Class 10 Mathematics Notes Chapter 11 (Chapter 11) – Examplar Problems (English) Book

Alright class, let's focus on Chapter 11: Areas Related to Circles from the NCERT Exemplar. This chapter is crucial not just for your board exams but also forms the basis for geometry questions in various government exams. We'll break down the concepts and formulas methodically.

Chapter 11: Areas Related to Circles - Detailed Notes

1. Review of Circle Basics:

• Circle: A collection of all points in a plane that are at a fixed distance (radius) from a fixed point (center).
• Radius (r): The fixed distance from the center to any point on the circle.
• Diameter (d): The distance across the circle passing through the center (d = 2r).
• Chord: A line segment joining any two points on the circle. The diameter is the longest chord.
• Circumference (C): The perimeter or 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: Usually taken as 22/7 or 3.14 (use the value specified in the question; if not specified, use 22/7 unless the numbers suggest 3.14 is more convenient).

2. Sector of a Circle:

• Definition: The region enclosed by two radii of the circle and their corresponding arc. Imagine a slice of pizza.
• Angle of the Sector (θ): The angle formed between the two radii at the center. Measured in degrees.
• Minor Arc/Sector: The arc/sector corresponding to the smaller angle (< 180°).
• Major Arc/Sector: The arc/sector corresponding to the reflex angle (> 180°).
• Length of an Arc of a Sector: The length of the curved part of the sector.
  • Formula: Length of Arc = (θ / 360°) × 2πr
• Area of a Sector: The space enclosed by the sector.
  • Formula 1 (using angle): Area of Sector = (θ / 360°) × πr²
  • Formula 2 (using arc length 'l'): Area of Sector = (1/2) × l × r
  • Area of Major Sector: Area of Circle - Area of Minor Sector = ( (360° - θ) / 360° ) × πr²
• Special Cases:
  • Semicircle: A sector with θ = 180°. Area = (180°/360°) × πr² = (1/2)πr². Arc Length = πr.
  • Quadrant: A sector with θ = 90°. Area = (90°/360°) × πr² = (1/4)πr². Arc Length = (1/2)πr.

3. Segment of a Circle:

• Definition: The region enclosed between a chord and its corresponding arc.
• Minor Segment: The region enclosed by a chord and the minor arc.
• Major Segment: The region enclosed by a chord and the major arc.
• Area of a Minor Segment:
  • Formula: Area of Minor 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(ΔOAB)
    • Where O is the center, A and B are points on the circle defining the chord AB.
    • Finding Area(ΔOAB):
      • If θ is given: Area(ΔOAB) = (1/2) × r² × sin θ
      • If ΔOAB is a right-angled triangle (θ = 90°): Area = (1/2) × base × height = (1/2) × r × r = (1/2)r²
      • If ΔOAB is an equilateral triangle (θ = 60°): Area = (√3 / 4) × side² = (√3 / 4) × r²
• Area of a Major Segment:
  • Formula: Area of Major Segment = Area of the Circle - Area of the Minor Segment.
  • Area of Major Segment = πr² - [Area of Minor Segment]

4. Areas of Combinations of Plane Figures:

• This section involves problems where areas of regions formed by combining circles with other shapes (squares, rectangles, triangles, other circles) need to be calculated.
• Strategy:
  1. Identify the Shapes: Break down the complex figure into basic geometric shapes (circles, sectors, segments, triangles, squares, rectangles, etc.).
  2. Identify the Required Region: Clearly understand which area needs to be calculated (e.g., shaded region, unshaded region, area of a track).
  3. Formulate a Plan: Determine whether you need to add or subtract the areas of the identified basic shapes to get the required area.
  4. Apply Formulas: Use the correct formulas for each shape.
  5. Calculate: Perform the calculations carefully, paying attention to units and the value of π.
• Common Examples:
  • Area of rings (region between concentric circles): Area = π(R² - r²), where R is the outer radius and r is the inner radius.
  • Area of shaded regions formed by intersecting circles or circles inscribed/circumscribed in polygons.
  • Area of flower beds, tracks, logos, etc.

Key Points for Exam Preparation:

• Memorize Formulas: Be thorough with all the formulas for circumference, area, arc length, sector area, and segment area.
• Understand Concepts: Don't just memorize; understand the difference between sector and segment, minor and major arcs/sectors/segments.
• Visualize: Draw diagrams whenever possible to visualize the problem and the region whose area needs to be calculated.
• Units: Always mention the correct units (cm, m for length; cm², m² for area). Ensure consistency throughout the calculation.
• Accuracy: Be careful with calculations, especially involving π and square roots.
• Practice: Solve a variety of problems from the NCERT textbook and Exemplar, focusing on combination figures.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Chapter 11 concepts for your practice:

1. If the area of a circle is 154 cm², then its circumference is:
   (a) 11 cm
   (b) 22 cm
   (c) 44 cm
   (d) 55 cm

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) 142/7 cm²
   (c) 152/7 cm²
   (d) 162/7 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°) × πR
   (b) (θ / 360°) × 2πR
   (c) (θ / 180°) × 2πR
   (d) (θ / 360°) × πR²

4. The area of a quadrant 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 chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segment (approx.) is: (Use π = 3.14)
   (a) 28.5 cm²
   (b) 30.5 cm²
   (c) 32.5 cm²
   (d) 34.5 cm²

6. 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

7. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is: (Use π = 22/7)
   (a) 22:7
   (b) 14:11
   (c) 7:22
   (d) 11:14

8. The area of the ring between two concentric circles whose circumferences are 88 cm and 132 cm is: (Use π = 22/7)
   (a) 770 cm²
   (b) 660 cm²
   (c) 550 cm²
   (d) 440 cm²

9. The area of a sector is 1/12th the area of the circle. Find the angle of the sector.
   (a) 30°
   (b) 45°
   (c) 60°
   (d) 90°

10. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is: (Use π = 22/7)
    (a) 11 cm
    (b) 22 cm
    (c) 33 cm
    (d) 44 cm

Answers to MCQs:

1. (c) 44 cm
2. (a) 132/7 cm²
3. (b) (θ / 360°) × 2πR
4. (a) 77/8 cm²
5. (a) 28.5 cm²
6. (a) r² sq. units
7. (b) 14:11
8. (a) 770 cm²
9. (a) 30°
10. (b) 22 cm

Study these notes thoroughly and practice the problems. Understanding the application of formulas in different scenarios is key. Good luck with your preparation!