1. Basic Circle Formulas
• Circumference: C = 2pr = pd
• Area of circle: A = pr²
• Diameter: d = 2r
• Value of p: p ˜ 22/7 or 3.14
2. Area of Sector of a Circle
Using Angle (in degrees)
Area of sector = (?/360°) × pr²
Using Angle (in radians)
Area of sector = (1/2) × r² × ?
Length of Arc
Arc length = (?/360°) × 2pr
3. Area of Segment of a Circle
Area of segment = Area of sector - Area of triangle
= (?/360°) × pr² - (1/2)r² sin ?
For minor segment: Use above formula
For major segment: Area of circle - Area of minor segment
4. Special Cases
Semicircle
? = 180°
Area = (1/2)pr²
Perimeter = pr + 2r
Quadrant
? = 90°
Area = (1/4)pr²
Perimeter = (pr/2) + 2r
5. Perimeter Formulas
• Sector: Perimeter = 2r + Arc length
• Segment: Perimeter = Chord length + Arc length
• Semicircle: Perimeter = pr + 2r
• Quadrant: Perimeter = (pr/2) + 2r
6. Angle Subtended by Arc
• At center: Angle = ? (given)
• At circumference: Angle = ?/2 (for same arc)
• Length of arc ? Angle at center
7. Area of Combinations of Plane Figures
Circle + Square/Rectangle/Triangle
- Calculate area of each shape separately
- Add or subtract based on overlap
- For shaded regions: Total area - Unshaded area
8. Common Problem Types
- Sector problems: Find area, perimeter, arc length
- Segment problems: Find area of segment
- Shaded region problems: Area between circles/combinations
- Path/walkway problems: Area of circular path
- Clock problems: Distance covered by clock hands
9. Important Formulas Summary
1. Area of circular ring:
A = p(R² - r²) where R = outer radius, r = inner radius
2. Width of circular path:
Width = R - r
3. Chord length:
Chord = 2r sin(?/2)
4. Area of triangle in segment:
Area = (1/2)r² sin ?
10. Quick Reference Table
11. Example Problems (NCERT Style)
Problem 1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Area = (?/360°) × pr²
= (60/360) × (22/7) × 6²
= (1/6) × (22/7) × 36
= (22 × 6)/7 = 132/7 ˜ 18.86 cm²
Problem 2: Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60°.
Solution:
Area of sector = (60/360) × p × 12² = (1/6) × p × 144 = 24p
Area of triangle = (1/2) × 12² × sin 60° = (1/2) × 144 × (v3/2) = 36v3
Area of segment = 24p - 36v3 ˜ 75.36 - 62.35 = 13.01 cm²
Problem 3: The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?
Solution:
Diameter = 80 cm, Radius = 40 cm
Circumference = 2pr = 2 × (22/7) × 40 = (1760/7) cm
Speed = 66 km/hr = 66 × (100000/60) cm/min = 110000 cm/min
Distance in 10 min = 110000 × 10 = 1100000 cm
Number of revolutions = Distance / Circumference
= 1100000 ÷ (1760/7) = 1100000 × (7/1760) = 4375 revolutions
12. Important Points to Remember
- p = 22/7 (approximately) or use p symbol if exact answer needed
- Always include units (cm², m², etc.)
- For segment area: Area = Area of sector - Area of triangle
- Major segment area = Circle area - Minor segment area
- In combinations, break into basic shapes
- For circular paths: Use p(R² - r²)
- Arc length = (?/360°) × 2pr
- Perimeter of sector = 2r + arc length
13. Conversion Factors
• 1 m = 100 cm
• 1 km = 1000 m
• 1 hectare = 10000 m²
• p ˜ 22/7 ˜ 3.14
• v2 ˜ 1.414, v3 ˜ 1.732