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IB Mathematics: Applications and Interpretation SL · 12.10: Geometry & Trigonometry · 12.10.10

Arc length and sector area Problem set A One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. A sector has radius 7.2 cm and angle 115° at the centre. Find the length of its arc.

    115°7.2 cmOAB
    • \(s=\frac{115}{360}\times2\pi\times7.2=14.45\)
    • 14.5 cm
  2. A sector has radius 11 cm and angle 72° at the centre. Find its area.

    72°11 cmOAB
    • \(A=\frac{72}{360}\times\pi\times11^2=76.03\)
    • 76.0 cm²
  3. An arc of length 15 cm subtends an angle of 70° at the centre of a circle. Find the radius of the circle.

    70°r15 cmOAB
    • \(15=\frac{70}{360}\times2\pi r\)
    • \(r=\frac{15\times360}{70\times2\pi}=12.28\)
    • r = 12.3 cm
  4. A sector of radius 7 cm has area 40 cm². Find the angle at the centre.

    θ7 cmOAB
    • \(40=\frac{\theta}{360}\times\pi\times7^2\)
    • \(\theta=\frac{40\times360}{\pi\times7^2}=93.54\)
    • θ = 93.5°

In context

  1. A designer cuts a segment from a circular metal disc of radius 9 cm. The straight edge AB subtends an angle of 75° at the centre O. Find

    75°9 cmOAB
    1. (a) the perimeter of the segment

      • \(\text{arc}=\frac{75}{360}\times2\pi\times9=11.78\)
      • \(AB^2=9^2+9^2-2\times9\times9\cos75^\circ,\; AB=10.96\)
      • \(\text{perimeter}=11.78+10.96=22.7\) cm
    2. (b) the area of the segment

      • \(\text{sector}=\frac{75}{360}\times\pi\times9^2=53.01\)
      • \(\text{triangle}=\frac{1}{2}\times9^2\sin75^\circ=39.12\)
      • \(\text{segment}=53.01-39.12=13.9\) cm²
  2. A lawn sprinkler sprays water up to 6.5 m and sweeps through 130°. Find

    130°6.5 mOAB
    1. (a) the area of the region

      • \(A=\frac{130}{360}\times\pi\times6.5^2=47.93\) m²
    2. (b) the total length of its edge (the arc and the two straight sides)

      • \(s=\frac{130}{360}\times2\pi\times6.5=14.75\)
      • \(\text{edge}=s+2r=14.75+13=27.7\) m
  3. A horse is tied by an 8 m rope to the outside corner of a barn that is 11 m by 14 m. The horse can graze only outside the barn. Find the area it can reach.

    270°8 mX
    • The shed fills a right angle at the corner, so the rope swings through 360° − 90° = 270°.
    • \(A=\frac{270}{360}\times\pi\times8^2=\frac{3}{4}\pi\times64=150.8\)
    • 151 m² (8 m is shorter than both walls, so the rope never wraps round the shed)

Stretch

  1. A circular pond has radius 10 m. A straight path AB of length 14 m cuts off a segment of the pond. Find

    θ10 mOAB
    1. (a) the angle AOB at the centre

      • \(\sin\left(\frac{\theta}{2}\right)=\frac{7}{10}=0.7\)
      • \(\frac{\theta}{2}=44.43^\circ,\; \theta=88.9^\circ\)
    2. (b) the area of the segment

      • \(\text{sector}=\frac{88.85}{360}\times\pi\times10^2=77.54\)
      • \(\text{triangle}=\frac{1}{2}\times10^2\times\sin88.85^\circ=49.99\)
      • \(\text{segment}=27.5\) m²

jmaths.xyz · AI SL · 12.10.10 Arc length and sector area