April 10, 2026
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Chapter 11: Areas Related to Circles
Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)
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Exercise 11.1 (Page 227)
13 marks
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Answer:
Area of sector = (θ/360°) × πr² = (60/360) × π × 6² = (1/6) × π × 36 = 6π cm².
Area = 6π cm² ≈ 18.85 cm².
Area of sector = (θ/360°) × πr² = (60/360) × π × 6² = (1/6) × π × 36 = 6π cm².
Area = 6π cm² ≈ 18.85 cm².
23 marks
Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer:
Circumference = 2πr = 22 → r = 22/(2π) = 11/π cm.
Area of quadrant = (1/4)πr² = (1/4)π × (121/π²) = 121/(4π) = (121×7)/(4×22) = 847/88 ≈ 9.625 cm².
Area = 121/(4π) cm² ≈ 9.625 cm².
Circumference = 2πr = 22 → r = 22/(2π) = 11/π cm.
Area of quadrant = (1/4)πr² = (1/4)π × (121/π²) = 121/(4π) = (121×7)/(4×22) = 847/88 ≈ 9.625 cm².
Area = 121/(4π) cm² ≈ 9.625 cm².
33 marks
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer:
In 60 minutes, minute hand sweeps 360°. In 5 minutes, angle = (5/60)×360° = 30°.
Area swept = (30/360) × π × 14² = (1/12) × π × 196 = (49π)/3 cm² ≈ 51.31 cm².
Area = (49π)/3 cm².
In 60 minutes, minute hand sweeps 360°. In 5 minutes, angle = (5/60)×360° = 30°.
Area swept = (30/360) × π × 14² = (1/12) × π × 196 = (49π)/3 cm² ≈ 51.31 cm².
Area = (49π)/3 cm².
43 marks
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 π = 3.14)
(i) minor segment (ii) major sector (Use π = 3.14)
Answer:
r = 10 cm, θ = 90°.
(i) Area of sector = (90/360) × 3.14 × 100 = 78.5 cm².
Area of triangle = (1/2)×10×10 = 50 cm².
Minor segment = 78.5 - 50 = 28.5 cm².
(ii) Major sector = total area - minor sector = 3.14×100 - 78.5 = 314 - 78.5 = 235.5 cm².
Minor segment = 28.5 cm², Major sector = 235.5 cm².
r = 10 cm, θ = 90°.
(i) Area of sector = (90/360) × 3.14 × 100 = 78.5 cm².
Area of triangle = (1/2)×10×10 = 50 cm².
Minor segment = 78.5 - 50 = 28.5 cm².
(ii) Major sector = total area - minor sector = 3.14×100 - 78.5 = 314 - 78.5 = 235.5 cm².
Minor segment = 28.5 cm², Major sector = 235.5 cm².
53 marks
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc (ii) area of the sector (iii) area of the segment formed by the corresponding chord.
(i) the length of the arc (ii) area of the sector (iii) area of the segment formed by the corresponding chord.
Answer:
r = 21 cm, θ = 60°.
(i) Arc length = (θ/360)×2πr = (60/360)×2×(22/7)×21 = (1/6)×2×22×3 = 22 cm.
(ii) Sector area = (60/360)×(22/7)×21² = (1/6)×(22/7)×441 = (1/6)×22×63 = 231 cm².
(iii) Area of equilateral triangle (since θ=60°) = (√3/4)×21² = (√3/4)×441 = (441√3)/4 ≈ 190.96 cm².
Segment = 231 - 190.96 = 40.04 cm².
Arc length = 22 cm, Sector area = 231 cm², Segment ≈ 40.04 cm².
r = 21 cm, θ = 60°.
(i) Arc length = (θ/360)×2πr = (60/360)×2×(22/7)×21 = (1/6)×2×22×3 = 22 cm.
(ii) Sector area = (60/360)×(22/7)×21² = (1/6)×(22/7)×441 = (1/6)×22×63 = 231 cm².
(iii) Area of equilateral triangle (since θ=60°) = (√3/4)×21² = (√3/4)×441 = (441√3)/4 ≈ 190.96 cm².
Segment = 231 - 190.96 = 40.04 cm².
