Topic Review — Circles — Solutions

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1. Parts of a circle and circumference

   1. Parts: (i) radius   (ii) diameter   (iii) arc
   2. Circumference, r = 7 cm: C = 2πr = 14π cm ≈ 43.96 cm
   3. Circumference, d = 15 m: C = πd = 15π ≈ 47.12 m
   4. Circumference 62.8 cm — radius: r = C ÷ (2π) = 62.8 ÷ 6.28 = 10 cm
   5. 5 laps of d = 80 m track: One lap = π × 80 ≈ 251.2 m; 5 laps ≈ 1 257 m
   6. Perimeter of semicircle d = 12 cm: Arc = πr = 6π ≈ 18.85 cm; Perimeter = 18.85 + 12 ≈ 30.85 cm

2. Area of a circle

   1. Area, r = 5 cm: A = πr² = 25π cm² ≈ 78.50 cm²
   2. Area, d = 18 m (r = 9 m): A = π × 81 ≈ 254.47 m²
   3. Area 154 cm² — radius (use π = 22/7): r² = 154 × 7/22 = 49; r = 7 cm
   4. Area, r = 3.5 m: A = π × 12.25 ≈ 38.48 m²
   5. Area of semicircle, d = 20 cm (r = 10 cm): ½ × π × 100 = 50π ≈ 157 cm²
   6. Area 200π mm² — radius: r² = 200; r = 10√2 mm

3. Arc length and sector area

   1. Arc length r = 9 cm, θ = 120°: l = (120/360) × 2π × 9 = (1/3) × 18π = 6π cm
   2. Arc length r = 6 m, θ = 270°: l = (270/360) × 2π × 6 = (3/4) × 12π = 9π ≈ 28.27 m
   3. Sector area r = 10 cm, θ = 72°: A = (72/360) × π × 100 = (1/5) × 100π = 20π cm²
   4. Sector area r = 8 m, θ = 45°: A = (45/360) × π × 64 = (1/8) × 64π = 8π ≈ 25.13 m²
   5. Arc = 10π cm, r = 15 cm — angle: l = (θ/360) × 2π × 15; 10π = (θ/360) × 30π; θ = 120°
   6. Area = 24π cm², r = 12 cm — angle: 24π = (θ/360) × π × 144; θ/360 = 24/144 = 1/6; θ = 60°

4. Mixed circle calculations

   1. Quadrant, r = 14 cm:
      1. Arc length = (90/360) × 2π × 14 = 7π ≈ 21.99 cm
      2. Perimeter = 7π + 14 + 14 ≈ 49.99 cm
      3. Area = (1/4) × π × 196 = 49π ≈ 153.94 cm²
   2. Circle in 10 cm square: Square area = 100 cm²; Circle area = π × 25 = 25π ≈ 78.54 cm²; Uncovered area ≈ 21.46 cm²
   3. Ring between r = 5 cm and r = 8 cm: Area = π × 64 − π × 25 = 39π ≈ 122.52 cm²
   4. Area equals circumference numerically: πr² = 2πr; r = 2 cm

5. Circle problem solving

   1. Pizza d = 32 cm, 8 equal slices (θ = 45°, r = 16 cm):
      1. Arc = (45/360) × 2π × 16 = 4π ≈ 12.57 cm
      2. Area = (45/360) × π × 256 = 32π ≈ 100.53 cm²
   2. Sprinklers: Sprinkler 1: π × 36 = 36π m²; Sprinkler 2: π × 16 = 16π m²; Total = 52π m²
   3. Bike wheel d = 70 cm for 1 km: Circumference = π × 70 ≈ 219.8 cm = 2.198 m; Rotations = 1000 ÷ 2.198 ≈ 455 complete rotations
   4. Sector r = 12 cm, θ = 150° vs “circle divided by 2.5”: Actual sector area = (150/360) × π × 144 = 60π cm². Circle area = 144π; 144π ÷ 2.5 = 57.6π cm². These are NOT equal. The student is incorrect; 150/360 = 5/12, not 1/2.5 = 2/5.