Class 10 Mathematics Notes Chapter 12 (Areas related to circles) – Mathematics Book

Alright class, let's focus on Chapter 12, 'Areas Related to Circles'. This is an important chapter, not just for your board exams but also as concepts from here frequently appear in various government competitive examinations. We'll break down the key ideas and formulas you need to master.

Chapter 12: Areas Related to Circles - Detailed Notes

1. Recap: Circle Basics

• Circle: A collection of all points in a plane that are at a fixed distance (radius) from a fixed point (centre).
• Radius (r): The fixed distance from the centre to any point on the circle.
• Diameter (d): The distance across the circle passing through the centre. d = 2r.
• Circumference (C): The distance around the circle.
  • Formula: C = 2πr or C = πd
• Area (A): The region enclosed by the circle.
  • Formula: A = πr²
• Value of π (Pi): A constant ratio of a circle's circumference to its diameter. Commonly used approximations are 22/7 or 3.14. Always use the value specified in the question, or 22/7 if not specified, unless 3.14 makes calculations significantly easier.

2. Sector of a Circle

• Definition: The region enclosed by two radii of a circle and their corresponding arc. Think of it like a slice of pizza.
• Angle of the Sector (θ): The angle formed by the two radii at the centre. Measured in degrees.
• Arc: A part of the circumference of the circle.
  • Minor Arc: The shorter arc connecting the endpoints of the two radii.
  • Major Arc: The longer arc connecting the endpoints of the two radii.
• Length of an Arc of a Sector: The length of the arc corresponding to the angle θ.
  • Formula: Length of Arc = (θ / 360°) × 2πr
  • Think: It's a fraction (θ/360°) of the total circumference.
• Area of a Sector: The area of the region bounded by the two radii and the arc.
  • Formula: Area of Sector = (θ / 360°) × πr²
  • Think: It's a fraction (θ/360°) of the total area of the circle.
  • Alternative Formula (using arc length 'l'): Area of Sector = (1/2) × l × r (where l is the length of the arc of that sector)
• Major and Minor Sectors:
  • Minor Sector: The sector with the smaller angle (θ < 180°).
  • Major Sector: The sector with the larger angle (360° - θ).
  • Area of Major Sector = Area of Circle - Area of Minor Sector = ((360° - θ) / 360°) × πr²

3. Segment of a Circle

• Definition: The region enclosed between a chord and its corresponding arc.
• Chord: A line segment connecting any two points on the circle.
• Minor Segment: The region bounded by a chord and the minor arc.
• Major Segment: The region bounded by a chord and the major arc.
• Area of a Segment:
  • Formula: Area of Segment = Area of the corresponding Sector - Area of the corresponding Triangle (formed by the two radii and the chord).
  • Area of Minor Segment = [ (θ / 360°) × πr² ] - [ Area of ΔOAB ] (where O is the centre, A and B are points on the circle, and θ is the angle ∠AOB).
  • Finding Area of Triangle (ΔOAB):
    • If θ = 60°, ΔOAB is equilateral (Area = (√3/4)r²).
    • If θ = 90°, ΔOAB is a right-angled isosceles triangle (Area = (1/2) × base × height = (1/2) × r × r = (1/2)r²).
    • For other angles, use the formula: Area = (1/2)r² sin(θ) or use basic trigonometry to find the base and height if sin(θ) is not readily known.
• Area of Major Segment:
  • Formula: Area of Major Segment = Area of Circle - Area of Minor Segment.

4. Areas of Combinations of Plane Figures

• This is a very common type of question in competitive exams.
• Strategy:
  1. Identify the basic geometric shapes involved (circles, squares, triangles, rectangles, sectors, segments, etc.).
  2. Visualize the required area (often a shaded region).
  3. Determine the relationship: Is the required area obtained by adding areas of simpler shapes or by subtracting the area of one shape from another?
  4. Calculate the dimensions (radius, side length, base, height, angle θ) needed for each shape, often using given information or geometric properties (like Pythagoras theorem).
  5. Calculate the areas of the individual shapes using the appropriate formulas.
  6. Add or subtract the areas as determined in step 3 to find the final required area.
• Common Combinations:
  • Circle inscribed in a square (diameter = side of square).
  • Square inscribed in a circle (diagonal of square = diameter of circle).
  • Circles touching each other externally or internally.
  • Area of pathways or rings (Area of outer circle - Area of inner circle = π(R² - r²)).
  • Designs involving sectors or segments cut from squares or rectangles.

