Class 10 Mathematics
Chapter 11: Areas Related to Circles
Exercise 11.1 – Stepwise Solutions
Note: Unless mentioned otherwise, take π = 22/7.
Q1) Area of a sector (r = 6 cm, angle = 60°)
Formula: Area of sector = (θ/360) × πr²
Area = (60/360) × π × 6²
= (1/6) × π × 36
= 6π
= 6 × 22/7 = 132/7 cm²
Answer: 132/7 cm² ≈ 18.86 cm²
Q2) Area of a quadrant when circumference = 22 cm
Step 1: Find radius using circumference 2πr = 22
r = 22/(2π) = 11/π
Using π = 22/7:
r = 11 ÷ (22/7) = 11 × 7/22 = 3.5 cm
Step 2: Quadrant area = (1/4)πr²
Area = (1/4) × (22/7) × (3.5)²
= (1/4) × (22/7) × 12.25
= 9.625 cm²
Answer: 9.625 cm²
Q3) Area swept by minute hand (length 14 cm) in 5 minutes
Step 1: Angle in 5 minutes = (5/60) × 360° = 30°
Step 2: Sector area = (30/360) × π × 14²
= (1/12) × π × 196
= 49π/3
= (49/3) × (22/7) = 154/3 cm²
Answer: 154/3 cm² ≈ 51.33 cm²
Q4) Circle radius 10 cm, chord subtends 90° at centre (Use π = 3.14)
(i) Area of minor segment
Step 1: Area of 90° sector = (90/360) × π × 10²
= (1/4) × 3.14 × 100 = 78.5 cm²
Step 2: Area of triangle (two radii with included angle 90°)
Area = (1/2)r²sin90° = (1/2) × 100 × 1 = 50 cm²
Minor segment = 78.5 − 50 = 28.5 cm²
(ii) Area of major sector (270°)
Area = (270/360) × 3.14 × 100
= (3/4) × 314 = 235.5 cm²
Q5) Circle radius 21 cm, arc angle 60°
(i) Length of the arc
Formula: Arc length = (θ/360) × 2πr
= (60/360) × 2π × 21
= (1/6) × 42π = 7π
= 7 × 22/7 = 22 cm
(ii) Area of the sector
Area = (60/360) × π × 21²
= (1/6) × π × 441
= 441π/6
= 441 × (22/7) ÷ 6 = 1386 ÷ 6 = 231 cm²
(iii) Area of the segment (minor)
Step 1: Triangle formed has sides 21, 21 with included angle 60° ⇒ equilateral triangle of side 21.
Area of triangle = (√3/4) × 21² = (√3/4) × 441 = 110.25√3
Step 2: Segment area = Sector area − Triangle area
= 231 − 110.25√3 cm²
Answer: 231 − 110.25√3 cm² (≈ 40.1 cm² if √3 ≈ 1.732)
Q6) Circle radius 15 cm, chord angle 60° (Use π = 3.14, √3 = 1.73)
Step 1: Area of 60° sector = (60/360) × 3.14 × 15²
= (1/6) × 3.14 × 225
= (1/6) × 706.5 = 117.75 cm²
Step 2: Area of triangle = (1/2)r²sin60°
sin60° = √3/2 = 1.73/2 = 0.865
Triangle area = (1/2) × 225 × 0.865
= 112.5 × 0.865 = 97.3125 cm²
Minor segment = 117.75 − 97.3125 = 20.4375 cm² ≈ 20.44 cm²
Step 3: Area of circle = 3.14 × 225 = 706.5 cm²
Major segment = 706.5 − 20.4375 = 686.0625 cm² ≈ 686.06 cm²
Q7) Circle radius 12 cm, chord angle 120° (Use π = 3.14, √3 = 1.73)
Step 1: Area of 120° sector = (120/360) × 3.14 × 12²
= (1/3) × 3.14 × 144
= (1/3) × 452.16 = 150.72 cm²
Step 2: Area of triangle = (1/2)r²sin120°
sin120° = sin60° = √3/2 = 1.73/2 = 0.865
Triangle area = (1/2) × 144 × 0.865
= 72 × 0.865 = 62.28 cm²
Segment area = 150.72 − 62.28 = 88.44 cm²
Q8) Horse tied at corner of square field (side 15 m) with rope 5 m
