Areas Related to Circles — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Maths · Ch 11Class 10 · CBSE 2026–27Medium weightage

Areas Related to Circles

What the board actually asks: almost every mark here comes from three moves — the length of an arc, the area of a sector (the fraction θ/360 of the circle), and the area of a segment (that sector minus triangle OAB). Get the central angle and the value of π right, and know when to leave √3 in surd form. The 2026-27 chapter is trimmed to sectors and segments only — combinations of plane figures are no longer in it.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

A slice of the circle — measured by its angle.

Every result in this chapter is one idea: a part of a circle is the fraction $\frac{θ}{360}$ of the whole, where $θ$ is the angle it makes at the centre. Take that fraction of the circumference $2 π r$ and you get an arc length; take it of the area $π r^{2}$ and you get a sector; cut a chord across and the leftover piece is a segment, found by subtracting triangle $O A B$ from the sector.

Formulae

Circumference of a circle

C = 2 π r

Tip: use π = 22/7 when r is a multiple of 7, unless the question gives a value of π.

Area of a circle

A = π r^{2}

Length of an Arc

Length of an arc (angle

θ

)

ℓ = \frac{θ}{360} \times 2 π r

Tip: θ/360 of the circumference.

Area of a Sector

Area of a sector (angle

θ

)

\frac{θ}{360} \times π r^{2}

Tip: θ/360 of the area.

Area of a Segment

Area of a segment

\frac{θ}{360} \times π r^{2} - area of △ O A B

Tip: Sector − triangle OAB.

Perimeter of a Sector

Perimeter = ℓ + 2 r

: the arc length plus the two radii.

Major Sector and Major Segment

Angle of the major sector

= 36 0^{\circ} - θ

. Area of major sector

= π r^{2} -

area of minor sector. Area of major segment

= π r^{2} -

area of minor segment.

Area of △OAB for 60°, 90°, 120°

For

12 0^{\circ}

: draw

O M ⊥ A B

;

∠ A O M = 6 0^{\circ}

, so

O M = \frac{r}{2}

and

A M = \frac{3}{2} r

. For

6 0^{\circ}

say that

△ O A B

is equilateral; for

9 0^{\circ}

the two sides are both

r

. Values to check against:

6 0^{\circ}

\frac{3}{4} r^{2}

;

9 0^{\circ}

\frac{1}{2} r^{2}

;

12 0^{\circ}

\frac{3}{4} r^{2}

Shortcuts & memory tricks →

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Sector of a circle

NCERT verbatim — memorise this exact wording· Ch 11,

“the portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle”

What it means: A 'pizza slice' of the disc — the region boxed in by two radii and the arc between them, with its pointed corner at the centre

O

Segment of a circle

NCERT verbatim — memorise this exact wording· Ch 11,

“the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle”

What it means: The region cut off by a straight chord and the arc it subtends — like the slice of melon left after one straight cut. No radius is involved.

Area of a sector

NCERT verbatim — memorise this exact wording· Ch 11,

“Area of a sector of a circle with radius r and angle with degree measure θ is (θ/360) × πr²”

What it means: Take the fraction

\frac{θ}{360}

of the whole circle's area

π r^{2}

. Convert the central angle to degrees first.

Length of an arc of a sector

NCERT verbatim — memorise this exact wording· Ch 11,

“Length of an arc of a sector of a circle with radius r and angle with degree measure θ is (θ/360) × 2πr”

What it means: The same fraction

\frac{θ}{360}

, but of the whole circumference

2 π r

— the curved boundary of the sector, measured in length (cm), not area.

Area of a segment

NCERT verbatim — memorise this exact wording· Ch 11,

“Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle.”

What it means: There is no direct formula — always subtract: segment

=

its sector

-

triangle

O A B

. Finding the triangle (the chord and the perpendicular height) is the hard part.

Key terms — NCERT verbatim (state word-for-word)

Minor & major sector: OAPB is called the minor sector and OAQB is called the major sector. (Ch 11, )
Minor & major segment: APB is called the minor segment and AQB is called the major segment. (Ch 11, )
Angle of the sector: ∠ AOB is called the angle of the sector. (Ch 11, )
'Segment' / 'sector' mean the minor one: When we write 'segment' and 'sector' we will mean the 'minor segment' and the 'minor sector' respectively, unless stated otherwise. (Ch 11, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Sector — minor and major

Two radii plus the arc between them cut the disc into two sectors. The smaller one (shaded,

O A P B

) is the minor sector and carries the angle

θ = ∠ A O B

; the larger one (

O A QB

) is the major sector with angle

36 0^{\circ} - θ

NCERT Fig 11.1Sector of a circle · shaded OAPB is the minor sector (angle ∠AOB), OAQB the major sector.

