Solutions — Applications of Circle Geometry

1. Arc length. Fluency

   • (a) r=6, θ=60°: (60/360)×2π(6) = ⅙×12π = 2π ≈ 6.28 cm.
   • (b) r=10, θ=90°: ¼×2π(10) = 5π ≈ 15.71 m.
   • (c) r=5, θ=216°: (216/360)×2π(5) = 0.6×10π = 6π ≈ 18.85 cm.
   • (d) Arc=12π, r=9. Find θ: (θ/360)×2π(9) = 12π ⇒ 18πθ/360 = 12π ⇒ θ = 12×360/18 = 240°.

2. Sector area. Fluency

   • (a) r=8, θ=45°: (45/360)×π(64) = ⅛×64π = 8π ≈ 25.13 cm².
   • (b) r=12, θ=120°: ⅓×π(144) = 48π ≈ 150.80 m².
   • (c) r=5, θ=300°: (300/360)×π(25) = ⅚×25π = (125/6)π ≈ 65.45 cm².
   • (d) Area=30π, θ=120°. Find r: ⅓πr² = 30π ⇒ r² = 90 ⇒ r = 3√10 ≈ 9.49 cm.

3. Segment area. Fluency

   • (a) r=10, θ=90°: Sector = ¼π(100) = 25π. Triangle = ½(10)(10) = 50. Segment = 25π−50 ≈ 28.54 cm².
   • (b) r=6, θ=60° (equilateral triangle): Sector = ⅙π(36) = 6π. Triangle (equilateral, side=6): ½(6)²sin60° = 18×(√3/2) = 9√3. Segment = 6π−9√3 ≈ 18.85−15.59 = 3.26 cm².
   • (c) r=8, θ=120°: Sector = ⅓π(64) = (64/3)π. Triangle = ½(8)²sin120° = 32×(√3/2) = 16√3. Segment = (64/3)π−16√3 ≈ 67.02−27.71 = 39.31 cm².
   • (d) r=5, θ=144°. Major segment: Minor sector = (144/360)π(25) = 10π. Triangle = ½(25)sin144° = 12.5×0.5878 ≈ 7.35. Minor segment = 10π−7.35 ≈ 24.07. Major segment = π(25)−24.07 = 78.54−24.07 ≈ 54.47 cm².

4. Perimeter of sectors and segments. Fluency

   • (a) Sector r=7, θ=80°: Arc = (80/360)×14π = (28/9)π ≈ 9.77 cm. Perimeter = 9.77+7+7 ≈ 23.77 cm.
   • (b) Sector r=12, θ=150°: Arc = (150/360)×24π = 10π ≈ 31.42 m. Perimeter = 31.42+12+12 ≈ 55.42 m.
   • (c) Segment r=9, θ=120°: Arc = (120/360)×18π = 6π ≈ 18.85 cm. Chord = 2×9×sin60° = 18×(√3/2) = 9√3 ≈ 15.59 cm. Perimeter = 18.85+15.59 ≈ 34.44 cm.
   • (d) Semicircle d=20 (r=10): Perimeter = πr + 2r = 10π+20 ≈ 51.42 cm. Area = ½πr² = 50π ≈ 157.08 cm².

5. Sector and segment from a diagram. Understanding

   • (a) Arc AB (r=12, θ=60°): (60/360)×2π(12) = ⅙×24π = 4π ≈ 12.57 cm.
   • (b) Chord AB (θ=60°, r=12): Triangle OAB has OA=OB=12 and angle 60°, so it is equilateral. AB = 12 cm.
   • (c) Sector area: (60/360)×π(144) = ⅙×144π = 24π ≈ 75.40 cm².
   • (d) Segment area: Triangle OAB (equilateral, side 12): area = (√3/4)×144 = 36√3 ≈ 62.35 cm². Segment = 24π−36√3 ≈ 75.40−62.35 = 13.05 cm².

