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CBSE Class 10 Maths Chapter 11 Areas Related to Circles Model Questions - 3 Marks - Part 1

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SECTION C — Short Answer Type Questions [3 Marks Each]

1. 3 Marks Q1. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°. Also, find the length of the corresponding arc of the sector. (Use π = ²²⁄₇).
   Answer:
   

   Given: Radius (r) = 6 cm, Angle (θ) = 60°

   1. Area of Sector:
      Area = (θ / 360°) × πr2
Area = (60° / 360°) × (22 / 7) × 6 × 6 = (1 / 6) × (22 / 7) × 36 = 132 / 7 cm2 ≈ 18.86 cm2
   2. Length of Arc:
      Length = (θ / 360°) × 2πr
Length = (1 / 6) × 2 × (22 / 7) × 6 = 44 / 7 cm ≈ 6.28 cm

   Result: Area of sector = ¹³²⁄₇ cm² (18.86 cm²), Arc length = ⁴⁴⁄₇ cm (6.28 cm)

2. 3 Marks Q2. A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding minor segment of the circle. (Use π = 3.14 and √3 = 1.732).
   Answer:
   

   Given: Radius (r) = 12 cm, Angle (θ) = 120°

   1. Area of Sector:
      Area = (120° / 360°) × 3.14 × 12 × 12 = (1 / 3) × 3.14 × 144 = 150.72 cm2
   2. Area of Triangle:
      Area = ½ × r2 × sin(120°) = ½ × 144 × (√3 / 2) = 36√3 = 36 × 1.732 = 62.352 cm2
   3. Area of Minor Segment:
      Area = Area of Sector − Area of Triangle = 150.72 − 62.352 = 88.368 cm2

   Result: Area of minor segment = 88.37 cm²

3. 3 Marks Q3. The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.
   Answer:
   

   Given: Radius (r) = 5.6 cm, Perimeter (P) = 27.2 cm

   1. Find Arc Length (l):
      Perimeter = 2r + l ⇒ 27.2 = 2(5.6) + l
l = 27.2 − 11.2 = 16 cm
   2. Area of Sector:
      Area = ½ × l × r = ½ × 16 × 5.6 = 44.8 cm2

   Result: Area of sector = 44.8 cm²

4. 3 Marks Q4. A horse is tied to a peg at one corner of a square grassy field of side 15 m by means of a 5 m long rope. Find the increase in the grazing area if the rope length is increased to 10 m. (Use π = 3.14).
   Answer:
   

   Corner angle of square (θ) = 90°

   1. Initial Grazing Area (r = 5 m):
      Area1 = (90° / 360°) × π × 52 = (1 / 4) × 3.14 × 25 = 19.625 m2
   2. New Grazing Area (r = 10 m):
      Area2 = (90° / 360°) × π × 102 = (1 / 4) × 3.14 × 100 = 78.5 m2
   3. Increase in Area:
      Increase = Area2 − Area1 = 78.5 − 19.625 = 58.875 m2

   Result: Increase in grazing area = 58.875 m²

5. 3 Marks Q5. The minute hand of a clock is 14 cm long. Find the area swept by the minute hand in 5 minutes. Also find the distance moved by the tip of the minute hand in this time interval.
   Answer:
   

   Given: Radius (r) = 14 cm

   1. Angle in 5 minutes (θ):
      θ = (360° / 60) × 5 = 30°
   2. Area Swept:
      Area = (30° / 360°) × (22 / 7) × 14 × 14 = (1 / 12) × (22 / 7) × 196 = 154 / 3 cm2 ≈ 51.33 cm2
   3. Distance Moved by Tip (Arc Length):
      Distance = (30° / 360°) × 2 × (22 / 7) × 14 = 22 / 3 cm ≈ 7.33 cm

   Result: Area swept = ¹⁵⁴⁄₃ cm², Distance moved = ²²⁄₃ cm

6. 3 Marks Q6. Find the area of a segment of a circle of radius 21 cm subtending an angle of 60° at the center. (Use √3 = 1.732).
   Answer:
   

