Area of a segment Quiz (Grade 7)

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Score: 0 / 10

Area of a segment — Notes & Worked Examples Show / Hide

Explanation:

A segment is the region enclosed between a chord and an arc of a circle. To find the area of a minor segment with radius r and central angle θ, subtract the area of the isosceles triangle formed by the two radii and the chord from the area of the sector:

Area of segment = Area of sector − Area of triangle = (θ360 × πr²) − (12r² sin θ). To find the perimeter of a segment, add the arc length (θ360 × 2πr) to the chord length (found using the Cosine Rule).

Example 1: Find the area of the minor segment of a circle with radius 10 cm and central angle 72° (3 s.f.).

Solution:

Area of sector = 72360 × π × 10² = 20π = 62.8318... cm²

Area of triangle = 12 × 10² × sin 72° = 47.5528... cm²

Segment area = 62.8318... − 47.5528... = 15.2790...

Answer: 15.3 cm² (3 s.f.)

Example 2: Find the exact area of a minor segment with radius 6 cm and central angle 60°.

Solution:

Area of sector = 60360 × π × 6² = 6π cm²

Area of triangle = 12 × 6² × sin 60° = 18 × √32 = 9√3 cm²

Subtract triangle from sector:

Answer: (6π − 9√3) cm²

One Page Guide (PDF) Area of a segment Worksheets

Question 1

[2 marks]

The diagram shows a sector OAB of a circle with radius 12 cm and angle 60°.
Calculate the area of the shaded segment correct to 2 decimal places.

cm²

Worked Solution & Answer

(The area of a sector is (angle ÷ 360°) × πr²)
Area = 60360π(12²) - 12(12²)\sin(60°) = 13.04 cm² B1

Answer: 13.04 cm² cm²

Question 2

[2 marks]

The diagram shows a sector OAB of a circle with radius 8 cm and angle 60°.
Calculate the area of the shaded segment correct to 2 decimal places.

cm²

Worked Solution & Answer

(The area of a sector is (angle ÷ 360°) × πr²)
Area = 60360π(8²) - 12(8²)\sin(60°) = 5.8 cm² B1

Answer: 5.8 cm² cm²

Question 3 Specimen Set 2 2H Q7

[3 marks]

The arc ABC is a quarter of a circle with centre O and radius 4.8 cm.
AC is a chord of the circle.
Work out the area of the shaded segment.
Give your answer correct to 3 significant figures.

cm²

Worked Solution & Answer

(The area of a circle is πr²)
Area of quarter circle = ¼ × π × 4.8² = 5.76π ≈ 18.0956 cm² M1
Area of triangle OAC = ½ × 4.8 × 4.8 = 11.52 cm² M1
Shaded segment = 18.0956 – 11.52 = 6.58 cm² (3 s.f.). A1

Answer: 6.58 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[4 marks]

The diagram shows a sector OPQR of a circle, centre O and radius 8 cm.

OPR is a triangle.
Work out the area of the shaded segment PQR.
Give your answer correct to 3 significant figures.

cm²

Worked Solution & Answer

(The area of a triangle is ½ × base × height)
Area of sector = ¼ × π × 8² = 16π ≈ 50.265 cm² M2
Area of triangle OPR = ½ × 8 × 8 = 32 cm² M1
Segment = 50.265 – 32 = 18.3 cm² (3 s.f.) A1

Answer: 18.3 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 5 IGCSE Jan 2022 2F Q5

[4 marks]

The diagram shows points A and B marked on a grid of squares.

(a) On the grid, draw the line with equation y = -2 [1 mark]

(b) Find the coordinates of M [2 marks]

d =

Worked Solution & Answer

(a)

Horizontal line through (0, –2) B1

Answer (a): Horizontal line drawn at y = −2

(b)

((–3 + 1)/2, (1 + 3)/2) = (–1, 2) M1 A1

Answer (b): (-1, 2)

(c)

AB² = 4² + 2² = 20
for BD = AB with x_D = 5 (4 units right of B), y_D = 3 – 2 = 1 ⇒ d = 1 B1

Answer (c): d = 1

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 IGCSE Jan 2023 1H Q21

[5 marks]

The diagram shows the cross section of a circular water pipe.

