Cambridge IGCSE Mathematics 0580 · Extended · Syllabus 2025-2027

Circles Worked ExamplesIGCSE Maths 0580

Exam-style worked examples on circumference and area, arcs and sectors, perimeter of a sector (which is not just the arc), the segment found by subtracting a triangle, and working backwards from an area to a radius. Diagram-based items add a sector angle in exact form, major arcs and sectors in terms of π, the perimeter of a major sector, water lying in a pipe (a minor segment), a plate with semicircles cut from its ends, the leaf between two quarter circles, and a belt round two pulleys.

This page has 13 exam-style IGCSE Maths 0580 circles worked examples — 1 Core, 9 Extended and 3 Challenge, 7 with diagrams. Each one shows the full method step by step, the final answer in the form a marker expects, and an examiner tip on where marks are commonly lost.

Example 1CoreCircumference and area[3 marks]

Circumference and area

A circle has radius $7\text{ cm}$ .

Calculate (a) its circumference, (b) its area, each correct to $3$ significant figures.

Example 2ExtendedArcs and sectors[4 marks]

Arc length and sector area

A sector of a circle has radius $9\text{ cm}$ and angle $60^{\circ}$ .

Calculate (a) the arc length, (b) the sector area, each correct to $3$ significant figures.

Example 3ExtendedArcs and sectors[3 marks]

Perimeter of a sector

Calculate the perimeter of the sector in the previous example (radius $9\text{ cm}$ , angle $60^{\circ}$ ), correct to $3$ significant figures.

Example 4ChallengeSegments[5 marks]

Area of a segment

A circle has radius $9\text{ cm}$ . A chord subtends an angle of $60^{\circ}$ at the centre.

Calculate the area of the minor segment, correct to $3$ significant figures.

Example 5ExtendedWorking backwards[3 marks]

Working backwards to the radius

A circle has area $78.5\text{ cm}^2$ .

Calculate its radius, correct to $2$ significant figures.

Example 6ExtendedCircumference and area[4 marks]

Circles in context

A circular pond of radius $4\text{ m}$ is surrounded by a path $1\text{ m}$ wide.

Calculate the area of the path, correct to $3$ significant figures.

Example 7ExtendedWorking backwards from a sector area (exact form)[5 marks]

Sector angle from an exact area

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius $9\text{ cm}$ . The area of the sector is $27\pi\text{ cm}^2$ .

(a) Find the angle $\theta$ .

(b) Find the perimeter of the sector. Give your answer in the form $a + b\pi$ .

NOT TO SCALE

Example 8ExtendedShaded region between two arcs[8 marks]

The leaf between two quarter circles

$ABCD$ is a square of side $10\text{ cm}$ . One arc from $B$ to $D$ is drawn with centre $A$ , and another arc from $B$ to $D$ is drawn with centre $C$ , each with radius $10\text{ cm}$ . The two arcs enclose the shaded leaf shape.

(a) Show that the area between the arc centred at $A$ and the chord $BD$ is $(25\pi - 50)\text{ cm}^2$ .

(b) Find the shaded area, correct to $3$ significant figures.

(c) Find the perimeter of the shaded shape, giving your answer in terms of $\pi$ .

(d) Find the total unshaded area inside the square, correct to $3$ significant figures.

NOT TO SCALE

Example 9ChallengeTangents, arcs and a belt round two circles[8 marks]

A belt round two pulleys

Two pulleys, with centres $P$ and $Q$ , have radii $12\text{ cm}$ and $3\text{ cm}$ . The distance $PQ$ is $18\text{ cm}$ . A tight belt passes round both pulleys, as shown. The straight parts of the belt are tangents to both circles.

(a) By drawing the radii to the points where one straight part of the belt touches the pulleys, show that each straight part has length $9\sqrt{3}\text{ cm}$ .

(b) Show that the belt touches the larger pulley through an angle of $240^{\circ}$ .

(c) Calculate the total length of the belt, correct to $3$ significant figures.

