Areas Related to Circles — Detailed Notes

Chapter 11 | CBSE Class 10 Mathematics (NCERT)

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Areas Related to Circles — Decision Map

Sector and Segment of a Circle

Areas Related to Circles — Formula Strip

11.1 Areas of Sector and Segment of a Circle

Definitions (recall)

Sector: part of the circular region enclosed by two radii and the corresponding arc.
Angle at the centre is the angle of the sector.
Minor sector angle $= θ$ ; major sector angle $= 36 0^{\circ} - θ$ .
Segment: part of the circular region enclosed between a chord and the corresponding arc.
Minor segment and major segment are formed by the same chord.
Remark: Unless stated otherwise, “sector” and “segment” mean the minor ones.

Area of a sector

Full circle (angle $36 0^{\circ}$ ) has area $π r^{2}$ . By unitary method:

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

where $θ$ is in degrees and $r$ is the radius.

Length of an arc

Full circumference $= 2 π r$ . Same unitary idea:

$Length of an arc of a sector of angle θ = \frac{θ}{360} \cdot 2 π r$

Area of a segment

For minor segment APB with sector OAPB and $△ O A B$ :

$Area of segment APB = Area of sector OAPB - Area of △ O A B$

Also:

$Area of major sector Area of major segment = π r^{2} - Area of minor sector = π r^{2} - Area of minor segment$

Alternatively, major sector area $= \frac{36 0 ^{\circ} - θ}{360} π r^{2}$ .

Better Method / Strategy — Sector vs Segment
Ask: does the region include the two radii (sector) or only the chord + arc (segment)?
For segment: always sector − triangle. Never stop at sector alone unless asked.

Worked idea A — Sector and major sector (NCERT Ex. 1 style)

Radius $r = 4$ cm, $θ = 3 0^{\circ}$ , $π = 3.14$ .

$Sector area = \frac{30}{360} \times 3.14 \times 4^{2} = \frac{1}{12} \times 50.24 = 4.19 cm^{2} (approx.)$

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

(Or use $\frac{330}{360} π r^{2}$ .)

Worked idea B — Segment with $θ = 12 0^{\circ}$ (NCERT Ex. 2 style)

$r = 21$ cm, $∠ A O B = 12 0^{\circ}$ , $π = \frac{22}{7}$ .

Step 1 — Sector: $\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = 462 cm^{2}$

Step 2 — Triangle: Draw $O M ⊥ A B$ . Then $∠ A O M = 6 0^{\circ}$ .

$O M = 21 cos 6 0^{\circ} = \frac{21}{2}, A M = 21 sin 6 0^{\circ} = \frac{21 3}{2}$

$Area of △ O A B = \frac{1}{2} \times A B \times O M = \frac{1}{2} \times 213 \times \frac{21}{2} = \frac{441 3}{4}$

Step 3 — Segment: $462 - \frac{441 3}{4} = \frac{21}{4} (88 - 213) cm^{2}$

Better Method / Strategy — Finding $△$ area

$θ$ Quick approach

$6 0^{\circ}$ $△ O A B$ equilateral $\Rightarrow$ area $= \frac{3}{4} r^{2}$

$9 0^{\circ}$ Right isosceles $\Rightarrow$ area $= \frac{1}{2} r^{2}$

$12 0^{\circ}$ Split to $6 0^{\circ}$ ; use $sin / cos$ or formula $\frac{1}{2} r^{2} sin θ$

General: area of $△ O A B = \frac{1}{2} r^{2} sin θ$ (when $θ$ is central angle).

Application patterns (exam favourites)

Minute hand: length $=$ radius; in $t$ minutes, angle $= 6 t^{\circ}$ (since $36 0^{\circ}$ in $60$ min). Area swept $=$ sector of that angle.
Quadrant: $θ = 9 0^{\circ}$ ; if circumference given, find $r$ first from $2 π r$ .
Grazing at a corner of a square field: rope length $r$ at a corner $\Rightarrow$ quarter-circle sector of radius $r$ .
Wipers / lighthouse: sector area with given blade length (radius) and sweep angle.
Equal sectors: $n$ equal ribs $\Rightarrow$ each central angle $= \frac{36 0 ^{\circ}}{n}$ .

Worked idea C — Clock / quadrant / arc (board patterns)

Minute hand $14$ cm in $5$ minutes: $θ = 3 0^{\circ}$ . $Area swept = \frac{30}{360} π (14)^{2} = \frac{1}{12} \cdot \frac{22}{7} \cdot 196 = \frac{154}{3} cm^{2}$

Quadrant from circumference $22$ cm: $2 π r = 22 \Rightarrow r = 3.5$ ; quadrant $= \frac{1}{4} π r^{2} = \frac{77}{8}$ cm² (with $π = \frac{22}{7}$ ).

Arc with $r = 21$ , $θ = 6 0^{\circ}$ : length $= \frac{60}{360} \cdot 2 π \cdot 21 = 22$ cm; sector $= \frac{1}{6} π \cdot 441 = 231$ cm².

Worked idea D — Grazing and design sectors

Horse at corner, rope $5$ m, square field: grazing inside field $= \frac{1}{4} π (5)^{2}$ . If rope becomes $10$ m, new area $= \frac{1}{4} π (10)^{2}$ ; increase $= \frac{1}{4} π (100 - 25) = \frac{75 π}{4}$ .

Umbrella with $8$ equal ribs, $r = 45$ : each sector angle $= 4 5^{\circ}$ ; area between consecutive ribs $= \frac{45}{360} π (45)^{2} = \frac{1}{8} π \cdot 2025$ .

Lighthouse sector $8 0^{\circ}$ , range $16.5$ km ( $π = 3.14$ ): area $= \frac{80}{360} \cdot 3.14 \cdot (16.5)^{2}$ .

Better Method / Strategy — Applications

Identify $r$ and $θ$ from the story.

Decide sector vs segment vs major.

Substitute given $π$ .

If cost is asked, multiply area by rate.

NCERT summary points (must retain)

Arc length $= \frac{θ}{360} \cdot 2 π r$
Sector area $= \frac{θ}{360} π r^{2}$
Segment area $=$ corresponding sector $-$ corresponding triangle

Formula MCQ alert

Area of sector of angle $p$ (degrees), radius $R$ , is $\frac{p}{360} π R^{2}$ (not $\frac{p}{180}$ ).

Exam-Oriented Analysis

Question type	Typical marks	What examiners look for
Direct sector / arc	1–2	Correct formula + $π$
Segment (60°/90°/120°)	2–3	Sector − triangle steps
Major sector/segment	2	Whole − minor or $(360 - θ)$
Clock / wiper / grazing	2–4	Correct $θ$ from context
Formula MCQ	1	$\frac{θ}{360} π r^{2}$

Common mistakes

Using $θ$ in radians formulas (NCERT Class 10 uses degrees).
Computing sector when segment is asked (forgot $-$ triangle).
Wrong $π$ value when question specifies $22/7$ or $3.14$ .
For $12 0^{\circ}$ , incorrect height/base of triangle.
Mixing major segment with major sector.
Forgetting that default “segment” means minor.

Scoring tips

Write formula, substitute, simplify; box final answer with units.
For major parts, either whole $-$ minor or $(360 - θ)$ formula — pick one and stay consistent.
In applications, first find $θ$ , then apply sector formula.
Show triangle area calculation clearly (especially for $6 0^{\circ}$ / $12 0^{\circ}$ ).
Keep intermediate answers exact (e.g. $\frac{154}{3}$ ) unless asked to approximate.