Chapter 11 : Areas Related to Circles Class 10 Maths NCERT Solutions

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Contents

Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 11: Areas Related to Circles — Exercise 11.1

Ever wondered how much pizza you’re actually getting in a “slice” or how much ground a windshield wiper clears? This exercise teaches you the math behind sectors and segments—the “slices” and “crusts” of the geometry world.

Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60 degrees.

  • The Logic: A sector is just a fraction of the total circle. Since a full circle is 360 degrees, a 60-degree sector is like taking a specific “slice.” We take the area of the whole circle (pi * r^2) and multiply it by the fraction (theta / 360).

  • The Steps:

    • Radius (r) = 6 cm, Angle (theta) = 60 degrees.

    • Formula: (theta / 360) * pi * r^2

    • Area = (60 / 360) * (22 / 7) * 6 * 6

    • Area = (1 / 6) * (22 / 7) * 36

    • Area = 22 * 6 / 7 = 132 / 7

  • The Result: The area of the sector is 132/7 cm^2 (or approx 18.86 cm^2).

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.

  • The Logic: A quadrant is exactly one-fourth (1/4) of a circle, which means the angle is 90 degrees. First, we need to “reverse-engineer” the radius from the circumference before we can find the area.

  • The Steps:

    1. Find the Radius: Circumference = 2 * pi * r = 22.

      • 2 * (22 / 7) * r = 22

      • r = 7 / 2 = 3.5 cm.

    2. Find the Area: Area of quadrant = (1 / 4) * pi * r^2

      • Area = (1 / 4) * (22 / 7) * (7 / 2) * (7 / 2)

      • Area = (1 * 11 * 7) / (2 * 4) = 77 / 8

  • The Result: The area of the quadrant is 77/8 cm^2 (or 9.625 cm^2).

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

  • The Logic: Think of the minute hand as the radius of a circle. In 60 minutes, it covers a full 360 degrees. So, in 5 minutes, we need to find what “slice” of the circle it covers.

  • Human Tip: 1 minute = 360 / 60 = 6 degrees. So, 5 minutes = 30 degrees!

  • The Steps:

    • Radius (r) = 14 cm, Angle (theta) = 30 degrees.

    • Area = (30 / 360) * (22 / 7) * 14 * 14

    • Area = (1 / 12) * 22 * 2 * 14

    • Area = (11 * 14) / 3 = 154 / 3

  • The Result: The area swept in 5 minutes is 154/3 cm^2 (or approx 51.33 cm^2).

Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use pi = 3.14)

  • The Logic: * Minor Segment: This is the area of the sector minus the area of the triangle formed by the chord and radii.

    • Major Sector: This is the “rest” of the circle. You can find it by using (360 – 90) = 270 degrees.

  • The Steps:

    1. Minor Sector Area: (90 / 360) * 3.14 * 10 * 10 = 0.25 * 314 = 78.5 cm^2.

    2. Triangle Area: (1 / 2) * base * height = (1 / 2) * 10 * 10 = 50 cm^2.

    3. Minor Segment: 78.5 – 50 = 28.5 cm^2.

    4. Major Sector Area: (270 / 360) * 3.14 * 10 * 10 = 0.75 * 314 = 235.5 cm^2.

  • The Result: (i) Minor segment = 28.5 cm^2, (ii) Major sector = 235.5 cm^2.

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60 degrees at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord.

  • The Logic: * Arc Length: Just a fraction of the circumference (2 * pi * r).

    • Segment Area: Area of sector – Area of equilateral triangle (since the angle is 60 and sides are equal, it must be equilateral!).

  • Pro-Tip: Area of an equilateral triangle = (root 3 / 4) * side^2.

  • The Steps:

    1. Arc Length: (60 / 360) * 2 * (22 / 7) * 21 = (1 / 6) * 2 * 22 * 3 = 22 cm.

    2. Sector Area: (60 / 360) * (22 / 7) * 21 * 21 = (1 / 6) * 22 * 3 * 21 = 231 cm^2.

    3. Triangle Area: (root 3 / 4) * 21 * 21 = (441 * root 3) / 4.

    4. Segment Area: 231 – (441 * root 3) / 4.

  • The Result: (i) 22 cm, (ii) 231 cm^2, (iii) [231 – (441 * root 3) / 4] cm^2.

Q6. A chord of a circle of radius 15 cm subtends an angle of 60 degrees at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use pi = 3.14 and root 3 = 1.73)

  • The Logic: Since the angle is 60° and the two radii are equal, the triangle formed is equilateral. To find the Minor Segment, we take the sector area and subtract the equilateral triangle area. For the Major Segment, we subtract the minor segment area from the total circle area.

  • The Steps:

    1. Sector Area: (60 / 360) * 3.14 * 15 * 15 = (1 / 6) * 706.5 = 117.75 cm^2.

    2. Triangle Area: (root 3 / 4) * 15 * 15 = (1.73 / 4) * 225 = 97.3125 cm^2.

    3. Minor Segment: 117.75 – 97.3125 = 20.4375 cm^2.

    4. Major Segment: (3.14 * 15 * 15) – 20.4375 = 706.5 – 20.4375 = 686.0625 cm^2.

  • The Result: Minor Segment = 20.4375 cm^2; Major Segment = 686.0625 cm^2.