NOT TO SCALE

1
(a) Let the upper straight part touch the large pulley at $S$ and the small pulley at $T$ . A tangent is perpendicular to the radius at the point of contact, so $PS$ and $QT$ are both perpendicular to $ST$ , and so $PS$ is parallel to $QT$ . Draw the line from $Q$ parallel to $ST$ , meeting $PS$ at $N$ . Then $QN = ST$ and $NS = QT = 3$ .
$PN = 12 - 3 = 9\text{ cm}$
2
Triangle $PNQ$ has a right angle at $N$ and hypotenuse $PQ = 18$ .
$ST = QN = \sqrt{18^{2} - 9^{2}} = \sqrt{243} = 9\sqrt{3}\text{ cm}$
3
(b) Use the same right-angled triangle to find angle $SPQ$ .
$\cos \angle SPQ = \frac{PN}{PQ} = \frac{9}{18} = \frac{1}{2} \implies \angle SPQ = 60^{\circ}$
4
By symmetry the lower straight part makes another $60^{\circ}$ on the other side of $PQ$ . The belt is off the large pulley across those $120^{\circ}$ facing $Q$ , and in contact everywhere else.
$360^{\circ} - 2 \times 60^{\circ} = 240^{\circ}$
5
(c) For the small pulley: $PS \parallel QT$ , so angle $PQT$ and angle $SPQ$ are co-interior and add to $180^{\circ}$ . The belt is off the small pulley across $2 \times 120^{\circ}$ facing $P$ .
$\angle PQT = 120^{\circ} \implies \text{contact angle} = 360^{\circ} - 240^{\circ} = 120^{\circ}$
6
Find the two curved lengths as fractions of each circumference.
$\tfrac{240}{360} \times 2\pi(12) = 16\pi, \qquad \tfrac{120}{360} \times 2\pi(3) = 2\pi$
7
Add the two straight parts and the two arcs.
$2 \times 9\sqrt{3} + 16\pi + 2\pi = 18\sqrt{3} + 18\pi = 87.72\ldots\text{ cm}$

Answer

(a) $\sqrt{18^2 - 9^2} = \sqrt{243} = 9\sqrt{3}$ (b) $\cos^{-1}\tfrac{9}{18} = 60^{\circ}$ , so $360^{\circ} - 120^{\circ} = 240^{\circ}$ (c) $87.7\text{ cm}$

Examiner tip

Everything rests on one fact: the tangent is perpendicular to the radius, which makes $PS$ parallel to $QT$ and creates the right-angled triangle. Using $12 + 3$ instead of $12 - 3$ for $PN$ is the triangle for a crossed belt, not this one. The belt does not cover half of each pulley — assuming two semicircles gives $15\pi + 18\sqrt{3}$ , which is too small.

Example 10ExtendedMajor arc length and major sector area, exact form, non-calculator[5 marks]

Major arc and major sector in terms of π

The diagram shows a circle with centre $O$ and radius $8\text{ cm}$ . Angle $AOB = 135^{\circ}$ . The major sector $AOB$ is shaded.

(a) The major arc $AB$ has length $n\pi\text{ cm}$ , where $n$ is a whole number. Work out $n$ . [2]

(b) Work out the exact area of the shaded sector, as a multiple of $\pi$ . [3]

Do not use a calculator.

NOT TO SCALE

Example 11ExtendedMajor arc plus two radii[4 marks]

Perimeter of a major sector

A badge is cut from a circle of radius $6.4\text{ cm}$ by removing a slice with an angle of $118^{\circ}$ at the centre, leaving the major sector shown.

A strip of ribbon is glued all the way round the edge of the badge. Work out the length of ribbon needed.

NOT TO SCALE

Example 12ChallengeMinor segment area via the angle at the centre, then a capacity[7 marks]

Water lying in a horizontal pipe

A drainage pipe lies horizontally. Its inside is a circle of radius $25\text{ cm}$ with centre $O$ , and the pipe is $3\text{ m}$ long.

Water lies along the bottom of the pipe and is $10\text{ cm}$ deep at its deepest point, as shown in the cross-section.

