CBSE & State Board Question Bank | 9th - 12th

Deepa Maths Academy • Global Examination Portal

CBSE Class 10 Maths Chapter 11 Areas Related to Circles Model Questions - 2 Marks - Part 2

Secure University-Grade Repository for Model Assessments, Board Examinations, and Step-by-Step Solutions.

Portal ID: DMA-EXAM-2026 • Security: Encrypted • Status: Active Examination

SECTION B — Very Short Answer Type Questions [2 Marks Each]

26. 2 Marks Q26. A chord of a circle of radius 14 cm subtends an angle of 60° at the center. Find the area of the corresponding minor segment of the circle.
    Answer:
    

    Given: Radius (r) = 14 cm, Central angle (θ) = 60°

    1. Area of Sector:
       Area = (60° / 360°) × (22 / 7) × 142 = (1 / 6) × 616 = 308 / 3 cm2 ≈ 102.67 cm2
    2. Area of Equilateral Triangle (θ = 60°):
       Area = (√3 / 4) × 142 = 49√3 cm2 ≈ 49 × 1.732 = 84.87 cm2
    3. Area of Minor Segment:
       Area = 102.67 − 84.87 = 17.80 cm2

    Result: Area of minor segment = (³⁰⁸⁄₃ − 49√3) cm² ≈ 17.80 cm²

27. 2 Marks Q27. Find the area of the corresponding major segment of a circle of radius 12 cm if the minor segment area for a 120° angle is calculated or given as 88.44 cm² (or use standard formulas).
    Answer:
    

    Given: Radius (r) = 12 cm, Minor segment area = 88.44 cm²

    1. Total Area of Circle:
       Area = πr2 = 3.14 × 122 = 3.14 × 144 = 452.16 cm2
    2. Area of Major Segment:
       Area = Area of Circle − Area of Minor Segment
Area = 452.16 − 88.44 = 363.72 cm2

    Result: Area of major segment = 363.72 cm²

28. 2 Marks Q28. A chord AB of a circle of radius 10 cm makes an angle of 90° at the center of the circle. Find the area of the minor segment.
    Answer:
    

    Given: Radius (r) = 10 cm, θ = 90°

    1. Area of Minor Sector:
       Area = (90° / 360°) × 3.14 × 102 = (1 / 4) × 314 = 78.5 cm2
    2. Area of Right Triangle AOB:
       Area = ½ × 10 × 10 = 50 cm2
    3. Area of Minor Segment:
       Area = 78.5 − 50 = 28.5 cm2

    Result: Area of minor segment = 28.5 cm²

29. 2 Marks Q29. Find the area of a segment of a circle of radius 21 cm subtending an angle of 120° at the center. (Use sin 120° = ^√3⁄₂).
    Answer:
    

    Given: Radius (r) = 21 cm, θ = 120°

    1. Area of Sector:
       Area = (120° / 360°) × (22 / 7) × 212 = (1 / 3) × 1386 = 462 cm2
    2. Area of Triangle:
       Area = ½ × r2 × sin(120°) = ½ × 441 × (√3 / 2) = 441√3 / 4 cm2 ≈ 190.95 cm2
    3. Area of Segment:
       Area = 462 − 190.95 = 271.05 cm2

    Result: Area of segment = (462 − ^441√3⁄₄) cm² ≈ 271.05 cm²

30. 2 Marks Q30. A chord subtends an angle of 60° at the center of a circle of radius 10 cm. Find the area of the corresponding minor segment. (Use π = 3.14 and √3 = 1.73).
    Answer:
    

    Given: Radius (r) = 10 cm, θ = 60°

    1. Area of Sector:
       Area = (60° / 360°) × 3.14 × 102 = (1 / 6) × 314 = 52.33 cm2
    2. Area of Triangle:
       Area = (√3 / 4) × 102 = 25 × 1.73 = 43.25 cm2
    3. Area of Minor Segment:
       Area = 52.33 − 43.25 = 9.08 cm2

    Result: Area of minor segment = 9.08 cm²

31. 2 Marks Q31. The perimeter of a square circumscribing a circle is 68 cm. Find the area of the circle.
    Answer:
    

    1. Find Side of Square:
       Side (a) = Perimeter / 4 = 68 / 4 = 17 cm
    2. Find Radius of Inscribed Circle:
       Diameter (d) = Side = 17 cm ⇒ Radius (r) = 17 / 2 = 8.5 cm
    3. Area of Circle:
       Area = πr2 = (22 / 7) × (17 / 2)2 = 3179 / 14 ≈ 227.07 cm2

    Result: Area of circle = ³¹⁷⁹⁄₁₄ cm² (227.07 cm²)

32. 2 Marks Q32. Find the area of the shaded region in a circle of radius 6 cm where a sector of angle 60° is removed.
    Answer:
    

    Given: Radius (r) = 6 cm, Removed angle = 60° ⇒ Remaining angle = 300°

    1. Area of Remaining Major Sector:
       Area = (300° / 360°) × πr2 = (5 / 6) × (22 / 7) × 36
Area = (5 / 6) × (792 / 7) = 660 / 7 cm2 ≈ 94.29 cm2

