes on Mensuration for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Circles, Arcs and Sectors — Cambridge IGCSE 0580 Maths Extended (2026)

Circumference, area, arc length, sector area and segment area. The fraction $\dfrac{\theta}{360}$ is the master tool — apply it to whichever full-circle quantity you want.

At a glance

Circumference: $C = 2\pi r$ or $\pi d$ .
Area of a circle: $A = \pi r^{2}$ .
Arc length: $L = \dfrac{\theta}{360} \times 2\pi r$ .
Sector area: $A_{\text{sector}} = \dfrac{\theta}{360} \times \pi r^{2}$ .
Segment area: sector area minus triangle area.
Use the EXACT $\pi$ button — round only at the end.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E6.3 — Calculate circumference and area of circles.
E6.4 — Calculate arc length and sector area.
E6.4 — Calculate area of a segment of a circle.

Circumference and area of a circle

$C = 2\pi r$ , $A = \pi r^{2}$ . Identify radius vs diameter first.

Formulae. $C = 2\pi r = \pi d, \qquad A = \pi r^{2}.$

The radius $r$ is half the diameter $d$ . Always check which the question gives you.

The diameter is twice the radius; circumference C = 2πr and area A = πr².

Worked. Circle radius $7\,\text{cm}$ . Find $C$ and $A$ to 3 s.f.

$C = 2\pi \times 7 = 14\pi \approx 44.0\,\text{cm}$ .
$A = \pi \times 7^{2} = 49\pi \approx 154\,\text{cm}^{2}$ .

Worked (diameter given). Diameter $20\,\text{cm}$ → radius $10\,\text{cm}$ .

$C = \pi \times 20 = 20\pi \approx 62.8\,\text{cm}$ .
$A = \pi \times 100 = 100\pi \approx 314\,\text{cm}^{2}$ .

Calculator $\pi$ button. Use the calculator's $\pi$ key — don't type $3.14$ unless the question says to.

$C = 2\pi r = \pi d$ .
$A = \pi r^{2}$ .
Radius $=$ diameter $\div 2$ .
Use exact $\pi$ , round at the end.

Arc length and sector area

Multiply the full-circle quantity by $\dfrac{\theta}{360}$ to scale to a sector of angle $\theta$ .

A sector is the pie-slice region of a circle bounded by two radii and an arc. The angle at the centre is $\theta$ (in degrees).

Arc length. $L = \dfrac{\theta}{360} \times 2\pi r.$

Sector area. $A_{\text{sector}} = \dfrac{\theta}{360} \times \pi r^{2}.$

The pattern: $\dfrac{\theta}{360}$ is the FRACTION of a full revolution, so it scales any full-circle quantity to its sector value.

Worked. Sector with radius $9\,\text{cm}$ and angle $80\degree$ .

Arc: $L = \dfrac{80}{360} \times 2\pi \times 9 = \dfrac{2}{9} \times 18\pi = 4\pi \approx 12.6\,\text{cm}$ .
Area: $A = \dfrac{80}{360} \times \pi \times 81 = \dfrac{2}{9} \times 81\pi = 18\pi \approx 56.5\,\text{cm}^{2}$ .

Perimeter of a sector. Two radii plus the arc: $P_{\text{sector}} = 2r + L.$

A common Cambridge trip-up: students forget the two radii.

The fraction θ÷360 scales any full-circle quantity down to the sector.

A sector's share of the circle is θ ÷ 360 — a quarter, half or three-quarters for 90°, 180°, 270°.

Arc: $L = \dfrac{\theta}{360} \times 2\pi r$ .
Sector area: $\dfrac{\theta}{360} \times \pi r^{2}$ .
Sector perimeter: $2r + L$ (two radii + arc).

Segment of a circle

Segment area $=$ sector area $-$ triangle area. Use $\dfrac{1}{2}r^{2}\sin\theta$ for the triangle.

A segment is the region between a chord and the arc it cuts off — a sector with the central triangle sliced away.

Formula. $A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} = \dfrac{\theta}{360}\pi r^{2} - \dfrac{1}{2}r^{2}\sin\theta.$

The triangle formed by the two radii and the chord has area $\dfrac{1}{2}r^{2}\sin\theta$ (see Trigonometry — area rule).

Slice the central triangle off the sector and the curved segment is what remains.

Worked. Segment in a circle radius $10\,\text{cm}$ , central angle $60\degree$ .