(a) Show that the angle at $O$ between the two radii drawn to the edges of the water surface is $106.3^{\circ}$ , correct to $1$ decimal place. [3]

(b) Work out how many litres of water are in the pipe. [4]

NOT TO SCALE

1
(a) The lowest point of the circle is $25\text{ cm}$ below $O$ and the water is $10\text{ cm}$ deep, so the water surface is $25 - 10 = 15\text{ cm}$ below $O$ . The vertical from $O$ to the surface bisects the surface, making a right-angled triangle with hypotenuse $25$ (a radius).
$\cos\alpha = \frac{15}{25} = 0.6 \implies \alpha = 53.130\ldots^{\circ}$
2
The angle at $O$ is twice the half-angle; write down more figures than the value being shown.
$2\alpha = 106.260\ldots^{\circ} = 106.3^{\circ} \text{ (1 d.p.)}$
3
(b) The water's cross-section is a minor segment: the sector minus the triangle above the water. Sector first, with the unrounded angle.
$\frac{106.260\ldots}{360} \times \pi \times 25^{2} = 579.55\ldots\text{ cm}^2$
4
The half-width of the water surface is $\sqrt{25^2 - 15^2} = 20\text{ cm}$ , so the triangle has base $40$ and height $15$ .
$\tfrac{1}{2} \times 40 \times 15 = 300\text{ cm}^2$
5
Subtract to get the segment.
$579.55\ldots - 300 = 279.55\ldots\text{ cm}^2$
6
The water is a prism $3\text{ m} = 300\text{ cm}$ long. Then divide by $1000$ for litres.
$279.55\ldots \times 300 = 83\,867.8\ldots\text{ cm}^3 = 83.867\ldots\text{ litres}$

Answer

(a) $\cos\alpha = \tfrac{15}{25}$ , $2\alpha = 106.26\ldots^{\circ} = 106.3^{\circ}$ (b) $83.9$ litres

Examiner tip

The shaded region is not a sector, so an angle is needed before any formula applies: on a segment question the March 2025 examiner report noted that "A significant number of candidates did not recognise the need to calculate an angle to work with sectors and triangles". The same report found that "some lost accuracy by rounding prematurely" — use the unrounded angle, not $106.3^{\circ}$ , in the sector. Convert the length to centimetres before multiplying.

Example 13ExtendedRectangle minus semicircles: area in terms of π and a curved perimeter[5 marks]

A plate with semicircles cut from its ends

A metal plate is made from a rectangle $30\text{ cm}$ long and $16\text{ cm}$ wide. A semicircle with diameter $16\text{ cm}$ is cut out of each short end, as shown.

(a) Find the exact area of the plate, in the form $a - b\pi$ . [2]

(b) Calculate the perimeter of the plate. [3]

NOT TO SCALE

These worked examples are written by Tutopiya to cover the same syllabus content as the official Cambridge IGCSE 0580 specification (2025-2027). They are exam-style questions, not reproductions of past paper questions.

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Circles Worked ExamplesIGCSE Maths 0580

Example 1CoreCircumference and area[3 marks]

Circumference and area

A circle has radius $7\text{ cm}$ .

Calculate (a) its circumference, (b) its area, each correct to $3$ significant figures.

Example 2ExtendedArcs and sectors[4 marks]

Arc length and sector area

A sector of a circle has radius $9\text{ cm}$ and angle $60^{\circ}$ .

Calculate (a) the arc length, (b) the sector area, each correct to $3$ significant figures.

Example 3ExtendedArcs and sectors[3 marks]

Perimeter of a sector

Calculate the perimeter of the sector in the previous example (radius $9\text{ cm}$ , angle $60^{\circ}$ ), correct to $3$ significant figures.

Example 4ChallengeSegments[5 marks]

Area of a segment

A circle has radius $9\text{ cm}$ . A chord subtends an angle of $60^{\circ}$ at the centre.

Calculate the area of the minor segment, correct to $3$ significant figures.

Example 5ExtendedWorking backwards[3 marks]

Working backwards to the radius

A circle has area $78.5\text{ cm}^2$ .

Calculate its radius, correct to $2$ significant figures.

Example 6ExtendedCircumference and area[4 marks]

Circles in context

A circular pond of radius $4\text{ m}$ is surrounded by a path $1\text{ m}$ wide.

Calculate the area of the path, correct to $3$ significant figures.

Example 7ExtendedWorking backwards from a sector area (exact form)[5 marks]

Sector angle from an exact area

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius $9\text{ cm}$ . The area of the sector is $27\pi\text{ cm}^2$ .

