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Area of a segment

Free Printable Area of a segment PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 7)

Download free area of a segment worksheets (PDF), GCSE maths past paper questions and worked solutions for Area of a segment (7U). This is a Grade 7 geometry topic from the GCSE questions by topic collection for Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

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Free Area of a segment PDF Worksheets & Answers

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GCSE Maths Grade 7 Revision Notes & Method

Guide to Area of a segment: Explanation, Diagrams & Worked Examples

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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.).

O 72° 10 cm 10 cm Minor segment

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
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Area of a segment Questions & Answers

Past Paper Example 1 IGCSE Jan 2023 1H Q21 5 marks
The diagram shows the cross section of a circular water pipe.
72°4.8 cmOABCDiagram NOTaccurately drawn
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.
Worked Solution & Mark Scheme (5 marks)
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
Past Paper Example 2 GCSE Nov 2021 3F Q27 4 marks
The diagram shows a sector OPQR of a circle, centre O and radius 8 cm.
OPRQ8 cm8 cm
OPR is a triangle.
Work out the area of the shaded segment PQR.
Give your answer correct to 3 significant figures.
Worked Solution & Mark Scheme (4 marks)
(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
Past Paper Example 3 Specimen Set 2 2H Q7 3 marks
ABCO4.8 cm
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.
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.
Worked Solution & Mark Scheme (3 marks)
(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

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