By the end of this chapter you'll be able to…

  • 1Calculate area and circumference of circle
  • 2Find area of sector and segment
  • 3Calculate arc length
  • 4Solve problems with combined shapes
💡
Why this chapter matters
Circle calculations used in engineering, design, architecture. Foundation for higher mathematics.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Areas Related to Circles — Class 10 Mathematics

"Pi (π) — the mathematics of circles, an irrational number with infinite digits, present everywhere."

1. About the Chapter

This chapter expands circle-area calculations to include:

  • Sector (pie-slice region)
  • Segment (region between chord and arc)
  • Combinations of plane figures (composite shapes)

Foundation for engineering, design, daily calculations.


2. Basic Circle Formulas (Recap)

Circumference

C = 2πr (where r = radius) C = πd (where d = diameter = 2r)

Area

A = πr²

Value of π

  • π ≈ 3.14159...
  • For class use: π = 22/7 (when problems give r as multiple of 7)
  • Or: π = 3.14 (decimal approximation)

3. Sector of a Circle

Definition

A sector is a region between two radii and an arc — like a pie slice.

Angle of Sector

The angle between the two radii (let it be θ).

Sector Area

Area of sector = (θ/360°) × πr²

(Since full circle is 360°, sector at angle θ is fraction θ/360°.)

Length of Arc

Arc length = (θ/360°) × 2πr

(Fraction of total circumference.)

Example

A sector with radius 14 cm and angle 90° (quarter circle).

  • Area = (90/360) × 22/7 × 14² = (1/4) × 22/7 × 196 = 154 cm²
  • Arc length = (90/360) × 2 × 22/7 × 14 = (1/4) × 88 = 22 cm

4. Segment of a Circle

Definition

A segment is a region between a CHORD and the corresponding ARC.

Types

  • Minor segment: smaller region (less than half circle)
  • Major segment: larger region (more than half circle)

Segment Area

Area of MINOR segment = Area of sector − Area of triangle

(The triangle is formed by the two radii and the chord.)

For sector angle θ: Area of segment = (θ/360°) × πr² − (1/2) r² sin θ

(The second term is area of triangle using ½ × base × height, or for any triangle, ½ ab sin C.)

Example

Find area of segment of circle (radius 10, angle 90°).

  • Sector area = (90/360) × π × 100 = 25π ≈ 78.5 cm²
  • Triangle area = (1/2)(10)(10) sin 90° = 50 cm²
  • Segment area = 78.5 − 50 = 28.5 cm²

Major Segment

  • Major segment = Area of circle − Minor segment
  • = πr² − Minor segment

5. Combinations of Plane Figures

Strategy

For complex shapes:

  1. DIVIDE into known simple shapes (rectangle, circle, triangle, etc.)
  2. CALCULATE each part's area
  3. ADD (for combined area) or SUBTRACT (for cut-out shapes)

Example 1: Park

A rectangular park 50 m × 30 m has a semicircular flower bed of radius 10 m at one end.

  • Park area = 50 × 30 = 1500 m²
  • Flower bed area = (1/2)π(10)² = 50π ≈ 157 m²
  • Lawn area = 1500 − 157 = 1343 m²

Example 2: Window

A rectangular window 1.5 m wide with a semicircular top of radius 0.75 m. Find area for glass needed.

  • Rectangular part: assume height = 2 m. Area = 1.5 × 2 = 3 m²
  • Semicircle: (1/2)π(0.75)² ≈ 0.88 m²
  • Total area ≈ 3.88 m²

Example 3: Hollow Pipe

A pipe has outer radius 5 cm, inner radius 3 cm. Cross-section area?

  • Outer circle area = π(5)² = 25π
  • Inner circle area = π(3)² = 9π
  • Cross-section = 25π − 9π = 16π ≈ 50.27 cm²

6. Worked Examples

Example 1: Find Arc Length

A circle has radius 21 cm. Find arc length subtending 60° at centre.

  • Arc length = (60/360) × 2 × 22/7 × 21
  • = (1/6) × 132 = 22 cm

Example 2: Find Sector Angle

Sector of radius 10 has arc length of 5.6π. Find angle.

  • 5.6π = (θ/360) × 2π × 10
  • θ = (5.6 × 360) / 20 = 100.8°

Example 3: Cut from Square

A square of side 14 cm has 4 semicircles drawn on its sides. Find the area of the shape formed.

  • Square area = 14² = 196 cm²
  • 4 semicircles = 2 full circles of radius 7
  • Total = 2 × π × 49 = 98π ≈ 308 cm²
  • Net area = 196 + 308 = 504 cm² (if added) OR 196 − 308 = NEGATIVE (so different setup)

(Actual setup depends — if circles are added outside or are cut from corners.)

Example 4: Two Wheels

A wheel has diameter 56 cm. How many revolutions for 11 km?