   Given: Radius (r) = 21 cm, θ = 60°

   1. Area of Sector:
      Areasector = (60° / 360°) × (22 / 7) × 212 = (1 / 6) × 22 × 63 = 231 cm2
   2. Area of Equilateral Triangle (θ = 60°):
      Areatriangle = (√3 / 4) × r2 = (1.732 / 4) × 441 = 190.953 cm2
   3. Area of Segment:
      Area = 231 − 190.953 = 40.047 cm2

   Result: Area of segment = 40.05 cm² (or 231 − ^441√3⁄₄ cm²)

7. 3 Marks Q7. An arc of length 22 cm subtends an angle of 50° at the center of a circle. Find the radius of the circle and the area of the corresponding sector.
   Answer:
   

   Given: Arc length (l) = 22 cm, θ = 50°

   1. Find Radius (r):
      l = (θ / 360°) × 2πr ⇒ 22 = (50° / 360°) × 2 × (22 / 7) × r
r = (22 × 360 × 7) / (50 × 44) = 50.4 cm
   2. Area of Sector:
      Area = ½ × l × r = ½ × 22 × 50.4 = 554.4 cm2

   Result: Radius = 50.4 cm, Area of sector = 554.4 cm²

8. 3 Marks Q8. The wheel of a motorcycle is of radius 35 cm. How many revolutions per minute must the wheel make so that the speed of the motorcycle is kept at 66 km/h?
   Answer:
   

   Given: Radius (r) = 35 cm, Speed = 66 km/h

   1. Circumference of Wheel:
      C = 2πr = 2 × (22 / 7) × 35 = 220 cm = 2.2 m
   2. Distance covered in 1 minute:
      Distance = (66 × 1000 m) / 60 min = 1100 m/min
   3. Revolutions per Minute (RPM):
      RPM = Distance per min / Circumference = 1100 / 2.2 = 500

   Result: 500 revolutions per minute

9. 3 Marks Q9. A chord of a circle of radius 10 cm subtends a right angle at the center. Find the areas of the corresponding: (i) minor sector, and (ii) major sector. (Use π = 3.14).
   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 Major Sector:
      Area = Total Circle Area − Minor Sector Area
Area = (3.14 × 100) − 78.5 = 314 − 78.5 = 235.5 cm2

   Result: (i) Minor sector = 78.5 cm², (ii) Major sector = 235.5 cm²

10. 3 Marks Q10. The difference between the circumference and radius of a circle is 37 cm. Using this relation, calculate the area of the circle.
    Answer:
    

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

    Result: Area of circle = 154 cm²

11. 3 Marks Q11. A round table cover has six equal designs as shown in the given circular frame. If the radius of the cover is 28 cm, find the total cost of making the designs at the rate of ₹0.35 per cm². (Use √3 = 1.7).
    Answer:
    

    Each sector angle (θ) = 360° / 6 = 60°

    1. Area of 1 Sector:
       Area = (60° / 360°) × (22 / 7) × 282 = (1 / 6) × 2464 = 410.67 cm2
    2. Area of 1 Equilateral Triangle:
       Area = (√3 / 4) × 282 = (1.7 / 4) × 784 = 333.2 cm2
    3. Area of 6 Designs:
       Total Area = 6 × (410.67 − 333.2) = 6 × 77.47 = 464.8 cm2
    4. Total Cost:
       Cost = 464.8 × 0.35 = ₹162.68

    Result: Total cost = ₹162.68

12. 3 Marks Q12. Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector. (Use π = 3.14).
    Answer:
    

    Given: r = 4 cm, θ = 30°

    1. Area of Sector:
       Area = (30° / 360°) × 3.14 × 42 = (1 / 12) × 50.24 = 4.19 cm2
    2. Area of Major Sector:
       Area = (330° / 360°) × 3.14 × 16 = 46.05 cm2

    Result: Sector area = 4.19 cm², Major sector area = 46.05 cm²

13. 3 Marks Q13. Find the area of the shaded region in a square of side 10 cm, where semicircles are drawn with each side as diameter. (Use π = 3.14).
    Answer:
    