(a) Find the angle $\theta$ .

(b) Find the perimeter of the sector. Give your answer in the form $a + b\pi$ .

NOT TO SCALE

Example 8ExtendedShaded region between two arcs[8 marks]

The leaf between two quarter circles

(a) Show that the area between the arc centred at $A$ and the chord $BD$ is $(25\pi - 50)\text{ cm}^2$ .

(b) Find the shaded area, correct to $3$ significant figures.

(c) Find the perimeter of the shaded shape, giving your answer in terms of $\pi$ .

(d) Find the total unshaded area inside the square, correct to $3$ significant figures.

NOT TO SCALE

Example 9ChallengeTangents, arcs and a belt round two circles[8 marks]

A belt round two pulleys

(a) By drawing the radii to the points where one straight part of the belt touches the pulleys, show that each straight part has length $9\sqrt{3}\text{ cm}$ .

(b) Show that the belt touches the larger pulley through an angle of $240^{\circ}$ .

(c) Calculate the total length of the belt, correct to $3$ significant figures.

NOT TO SCALE

1
(a) Let the upper straight part touch the large pulley at $S$ and the small pulley at $T$ . A tangent is perpendicular to the radius at the point of contact, so $PS$ and $QT$ are both perpendicular to $ST$ , and so $PS$ is parallel to $QT$ . Draw the line from $Q$ parallel to $ST$ , meeting $PS$ at $N$ . Then $QN = ST$ and $NS = QT = 3$ .
$PN = 12 - 3 = 9\text{ cm}$
2
Triangle $PNQ$ has a right angle at $N$ and hypotenuse $PQ = 18$ .
$ST = QN = \sqrt{18^{2} - 9^{2}} = \sqrt{243} = 9\sqrt{3}\text{ cm}$
3
(b) Use the same right-angled triangle to find angle $SPQ$ .
$\cos \angle SPQ = \frac{PN}{PQ} = \frac{9}{18} = \frac{1}{2} \implies \angle SPQ = 60^{\circ}$
4
By symmetry the lower straight part makes another $60^{\circ}$ on the other side of $PQ$ . The belt is off the large pulley across those $120^{\circ}$ facing $Q$ , and in contact everywhere else.
$360^{\circ} - 2 \times 60^{\circ} = 240^{\circ}$
5
(c) For the small pulley: $PS \parallel QT$ , so angle $PQT$ and angle $SPQ$ are co-interior and add to $180^{\circ}$ . The belt is off the small pulley across $2 \times 120^{\circ}$ facing $P$ .
$\angle PQT = 120^{\circ} \implies \text{contact angle} = 360^{\circ} - 240^{\circ} = 120^{\circ}$
6
Find the two curved lengths as fractions of each circumference.
$\tfrac{240}{360} \times 2\pi(12) = 16\pi, \qquad \tfrac{120}{360} \times 2\pi(3) = 2\pi$
7
Add the two straight parts and the two arcs.
$2 \times 9\sqrt{3} + 16\pi + 2\pi = 18\sqrt{3} + 18\pi = 87.72\ldots\text{ cm}$

Answer

(a) $\sqrt{18^2 - 9^2} = \sqrt{243} = 9\sqrt{3}$ (b) $\cos^{-1}\tfrac{9}{18} = 60^{\circ}$ , so $360^{\circ} - 120^{\circ} = 240^{\circ}$ (c) $87.7\text{ cm}$

Examiner tip

Example 10ExtendedMajor arc length and major sector area, exact form, non-calculator[5 marks]

Major arc and major sector in terms of π

The diagram shows a circle with centre $O$ and radius $8\text{ cm}$ . Angle $AOB = 135^{\circ}$ . The major sector $AOB$ is shaded.

(a) The major arc $AB$ has length $n\pi\text{ cm}$ , where $n$ is a whole number. Work out $n$ . [2]

(b) Work out the exact area of the shaded sector, as a multiple of $\pi$ . [3]

Do not use a calculator.

NOT TO SCALE

Example 11ExtendedMajor arc plus two radii[4 marks]

Perimeter of a major sector

A badge is cut from a circle of radius $6.4\text{ cm}$ by removing a slice with an angle of $118^{\circ}$ at the centre, leaving the major sector shown.