Q7. A chord of a circle of radius 12 cm subtends an angle of 120 degrees at the centre. Find the area of the corresponding segment of the circle. (Use pi = 3.14 and root 3 = 1.73)

  • The Logic: This is a tricky one because the angle is 120°. You can’t use the equilateral triangle shortcut here. You have to find the area of the triangle using the general formula: (1/2) * r^2 * sin(theta) or by splitting it into two 30-60-90 triangles.

  • The Steps:

    1. Sector Area: (120 / 360) * 3.14 * 12 * 12 = (1 / 3) * 3.14 * 144 = 150.72 cm^2.

    2. Triangle Area: (1 / 2) * 12 * 12 * sin(120°). Since sin 120° = root 3 / 2:

      • Area = 72 * (1.73 / 2) = 36 * 1.73 = 62.28 cm^2.

    3. Segment Area: 150.72 – 62.28 = 88.44 cm^2.

  • The Result: The area of the segment is 88.44 cm^2.

Q8. A horse is tied to a tether at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use pi = 3.14)

  • The Logic: The horse is at the corner of a square, so the grazing area is a quadrant (90° sector). The field size (15 m) is just a distraction—the only thing that matters is the length of the rope (the radius).

     

  • The Steps:

    1. Initial Area (r = 5): (90 / 360) * 3.14 * 5 * 5 = (1 / 4) * 78.5 = 19.625 m^2.

    2. New Area (r = 10): (90 / 360) * 3.14 * 10 * 10 = (1 / 4) * 314 = 78.5 m^2.

    3. Increase: 78.5 – 19.625 = 58.875 m^2.

  • The Result: (i) 19.625 m^2, (ii) 58.875 m^2.

Q9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

  • The Logic: Total wire = Circumference + (5 * Diameter). For the sector area, the angle is 360 / 10 = 36°.

  • The Steps:

    1. Total Wire: [pi * D] + [5 * D] = (22 / 7 * 35) + (5 * 35) = 110 + 175 = 285 mm.

    2. Sector Area: (36 / 360) * (22 / 7) * (35 / 2) * (35 / 2)

      • Area = (1 / 10) * (22 / 7) * (1225 / 4) = 385 / 4 = 96.25 mm^2.

  • The Result: (i) 285 mm, (ii) 96.25 mm^2.

Q10. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

  • The Logic: If there are 8 ribs, the circle is divided into 8 equal sectors. Angle = 360 / 8 = 45°.

  • The Steps:

    • Radius (r) = 45 cm.

    • Area = (1 / 8) * (22 / 7) * 45 * 45

    • Area = (11 / 4) * (2025 / 7) = 22275 / 28

  • The Result: The area between two ribs is 22275/28 cm^2 (approx 795.53 cm^2).

Q11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115 degrees. Find the total area cleaned at each sweep of the blades.

  • The Logic: Each wiper creates a sector. Since there are two wipers and they don’t overlap, we simply find the area of one sector and multiply it by 2.

  • The Steps:

    • Radius (r) = 25 cm, Angle (theta) = 115 degrees.

    • Area of one sector = (theta / 360) * pi * r^2

    • Total Area = 2 * (115 / 360) * (22 / 7) * 25 * 25

    • Total Area = 2 * (23 / 72) * (22 / 7) * 625

    • Total Area = (23 * 11 * 625) / (18 * 7) = 158125 / 126

  • The Result: The total area cleaned is 158125/126 cm^2 (approx 1254.96 cm^2).

Q12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80 degrees to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use pi = 3.14)

  • The Logic: The lighthouse light acts as a giant sector on the surface of the water. The distance the light travels (16.5 km) is our radius.

  • The Steps:

    • Radius (r) = 16.5 km, Angle (theta) = 80 degrees.

    • Area = (80 / 360) * 3.14 * 16.5 * 16.5

    • Area = (2 / 9) * 3.14 * 272.25

    • Area = 2 * 3.14 * 30.25 = 189.97

  • The Result: The area of the sea warned is 189.97 km^2.

Q13. A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm^2. (Use root 3 = 1.7)

  • The Logic: The “designs” are actually six equal segments.

    1. Find the area of one sector (Angle = 360 / 6 = 60 degrees).

    2. Find the area of the equilateral triangle (since angle is 60).

    3. Subtract triangle from sector to get one segment.

    4. Multiply by 6 for total design area, then multiply by the cost.

  • The Steps:

    1. One Sector Area: (60 / 360) * (22 / 7) * 28 * 28 = (1 / 6) * 22 * 4 * 28 = 1232 / 3 = 410.67 cm^2.

    2. One Triangle Area: (root 3 / 4) * side^2 = (1.7 / 4) * 28 * 28 = 1.7 * 7 * 28 = 333.2 cm^2.

    3. One Segment Area: 410.67 – 333.2 = 77.47 cm^2.

    4. Total Design Area: 6 * 77.47 = 464.82 cm^2.

    5. Total Cost: 464.82 * 0.35 = 162.687.

  • The Result: The total cost of making the designs is Rs 162.68 (approx).

Q14. Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is:

(A) (p / 180) * 2piR (B) (p / 180) * piR^2 (C) (p / 360) * 2piR (D) (p / 720) * 2piR^2

  • The Logic: The standard formula is (p / 360) * pi * R^2. We need to find which option matches this.

  • The Steps:

    • Option (D) says (p / 720) * 2 * pi * R^2.

    • If we simplify 2 / 720, we get 1 / 360.

    • So, (p / 720) * 2 * pi * R^2 = (p / 360) * pi * R^2.

  • The Result: (D) (p/720) * 2piR^2

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