    Side of square = 10 cm, Radius of semicircles (r) = 5 cm

    1. Area of Unshaded Regions (2 pairs of semicircles):
       Area(I + III) = Area of Square − Area of 2 Semicircles
Area(I + III) = 102 − [2 × (½ × 3.14 × 52)] = 100 − 78.5 = 21.5 cm2
Similarly, Area(II + IV) = 21.5 cm2
    2. Area of Shaded Region:
       Area = Area of Square − Total Unshaded Area
Area = 100 − (21.5 + 21.5) = 100 − 43 = 57 cm2

    Result: Area of shaded region = 57 cm²

14. 3 Marks Q14. From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut. Find the area of the remaining portion of the square.
    Answer:
    

    1. Area of Square: 4 × 4 = 16 cm2
    2. Area of 4 Quadrants:
       4 × (¼ × π × 12) = π cm2
    3. Area of Center Circle (r = 1 cm):
       π × 12 = π cm2
    4. Remaining Area:
       Area = 16 − 2π = 16 − 2(22 / 7) = 16 − 44 / 7 = 68 / 7 cm2 ≈ 9.71 cm2

    Result: Remaining area = ⁶⁸⁄₇ cm² (9.71 cm²)

15. 3 Marks Q15. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design (shaded region).
    Answer:
    

    Radius of circle (R) = 32 cm

    1. Area of Circle:
       Area = πR2 = (22 / 7) × 322 = 22528 / 7 cm2
    2. Side of Equilateral Triangle (a):
       R = a / √3 ⇒ a = 32√3 cm
    3. Area of Equilateral Triangle:
       Area = (√3 / 4) × a2 = (√3 / 4) × (32√3)2 = 768√3 cm2
    4. Area of Design:
       Area = (22528 / 7 − 768√3) cm2

    Result: Area of design = (²²⁵²⁸⁄₇ − 768√3) cm²

16. 3 Marks Q16. Find the area of the shaded region where a circular arc of radius 6 cm has been drawn with vertex A of an equilateral triangle ABC of side 12 cm as center.
    Answer:
    

    1. Area of Equilateral Triangle (side = 12 cm):
       Area = (√3 / 4) × 122 = 36√3 cm2
    2. Area of Major Sector (r = 6 cm, θ = 360° − 60° = 300°):
       Area = (300° / 360°) × (22 / 7) × 62 = (5 / 6) × (22 / 7) × 36 = 660 / 7 cm2
    3. Total Shaded Area:
       Area = (36√3 + 660 / 7) cm2

    Result: Shaded area = (36√3 + ⁶⁶⁰⁄₇) cm²

17. 3 Marks Q17. A rectangular park is 60 m long and 40 m wide. Two cross paths each 2 m wide run at right angles through the center of the park parallel to its sides. Find the remaining area of the park.
    Answer:
    

    1. Total Park Area: 60 × 40 = 2400 m2
    2. Area of Cross Paths:
       Area = (60 × 2) + (40 × 2) − (2 × 2) = 120 + 80 − 4 = 196 m2
    3. Remaining Area:
       Area = 2400 − 196 = 2204 m2
Alternatively: (60 − 2) × (40 − 2) = 58 × 38 = 2204 m2

    Result: Remaining area = 2204 m²

18. 3 Marks Q18. Two circular pieces of radii r₁ = 8 cm and r₂ = 6 cm are combined. Find the radius and area of a new circle whose area is equal to the sum of the areas of the two given circles.
    Answer:
    

    1. Find New Radius (R):
       πR2 = πr12 + πr22 ⇒ R2 = 82 + 62 = 64 + 36 = 100
R = 10 cm
    2. Area of New Circle:
       Area = πR2 = (22 / 7) × 100 = 2200 / 7 cm2 ≈ 314.29 cm2

    Result: New radius = 10 cm, Area = 314.29 cm²

19. 3 Marks Q19. A pathway of uniform width 3.5 m runs around a circular garden whose radius is 10.5 m. Find the total cost of paving the path at ₹10 per m².
    Answer:
    