A strip of ribbon is glued all the way round the edge of the badge. Work out the length of ribbon needed.

NOT TO SCALE

Example 12ChallengeMinor segment area via the angle at the centre, then a capacity[7 marks]

Water lying in a horizontal pipe

A drainage pipe lies horizontally. Its inside is a circle of radius $25\text{ cm}$ with centre $O$ , and the pipe is $3\text{ m}$ long.

Water lies along the bottom of the pipe and is $10\text{ cm}$ deep at its deepest point, as shown in the cross-section.

(a) Show that the angle at $O$ between the two radii drawn to the edges of the water surface is $106.3^{\circ}$ , correct to $1$ decimal place. [3]

(b) Work out how many litres of water are in the pipe. [4]

NOT TO SCALE

1
(a) The lowest point of the circle is $25\text{ cm}$ below $O$ and the water is $10\text{ cm}$ deep, so the water surface is $25 - 10 = 15\text{ cm}$ below $O$ . The vertical from $O$ to the surface bisects the surface, making a right-angled triangle with hypotenuse $25$ (a radius).
$\cos\alpha = \frac{15}{25} = 0.6 \implies \alpha = 53.130\ldots^{\circ}$
2
The angle at $O$ is twice the half-angle; write down more figures than the value being shown.
$2\alpha = 106.260\ldots^{\circ} = 106.3^{\circ} \text{ (1 d.p.)}$
3
(b) The water's cross-section is a minor segment: the sector minus the triangle above the water. Sector first, with the unrounded angle.
$\frac{106.260\ldots}{360} \times \pi \times 25^{2} = 579.55\ldots\text{ cm}^2$
4
The half-width of the water surface is $\sqrt{25^2 - 15^2} = 20\text{ cm}$ , so the triangle has base $40$ and height $15$ .
$\tfrac{1}{2} \times 40 \times 15 = 300\text{ cm}^2$
5
Subtract to get the segment.
$579.55\ldots - 300 = 279.55\ldots\text{ cm}^2$
6
The water is a prism $3\text{ m} = 300\text{ cm}$ long. Then divide by $1000$ for litres.
$279.55\ldots \times 300 = 83\,867.8\ldots\text{ cm}^3 = 83.867\ldots\text{ litres}$

Answer

(a) $\cos\alpha = \tfrac{15}{25}$ , $2\alpha = 106.26\ldots^{\circ} = 106.3^{\circ}$ (b) $83.9$ litres

Examiner tip

Example 13ExtendedRectangle minus semicircles: area in terms of π and a curved perimeter[5 marks]

A plate with semicircles cut from its ends

A metal plate is made from a rectangle $30\text{ cm}$ long and $16\text{ cm}$ wide. A semicircle with diameter $16\text{ cm}$ is cut out of each short end, as shown.

(a) Find the exact area of the plate, in the form $a - b\pi$ . [2]

(b) Calculate the perimeter of the plate. [3]

NOT TO SCALE

Circles Worked ExamplesIGCSE Maths 0580

Circumference and area

Arc length and sector area

Perimeter of a sector

Area of a segment

Working backwards to the radius

Circles in context

Sector angle from an exact area

The leaf between two quarter circles

A belt round two pulleys

Major arc and major sector in terms of π

Perimeter of a major sector

Water lying in a horizontal pipe

A plate with semicircles cut from its ends

More IGCSE Maths 0580 worked examples

Mensuration

Number

Algebra

Functions

Coordinate Geometry

Geometry

Vectors and Transfomations

Trigonometry

Sets

Probability

Statistics

Calculator Skills

Circles Worked ExamplesIGCSE Maths 0580

Circumference and area

Arc length and sector area

Perimeter of a sector

Area of a segment

Working backwards to the radius

Circles in context

Sector angle from an exact area

The leaf between two quarter circles

A belt round two pulleys

Major arc and major sector in terms of π

Perimeter of a major sector

Water lying in a horizontal pipe

A plate with semicircles cut from its ends

More IGCSE Maths 0580 worked examples

Mensuration

Number

Algebra

Functions

Coordinate Geometry

Geometry

Vectors and Transfomations

Trigonometry

Sets

Probability

Statistics

Calculator Skills