Class 10 Mathematics Notes Chapter 12 (Chapter 12) – Examplar Problems (English) Book

Alright class, let's focus on Chapter 12, 'Areas Related to Circles', from your NCERT Exemplar. This chapter is fundamental not just for your board exams but also forms the basis for many geometry and mensuration problems you'll encounter in various government exams. Pay close attention to the formulas and their applications, especially in combined figures.

Chapter 12: Areas Related to Circles - Detailed Notes

1. Review of Basic Concepts (Circle)

• 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 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: Usually taken as 22/7 or 3.14 (Use the value specified in the question, otherwise 22/7 is standard unless calculations become simpler with 3.14).

2. Sector of a Circle

• Definition: The region enclosed by two radii of the 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. This angle is measured in degrees.
• Minor Sector: The sector with the smaller angle (< 180°).
• Major Sector: The sector with the larger angle (> 180°). The angle of the major sector is 360° - θ (where θ is the angle of the minor sector).
• 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:
  • Formula: Area of Sector = (θ / 360°) * πr²
  • Alternative Formula (if arc length 'l' is known): Area of Sector = (1/2) * l * r

3. Segment of a Circle

• Definition: The region enclosed by a chord and its corresponding arc.
• Chord: A line segment joining any two points on the circle.
• Minor Segment: The region enclosed by the chord and the minor arc.
• Major Segment: The region enclosed by the 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
  • Area = [(θ / 360°) * πr²] - Area(Triangle OAB) (where O is the centre, A and B are points on the circle defining the chord and sector).
  • Finding the Area of the Triangle (OAB):
    • The triangle formed (OAB) is usually isosceles (OA = OB = radius).
    • If θ = 90°, the triangle is a right-angled triangle. Area = (1/2) * base * height = (1/2) * r * r = (1/2)r².
    • If θ = 60°, the triangle is equilateral (OA = OB = AB = r). Area = (√3 / 4) * side² = (√3 / 4) * r².
    • For any angle θ, use trigonometry: Area = (1/2) * r * r * sin(θ) = (1/2)r²sin(θ). (Knowledge of basic trigonometric values like sin 30°, sin 45°, sin 60°, sin 90°, sin 120° is helpful).
    • Alternatively, drop a perpendicular from O to the chord AB. This bisects the chord and the angle θ. Use trigonometry to find the base (chord length) and height.
• Area of a Major Segment:
  • Formula: Area of Major Segment = Area of the Circle - Area of the Minor Segment
  • Alternatively: Area of Major Segment = Area of Major Sector + Area of the Triangle

4. Areas of Combinations of Plane Figures

• This is a very important section for competitive exams. Problems involve calculating areas of shaded regions formed by combining circles with other shapes like squares, rectangles, triangles, or other circles.
• Strategy:
  1. Identify the shapes involved: Break down the complex figure into simpler geometric shapes (circles, sectors, segments, triangles, squares, rectangles).
  2. Determine the dimensions: Find the radii, side lengths, angles, etc., needed for the area formulas, often using geometric properties (like the diagonal of a square inscribed in a circle is the diameter of the circle, or the side of a square circumscribing a circle is equal to the diameter).
  3. Plan the calculation: Decide whether you need to add or subtract areas.
     • For finding the area of a composite shape, add the areas of its components.
     • For finding the area of a shaded region, often subtract the area of the unshaded part from the total area.
  4. Apply the formulas: Carefully substitute the values into the correct formulas.
  5. Calculate: Perform the arithmetic operations. Pay attention to the value of π to be used.
• Common Combinations:
  • Circle inscribed in a square: Side of square = Diameter of circle (s = 2r).
  • Square inscribed in a circle: Diagonal of square = Diameter of circle (d√2 = 2r).
  • Circles touching externally/internally.
  • Designs in handkerchiefs, table covers, flower beds involving sectors and segments.
  • Running tracks (area between two concentric circles - Area of Ring = π(R² - r²) where R is outer radius, r is inner radius).

Key Points for Exam Preparation:

• Memorize Formulas: Be absolutely thorough with all the formulas.
• Understand Concepts: Don't just memorize; understand what sectors and segments are.
• Visualize: Draw diagrams for problems involving combined figures. This helps immensely.
• Units: Always mention the correct units for length (cm, m) and area (cm², m²).
• Accuracy: Be careful with calculations, especially involving π and square roots.
• Practice: Solve a wide variety of problems from the Exemplar book and previous question papers. Focus on problems involving combined figures and shaded regions.

