Class 10 Notes - Areas Related to Circles
Introduction
Circles are among the most common shapes we see in our surroundings—wheels, coins, plates, and clocks are all circular. In earlier classes, students learned about the basic properties of circles—radius, diameter, circumference, and area. In Class X, we expand on this understanding by exploring how to calculate the areas of different segments and sectors of a circle, as well as how to apply these in solving real-world problems.
Learning Objectives
- Understand and apply the formulas for area and circumference of a circle
- Calculate the area of sectors and segments
- Solve problems based on circular regions and composite shapes
- Use π in calculations both as 22/7 and 3.14
1. Basic Terms and Review
- Radius (r): Distance from center to the boundary
- Diameter (d): 2 × radius
- Circumference: 2πr
- Area of Circle: πr²
2. Circumference and Area of a Circle
Circumference: 2πr = πd
Area: πr²
Note on π: Use 22/7 if radius/diameter is divisible by 7; otherwise use 3.14.
3. Sector of a Circle
A sector is a “slice” of the circle formed by two radii and the arc.
Area of Sector: (θ/360) × πr²
Length of Arc: (θ/360) × 2πr
4. Segment of a Circle
A segment is the region between a chord and the arc.
Area of Segment: Area of sector − Area of triangle
Area of triangle: (1/2) × r² × sin(θ)
5. Applications in Real Life
- Roundabouts in roads
- Land area under circular ponds
- Engineering and gear design
- Pie charts
6. Composite Figures Involving Circles
Break the figure into known shapes. Use addition or subtraction of areas as required.
- Semicircle with rectangle
- Quadrant inside a square
- Circular flower beds with paths
7. Important Formulas Summary
| Shape | Formula |
|---|
| Area of Circle | πr² |
| Circumference | 2πr |
| Area of Sector | (θ/360) × πr² |
| Length of Arc | (θ/360) × 2πr |
| Area of Segment | Sector area − Triangle area |
| Semicircle Area | (1/2) × πr² |
| Quadrant Area | (1/4) × πr² |
8. Example Problems
Example 1: Sector Area
Radius = 7 cm, Angle = 90°
Area = (90/360) × π × 7² = (1/4) × 22/7 × 49 = 38.5 cm²
Example 2: Segment Area
Radius = 14 cm, Angle = 60°
Sector area = (1/6) × 22/7 × 196 = 61.33 cm²
Triangle area ≈ 42.87 cm²
Segment area ≈ 18.46 cm²
Example 3: Composite Figure
Rectangle: 14×7 cm, semicircle cut from one side
Rectangle area = 98 cm²
Semicircle area = (1/2) × π × 3.5² ≈ 19.25 cm²
Remaining = 78.75 cm²
9. Real-Life Word Problems
Example: Circular Path
Garden radius = 10 m, Path = 1 m wide
Outer radius = 11 m
Area of path = π(11² − 10²) = 66.13 m²
Example: Wheel Rotation
Radius = 0.7 m, 100 revolutions
Circumference = 4.4 m
Distance = 100 × 4.4 = 440 m
10. Tips for Solving Area Problems
- Sketch the diagram if not provided
- Check if radius is divisible by 7 for using 22/7
- Ensure units are consistent
- Use the correct formulas carefully
11. Common Mistakes to Avoid
- Confusing arc length with area
- Forgetting to square radius
- Incorrect subtraction in segment area
- Using wrong π approximation
12. Practice Exercises
Level 1: Basic
- Find the area and circumference of a circle with radius 14 cm
- A wheel of radius 28 cm makes one full turn. Find the distance covered
Level 2: Moderate
- Find the area of a quadrant of radius 21 cm
- A sector has angle 60° and radius 6 cm. Find area and arc length
Level 3: Advanced
- Circle of radius 10 cm, sector with angle 120°. Find segment area
- Square of side 14 cm with 4 quadrant cutouts of radius 7 cm. Find area left
13. Case-Based Questions
Case 1: Playground Design
Circular garden of radius 35 m, path width = 2 m
Find area of the path.
Case 2: Clock Face Design
Minute hand = 10 cm
Find area swept in 30 minutes.
14. Summary
- Use the correct formula based on the part of the circle (full, half, quarter, segment, etc.)
- Composite problems may involve both addition and subtraction of areas
- Keep units consistent and cross-check your answers