OABC is a sector of the circle, centre O
The shaded region in the diagram represents the water flowing in the pipe.
The water flows at 14 cm/s in the pipe.
Work out the volume of water that has flowed through the pipe in 3 minutes.
Give your answer in cm³ correct to 3 significant figures.

cm³

Worked Solution & Answer

Area of sector OABC = (72/360) × π × 4.8² = 14.47645... cm² M1
Area of triangle OAC = ½ × 4.8² × sin 72° = 10.95617... cm² M1
Area of shaded segment = 14.47645... − 10.95617... = 3.52028... cm² M1
In 3 minutes (180 seconds), length of flow = 14 × 180 = 2520 cm M1
Volume = 3.52028... × 2520 = 8871.1... ≈ 8870 cm³ (3 s.f.). A1

Answer: 8870 cm³

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 IGCSE Jan 2023 2H Q14

[5 marks]

A, B, C and D are points on a circle, centre O
EBF is the tangent to the circle at B

(a) Work out the size of angle DCB [1 mark]

(b) Give a reason for your answer to (a)(i) [1 mark]

(b) Work out the size of angle ADO [3 marks]

Worked Solution & Answer

(a)

180° − 40° = 140° (opposite angles of a cyclic quadrilateral sum to 180°) B1

Answer (a): 140 °

(b)

Opposite angles of a cyclic quadrilateral sum to 180°

Answer (b): Opposite angles of a cyclic quadrilateral sum to 180°

(b)

Angle DOB = 2 × 40° = 80° (angle at centre) M1
Since OD = OB, angle ODB = (180° − 80°)/2 = 50° M1
By alternate segment theorem, angle ADB = angle ABF = 66°
Therefore angle ADO = 66° − 50° = 16° A1

Answer (b): 16 °

Question Paper (PDF) • Mark Scheme (PDF)

Question 8 IGCSE Jun 2018 2F Q9

[5 marks]

R and S are points on a circle with centre O.

(a) On the diagram above, shade a segment of the circle. [1 mark]

(b) Write down the mathematical name of the straight line RS. [1 mark]

(c) QT is a tangent to the circle.
Angle OPQ = 18°
Work out the size of angle PQT.
Give a reason for each stage of your working. [3 marks]

Worked Solution & Answer

(a)

Region between chord RS and arc RS shaded

Answer (a): Region between chord RS and arc RS shaded

(b)

Chord

Answer (b): Chord

(c)

Angle OQP = 18° (base angles of isosceles triangle OPQ are equal as OP = OQ) M1
Angle OQT = 90° (angle between tangent and radius is 90°) M1
Angle PQT = 90° – 18° = 72° A1

Answer (c): 72 °

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 IGCSE Jun 2024 1FR Q9

[5 marks]

The diagram shows three points, A, B and C, marked on a grid.

(a) Write down the coordinates of A [1 mark]

(b) Find the coordinates of the midpoint of AB [2 marks]

Worked Solution & Answer

(a)

A = (−4, 2) B1

Answer (a): −4, 2

(b)

Midpoint = ((−4+2)/2, (2+4)/2) = (−1, 3) M1 A1

Answer (b): −1, 3

(c)

D = A + (C − B) = (−4 + 1, 2 − 3) = (−3, −1) M1 A1

Answer (c): −3, −1

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 IGCSE Jun 2025 1HR Q25

[6 marks]

BAC is a sector of a circle, centre A
BCD is a sector of a circle, centre C

Angle BAC = 40°
Angle BCD = 130°
Area of shaded segment = 28 cm²
Find the length of the arc BD
Give your answer correct to 3 significant figures.

Worked Solution & Answer

(The area of a sector is (angle ÷ 360°) × πr²)
Area of segment = (40/360)πr² − (1/2)r² sin 40° = 28 ⇒ r²(0.349066 − 0.321394) = 28 ⇒ r² = 1011.86 ⇒ r = 31.81 cm M3
Chord BC = 2 × 31.81 × sin 20° = 21.759 cm M2
Arc BD = (130/360) × 2π × 21.759 = 49.37... = 49.4 cm (3 s.f.). A1

Answer: 49.4 cm

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 41 marks)

Area of a segment Worksheets Next: Quad. sim. equations →

Grade 7 Quizzes

19 Topics

7A - Surds	7B - Bounds
7C - Proportion (harder)	7D - Algebraic fractions
7E - The quadratic formula	7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs	7H - Rearranging (harder)
7J - Functions (harder)	7K - Iteration
7L - Area of any triangle	7M - The sine rule
7N - The cosine rule	7P - Congruent triangles
7Q - 3D pythagoras / trig.	7R - Histograms
7S - Conditional probability	7T - Exact trig. values (harder)
7U - Area of a segment