Sector area: $\dfrac{60}{360}\pi \times 100 = \dfrac{50\pi}{3} \approx 52.36\,\text{cm}^{2}$ .
Triangle area: $\dfrac{1}{2} \times 100 \times \sin 60\degree = 50 \times \dfrac{\sqrt{3}}{2} \approx 43.30\,\text{cm}^{2}$ .
Segment: $52.36 - 43.30 \approx 9.06\,\text{cm}^{2}$ .

Major vs minor segment. The chord splits the circle into a minor (small) and major (large) segment. The formula above gives the minor segment for $\theta < 180\degree$ .

Segment $=$ sector $-$ triangle.
Triangle area: $\dfrac{1}{2}r^{2}\sin\theta$ .
Minor segment: $\theta < 180\degree$ .

Units, rounding and exact answers

Keep $\pi$ exact through the calculation. Round only at the end (3 s.f. unless told otherwise).

Exact vs decimal answers.

"Give your answer in terms of $\pi$ " → leave $\pi$ in the answer (e.g. $4\pi\,\text{cm}^{2}$ ).
"Give your answer to 3 significant figures" → use the calculator's $\pi$ button, round only at the end.

Common error. Rounding $\pi$ to $3.14$ early in a multi-step calculation. This compounds errors and can lose marks.

Units.

Lengths and arc lengths: linear units ( $\text{cm}, \text{m}$ ).
Areas: squared units ( $\text{cm}^{2}, \text{m}^{2}$ ).
Always include the unit in the final answer.

Quantity	Formula
Circumference	$C = 2\pi r = \pi d$
Area	$A = \pi r^{2}$
Arc length	$\tfrac{\theta}{360} \times 2\pi r$
Sector area	$\tfrac{\theta}{360} \times \pi r^{2}$
Sector perimeter	$2r + \text{arc}$
Segment area	sector $- \tfrac{1}{2}r^{2}\sin\theta$

Use the calculator $\pi$ key.
Round only at the end (3 s.f. typical).
Always include units.

Quick recap

$C = 2\pi r$ . $A = \pi r^{2}$ .
Arc length: $\dfrac{\theta}{360} \times 2\pi r$ .
Sector area: $\dfrac{\theta}{360} \times \pi r^{2}$ .
Sector perimeter: $2r + L$ .
Segment $=$ sector $-$ triangle ( $\dfrac{1}{2}r^{2}\sin\theta$ ).
Use exact $\pi$ — round at the end.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Circumference — perimeter of a circle: $C = 2\pi r$ .
Radius — distance from centre to circle: $r = d/2$ .
Sector — pie-slice region bounded by two radii and an arc.
Arc — portion of the circumference.
Segment — region between a chord and the arc it cuts off.

How it’s examined

Circle, arc and sector questions appear every Paper 4 (3-6 marks each), often combined with composite-shape questions. Examiner reports flag forgetting the two radii in sector perimeter as a frequent error. Exact- $\pi$ answers are increasingly preferred — read the question.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E6.3-6.4); 0580/42 Oct/Nov 2024 — Q11 (sector + segment); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Circles

Step-by-step solutions to past-paper-style questions on circles, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Circles

Almost every circles exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the circumference / area / arc length / sector area — for one circle or sector, using one formula.

How to approach it

Pick the formula and substitute: $C = 2\pi r$ , $A = \pi r^{2}$ , and scale by $\frac{\theta}{360}$ for an arc or sector. Halve the diameter to a radius first if a diameter is given.

Common trap

Squaring the diameter instead of the radius, or mixing the length and area formulae. Examiner reports also flag a calculator left in radian mode — 0580 always uses degrees.

Multi-step problem

Recognise it by

Several pieces combined — a segment (sector minus triangle), a sector perimeter (arc plus two radii), or a compound shape stitched from a rectangle and an arc.

How to approach it

Break the figure into standard parts, compute each (sector, triangle with $\frac{1}{2}ab\sin C$ , arc, straight edges), then add or subtract as the shape requires.

Common trap

Giving only the arc and omitting the two radii for a sector perimeter, or including an interior edge that the arc replaces. Examiner reports flag both.

Word problem

Recognise it by

A real-world object — a circular table, a running track — described in words, with no formula stated.

How to approach it

Translate the object into circles, sectors and rectangles, identify the radius (halving any diameter), then apply the circle formulae to each part.

Common trap

Using the diameter in $\pi r^{2}$ , or counting edges that are not part of the outer boundary. Examiner reports flag these in composite practical questions.

