1. Chapter at a Glance
A pizza slice, the region swept by a car's wiper, the area a grazing cow can reach when tied to a peg, the arc traced by a clock's minute hand — all of these are parts of circles. In this chapter we find the length of an arc, the area of a sector (a "pizza slice") and the area of a segment (the region between a chord and its arc). For board exams, segment problems use central angles of 60°, 90° and 120°.
- Perimeter and area of a circle (recap).
- Minor and major arcs, sectors and segments.
- Arc length = (θ/360°) × 2πr.
- Sector area = (θ/360°) × πr².
- Segment area = sector area − triangle area (θ = 60°, 90°, 120°).
- Real-life problems: clocks, wipers, grazing animals, umbrellas, brooches.
2. Recap: The Circle
| Quantity | Formula |
|---|---|
| Circumference | 2πr = πd |
| Area | πr² |
| Semicircle area | πr²/2 |
| Ring (annulus) area | π(R² − r²) |
Use π = 22/7 when the radius is a multiple of 7, and 3.14 otherwise (or as instructed).
3. Sectors and Arcs
A sector is the region between two radii and the arc between them. The smaller one is the minor sector, the larger the major sector. If the minor sector has angle θ, the major has (360° − θ).
- Arc length l = (θ/360) × 2πr
- Sector area = (θ/360) × πr² = ½ × l × r
Reason: a full turn (360°) gives the whole circumference and the whole area; an angle θ takes the fraction θ/360 of each.
Find the area of the sector of a circle of radius 4 cm with angle 30°. Also find the major sector's area (π = 3.14).
Solution: Minor = (30/360) × 3.14 × 16 = 4.19 cm² (approx). Major = 3.14 × 16 − 4.19 = 50.24 − 4.19 = 46.05 cm².
The minute hand of a clock is 14 cm long. Find the area swept in 5 minutes (π = 22/7).
Solution: 5 minutes = 30°. Area = (1/12) × (22/7) × 196 = 154/3 ≈ 51.33 cm².
4. Segments
Area of minor segment = area of sector OAB − area of △OAB
Area of major segment = πr² − area of minor segment.
| Central angle θ | Area of △OAB | Why |
|---|---|---|
| 60° | (√3/4) r² | △OAB is equilateral |
| 90° | ½ r² | Right isosceles triangle |
| 120° | (√3/4) r² | Height r/2, base r√3: ½ × r√3 × r/2 |
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the areas of the minor and major segments (π = 3.14).
Solution: Sector = ¼ × 314 = 78.5. Triangle = ½ × 100 = 50. Minor segment = 28.5 cm². Major = 314 − 28.5 = 285.5 cm².
Find the area of the segment of a circle of radius 21 cm cut off by a chord subtending 120° (π = 22/7, √3 = 1.73).
Solution: Sector = ⅓ × (22/7) × 441 = 462. Triangle = (√3/4) × 441 = 190.73. Segment = 271.27 cm².
5. Applications
A horse is tied to a peg at one corner of a square field of side 15 m with a 5 m rope. Find the area it can graze (π = 3.14).
Solution: A quarter circle of radius 5: ¼ × 3.14 × 25 = 19.625 m².
A car has two non-overlapping wipers, each 25 cm long, sweeping 115°. Find the total area cleaned (π = 22/7).
Solution: 2 × (115/360) × (22/7) × 625 = ≈ 1254.96 cm².
6. Solved Problems (Board Pattern)
In a circle of radius 21 cm, an arc subtends 60° at the centre. Find (i) the arc length, (ii) the sector area, (iii) the segment area (√3 = 1.73).
Solution: (i) (1/6) × 2 × (22/7) × 21 = 22 cm. (ii) (1/6) × (22/7) × 441 = 231 cm². (iii) Triangle = (1.73/4) × 441 = 190.73 → segment = 40.27 cm².
An umbrella has 8 ribs equally spaced. Treating it as a flat circle of radius 45 cm, find the area between two consecutive ribs (π = 22/7).
Solution: θ = 45°. Area = (1/8) × (22/7) × 2025 = ≈ 795.54 cm².
A lighthouse spreads light over a sector of 80° to a distance of 16.5 km. Find the sea area covered (π = 3.14).
Solution: (80/360) × 3.14 × 272.25 = ≈ 189.97 km².