(i) Grazing area = area of quarter circle of radius 5 m
Area = (1/4) × π × 5²
Using π = 3.14:
= (1/4) × 3.14 × 25 = 19.625 m²
Answer: 19.625 m²
(ii) Increase if rope is 10 m (Use π = 3.14)
New area = (1/4) × 3.14 × 10²
= (1/4) × 3.14 × 100 = 78.5 m²
Increase = 78.5 − 19.625 = 58.875 m²
Q9) Brooch: circle wire diameter 35 mm; plus 5 diameters (10 equal sectors)
(i) Total wire length
Circumference = πd = (22/7) × 35 = 110 mm
5 diameters = 5 × 35 = 175 mm
Total length = 110 + 175 = 285 mm
(ii) Area of each sector
Radius r = 35/2 = 17.5 mm
Area of circle = πr² = (22/7) × (17.5)²
(17.5)² = 306.25
Area = (22/7) × 306.25 = 962.5 mm²
Each of 10 equal sectors ⇒ area = 962.5/10 = 96.25 mm²
Q10) Umbrella: 8 equal ribs, radius = 45 cm
Angle between two consecutive ribs = 360°/8 = 45°
Area between two ribs = area of 45° sector
= (45/360) × π × 45²
= (1/8) × π × 2025 = 2025π/8
= (2025/8) × (22/7) = 44550/56 cm²
Answer: 44550/56 cm² ≈ 795.54 cm²
Q11) Two wipers (no overlap), each length 25 cm, sweep 115°
Area cleaned by 1 wiper = (115/360) × π × 25²
= (115/360) × π × 625
Total for 2 wipers:
= 2 × (115/360) × π × 625
= (230/360) × π × 625 = (23/36) × π × 625
= (14375π)/36
= (14375/36) × (22/7) = 316250/252 cm²
Answer: 316250/252 cm² ≈ 1254.96 cm²
Q12) Lighthouse warning sector: angle 80°, radius 16.5 km (Use π = 3.14)
Area = (80/360) × 3.14 × (16.5)²
(16.5)² = 272.25
Area = (2/9) × 3.14 × 272.25
3.14 × 272.25 = 854.865
Area = (2/9) × 854.865 = 1709.73/9
Answer: ≈ 189.97 km²
Q13) Round table cover with 6 designs, radius = 28 cm (Use √3 = 1.7)
Idea: Designs area = Area of circle − Area of inscribed regular hexagon
Step 1: Area of circle = πr²
= (22/7) × 28² = (22/7) × 784
= 22 × 112 = 2464 cm²
Step 2: For a regular hexagon inscribed in a circle, side a = radius = 28 cm
Area of regular hexagon = (3√3/2) a²
= (3 × 1.7 / 2) × 28²
= (5.1/2) × 784 = 2.55 × 784
= 1999.2 cm²
Step 3: Designs area = 2464 − 1999.2 = 464.8 cm²
Step 4: Cost at ₹0.35 per cm²
Cost = 0.35 × 464.8 = ₹162.68
Q14) MCQ: Correct formula for area of a sector (angle p°, radius R)
Known formula: Area of sector = (p/360) × πR²
Option (C): (p/720) × 2πR² = (p × 2πR²)/720 = pπR²/360
So option (C) matches.
Final Answers (Quick List)
- Q1: 132/7 cm² ≈ 18.86 cm²
- Q2: 9.625 cm²
- Q3: 154/3 cm² ≈ 51.33 cm²
- Q4: Minor segment = 28.5 cm²; Major sector = 235.5 cm²
- Q5: Arc = 22 cm; Sector area = 231 cm²; Segment = 231 − 110.25√3 cm²
- Q6: Minor segment ≈ 20.44 cm²; Major segment ≈ 686.06 cm²
- Q7: Segment ≈ 88.44 cm²
- Q8: 19.625 m²; Increase = 58.875 m²
- Q9: Wire = 285 mm; Each sector = 96.25 mm²
- Q10: ≈ 795.54 cm²
- Q11: ≈ 1254.96 cm²
- Q12: ≈ 189.97 km²
- Q13: Designs cost ≈ ₹162.68
- Q14: Option (C)