1Circumference and area of a circle
CoreDirect calculation• circle
Question
Find the circumference and area of a circle of radius $5$ cm. Give answers to 1 d.p.
Step-by-step solution
1. Step 1
  $C = 2\pi r$ .
  $C = 2\pi(5) = 10\pi \approx 31.4\,\text{cm}$
2. Step 2
  $A = \pi r^{2}$ .
  $A = \pi(25) = 25\pi \approx 78.5\,\text{cm}^{2}$
Answer
$C \approx 31.4\,\text{cm},\ A \approx 78.5\,\text{cm}^{2}$
2Arc length
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q13• arc length
Question
Find the length of an arc of a circle, radius $9$ cm, that subtends an angle of $80\degree$ at the centre.
Step-by-step solution
1. Step 1
  Arc length $= \dfrac{\theta}{360} \times 2\pi r$ .
  $L = \dfrac{80}{360} \times 2\pi(9) = \dfrac{2}{9} \times 18\pi = 4\pi$
Answer
$4\pi\,\text{cm} \approx 12.6\,\text{cm}$
3Sector area
ExtendedDirect calculation• sector
Question
Find the area of a sector of a circle, radius $6$ cm, with sector angle $120\degree$ .
Step-by-step solution
1. Step 1
  Sector area $= \dfrac{\theta}{360} \times \pi r^{2}$ .
  $A = \dfrac{120}{360} \times \pi(36) = 12\pi$
Answer
$12\pi\,\text{cm}^{2} \approx 37.7\,\text{cm}^{2}$
4Area of a segment
ExtendedMulti-step problem• segment
Question
A chord subtends an angle of $90\degree$ at the centre of a circle of radius $4$ cm. Find the area of the minor segment.
Step-by-step solution
1. Step 1
  Segment area = sector area − triangle area.
  $A = \dfrac{90}{360} \pi(16) - \dfrac{1}{2}(4)(4)\sin 90\degree$
2. Step 2
  Compute.
  $A = 4\pi - 8 \approx 4.57\,\text{cm}^{2}$
Answer
$4\pi - 8 \approx 4.57\,\text{cm}^{2}$
5Find area given the diameter
CoreWord problem• diameter, area
Question
A circular table has diameter $1.4$ m. Find the area of its top in $\text{m}^{2}$ , correct to 2 decimal places.
Step-by-step solution
1. Step 1
  Halve the diameter to get the radius.
  $r = \dfrac{1.4}{2} = 0.7\,\text{m}$
2. Step 2
  Apply $A = \pi r^{2}$ .
  $A = \pi(0.7)^{2} = 0.49\pi \approx 1.54\,\text{m}^{2}$
Answer
$A \approx 1.54\,\text{m}^{2}$
Examiner tip
The examiner report flags candidates who square the diameter instead of the radius. Always halve the diameter first.
6Perimeter of a sector
ExtendedMulti-step problem• Adapted from 0580/22 Oct/Nov 2023 Q14• sector, perimeter
Question
A sector has radius $10$ cm and sector angle $72\degree$ . Find the perimeter of the sector.
Step-by-step solution
1. Step 1
  Perimeter = arc + two radii.
2. Step 2
  Arc length first.
  $L = \dfrac{72}{360}\times 2\pi(10) = \dfrac{1}{5}\times 20\pi = 4\pi$
3. Step 3
  Add the two radii.
  $P = 4\pi + 2(10) = 4\pi + 20 \approx 32.6\,\text{cm}$
Answer
$P = 4\pi + 20 \approx 32.6\,\text{cm}$
Examiner tip
The examiner report flags candidates who give only the arc length and forget the two radii. Sector PERIMETER always = arc length + $2r$ .
7Area between two concentric circles (annulus)
ExtendedDirect calculation• annulus, area
Question
An annulus is formed by two concentric circles of radii $5$ cm and $8$ cm. Find the area of the annulus in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Outer area minus inner area.
  $A = \pi(8)^{2} - \pi(5)^{2} = 64\pi - 25\pi$
2. Step 2
  Subtract.
  $A = 39\pi\,\text{cm}^{2}$
Answer
$39\pi\,\text{cm}^{2} \approx 122.5\,\text{cm}^{2}$