Key Points for Problem Solving:

• Read Carefully: Understand what area is being asked for. Draw a diagram if not provided.
• Units: Ensure all measurements are in the same unit before calculating. The final area will be in square units (cm², m², etc.).
• Formula Accuracy: Memorize the formulas correctly.
• Calculation: Be careful with calculations, especially involving π and square roots. Simplify where possible.
• Visualization: Break down complex figures into simpler, known shapes.

This chapter essentially applies the basic circle formulas to parts of a circle (sectors and segments) and then combines these with other geometric shapes. Practice is key to mastering the combinations.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Chapter 12 concepts for your practice:

1. If the circumference and the area of a circle are numerically equal, then the diameter of the circle is:
   (a) π/2 units
   (b) 2 units
   (c) 4 units
   (d) π units

2. The area of a sector of a circle with radius 6 cm if the angle of the sector is 60° is: (Use π = 22/7)
   (a) 132/7 cm²
   (b) 144/7 cm²
   (c) 154/7 cm²
   (d) 120/7 cm²

3. The length of the arc of a sector of angle θ (in degrees) of a circle with radius R is:
   (a) (θ / 180°) × 2πR
   (b) (θ / 360°) × πR²
   (c) (θ / 180°) × πR
   (d) (θ / 360°) × 2πR

4. The area of a quadrant (sector with angle 90°) of a circle whose circumference is 22 cm is: (Use π = 22/7)
   (a) 77/8 cm²
   (b) 77/4 cm²
   (c) 77/2 cm²
   (d) 77 cm²

5. A wire is bent to form a square enclosing an area of 121 cm². If the same wire is bent into the form of a circle, the area of the circle is: (Use π = 22/7)
   (a) 144 cm²
   (b) 154 cm²
   (c) 169 cm²
   (d) 176 cm²

6. The area of the segment of a circle of radius 14 cm corresponding to a central angle of 90° is: (Use π = 22/7)
   (a) 56 cm²
   (b) 154 cm²
   (c) 98 cm²
   (d) 112 cm²

7. The minute hand of a clock is 7 cm long. The area swept by the minute hand in 10 minutes is: (Use π = 22/7)
   (a) 77/3 cm²
   (b) 77 cm²
   (c) 154/3 cm²
   (d) 154 cm²

8. The area of the largest triangle that can be inscribed in a semi-circle of radius 'r' units is:
   (a) r² sq. units
   (b) (1/2)r² sq. units
   (c) 2r² sq. units
   (d) √2 r² sq. units

9. The area of a circular ring formed by two concentric circles whose radii are 5.7 cm and 4.3 cm respectively is: (Take π = 3.14)
   (a) 43.96 cm²
   (b) 44 cm²
   (c) 43.98 cm²
   (d) 45.5 cm²

10. The distance covered by a wheel of diameter 70 cm in 10 revolutions is: (Use π = 22/7)
    (a) 22 m
    (b) 220 m
    (c) 2.2 m
    (d) 2200 m

Answer Key:

1. (c) 4 units
2. (a) 132/7 cm²
3. (d) (θ / 360°) × 2πR
4. (a) 77/8 cm²
5. (b) 154 cm²
6. (a) 56 cm²
7. (c) 154/3 cm²
8. (a) r² sq. units
9. (a) 43.96 cm² (Calculation: π(R² - r²) = π(R-r)(R+r) = 3.14 * (5.7-4.3) * (5.7+4.3) = 3.14 * 1.4 * 10 = 43.96)
10. (a) 22 m (Calculation: Circumference = πd = (22/7)*70 = 220 cm. Distance = 10 * 220 cm = 2200 cm = 22 m)

Keep practicing these concepts and solving problems from the textbook and previous exam papers. Good luck!