Is 'areas of combinations of plane figures' still in the syllabus?

Not as part of this chapter. The rationalised Chapter 11 is Introduction → 11.1 Areas of Sector and Segment of a Circle → 11.2 Summary, and the three-point summary covers only arc length, sector area, and segment area. The composite-shapes technique (adding and subtracting simple regions) is on the chapter page in the appendix, marked non-examinable here, because it is a natural extension and still appears in general mensuration problems.

How many questions are in this chapter now?

Fourteen, all in the single Exercise 11.1. The older edition split sector/segment questions and combination-of-shapes questions across two exercises; the second one is gone.

How do I find the area of a segment if I'm only given the sector angle and radius?

Compute the sector area with (θ/360°)πr², then subtract the area of the triangle formed by the two radii and the chord — which is ½r²sin θ for any angle θ, not just special ones. For θ = 90°, this triangle is simply ½ × r × r, since sin 90° = 1.

Chapter 11 of 14

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Haryana (BSEH)Class 10 Mathematics★ Unit VI — Mensuration: 10 marks, shared with Surface Areas and Volumes · CBSE Class 10 Maths (Chapter 11)

Areas Related to Circles

Areas Related to Circles — the area of a sector and a segment of a circle, and the arc length formula, applied to clocks, wipers, brooches and lighthouses. CBSE Class 10 Maths Chapter 11, updated to the current single-exercise structure.

⏱ 20 min readEasy difficulty✓ Free · NCERT-aligned

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

1Find the area of a sector given its radius and angle
2Find the length of an arc given its radius and angle
3Find the area of a segment by subtracting the triangle from the sector
4Move between a minor sector/segment and its major counterpart using the full circle
5Break a real object — a clock, an umbrella, a table cover — into the sector it actually is

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Why this chapter matters

Every question in this chapter is one idea applied to a new picture: a fraction of the circle, where the fraction is the angle out of 360°. Clock hands, wipers, brooches, table covers — all of it is sector minus triangle, over and over.

Before you start — revise these

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

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Circles

What CBSE examines here (2026-27). The area of a sector and a segment of a circle, and the length of an arc — all three built from one idea: a fraction of the whole circle, where the fraction is the angle out of 360°. The chapter runs to one exercise, 11.1. Areas of combinations of plane figures were removed — the appendix at the end keeps the technique for reference, marked as background.

"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)
Arc length (the curved boundary of a sector)

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. 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: Combine a Sector and a Segment

A chord of a circle of radius 14 cm subtends an angle of 90° at the centre. Find the sum of the areas of the minor segment and the major sector.

Sector (90°) area = (90/360) × 22/7 × 14² = (1/4) × 22/7 × 196 = 154 cm²
Triangle area (right angle, both sides 14) = (1/2)(14)(14) = 98 cm²
Minor segment = 154 − 98 = 56 cm²
Major sector = full circle − minor sector = (22/7)(14²) − 154 = 616 − 154 = 462 cm²
Sum = 56 + 462 = 518 cm²

This is the pattern behind almost every question in this exercise: sector, subtract the triangle for the segment, subtract from the full circle for the major piece.

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

6. Common Mistakes

Confusing sector with segment
- Sector = between TWO RADII and arc.
- Segment = between CHORD and arc.
Wrong angle conversion
- Always use angle out of 360° in fraction.
Wrong π value
- Use 22/7 if numbers are multiples of 7; else 3.14.
Forgetting to convert units
- cm² vs m². 1 m² = 10,000 cm².
Using the wrong triangle formula for the segment
- The triangle cut off by the chord has two sides equal to the radius, so its area is ½r²sin θ — not ½ × base × height unless you have actually found the base and height separately.

7. 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

8. 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.

9. 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, and major sector/segment = full circle − minor sector/segment

Practice 15+ problems. This chapter builds on Chapter 10 (Circles) and feeds Chapter 12 (Surface Areas and Volumes).

Circles and their parts — the geometry that surrounds us.

Appendix — beyond the current syllabus

Not examinable in CBSE 2026-27. Areas of combinations of plane figures — composite shapes built by adding or subtracting circles, sectors, triangles and rectangles — were removed from this chapter, along with the exercise that practised them. The rationalised chapter is just 11.1 Areas of Sector and Segment of a Circle, followed by 11.2 Summary, and the summary's three points are the sector-area formula, the arc-length formula, and the segment-as-sector-minus-triangle relationship — nothing about composite shapes. The technique is kept here because it is a natural extension of what the chapter does teach, and it still turns up in mensuration problems elsewhere.

The method

Divide the composite shape into simple pieces you already know how to measure — circles, sectors, triangles, rectangles.
Decide add or subtract — is the piece part of the shape, or cut out of it?
Compute each piece and combine.

Example — a hollow pipe's cross-section

A pipe has outer radius 5 cm and inner radius 3 cm. Find the area of the cross-section (the annular ring between the two circles).

Outer circle area = π(5)² = 25π
Inner circle area (the hollow part, to be removed) = π(3)² = 9π
Cross-section = 25π − 9π = 16π ≈ 50.27 cm²

Example — a park with a flower bed

A rectangular park 50 m × 30 m has a semicircular flower bed of radius 10 m built into one end. Find the area of the park available for lawn.

Park area = 50 × 30 = 1500 m²
Flower bed area = ½π(10)² = 50π ≈ 157 m²
Lawn area = 1500 − 157 = 1343 m²

If you meet this in an older guidebook or PYQ, the method above is exactly right — it simply is not part of the current chapter's own exercise.

Key formulas & results

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

Area of a sector

(θ/360°) × πr²

θ is the angle at the centre, in degrees

Length of an arc

(θ/360°) × 2πr

The same fraction applied to the circumference instead of the area

Area of a segment

area of sector − area of the triangle OAB

The triangle has two sides equal to the radius and included angle θ

Area of that triangle

½ r² sin θ

Standard ½ab sin C with a = b = r

Major from minor

major sector = πr² − minor sector; major segment = πr² − minor segment

Both come from subtracting off the full circle

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Common mistakes & fixes

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

WATCH OUT

✗ Using ½ × base × height for the segment's triangle

✓ The triangle cut off by a chord has two known sides (both the radius) and a known included angle — that is ½r²sin θ, not a base-height calculation you have to derive separately.

WATCH OUT

✗ Forgetting the segment is sector minus triangle, not sector alone

✓ A sector always includes the straight edges to the centre; a segment is bounded only by the chord and the arc. Subtracting the triangle is what removes the straight edges.

WATCH OUT

✗ Mixing π = 22/7 and π = 3.14 in the same problem

✓ Use whichever the question specifies, and use only that one value throughout — switching partway changes the last digit of every answer.

WATCH OUT

✗ Not converting a real object into an angle first

✓ A clock's minute hand sweeping 5 minutes, an umbrella's 8 equally spaced ribs, a brooch's 10 equal sectors — each of these implies an angle (30°, 45°, 36°) that has to be worked out before the sector formula can be used at all.

NCERT exercises (with solutions)

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

Ex 11.1

Exercise 11.1

Area of sector and segment, arc length — clocks, wipers, brooches, table covers, and one MCQ

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Areas Related to Circles?

3 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

3 questions~2 min

5-minute revision

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

•Sector: region between two radii and the arc between them
•Segment: region between a chord and its arc — smaller (minor) or larger (major)
•Area of sector = (θ/360°)πr²; arc length = (θ/360°)2πr — the same fraction, two different circle quantities
•Segment area = sector area − triangle area, where the triangle is ½r²sin θ
•Major sector = full circle − minor sector; major segment = full circle − minor segment
•This chapter has a single exercise, 11.1, with 14 questions
•Areas of combinations of plane figures were removed from this chapter

Haryana (BSEH) marks blueprint

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

Typical chapter weightage: 6-8

Question type	Marks each	Typical count	What it tests
MCQ	1	2	Formulas
Short	2-3	1	Single shape
Long	5	1	Combinations

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.

Identify the angle first — many questions describe a real object and expect you to work out θ from it (a clock hand, evenly spaced ribs, a fraction of a full turn)

State whether you need the sector, the segment, or the difference between minor and major, before computing anything

For the segment's triangle, default to ½r²sin θ rather than hunting for a base and height

Keep the same value of π throughout one question

Going beyond the textbook

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

STRETCH

Brahmagupta formula for cyclic quadrilateral area

STRETCH

Heron's formula

STRETCH

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

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