8Perimeter of a compound shape containing an arc
ExtendedMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q8• compound shape, arc length, perimeter
Question
A shape consists of a rectangle $6$ cm by $4$ cm with a semicircle of diameter $4$ cm added to one of the short sides (so the rectangle is closed off and the semicircle bulges outward). Find the perimeter of the resulting shape.
Step-by-step solution
1. Step 1
  Identify the boundary: three sides of the rectangle (two long sides + one short side) plus the curved semicircle.
2. Step 2
  Length of the three rectangle sides.
  $6 + 4 + 6 = 16\,\text{cm}$
3. Step 3
  Semicircle of diameter $4$ has arc length $\dfrac{1}{2}\times \pi \times 4 = 2\pi$ cm.
4. Step 4
  Total perimeter.
  $P = 16 + 2\pi \approx 22.3\,\text{cm}$
Answer
$P = 16 + 2\pi \approx 22.3\,\text{cm}$
Examiner tip
The examiner report flags candidates who include the straight edge that the semicircle replaces — that edge is INTERIOR after the semicircle is attached, so it does not contribute to the perimeter.
9Area of a segment using a non-right angle
ChallengeMulti-step problem• Adapted from 0580/42 Feb/Mar 2024 Q11• segment, trigonometry
Question
A chord subtends an angle of $140\degree$ at the centre of a circle of radius $9$ cm. Find the area of the minor segment, giving your answer to 3 significant figures.
Step-by-step solution
1. Step 1
  Sector area.
  $A_{\text{sector}} = \dfrac{140}{360}\times \pi (9)^{2} = \dfrac{7}{18}\times 81\pi = \dfrac{63\pi}{2}$
2. Step 2
  Triangle area uses $\frac{1}{2}ab\sin C$ with $a = b = 9$ and $C = 140\degree$ .
  $A_{\text{triangle}} = \dfrac{1}{2}(9)(9)\sin 140\degree = \dfrac{81}{2}\sin 140\degree$
3. Step 3
  Segment = sector − triangle.
  $A = \dfrac{63\pi}{2} - \dfrac{81}{2}\sin 140\degree \approx 98.96 - 26.03 \approx 72.9\,\text{cm}^{2}$
Answer
$A \approx 72.9\,\text{cm}^{2}$
Examiner tip
The examiner report flags candidates who use $\frac{1}{2}\times\text{base}\times\text{height}$ for the triangle when no perpendicular height is given. Use $\frac{1}{2}ab\sin C$ with the two radii as $a$ and $b$ .
10Composite practical problem — running track
ChallengeWord problem• compound shape, real-world
Question
A running track is shaped like a rectangle with two semicircular ends. The straight sides are each $80$ m long and the radius of each semicircle is $30$ m. (a) Find the total perimeter. (b) Find the total enclosed area. Give answers to 1 d.p.
Step-by-step solution
1. Step 1
  Perimeter = two straights + the two semicircles, which together form a full circle of radius $30$ m.
  $P = 2(80) + 2\pi(30) = 160 + 60\pi$
2. Step 2
  Evaluate.
  $P \approx 160 + 188.50 \approx 348.5\,\text{m}$
3. Step 3
  Area = rectangle ( $80 \times 60$ ) + full circle of radius $30$ .
  $A = 80 \times 60 + \pi(30)^{2} = 4800 + 900\pi$
4. Step 4
  Evaluate.
  $A \approx 4800 + 2827.43 \approx 7627.4\,\text{m}^{2}$
Answer
(a) $P \approx 348.5\,\text{m}$ (b) $A \approx 7627.4\,\text{m}^{2}$
Examiner tip
The examiner report flags candidates who use the diameter ( $60$ m) instead of the radius ( $30$ m) in $\pi r^{2}$ , or who include the rectangle's short ends in the perimeter — those edges are replaced by the semicircles.