2-mark 'find the area of the sector' (Section A/B)

Segment — minor and major

A chord

A B

(not a radius) splits the disc into two segments. The piece between the chord and its arc is the minor segment (

A P B

, shaded); the rest is the major segment (

A QB

). A segment always needs the triangle

O A B

to measure it.

NCERT Fig 11.2Segment of a circle · shaded APB is the minor segment cut off by chord AB, AQB the major segment.

3-mark 'find the area of the segment' (Section C)

Length of an arc

The curved edge of a sector of angle

θ

is the fraction

\frac{θ}{360}

of the full circumference:

ℓ = \frac{θ}{360} \times 2 π r

. It is a length (cm), so never square it.

1-2 mark direct, or part (i) of a case-based problem

Area of a sector

The same fraction, taken of the area:

\frac{θ}{360} \times π r^{2}

. For the major sector the fast route is

π r^{2} - (minor sector)

, which avoids re-substituting

36 0^{\circ} - θ

2-3 mark direct 'find the area of the sector'

Area of a segment = sector − triangle

There is no direct formula. Area of the minor segment

= \frac{θ}{360} \times π r^{2} - area of △ O A B

. Get the triangle from a perpendicular dropped from

O

to the chord (using

cos

and

sin

of half the angle), or — when

θ = 6 0^{\circ}

— as an equilateral triangle of side

r

3-5 mark long answer (Section C/D)

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Ch 11, Example 1,

“Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14).”

Apply the sector-area formula.Area of the sector

= \frac{θ}{360} \times π r^{2} = \frac{30}{360} \times 3.14 \times 4^{2} = \frac{12.56}{3} cm^{2}

.Concept: fraction of the circle

1 mark

Evaluate the minor sector.

= 4.19 cm^{2}

(approx.).Concept: answer with unit

1 mark

Major sector = whole − minor.Area of the major sector

= π r^{2} - 4.19 = 3.14 \times 16 - 4.19 = 46.05 \approx 46.1 cm^{2}

(approx.).Concept: major = whole − minor

1 mark

Mark-scheme logic (authored): 3-mark shape: 1 for the sector formula with correct substitution, 1 for the minor-sector value

4.19 cm^{2}

, 1 for the major sector via

π r^{2} - minor

. NCERT also gives the alternative

\frac{360 - 30}{360} π r^{2}

route — either earns the last mark.

Ch 11,

“Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.”

Ch 11, Example 2,

“Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is 21 cm and ∠ AOB = 120°. (Use π = 22/7)”

Ch 11,

“The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.”

Ch 11,

“In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord”

Problem-solving strategies AUTHORED

Arc, sector or segment?All three start from the fraction

\frac{θ}{360}

. Multiply it by

2 π r

for an arc (length), by

π r^{2}

for a sector (area).

'Area of the segment'?There is no direct formula — compute sector

-

triangle

O A B

. Find the triangle separately.

60° at the centre?Two radii with a

6 0^{\circ}

angle make an equilateral triangle — its area is

\frac{3}{4} a^{2}

with

a = r

'Major sector / segment'?Major

=

whole

-

minor:

π r^{2} - (minor sector)

, or

π r^{2} - (minor segment)

. Or use the angle

36 0^{\circ} - θ

Which value of π?Use what the question states —

\frac{22}{7}

by default in this exercise,

3.14

when told. Keep

3

in surd form unless a value is given.

Common mistakes that cost marks

Where students lose easy marks

Reaching for a 'segment formula' — there isn't one. A segment is always sector

-

triangle

O A B

Forgetting to turn a time / fraction into a central angle — 5 minutes on a clock is

3 0^{\circ}

, not

5^{\circ}

Mixing up arc length (

\times 2 π r

) with sector area (

\times π r^{2}

) — check the unit: cm for an arc, cm

^{2}

for an area.

Rounding π early or dropping the unit — state cm

^{2}

and keep

3

exact unless a numerical value is supplied.

Formulae

Circumference of a circle

C = 2 π r

Tip: use π = 22/7 when r is a multiple of 7, unless the question gives a value of π.