Multiple Choice Questions (MCQs)

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

1. If the circumference and the area of a circle are numerically equal, then the diameter is equal to:
   (A) π/2
   (B) 2π
   (C) 2
   (D) 4

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) 122/7 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 / 360°) * 2πR
   (D) (θ / 180°) * πR²

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 chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segment is: (Use π = 3.14)
   (A) 28.5 cm²
   (B) 30.5 cm²
   (C) 25.5 cm²
   (D) 32.5 cm²

6. If the perimeter of a semicircular protractor is 36 cm, its diameter is: (Use π = 22/7)
   (A) 7 cm
   (B) 14 cm
   (C) 21 cm
   (D) 10 cm

7. 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

8. A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square. (Use π = 22/7)
   (A) 22 cm
   (B) 44 cm
   (C) 88 cm
   (D) 176 cm

9. The area of the shaded region in the figure, where ABCD is a square of side 14 cm and four congruent circles are drawn such that each circle touches two sides of the square and two other circles, is:
   (Assume the circles are inside the square, centered at the corners of a smaller square within ABCD, touching the midpoints of the sides)
   (A) 42 cm²
   (B) 196 cm²
   (C) 154 cm²
   (D) 48 cm²

10. 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) 1430 cm²
    (D) 880 cm²

Answers to MCQs:

1. (D) 4 (Given 2πr = πr², so 2 = r. Diameter = 2r = 4)
2. (A) 132/7 cm² (Area = (60/360) * (22/7) * 6 * 6 = (1/6) * (22/7) * 36 = 132/7)
3. (C) (θ / 360°) * 2πR (Direct formula)
4. (A) 77/8 cm² (Circumference = 2πr = 22 => 2 * (22/7) * r = 22 => r = 7/2 cm. Area of Quadrant = (90/360) * πr² = (1/4) * (22/7) * (7/2)² = (1/4) * (22/7) * (49/4) = 77/8)
5. (A) 28.5 cm² (Area of Sector = (90/360) * 3.14 * 10² = (1/4) * 314 = 78.5 cm². Area of Triangle = (1/2) * base * height = (1/2) * 10 * 10 = 50 cm². Area of Segment = 78.5 - 50 = 28.5 cm²)
6. (B) 14 cm (Perimeter = Circumference of semi-circle + diameter = πr + 2r = r(π + 2). 36 = r(22/7 + 2) = r(22/7 + 14/7) = r(36/7). So, r = 7 cm. Diameter = 2r = 14 cm)
7. (A) r² sq. units (The largest triangle will have the diameter as its base (2r) and the height equal to the radius (r). Area = (1/2) * base * height = (1/2) * (2r) * r = r²)
8. (B) 44 cm (Length of wire = Circumference = 2 * (22/7) * 28 = 176 cm. Perimeter of square = 4 * side = 176 cm. Side = 176 / 4 = 44 cm)
9. (A) 42 cm² (Side of square = 14 cm. Area of square = 14² = 196 cm². Each circle touches two sides, so the radius of each circle is 14/4 = 3.5 cm? No, if four circles are placed such that they touch sides and each other, the radius of each circle must be 14/4 = 3.5 cm. This is incorrect setup description. Assuming 4 quadrants are cut from corners, each of radius 7 cm, or one circle in the middle of radius 7cm. Let's assume the question implies 4 quadrants of radius 7cm at each corner are removed, or a circle of diameter 14cm is inscribed. Let's re-read the exemplar problem style. A common problem is a square with 4 circles touching at the corners, radius = side/4 is incorrect. The standard problem involves four circles at the corners, each with radius = side/2 = 7cm, but only the quadrants are inside the square. Area of 4 quadrants = Area of 1 circle = πr² = (22/7)*7² = 154 cm². Area of shaded region (Square - Circle) = 196 - 154 = 42 cm². Let's assume this interpretation.)
10. (A) 770 cm² (Outer: 2πR = 132 => R = 132 * 7 / (2 * 22) = 21 cm. Inner: 2πr = 88 => r = 88 * 7 / (2 * 22) = 14 cm. Area of Ring = π(R² - r²) = (22/7) * (21² - 14²) = (22/7) * (21-14) * (21+14) = (22/7) * 7 * 35 = 22 * 35 = 770 cm²)

Make sure you understand the reasoning behind each answer, especially for the combination figures. Good luck with your preparation!