Key Formulae — Circles

The formulae you need to memorise for circles on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Circumference
$C = 2\pi r = \pi d$
$r$
radius
$d$
diameter $= 2r$
When to use
Distance around any circle.
Area of a circle
$A = \pi r^{2}$
$r$
radius
When to use
Area enclosed by any circle.
Arc length
$L = \dfrac{\theta}{360} \times 2\pi r$
$\theta$
sector angle in degrees
When to use
Length of a portion of the circumference.
Sector area
$A = \dfrac{\theta}{360} \times \pi r^{2}$
$\theta$
sector angle in degrees
When to use
Area of a 'pie-slice' of a circle.

Key Definitions and Keywords — Circles

Definitions to memorise and the exact keywords mark schemes credit for circles answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Radius
Examiner keyword
Distance from the centre to the edge of the circle.
Diameter
Examiner keyword
Distance across a circle through its centre $= 2r$ .
Arc
Examiner keyword
A portion of the circumference. Minor arc = shorter; major arc = longer.
Sector
Examiner keyword
A 'pie slice' bounded by two radii and an arc.
Segment
Region bounded by a chord and an arc.

Common Mistakes and Misconceptions — Circles

The traps other students keep falling into on circles questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using diameter instead of radius (or vice versa)
Why it happens
Reading the question too quickly.
How to avoid it
Note: $C = \pi d = 2\pi r$ . If the diagram gives diameter, halve it before using $A = \pi r^{2}$ .
✕Using radians in the $\theta/360$ formula
0580/42 — recurring
Why it happens
Calculator left in radian mode.
How to avoid it
0580 always uses degrees. Set the calculator to DEG mode at the start of the paper.
✕Calling a sector a segment
Why it happens
Mixing up the two regions.
How to avoid it
Sector = two RADII + arc. Segment = one CHORD + arc.
✕Using $\pi r^{2}$ for arc length
Why it happens
Mixing up area and length formulae.
How to avoid it
Length comes from circumference $2\pi r$ . Area comes from $\pi r^{2}$ .

Circles — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

A sector is the pie-slice region of a circle bounded by two radii and an arc. The angle at the centre is $\theta$ (in degrees).

Arc length. $L = \dfrac{\theta}{360} \times 2\pi r.$

Sector area. $A_{\text{sector}} = \dfrac{\theta}{360} \times \pi r^{2}.$

The pattern: $\dfrac{\theta}{360}$ is the FRACTION of a full revolution, so it scales any full-circle quantity to its sector value.

Worked. Sector with radius $9\,\text{cm}$ and angle $80\degree$ .

Arc: $L = \dfrac{80}{360} \times 2\pi \times 9 = \dfrac{2}{9} \times 18\pi = 4\pi \approx 12.6\,\text{cm}$ .
Area: $A = \dfrac{80}{360} \times \pi \times 81 = \dfrac{2}{9} \times 81\pi = 18\pi \approx 56.5\,\text{cm}^{2}$ .

Perimeter of a sector. Two radii plus the arc: $P_{\text{sector}} = 2r + L.$

A common Cambridge trip-up: students forget the two radii.

The fraction θ÷360 scales any full-circle quantity down to the sector.

A sector's share of the circle is θ ÷ 360 — a quarter, half or three-quarters for 90°, 180°, 270°.

Quantity

Formula

Circumference

C = 2\pi r = \pi d

Area

A = \pi r^{2}

Arc length

\tfrac{\theta}{360} \times 2\pi r

Sector area

\tfrac{\theta}{360} \times \pi r^{2}

Sector perimeter

2r + \text{arc}

Segment area

sector

- \tfrac{1}{2}r^{2}\sin\theta

Circles

Short Notes - Circles

Detailed Study Notes

At a glance

What you’ll learn

Circumference and area of a circle

Arc length and sector area

Segment of a circle

Units, rounding and exact answers

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Circles

1Circumference and area of a circle

2Arc length

3Sector area

4Area of a segment

5Find area given the diameter

6Perimeter of a sector

7Area between two concentric circles (annulus)

8Perimeter of a compound shape containing an arc

9Area of a segment using a non-right angle

10Composite practical problem — running track

Circumference

Area of a circle

Arc length

Sector area

Radius

Diameter

Arc

Sector

Segment

✕Using diameter instead of radius (or vice versa)

✕Using radians in the θ/360\theta/360θ/360 formula

✕Calling a sector a segment

✕Using πr2\pi r^{2}πr2 for arc length

Circles — frequently asked questions

Circles

Short Notes - Circles

Detailed Study Notes

At a glance

What you’ll learn

Circumference and area of a circle

Arc length and sector area

Segment of a circle

Units, rounding and exact answers

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Circles

1Circumference and area of a circle

2Arc length

3Sector area

4Area of a segment

5Find area given the diameter

6Perimeter of a sector

7Area between two concentric circles (annulus)

8Perimeter of a compound shape containing an arc

9Area of a segment using a non-right angle

10Composite practical problem — running track

Circumference

Area of a circle

Arc length

Sector area

Radius

Diameter

Arc

Sector

Segment

✕Using diameter instead of radius (or vice versa)

✕Using radians in the θ/360\theta/360θ/360 formula

✕Calling a sector a segment

✕Using πr2\pi r^{2}πr2 for arc length

Circles — frequently asked questions

✕Using radians in the $\theta/360$ formula

✕Using $\pi r^{2}$ for arc length

✕Using radians in the $\theta/360$ formula

✕Using $\pi r^{2}$ for arc length