Area of a circle

A = π r^{2}

Length of an Arc

Length of an arc (angle

θ

)

ℓ = \frac{θ}{360} \times 2 π r

Tip: θ/360 of the circumference.

Area of a Sector

Area of a sector (angle

θ

)

\frac{θ}{360} \times π r^{2}

Tip: θ/360 of the area.

Area of a Segment

Area of a segment

\frac{θ}{360} \times π r^{2} - area of △ O A B

Tip: Sector − triangle OAB.

Perimeter of a Sector

Perimeter = ℓ + 2 r

: the arc length plus the two radii.

Major Sector and Major Segment

Angle of the major sector

= 36 0^{\circ} - θ

. Area of major sector

= π r^{2} -

area of minor sector. Area of major segment

= π r^{2} -

area of minor segment.

Area of △OAB for 60°, 90°, 120°

For

12 0^{\circ}

: draw

O M ⊥ A B

;

∠ A O M = 6 0^{\circ}

, so

O M = \frac{r}{2}

and

A M = \frac{3}{2} r

. For

6 0^{\circ}

say that

△ O A B

is equilateral; for

9 0^{\circ}

the two sides are both

r

. Values to check against:

6 0^{\circ}

\frac{3}{4} r^{2}

;

9 0^{\circ}

\frac{1}{2} r^{2}

;

12 0^{\circ}

\frac{3}{4} r^{2}

Shortcuts & Tricks

Sector area = ½ × arc length × radius.

MCQ / check only: don't write this as working

Area one-twelfth of the circle ⇒ θ = 30°.

MCQ / check only: don't write this as working

Draw OAB, label θ, then subtract.

Safe in written answers

Sector perimeter includes the two radii.

MCQ / check only: don't write this as working

Same θ: arc ∝ r, area ∝ r².

MCQ / check only: don't write this as working

Memory Tricks

Kitna hissa? θ/360 — lambai 2πr ki, area πr² ki.

arc and sector. θ/360 × 2πr; θ/360 × πr².

Segment = Sector minus Tikona (triangle).

segment = sector − △OAB.

Major = poora gola minus minor.

πr² − minor.

Mindmap

- - A part of a circle = θ/360 of the whole
  - θ = angle at the centre (in degrees)
  - × 2πr → length · × πr² → area
- - Minor sector = angle θ (shaded)
  - Major sector = angle 360° − θ
  - Area = (θ/360) × πr²
- - ℓ = (θ/360) × 2πr
  - It is a length (cm) — never square it
- - No direct formula
  - Segment = sector − triangle OAB
  - Minor / major segment about chord AB
- - Drop OM ⊥ AB → cos, sin of half-angle
  - θ = 60° ⇒ equilateral, area (√3/4)r²
- - Segment ≠ a single formula
  - Convert time/fraction → central angle
  - cm for arc, cm² for area
  - Use the stated π; keep √3 exact

Where the sector, arc and segment formulae come from

NCERT derives all three results by the Unitary Method (pp.155-156): scale the whole circle down by the fraction of the angle at the centre. This is the reasoning behind the formulae in the formula strip — worth understanding, not just memorising.

A full circle subtends 360° at the centre, and the area of the circular region is πr².

area of circle = π r^{2}

Unitary Method: when the angle at the centre is 360° the sector area is πr², so when the angle is 1° the sector area is πr²/360.

\frac{π r ^{2}}{360}

Therefore for an angle θ at the centre, the sector area is θ times the 1° value.

Area of the sector of angle θ = \frac{θ}{360} \times π r^{2}

This is the area-of-a-sector formula (NCERT p.155).

The same Unitary-Method argument on the whole circumference 2πr (which corresponds to 360°) gives the arc length.

length of an arc of angle θ = \frac{θ}{360} \times 2 π r

The minor segment is what remains of the sector after triangle OAB is removed, so its area is the sector minus that triangle.

Area of segment A P B = \frac{θ}{360} \times π r^{2} - area of △ O A B

NCERT p.156 — segment = sector − corresponding triangle.

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

1 markSection A

A regular pentagon is inscribed in a circle with centre O, of radius 5 cm. What is the area of the shaded part of the circle?