  • Circumference = π × 56 = (22/7) × 56 = 176 cm = 1.76 m
  • Distance = 11 km = 11,000 m
  • Revolutions = 11,000 / 1.76 = 6,250

7. Common Mistakes

  1. Confusing sector with segment

    • Sector = between TWO RADII and arc.
    • Segment = between CHORD and arc.
  2. Wrong angle conversion

    • Always use angle out of 360° in fraction.
  3. Wrong π value

    • Use 22/7 if numbers are multiples of 7; else 3.14.
  4. Forgetting to convert units

    • cm² vs m². 1 m² = 10,000 cm².
  5. Missing parts in combinations

    • Check if shape adds or subtracts. Re-read problem.

8. Real-World Applications

Architecture

  • Domed buildings: surface area
  • Stadiums: layout calculations
  • Indian temples use circle geometry

Engineering

  • Wheel/cog calculations
  • Pipe cross-sections
  • Pizza box, soup can design

Design

  • Logos, watch faces
  • Sport fields (cricket field with rope, football pitch curves)

Indian Use

  • Indian Rangoli patterns use sectors and segments
  • Sundials use sector geometry
  • Cricket boundary calculations

9. Indian Context

π Approximations Through Indian History

  • Aryabhata (5th c.): π ≈ 3.1416 (very accurate)
  • Bhaskara II: refined π calculations
  • Madhava (14th c.): infinite series for π — 200 years before Newton!

This made India a global leader in circle geometry.


10. Conclusion

Areas related to circles bring together:

  • Pi (π)
  • Geometry
  • Algebra

Master:

  • Sector area: (θ/360°) × πr²
  • Arc length: (θ/360°) × 2πr
  • Segment = sector − triangle
  • Combinations: ADD/SUBTRACT simple shapes

Practice 15+ problems. This chapter combines Chapter 10 (circles), Chapter 12 (volumes), and basic geometry.

Circles and their parts — the geometry that surrounds us.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Circumference
C = 2πr
Area of circle
A = πr²
Sector area
A_sector = (θ/360°) × πr²
θ in degrees
Arc length
L = (θ/360°) × 2πr
Segment area
A_segment = sector area − triangle area
Minor segment
π values
22/7 or 3.14
Use 22/7 when r is multiple of 7
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Confusing sector and segment
SECTOR is bounded by 2 radii + arc (pie slice). SEGMENT is bounded by chord + arc.
WATCH OUT
Wrong fraction for sector
Use θ/360°, not θ/180°.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Area
Find area of circle with radius 7 cm (use π = 22/7).
Show solution
✦ Answer: A = πr² = (22/7) × 49 = 154 cm².
Q2MEDIUM· Sector
Find area of sector with radius 14 cm and angle 60°.
Show solution
Step 1 — Apply formula. A_sector = (θ/360°) × πr² = (60/360) × (22/7) × 14² = (1/6) × (22/7) × 196 Step 2 — Compute. = (1/6) × (22 × 28) = (1/6) × 616 = 102.67 cm² Or: 308/3 cm² Step 3 — Alternative. = (22/7) × 196 / 6 = 22 × 28 / 6 = 616/6 = 102.67 cm² ✦ Answer: Area = 102.67 cm² (or 308/3 cm²).
Q3HARD· Combination
A rectangular ground is 100 m × 60 m with a semi-circular ring at one end (radius 30 m). Find total area.
Show solution
Step 1 — Rectangle area. Rectangle = 100 × 60 = 6000 m² Step 2 — Semi-circle area. Semicircle = (1/2) × π × 30² = (1/2) × (22/7) × 900 = 11 × 900 / 7 = 9900/7 ≈ 1414.29 m² Step 3 — Total area. Total = 6000 + 1414.29 = 7414.29 m² Step 4 — Express. Or: 6000 + 450π = 6000 + 450(22/7) = 6000 + 9900/7 = (42000 + 9900)/7 = 51900/7 ≈ 7414.29 m² ✦ Answer: Total area ≈ 7414.29 m² (or 51900/7 m²).

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • A_circle = πr²
  • C = 2πr
  • Sector area = (θ/360) × πr²
  • Arc length = (θ/360) × 2πr
  • Segment = sector − triangle
  • Combinations: add/subtract simple shapes
  • Aryabhata: π ≈ 3.1416 (5th century)
  • Madhava: infinite series for π

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8

Question typeMarks eachTypical countWhat it tests
MCQ12Formulas
Short2-31Single shape
Long51Combinations
Prep strategy
  • Memorise all formulas
  • Use 22/7 for clean numbers
  • Practice combination problems

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Engineering pipes

Cross-section areas calculated using circle formulas.

Stadium design

Cricket grounds, football fields use circle/sector geometry.

Indian Rangoli

Traditional art uses sectors, segments.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Memorise formulas
  2. Identify sector vs segment
  3. Break combinations into simple shapes
  4. Verify units

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Brahmagupta formula for cyclic quadrilateral area
  • Heron's formula
  • Inscribed and circumscribed shapes

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 10 BoardHigh
Maths OlympiadMedium
JEE FoundationMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Verified by the tuition.in editorial team
Last reviewed on 20 May 2026. Written and reviewed by subject-matter experts — read about our process.
Editorial process →
Header Logo