Areas Related to Circles – Class 10 Maths Chapter 11 Notes & NCERT PDF

Class 10MathematicsChapter 11NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 11 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 11: Areas Related to Circles

Introduction

You already know the area and circumference of a circle. In this chapter we find the areas of parts of a circular region — a sector and a segment — which appear in designs, clock hands, wiper blades and field shapes.

Key Concepts

Sector

The region enclosed by two radii and the arc between them. The angle between the radii is the angle of the sector.

Minor sector: the smaller region (angle θ); major sector: the rest (angle 360° – θ).

Segment

The region enclosed by a chord and the arc it cuts off.

Minor segment: between the chord and the minor arc; major segment: between the chord and the major arc.

Deriving the formula (unitary method)

A full circle (360°) has area πr². So a sector of angle 1° has area πr²/360, and a sector of angle θ has area (θ/360) × πr².

Similarly, the length of the arc of a sector of angle θ is (θ/360) × 2πr.

Key Formulas

Area of sector of angle θ = (θ/360°) × πr²

Length of arc of sector of angle θ = (θ/360°) × 2πr

Area of major sector = πr² – area of minor sector

Area of minor segment = area of sector – area of the triangle formed by the two radii and the chord

Area of major segment = πr² – area of minor segment

For triangle OAB with angle θ at the centre: if θ = 90°, area = r²/2; if θ = 60°, the triangle is equilateral with area (√3/4)r²; for θ = 120°, area = (1/2)r² sin 120° = (√3/4)r².

Methods Step by Step

Identify the centre, radius and the angle at the centre.

Find the sector area using θ/360 × πr².

For a segment, find the area of △OAB (use the special-angle facts or drop a perpendicular from O).

Subtract: segment = sector – triangle.

Use the value of π stated in the question (3.14 or 22/7).

Worked Examples (from the book)

Sector of radius 4 cm and angle 30° (π = 3.14). Area = (30/360) × 3.14 × 16 ≈ 4.19 cm². Major sector = 3.14 × 16 – 4.19 ≈ 46.05 cm².

Segment AYB, radius 21 cm, ∠AOB = 120° (π = 22/7). Sector OAYB = (120/360) × (22/7) × 21 × 21 = 462 cm². Area of △OAB = (441√3)/4 cm². Segment = 462 – 441√3/4 = (21/4)(88 – 21√3) cm².

Quadrant of a circle of radius 14 cm (π = 22/7): area = (90/360) × (22/7) × 196 = 154 cm².

Sector of angle 60° in a circle of radius 21 cm (π = 22/7): area = 231 cm², arc length = (60/360) × 2 × (22/7) × 21 = 22 cm.

Minor segment of a circle of radius 14 cm with a right angle at the centre: 154 – (1/2) × 14 × 14 = 56 cm².

Board Exam Focus

Sector/segment area questions with π = 22/7 or 3.14 (2–3 marks).

Real-life contexts: area swept by a minute hand, a car wiper, a horse tied to a peg (quadrant), a brooch or umbrella design.

Combination figures: shaded region = sum or difference of sectors, triangles, squares.

Formula Sheet

Circle: area πr², circumference 2πr

Sector: (θ/360)πr²; arc: (θ/360)2πr

Segment = sector – triangle

Equilateral triangle of side a: (√3/4)a²

Common Mistakes

Using the diameter instead of the radius.

Forgetting to subtract the triangle when finding a segment.

Mixing up arc length (2πr) with area (πr²).

Using a different value of π from the one the question states.

Treating the major segment as 'sector – triangle' — it is πr² – minor segment.

Writing answers without square units (cm²).

💡 Key Learning Points

✓Define sector, segment, minor/major sector and minor/major segment of a circle.
✓Derive and use area of a sector = (θ/360) × πr² and arc length = (θ/360) × 2πr.
✓Find the area of a segment as area of sector minus area of the corresponding triangle.
✓Find major sector and major segment areas by subtracting from the area of the circle.
✓Apply these results to real-life and combination-figure problems using the given value of π.

👨‍🏫 Teaching Tips

→Cut paper circles into sectors (pizza slices) to show that a sector's area is a fraction θ/360 of the whole circle.
→Fold a paper circle along a chord to show the minor and major segments.
→Use a clock face: find the area swept by the minute hand in 5 or 15 minutes.
→Always write 'segment = sector – triangle' as a boxed rule before solving.
→Practise special-angle triangles (60° equilateral, 90° right-angled) so the triangle part is quick.
→Colour-code shaded-region problems: add what is shaded, subtract what is not.

📋 Assessment Questions

Can the student find the area of a sector for any given angle and radius?
Can the student find the area of a minor segment for angles 60°, 90° and 120°?
Can the student find the arc length of a sector?
Can the student solve a real-life problem like the area swept by a minute hand?
Does the student use the value of π given and state square units?

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