Arc length = 22 cm, Sector area = 231 cm², Segment ≈ 40.04 cm².
63 marks
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
Answer:
r = 15 cm, θ = 60°.
Sector area = (60/360) × 3.14 × 225 = (1/6) × 706.5 = 117.75 cm².
Triangle area = (1/2)×15×15×sin60° = (225/2)×0.866 = 112.5 × 0.866 = 97.425 cm².
Minor segment = 117.75 - 97.425 = 20.325 cm².
Major segment = Total area - minor segment = 3.14×225 - 20.325 = 706.5 - 20.325 = 686.175 cm².
Minor segment ≈ 20.33 cm², Major segment ≈ 686.18 cm².
r = 15 cm, θ = 60°.
Sector area = (60/360) × 3.14 × 225 = (1/6) × 706.5 = 117.75 cm².
Triangle area = (1/2)×15×15×sin60° = (225/2)×0.866 = 112.5 × 0.866 = 97.425 cm².
Minor segment = 117.75 - 97.425 = 20.325 cm².
Major segment = Total area - minor segment = 3.14×225 - 20.325 = 706.5 - 20.325 = 686.175 cm².
Minor segment ≈ 20.33 cm², Major segment ≈ 686.18 cm².
73 marks
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Answer:
r = 12 cm, θ = 120°.
Sector area = (120/360) × 3.14 × 144 = (1/3) × 452.16 = 150.72 cm².
Triangle area = (1/2)×12×12×sin120° = 72 × (√3/2) = 72 × 0.865 = 62.28 cm².
Minor segment = 150.72 - 62.28 = 88.44 cm².
Area of corresponding segment = 88.44 cm².
r = 12 cm, θ = 120°.
Sector area = (120/360) × 3.14 × 144 = (1/3) × 452.16 = 150.72 cm².
Triangle area = (1/2)×12×12×sin120° = 72 × (√3/2) = 72 × 0.865 = 62.28 cm².
Minor segment = 150.72 - 62.28 = 88.44 cm².
Area of corresponding segment = 88.44 cm².
84 marks
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). 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 π = 3.14)
(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 π = 3.14)
Answer:
(i) Rope = 5 m, corner angle = 90° → quadrant area = (1/4) × π × 5² = (1/4)×3.14×25 = 19.625 m².
(ii) With rope 10 m, quadrant area = (1/4)×3.14×100 = 78.5 m².
Increase = 78.5 - 19.625 = 58.875 m².
Grazing area (5m) = 19.625 m², increase = 58.875 m².
(i) Rope = 5 m, corner angle = 90° → quadrant area = (1/4) × π × 5² = (1/4)×3.14×25 = 19.625 m².
(ii) With rope 10 m, quadrant area = (1/4)×3.14×100 = 78.5 m².
Increase = 78.5 - 19.625 = 58.875 m².
Grazing area (5m) = 19.625 m², increase = 58.875 m².
94 marks
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 as shown in Fig. 11.9. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Answer:
Diameter = 35 mm → radius = 17.5 mm.
(i) Circumference = 2πr = 2×(22/7)×17.5 = 110 mm. Length of 5 diameters = 5×35 = 175 mm.
Total wire = 110 + 175 = 285 mm.
(ii) Each sector angle = 360°/10 = 36°. Area of each sector = (36/360)×πr² = (1/10)×(22/7)×(17.5)² = (1/10)×(22/7)×306.25 = (22×306.25)/(70) = 96.25 mm².
Total wire = 285 mm, each sector area = 96.25 mm².
Diameter = 35 mm → radius = 17.5 mm.
(i) Circumference = 2πr = 2×(22/7)×17.5 = 110 mm. Length of 5 diameters = 5×35 = 175 mm.
Total wire = 110 + 175 = 285 mm.
(ii) Each sector angle = 360°/10 = 36°. Area of each sector = (36/360)×πr² = (1/10)×(22/7)×(17.5)² = (1/10)×(22/7)×306.25 = (22×306.25)/(70) = 96.25 mm².
Total wire = 285 mm, each sector area = 96.25 mm².