6. Inscribed and circumscribed circles. Understanding

   • (a) Equilateral triangle, side 6. Circumradius: R = a/√3 = 6/√3 = 2√3 ≈ 3.46 cm. (Or R = a/(√3) more precisely R = a√3/3.)
   • (b) Right triangle legs 6, 8 (hyp=10). Circumradius: For a right triangle, the circumradius = half the hypotenuse = 10/2 = 5 cm.
   • (c) Right triangle legs 5, 12 (hyp=13). Inradius: r = (5+12−13)/2 = 4/2 = 2 cm.
   • (d) Square side 10 inscribed in circle: Diagonal = 10√2. Circumradius = half diagonal = 5√2 ≈ 7.07 cm.

7. Mixed circle applications. Understanding

   • (a) Wiper r=45, θ=110°: A = (110/360)×π(2025) = (11/36)×2025π = 618.75π ≈ 1943.84 cm².
   • (b) Pizza r=15, 8 slices (θ=45° each): Arc = (45/360)×2π(15) = ⅛×30π = (15/4)π ≈ 11.78 cm.
   • (c) Semicircular lane r_in=36, r_out=37.22: Area = ½π(37.22²−36²) = ½π(1385.3−1296) = ½π(89.3) = 44.65π ≈ 140.25 m².
   • (d) Clock hand r=8, θ=90°: Tip distance = arc = ¼×2π(8) = 4π ≈ 12.57 cm. Area swept = ¼π(64) = 16π ≈ 50.27 cm².

8. Circles and triangles. Understanding

   • (a) r=5, central 120°. Chord: chord = 2r×sin(θ/2) = 2(5)sin60° = 10×(√3/2) = 5√3 ≈ 8.66 cm.
   • (b) Chord=8, central 90°. Radius: 8 = 2r×sin45° = 2r×(√2/2) = r√2 ⇒ r = 8/√2 = 4√2 ≈ 5.66 cm.
   • (c) Isosceles inscribed, R=10, apex 40°: The inscribed angle at apex = 40°. By Theorem 1, central angle = 80°. Base chord = 2R×sin(40°) = 20×0.6428 ≈ 12.86 cm.
   • (d) Chord=10, inscribed angle 30°: By the extended sine rule: chord/sin(θ) = 2R ⇒ 10/sin30° = 2R ⇒ 2R = 10/0.5 = 20 ⇒ R = 10 cm.

9. Arch design. Problem Solving

   • (a) Find r (sagitta = r − √(r²−1) = 0.5): r − √(r²−1) = 0.5 ⇒ √(r²−1) = r−0.5 ⇒ r²−1 = r²−r+0.25 ⇒ −1 = −r+0.25 ⇒ r = 1.25 m.
   • (b) Central angle θ: cos(θ/2) = (r−sagitta)/r = (1.25−0.5)/1.25 = 0.6 ⇒ θ/2 = cos²¹(0.6) ≈ 53.13° ⇒ θ ≈ 106.26°.
   • (c) Arc length: l = (106.26/360)×2π(1.25) ≈ 0.2952×7.854 ≈ 2.32 m.
   • (d) Segment area: Sector area = (106.26/360)×π(1.25)² ≈ 0.2952×4.909 ≈ 1.449 m². Triangle area = ½(1.25)²×sin106.26° ≈ ½(1.5625)(0.96) ≈ 0.75 m². Segment = 1.449−0.75 ≈ 0.70 m².

10. Gear wheels. Problem Solving

    • (a) Verify gears mesh and theorem: r_A + r_B = 8+6 = 14 cm = distance between centres. So the circles are externally tangent ✓. The radius-perpendicular-to-tangent theorem guarantees the common tangent at the contact point is perpendicular to the line of centres (since each radius to the contact point is perpendicular to the tangent, and both radii point in the same direction along the line of centres).
    • (b) Arc on Gear A (θ=45°, r=8): Arc = (45/360)×2π(8) = ⅛×16π = 2π ≈ 6.28 cm.
    • (c) Gear B rotation: Arc on B = 2π. (φ/360)×2π(6) = 2π ⇒ 12πφ/360 = 2π ⇒ φ = 360×2/12 = 60°.
    • (d) Gear ratio: In 1 rotation of A, arc = 2π(8) = 16π. B rotations = 16π/(2π×6) = 16/12 = 4/3 ≈ 1.33 rotations/sec. The ratio 4/3 is called the gear ratio (or transmission ratio). Since B is smaller, it spins faster.