    Inner radius (r) = 10.5 m, Outer radius (R) = 10.5 + 3.5 = 14 m

    1. Area of Pathway:
       Area = π(R2 − r2) = (22 / 7) × (142 − 10.52) = (22 / 7) × 85.75 = 269.5 m2
    2. Total Cost:
       Cost = 269.5 × 10 = ₹2695

    Result: Total paving cost = ₹2,695

20. 3 Marks Q20. 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, and (ii) the area of each sector of the brooch.
    Answer:
    

    Diameter (d) = 35 mm, Radius (r) = 35/2 mm

    1. Total Length of Silver Wire:
       Length = Circumference + 5 × Diameter = πd + 5d
Length = [(22 / 7) × 35] + (5 × 35) = 110 + 175 = 285 mm
    2. Area of Each Sector:
       Area = (1 / 10) × πr2 = (1 / 10) × (22 / 7) × (35 / 2)2 = 385 / 4 mm2 = 96.25 mm2

    Result: (i) Wire length = 285 mm, (ii) Sector area = ³⁸⁵⁄₄ mm² (96.25 mm²)

21. 3 Marks Q21. An umbrella has 8 ribs which are equally spaced. Assuming the 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 sectors = 8

    1. Angle of each sector (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)

22. 3 Marks Q22. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles, and compare its area.
    Answer:
    

    1. Find New Radius (R):
       2πR = 2πr1 + 2πr2 ⇒ R = r1 + r2 = 19 + 9 = 28 cm
    2. Area Comparison:
       Areanew = π × 282 = 784π cm2 = 2464 cm2
Sum of individual areas = π(192 + 92) = 442π cm2 = 1389.14 cm2

    Result: New radius = 28 cm; New area (784π cm²) is greater than the sum of the two areas (442π cm²) by 342π cm².

23. 3 Marks Q23. A square of diagonal 4√2 cm is inscribed in a circle. Find the area of the region lying outside the square but inside the circle.
    Answer:
    

    Diagonal of square (d) = Diameter of circle = 4√2 cm ⇒ Radius (r) = 2√2 cm

    1. Side of Square (a):
       a = d / √3 = (4√2) / √2 = 4 cm
    2. Area of Square: a2 = 16 cm2
    3. Area of Circle: πr2 = (22 / 7) × 8 = 176 / 7 cm2 ≈ 25.14 cm2
    4. Area Outside Square:
       Area = (176 / 7) − 16 = 64 / 7 cm2 ≈ 9.14 cm2

    Result: Required area = ⁶⁴⁄₇ cm² (9.14 cm²)

24. 3 Marks Q24. Find the area of the shaded region in a circle of radius 7 cm where a sector of angle 90° is combined with an external equilateral triangle of side 7 cm.
    Answer:
    

    1. Area of Sector (r = 7 cm, θ = 90°):
       Area = (90° / 360°) × (22 / 7) × 72 = 77 / 2 = 38.5 cm2
    2. Area of Equilateral Triangle (side = 7 cm):
       Area = (√3 / 4) × 72 = 49√3 / 4 cm2 ≈ 21.22 cm2
    3. Combined Area:
       Total Area = 38.5 + 21.22 = 59.72 cm2

    Result: Total area = (38.5 + ^49√3⁄₄) cm² ≈ 59.72 cm²

25. 3 Marks Q25. A chord AB of a circle of radius 14 cm subtends an angle of 90° at the center. Find the area of the minor segment and verify it using right triangle properties.
    Answer:
    

    Given: r = 14 cm, θ = 90°

    1. Area of Sector:
       Area = (90° / 360°) × (22 / 7) × 142 = (1 / 4) × 616 = 154 cm2
    2. Area of Right Triangle AOB:
       Area = ½ × base × height = ½ × 14 × 14 = 98 cm2
    3. Area of Minor Segment:
       Area = 154 − 98 = 56 cm2
    4. Verification: The triangle is isosceles right-angled at O; area verified via ½r2sin(90°) = 98 cm2, confirming segment area = 56 cm².

    Result: Area of minor segment = 56 cm²

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