104 marks
An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between two consecutive ribs of the umbrella.
Answer:
Radius = 45 cm, number of ribs = 8 → each sector angle = 360°/8 = 45°.
Area between two ribs = area of sector = (45/360) × π × 45² = (1/8) × (22/7) × 2025 = (22×2025)/(56) = (44550)/56 ≈ 795.54 cm².
Area ≈ 795.54 cm².
Radius = 45 cm, number of ribs = 8 → each sector angle = 360°/8 = 45°.
Area between two ribs = area of sector = (45/360) × π × 45² = (1/8) × (22/7) × 2025 = (22×2025)/(56) = (44550)/56 ≈ 795.54 cm².
Area ≈ 795.54 cm².
114 marks
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Answer:
For one wiper: sector area = (115/360) × π × 25² = (115/360) × (22/7) × 625.
Compute: (115×22×625)/(360×7) = (115×22×625)/(2520) = (1581250)/(2520) ≈ 627.48 cm².
Two wipers = 2 × 627.48 = 1254.96 cm².
Total area ≈ 1254.96 cm².
For one wiper: sector area = (115/360) × π × 25² = (115/360) × (22/7) × 625.
Compute: (115×22×625)/(360×7) = (115×22×625)/(2520) = (1581250)/(2520) ≈ 627.48 cm².
Two wipers = 2 × 627.48 = 1254.96 cm².
Total area ≈ 1254.96 cm².
124 marks
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Answer:
Area = (80/360) × 3.14 × (16.5)² = (2/9) × 3.14 × 272.25 = (2 × 3.14 × 272.25)/9 = (1709.73)/9 ≈ 189.97 km².
Area ≈ 189.97 km².
Area = (80/360) × 3.14 × (16.5)² = (2/9) × 3.14 × 272.25 = (2 × 3.14 × 272.25)/9 = (1709.73)/9 ≈ 189.97 km².
Area ≈ 189.97 km².
134 marks
A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use √3 = 1.7)
Answer:
Radius = 28 cm. Each sector angle = 360°/6 = 60°. Area of one sector = (60/360) × π × 28² = (1/6) × (22/7) × 784 = (22×784)/(42) = (17248)/42 ≈ 410.67 cm².
Area of equilateral triangle in sector = (√3/4)×28² = (1.7/4)×784 = (1.7×196) = 333.2 cm².
Area of one design (segment) = sector - triangle = 410.67 - 333.2 = 77.47 cm².
Total area of 6 designs = 6 × 77.47 = 464.82 cm².
Cost = 464.82 × 0.35 = ₹ 162.69.
Cost ≈ ₹ 162.69.
Radius = 28 cm. Each sector angle = 360°/6 = 60°. Area of one sector = (60/360) × π × 28² = (1/6) × (22/7) × 784 = (22×784)/(42) = (17248)/42 ≈ 410.67 cm².
Area of equilateral triangle in sector = (√3/4)×28² = (1.7/4)×784 = (1.7×196) = 333.2 cm².
Area of one design (segment) = sector - triangle = 410.67 - 333.2 = 77.47 cm².
Total area of 6 designs = 6 × 77.47 = 464.82 cm².
Cost = 464.82 × 0.35 = ₹ 162.69.
Cost ≈ ₹ 162.69.
144 marks
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 × 2πR (B) p/180 × πR² (C) p/360 × 2πR (D) p/720 × 2πR²
Area of a sector of angle p (in degrees) of a circle with radius R is
(A) p/180 × 2πR (B) p/180 × πR² (C) p/360 × 2πR (D) p/720 × 2πR²
Answer: (D) p/720 × 2πR² = (p/360) × πR² (the standard formula).
Option (D) simplifies to correct formula.
Option (D) simplifies to correct formula.
Key Concepts Summary
Formula 1Area of Circle
What is the area of a circle of radius r?
Answer: Area = πr².
Formula 2Area of Sector
What is the area of a sector of angle θ (in degrees) of a circle of radius r?
Answer: Area = (θ/360) × πr².
Formula 3Area of Segment
How do you find the area of a minor segment?
Answer: Area of minor segment = Area of sector - Area of corresponding triangle.
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