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Lesson 3 of 10

Measuring Space: Perimeter and Area · Lesson 3 of 10

Length of an Arc and Perimeter Puzzles

Even a slice of a circle has a boundary bill to pay.

Learning Objectives

• Derive arc length from circumference. • Understand semicircle and quarter-circle arc lengths. • Use central angle to calculate general arc length. • Apply arc length to athletics tracks. • Solve perimeter puzzles built from circular arcs.

The circumference of a circle is the total distance around its boundary, and for a circle of radius r, it is 2πr. An arc is only a part of this boundary, so its length must be some fraction of the full circumference.

That fraction is determined by the angle at the centre of the circle. A complete circle represents a turn of 360°. So, if an arc is formed by a central angle of θ°, then it represents θ/360 of the whole circle. This means the arc length is found by taking the same fraction of the total circumference. A larger central angle gives a longer arc, while a smaller central angle gives a shorter arc.

Semicircle and Quarter Circle

Semicircular arcLaTeX
Quarter-circle arcLaTeX

A semicircle corresponds to 180° out of 360°, or one-half of the circle. A quarter circle corresponds to 90° out of 360°, or one-fourth.

Arc fractions of a circle A full circle represents a 360 degree central angle, a semicircle represents 180 degrees, and a quarter circle represents 90 degrees. Full circle 360° O 360° = 1 whole circle Semicircle 180° O 180° = 1/2 circle Quarter circle 90° O 90° = 1/4 circle Arc fraction = central angle ÷ 360°
Arc fractions of a circle

General Arc-Length Formula

If an arc subtends θ° at the centre, it represents θ/360 of the full circumference.

Arc lengthLaTeX
Worked Example: 60° Arc

Problem
Find the length of a 60° arc in a circle of radius 3.5 cm using π≈22/7.

  1. 1.Full circumference=2×22/7×3.5=22 cm.
  2. 2.A 60° arc is 60/360=1/6 of the circle.
  3. 3.Arc length=22/6=11/3 cm.

The 400 m Track

A standard track can be modelled as two straight sections and two semicircular sections. The two semicircles together behave like one full circle. The outer lane has a larger effective radius, so it has a larger curved distance even though the straight sections are unchanged.

Why the Stagger Appears

Problem
Two adjacent lanes differ in radius by w. How much extra curved distance does the outer runner cover over two semicircles?

  1. 1.Two semicircles form one complete circle.
  2. 2.Inner curved distance=2πr.
  3. 3.Outer curved distance=2π(r+w).
  4. 4.Difference=2π(r+w)−2πr=2πw.
  5. 5.So the extra curved distance depends only on lane width w, not on the original radius.

Perimeter Puzzle: Two Overlapping Circles

If two equal circles of radius r each pass through the other's centre, the triangle formed by the two centres and either intersection point is equilateral. That gives central angles of 60° and lets us determine which arcs form the outer boundary.

Two congruent intersecting circles Two congruent circles with centres O1 and O2. The distance between their centres equals the radius, so each centre lies on the other circle. The circles intersect at C and D, and the corresponding central angles are 120 degrees. 120° 120° O₁ O₂ C D 120° arc CD of left circle 120° arc CD of right circle O₁O₂ = r, so each centre lies on the other circle
Two congruent intersecting circles
Perimeter Puzzle: Two Overlapping Circles

Problem
Two equal circles, each of radius r, pass through each other's centre. Find the perimeter of the outer boundary of the combined figure.

  1. 1.Let the centres of the two circles be A and B, and let their intersection points be C and D.
  2. 2.Because each circle passes through the other's centre, AB = r. Also, AC = BC = r because C lies on both circles.
  3. 3.Therefore triangle ABC is equilateral, so ∠CAB = ∠CBA = 60°.
  4. 4.Similarly, triangle ABD is equilateral, so ∠DAB = ∠DBA = 60°.
  5. 5.At centre A, the angle between AC and AD is therefore 60° + 60° = 120°. The same is true at centre B.
  6. 6.So the smaller arc CD of each circle measures 120°.
  7. 7.These 120° arcs lie inside the overlapping region and are not part of the outside perimeter.
  8. 8.Therefore the part of each circle that forms the outer boundary is the remaining 360° − 120° = 240° arc.
  9. 9.Length of a 240° arc = (240/360) × 2πr = 4πr/3.
  10. 10.There are two identical 240° outer arcs, one from each circle.
  11. 11.Hence total perimeter = 2 × 4πr/3 = 8πr/3.
Outer perimeterLaTeX

Perimeter Puzzle: Many Semicircles, Same Path Length

If one path from P to Q is a single semicircle and another path is made of several smaller semicircles whose diameters exactly partition PQ, both paths can have the same total length. The reason is that semicircle arc length is π times the radius, and the smaller radii add to the large radius.

Practice Problems

Practice Problems
  1. Find the length of a 120° arc in a circle of radius 6.3 m.
  2. Find the perimeter of a sector of radius 14 cm and angle 75°.
  3. A car tyre has diameter 56 cm. Find the distance travelled in one revolution.
  4. A tyre of diameter 56 cm rolls 10 km. Estimate the number of revolutions.
  5. Two adjacent track lanes differ in radius by 1.22 m. Find the extra curved distance in one lap.
  6. Explain algebraically why one large semicircle can have the same arc length as several smaller semicircles placed along the same diameter.

Key Takeaways

Key Takeaways

• Arc length is the same fraction of circumference as its central angle is of 360°. • A semicircular arc has length πr. • A quarter-circle arc has length πr/2. • Track stagger comes from larger curved radius. • Circular perimeter puzzles often simplify through central angles and additivity of radii.

Coming Next

Next, we move from boundary length to area and derive rectangle and parallelogram formulas.