    Result: Area of shaded region = ⁶⁶⁰⁄₇ cm² (94.29 cm²)

33. 2 Marks Q33. A circular arc of length 22 cm subtends an angle of 60° at the center. Find the area of the circle.
    Answer:
    

    Given: Arc length (l) = 22 cm, θ = 60°

    1. Find Radius (r):
       l = (θ / 360°) × 2πr ⇒ 22 = (60° / 360°) × 2 × (22 / 7) × r
22 = (1 / 6) × (44 / 7) × r ⇒ r = 21 cm
    2. Area of Circle:
       Area = πr2 = (22 / 7) × 21 × 21 = 1386 cm2

    Result: Area of circle = 1386 cm²

34. 2 Marks Q34. If the perimeter of a semicircular field is 36 m, find its diameter.
    Answer:
    

    1. Perimeter Formula:
       Perimeter = πr + 2r = r(π + 2)
    2. Solve for Radius (r):
       36 = r[(22 / 7) + 2] ⇒ 36 = r(36 / 7) ⇒ r = 7 m
    3. Find Diameter (d):
       d = 2r = 2 × 7 = 14 m

    Result: Diameter = 14 m

35. 2 Marks Q35. The cost of fencing a circular field at the rate of ₹24 per meter is ₹5280. Find the area of the field.
    Answer:
    

    1. Find Circumference:
       Circumference = Total Cost / Rate = 5280 / 24 = 220 m
    2. Find Radius (r):
       2πr = 220 ⇒ 2 × (22 / 7) × r = 220 ⇒ r = 35 m
    3. Area of Field:
       Area = πr2 = (22 / 7) × 35 × 35 = 3850 m2

    Result: Area of field = 3850 m²

36. 2 Marks Q36. Find the area of a sector of a circle of radius 28 cm and central angle 45°.
    Answer:
    

    Given: Radius (r) = 28 cm, θ = 45°

    1. Calculate Sector Area:
       Area = (θ / 360°) × πr2
Area = (45° / 360°) × (22 / 7) × 28 × 28 = (1 / 8) × 2464 = 308 cm2

    Result: Area of sector = 308 cm²

37. 2 Marks Q37. A pendulum of length 14 cm describes an arc of length 22 cm. Find the angle described by the pendulum at the center in degrees.
    Answer:
    

    Given: Radius (r) = 14 cm, Arc length (l) = 22 cm

    1. Formula: l = (θ / 360°) × 2πr
    2. Solve for θ:
       22 = (θ / 360°) × 2 × (22 / 7) × 14
22 = (θ / 360°) × 88 ⇒ θ = (22 × 360) / 88 = 90°

    Result: Angle = 90°

38. 2 Marks Q38. The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances traveled by their tips in 2 days.
    Answer:
    

    1. Short Hand (Hour hand, r = 4 cm):
       In 2 days (48 hrs), completes 4 revolutions.
       Distance1 = 4 × (2πr1) = 4 × 2 × (22 / 7) × 4 = 704 / 7 cm
    2. Long Hand (Minute hand, r = 6 cm):
       In 2 days (48 hrs), completes 48 revolutions.
       Distance2 = 48 × (2πr2) = 48 × 2 × (22 / 7) × 6 = 12672 / 7 cm
    3. Total Distance:
       Total = (704 + 12672) / 7 = 13376 / 7 cm ≈ 1910.86 cm

    Result: Total distance = ¹³³⁷⁶⁄₇ cm (1910.86 cm)

39. 2 Marks Q39. Find the area of the greatest circle that can be cut out from a square piece of cardboard of side 28 cm.
    Answer:
    

    1. Determine Circle Dimensions:
       Diameter (d) = Side of square = 28 cm ⇒ Radius (r) = 14 cm
    2. Calculate Area:
       Area = πr2 = (22 / 7) × 14 × 14 = 616 cm2

    Result: Area of greatest circle = 616 cm²

40. 2 Marks Q40. The area of a sector is ⁵⁄₁₈ of the area of the circle. Find the central angle of the sector.
    Answer:
    

    1. Equate Ratios:
       (θ / 360°) × πr2 = (5 / 18) × πr2
    2. Solve for θ:
       θ / 360° = 5 / 18 ⇒ θ = (5 / 18) × 360° = 100°

    Result: Central angle = 100°

41. 2 Marks Q41. If the radius of a circle is increased by 10%, find the percentage increase in its area.
    Answer:
    

    1. Original Area: A1 = πr2
    2. New Radius: r' = 1.1r
    3. New Area: A2 = π(1.1r)2 = 1.21πr2 = 1.21 A1
    4. Percentage Increase:
       Increase = [(1.21 A1 − A1) / A1] × 100% = 21%

    Result: Percentage increase in area = 21%

42. 2 Marks Q42. If the radius of a circle is decreased by 50%, find the percentage decrease in its area.
    Answer:
    