Regular pentagon inscribed in a circle with centre O; two triangles from O to pentagon sides and two segments of the circle shaded

(A) 2π cm^2
(B) 4π cm^2
(C) 5π cm^2
(D) 10π cm^2

Step-marked solution

[1 mark] The radii to the 5 vertices of the regular pentagon divide the circle into 5 congruent sectors, each of area (1/5)·π·5² = 5π cm². From the figure, the shaded part is one complete sector, plus the triangle of a second sector and the segment of a third sector; a triangle and a segment from congruent sectors together make one full sector. So shaded area = 2 × 5π = 10π cm².

1 markSection A

In the given figure, the area of the segment ACB is

Circle with centre O; radii OA and OB, each r, at right angles (90 degrees marked); segment ACB between chord AB and the arc shaded

(A) r^2/4 (π − 2)
(B) r^2/4 (π + 2)
(C) r^2/4 (π − 1)
(D) r^2/4 (π + 1)

Step-marked solution

[1 mark] From the figure, ∠AOB = 90° and OA = OB = r. Area of segment ACB = area of sector OACB − area of △OAB = (90°/360°)πr² − (1/2)·r·r = πr²/4 − r²/2 = (r²/4)(π − 2).

2 marksSection B

ABCD is a rhombus with side 3 cm. Two arcs are drawn from points A and C respectively such that the radius equals the side of the rhombus. If BD is a line of symmetry for the figure, then find the area of the shaded part of the figure in terms of π. OR Wasim made a model of Pac-Man with area 120π cm^2 and mouth angle 60° at the centre. Find the minimum length of ribbon needed for the entire boundary in terms of π.

Rhombus ABCD with side 3 cm and angle A marked 60 degrees; arcs from A and C meet at D and B with the lens between them shaded; BD dashed

Shaded circle with a sector cut out at the centre, its angle marked 60 degrees

Step-marked solution

[1 mark] From the figure, ∠DAB = 60° and the arc through D and B is drawn with centre A, radius 3 cm. Area of sector ABD = (60/360) × π × 3² = 3π/2 cm²; △ABD has AB = AD = 3 cm and ∠A = 60°, so it is equilateral: area = (√3/4) × 3² = 9√3/4 cm². [OR Let the radius be r cm; the model is a sector of angle 360° − 60° = 300°: 120π = (300/360) × π × r² ⇒ r² = 144 ⇒ r = 12 cm.]
[1 mark] BD is a line of symmetry, so the shaded part = 2 × (area of sector ABD − area of △ABD) = 2 × (3π/2 − 9√3/4) = (3π − 9√3/2) cm². [OR Ribbon length = arc length + 2 radii = (300/360) × 2 × π × 12 + 24 = (20π + 24) cm.]

Practise Areas Related to Circles competency questions

Read NCERT Chapter 11 on LazyTopper (jemh111)

Official NCERT chapter PDF (jemh111) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Mathematics, Ch 11 (jemh111.pdf, printed pages 154-160). One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Sector of a circle	verbatim	NCERT 2026-27 Ch11
Segment of a circle	verbatim	NCERT 2026-27 Ch11
Area of a sector	verbatim	NCERT 2026-27 Ch11 (Summary 2)
Length of an arc of a sector	verbatim	NCERT 2026-27 Ch11 (Summary 1)
Area of a segment	verbatim	NCERT 2026-27 Ch11 (Summary 3)
Minor & major sector	verbatim	NCERT 2026-27 Ch11
Minor & major segment	verbatim	NCERT 2026-27 Ch11
Angle of the sector	verbatim	NCERT 2026-27 Ch11
'Segment' / 'sector' mean the minor one	verbatim	NCERT 2026-27 Ch11 (Remark)
Area of a sector + major sector · NCERT Example 1	example	NCERT 2026-27 Ch11 Example 1
Area of a sector · NCERT Exercise 11.1 Q1	example	NCERT 2026-27 Ch11 Exercise 11.1 Q1
Area of a segment · NCERT Example 2	example	NCERT 2026-27 Ch11 Example 2 -158
Area swept by a minute hand · NCERT Exercise 11.1 Q3	example	NCERT 2026-27 Ch11 Exercise 11.1 Q3
Arc length + sector + segment · NCERT Exercise 11.1 Q5	example	NCERT 2026-27 Ch11 Exercise 11.1 Q5
fig_sector_111	ncert-figure	NCERT 2026-27 Ch11 Fig 11.1
fig_segment_112	ncert-figure	NCERT 2026-27 Ch11 Fig 11.2

Areas Related to Circles · spec-rendered note · NCERT Reprint 2026-27, Class 10 Mathematics, Ch 11 (jemh111.pdf, printed pages 154-160)

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