7. Master Formula Table
| Quantity | Formula |
|---|---|
| Arc length | (θ/360) × 2πr |
| Sector area | (θ/360) × πr² = ½lr |
| Major sector | πr² − minor sector |
| Sector perimeter | 2r + arc length |
| Minor segment | sector − ½ r² sin θ |
| Triangle (60° or 120°) | (√3/4) r² |
| Triangle (90°) | ½ r² |
| Clock: minute hand | 6° per minute |
8. Master Summary Table
| Region | How to find | Typical context |
|---|---|---|
| Sector | Fraction θ/360 of circle | Clock hands, wipers, slices |
| Arc | Fraction of circumference | Pendulum swing, track bend |
| Minor segment | Sector − triangle | Designs, brooches |
| Major segment | Circle − minor segment | Remaining region |
| Quarter circle | θ = 90° | Grazing at a corner |
Chart — sector area as a fraction of a circle:
9. Mind Map
- 2πr
- πr²
- (θ/360)πr²
- Arc (θ/360)2πr
- Sector − triangle
- 60°, 90°, 120°
- Clock, wiper
- Grazing, lighthouse
10. Tips and Tricks
- Convert time to angle first: minute hand 6°/min; hour hand 0.5°/min.
- Simplify θ/360 to a simple fraction (60° → 1/6, 120° → 1/3) before multiplying.
- For a 60° segment, the chord equals the radius (equilateral triangle).
- Confusing sector (two radii) with segment (chord).
- Using diameter in place of radius.
- Forgetting the two radii when asked for the perimeter of a sector.
11. Important One-liners
- A sector of angle θ is θ/360 of the circle.
- Minor + major sector = whole circle.
- Segment area = sector area − triangle area.
- A semicircle is a sector of 180°.
- The minute hand turns 360° in 60 minutes.
12. Exam-pattern Questions with Detailed Answers
The area of a sector of angle 90° of a circle of radius 14 cm is (π = 22/7):
- 154 cm²
- 616 cm²
- 44 cm²
- 308 cm²
Show answer
¼ × 616 = 154.
Answer: (A)
Find the length of the arc of a circle of radius 7 cm subtending 90° (π = 22/7).
Show answer
¼ × 2 × 22/7 × 7 = 11 cm.
Answer: 11 cm
Find the angle swept by the minute hand in 20 minutes.
Show answer
20 × 6° = 120°.
Answer: 120°
Find the perimeter of a sector of radius 10.5 cm and angle 60° (π = 22/7).
Show answer
Arc = (1/6) × 2 × 22/7 × 10.5 = 11. Perimeter = 21 + 11 = 32 cm.
Answer: 32 cm
The area of a sector of radius 6 cm is 6π cm². Find its angle.
Show answer
(θ/360) × 36π = 6π → θ = 60°.
Answer: 60°
A chord of a circle of radius 15 cm subtends 60° at the centre. Find the areas of the minor and major segments (π = 3.14, √3 = 1.73).
Show answer
Sector = (1/6) × 3.14 × 225 = 117.75. Triangle = (1.73/4) × 225 = 97.31. Minor = 20.44 cm². Major = 706.5 − 20.44 = 686.06 cm².
Answer: 20.44 cm² and 686.06 cm²
The perimeter of a sector of radius 5.2 cm is 16.4 cm. Find its area.
Show answer
Arc = 16.4 − 10.4 = 6. Area = ½ × 6 × 5.2 = 15.6 cm².
Answer: 15.6 cm²
A pendulum swings through 30° and its bob traces an arc of 8.8 cm. Find the pendulum's length (π = 22/7).
Show answer
(1/12) × 2 × (22/7) × r = 8.8 → r = 8.8 × 42/22 = 16.8 cm.
Answer: 16.8 cm
A brooch is a circle of diameter 35 mm made with silver wire, divided into 10 equal sectors by 5 diameters. Find the total wire needed and the area of each sector (π = 22/7).
Show answer
Wire = circumference + 5 diameters = 110 + 175 = 285 mm. Each sector = (1/10) × (22/7) × 17.5² = 96.25 mm².
Answer: 285 mm; 96.25 mm²
A round table cover has six equal designs (segments) of a circle of radius 28 cm, each with central angle 60°. Find the cost of making the designs at ₹0.35 per cm² (√3 = 1.7).
Show answer
One segment = (1/6)(22/7)(784) − (1.7/4)(784) = 410.67 − 333.2 = 77.47. Six = 464.8 cm². Cost = 464.8 × 0.35 ≈ ₹162.68.
Answer: ≈ ₹162.68
Find the difference between the areas of a sector of angle 120° and its corresponding major sector in a circle of radius 21 cm (π = 22/7).
Show answer
Circle = 1386. Minor = 462; major = 924. Difference = 462 cm².
Answer: 462 cm²
In a circle of radius 21 cm, an arc subtends 120° at the centre. Find the area of the major segment (π = 22/7, √3 = 1.73).
Show answer
Minor segment = 462 − 190.73 = 271.27. Major = 1386 − 271.27 = 1114.73 cm².
Answer: 1114.73 cm²
13. Quick Revision Summary
- Arc = (θ/360)2πr; sector = (θ/360)πr² = ½lr.
- Major sector = circle − minor sector.
- Segment = sector − triangle; triangle = (√3/4)r² for 60°/120°, ½r² for 90°.
- Minute hand: 6° per minute.
- Grazing at a corner of a square: quarter circle.