Basic Formulae
For a circle with radius r and central angle θ degrees:
| Quantity | Formula |
|---|---|
| Circumference | 2πr |
| Area of circle | πr² |
| Arc length (minor/major) | (θ/360) × 2πr |
| Area of sector | (θ/360) × πr² |
| Area of minor segment | Area of sector − Area of △OAB |
| Area of major segment | Area of circle − Area of minor segment |
Use π = 22/7 unless stated otherwise (or π = 3.14 if specified)
Sector of a Circle
A sector is the region between two radii and the arc connecting them.
Minor sector: smaller angle (< 180°) Major sector: larger angle (> 180°)
Worked examples:
Q1. Find the area of a sector with radius 7 cm and central angle 60°. Area = (60/360) × π × 7² = (1/6) × (22/7) × 49 = (1/6) × 154 = 25.67 cm²
Q2. A sector has area 77 cm² and radius 14 cm. Find the angle. 77 = (θ/360) × (22/7) × 196 77 = (θ/360) × 616 θ/360 = 77/616 = 1/8 θ = 45°
Segment of a Circle
A segment is the region between a chord and the arc.
Area of minor segment = Area of sector − Area of triangle OAB
Where triangle OAB has O as centre and A, B on circle.
Area of Triangle OAB
For common angles:
| θ | Area of △OAB |
|---|---|
| 60° | (√3/4)r² (equilateral triangle) |
| 90° | ½r² (right-angled isoceles) |
| 120° | (√3/4)r² (same as 60° — isoceles, half of equilateral) |
For general θ: Area of △OAB = ½ × r² × sin θ
Worked example:
Q1. Find area of minor segment with radius 6 cm and central angle 90°.
Area of sector = (90/360) × π × 36 = (1/4) × (22/7) × 36 = 28.28 cm²
Area of △OAB (right isoceles with legs 6 cm) = ½ × 6 × 6 = 18 cm²
Area of segment = 28.28 − 18 = 10.28 cm²
Combination of Plane Figures
Many problems involve finding areas of combined shapes (circles + rectangles, sectors + triangles, etc.)
Strategy: Identify each component shape, find area of each, then add or subtract as needed.
Type 1 — Flower/Petal design in a square:
Q. In a square of side 10 cm, a circle of maximum size is inscribed. Find the area of the regions outside the circle but inside the square.
Circle radius = 5 cm; Area = πr² = 78.57 cm² Square area = 100 cm² Shaded area = 100 − 78.57 = 21.43 cm²
Type 2 — Running track:
Q. A running track is made with two straight sections of 90 m each, and two semicircles of diameter 35 m at each end. Find the total area of the track and the total length of the track.
The inner boundary has two straight sides (90m each) and two semicircles (radius = 17.5 m, so one full circle). Length of inner boundary = 2(90) + 2π(17.5) = 180 + 110 = 290 m
Type 3 — Sector with triangle:
Q. A chord AB of a circle with radius 10 cm subtends 60° at the centre. Find the area of the minor segment.
Sector area = (60/360) × π × 100 = 100π/6 = 52.36 cm²
Since θ = 60° and OA = OB = 10 cm, △OAB is equilateral (all sides = 10 cm) Area of △OAB = (√3/4) × 100 = 43.30 cm²
Segment area = 52.36 − 43.30 = 9.06 cm²
Type 4 — Shaded region between two concentric circles:
Q. Two concentric circles have radii 7 cm and 14 cm. Find the area of the ring between them.
Area = π(14²) − π(7²) = π(196 − 49) = 147π = 147 × 22/7 = 462 cm²
Type 5 — Clock hands / Quadrant problems:
Q. Find the area swept by the minute hand of a clock of length 14 cm in 15 minutes.
In 60 minutes, the minute hand sweeps 360°. In 15 minutes, it sweeps 90°.
Area = (90/360) × π × 14² = (1/4) × (22/7) × 196 = 154 cm²
Board Exam Tips
- 5–6 marks from this chapter
- Always write the formula before substituting — examiners want to see the formula
- Use π = 22/7 with integer radii (7, 14, 21, 35) for exact answers; use π = 3.14 when asked
- Area of segment: sector minus triangle — never forget to subtract the triangle
- For 60° and 90° segments, know the triangle area formulae (√3/4 r² and ½r²)
- Combination figures: draw the figure, shade the required region, then set up Area = sum or difference
📚 RD Sharma Mathematics Class 10 — Areas Related to Circles
📝 Oswaal CBSE Maths Question Bank — Mensuration Board Questions