    1. Original Area: A1 = πr2
    2. New Radius: r' = 0.5r
    3. New Area: A2 = π(0.5r)2 = 0.25πr2 = 0.25 A1
    4. Percentage Decrease:
       Decrease = [(A1 − 0.25 A1) / A1] × 100% = 75%

    Result: Percentage decrease in area = 75%

43. 2 Marks Q43. A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. Find the radius of the circular base.
    Answer:
    

    1. Base Circumference:
       Width of sheet = Circumference of circular base = 22 cm
    2. Solve for Radius (r):
       2πr = 22 ⇒ 2 × (22 / 7) × r = 22 ⇒ r = 7 / 2 = 3.5 cm

    Result: Radius of circular base = 3.5 cm

44. 2 Marks Q44. Find the area of a square inscribed in a circle of radius 8 cm.
    Answer:
    

    1. Diagonal of Square (d):
       d = Diameter of circle = 2 × 8 = 16 cm
    2. Area of Square:
       Area = d2 / 2 = 162 / 2 = 256 / 2 = 128 cm2

    Result: Area of inscribed square = 128 cm²

45. 2 Marks Q45. A wire is looped in the form of a circle of radius 28 cm. It is rebent into a square form. Determine the length of the side of the square.
    Answer:
    

    1. Length of Wire (Circumference):
       C = 2πr = 2 × (22 / 7) × 28 = 176 cm
    2. Side of Square (a):
       Perimeter of square = 4a = 176 ⇒ a = 176 / 4 = 44 cm

    Result: Side of square = 44 cm

46. 2 Marks Q46. The difference between the circumference and radius of a circle is 37 cm. Find the area of the circle.
    Answer:
    

    1. Formulate Equation:
       2πr − r = 37 ⇒ r(2π − 1) = 37
    2. Solve for Radius (r):
       r[(44 / 7) − 1] = 37 ⇒ r(37 / 7) = 37 ⇒ r = 7 cm
    3. Area of Circle:
       Area = πr2 = (22 / 7) × 72 = 154 cm2

    Result: Area of circle = 154 cm²

47. 2 Marks Q47. A circular field has a diameter of 84 m. A road 3.5 m wide runs outside around it. Find the cost of gravelling the road at ₹20 per m².
    Answer:
    

    Inner radius (r) = 84 / 2 = 42 m, Outer radius (R) = 42 + 3.5 = 45.5 m

    1. Area of Road:
       Area = π(R2 − r2) = (22 / 7) × (45.5 − 42) × (45.5 + 42)
Area = (22 / 7) × 3.5 × 87.5 = 962.5 m2
    2. Total Cost:
       Cost = 962.5 × 20 = ₹19,250

    Result: Cost of gravelling = ₹19,250

48. 2 Marks Q48. Two circles touch externally. The sum of their areas is 130π cm² and the distance between their centers is 14 cm. If the radii are r₁ and r₂, find their values if r₁² + r₂² + 2r₁r₂ relations apply.
    Answer:
    

    1. Equations:
       r1 + r2 = 14
πr12 + πr22 = 130π ⇒ r12 + r22 = 130
    2. Solve for Radii:
       (r1 + r2)2 = r12 + r22 + 2r1r2 ⇒ 142 = 130 + 2r1r2
196 − 130 = 2r1r2 ⇒ r1r2 = 33
(r1 − r2)2 = 130 − 66 = 64 ⇒ r1 − r2 = 8
    3. Solve Simultaneous Equations:
       r1 = 11 cm, r2 = 3 cm

    Result: Radii are 11 cm and 3 cm

49. 2 Marks Q49. Find the area of a quadrant of a circle whose perimeter is 25 cm.
    Answer:
    

    1. Perimeter of Quadrant:
       Perimeter = ½πr + 2r = r(½π + 2) = r(11 / 7 + 2) = (25 / 7)r
    2. Solve for Radius (r):
       25 = (25 / 7)r ⇒ r = 7 cm
    3. Area of Quadrant:
       Area = ¼ × πr2 = ¼ × (22 / 7) × 49 = 77 / 2 = 38.5 cm2

    Result: Area of quadrant = 38.5 cm²

50. 2 Marks Q50. 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.
    Answer:
    

    Radius (r) = 45 cm, Total equal sectors = 8

    1. Central Angle (θ): θ = 360° / 8 = 45°
    2. Area between Consecutive Ribs:
       Area = (1 / 8) × πr2 = (1 / 8) × (22 / 7) × 452 = 22275 / 28 cm2 ≈ 795.54 cm2

    Result: Area = ²²²⁷⁵⁄₂₈ cm² (795.54 cm²)

Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :

English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.

தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.

மேலதிக விவரங்களுக்கு எங்களைத் தொடர்பு கொள்ளவும் → For further details, contact us →

Copyright © 2026 Deepa Maths Academy | Global Examination & Academic Portal. All Rights Reserved.

மாணவர்களுடன் பகிரவும்: