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Lesson 2 of 2

Areas Related to Circles · Lesson 2 of 2

Chapter Summary and Practice

Practice doesn't always make perfect, but it definitely makes you question all your life choices right before the exam.

Learning Objectives

• Recall the formulas for arc length, sector area and segment area. • Choose the correct formula from the information given. • Distinguish between minor and major regions in mixed problems. • Solve exam-style questions involving sectors, segments and circular applications. • Check units and reasonableness of final answers.

This chapter connects central angles with lengths and areas. Once the fraction θ/360 is recognised, the same idea gives both arc length and sector area, while segment questions add one extra step: subtract the corresponding triangle.

Sector of a Circle

A sector is enclosed by two radii and the corresponding arc. For a central angle θ°, the sector occupies θ/360 of the full circular region.

Area of a sectorLaTeX
Length of an arcLaTeX

Segment of a Circle

A segment is enclosed by a chord and its corresponding arc. The minor segment is obtained by removing triangle OAB from sector OAB.

Area of a minor segmentLaTeX

Major Regions

Area of major sectorLaTeX
Area of major segmentLaTeX

How to Choose the Right Method

Question asks forMain idea
Arc lengthUse θ/360 of the circumference 2πr
Sector areaUse θ/360 of the circle area πr²
Minor segmentSector area − triangle area
Major sectorCircle area − minor sector area
Major segmentCircle area − minor segment area
Area swept by a rotating armConvert the movement into a central angle, then use sector area
Equal divisions of a circleCentral angle = 360° ÷ number of equal parts
Common Mistakes

• Using diameter in place of radius in πr², • Using θ/180 instead of θ/360 for sector area, • Forgetting to subtract the triangle when finding a segment, • Confusing major sector with major segment, • Writing cm instead of cm² for area or cm² instead of cm for arc length,

Guided Practice

Guided Example 1: Sector Area

Problem
Find the area of a 60° sector of radius 6 cm. Take π = 22/7.

  1. 1.A = (θ/360)πr².
  2. 2.A = (60/360) × (22/7) × 6².
  3. 3.A = 132/7 cm² ≈ 18.86 cm².
Guided Example 2: Quadrant from Circumference

Problem
The circumference of a circle is 44 cm. Find the area of one quadrant. Take π = 22/7.

  1. 1.Circumference = 2πr = 44.
  2. 2.2 × (22/7) × r = 44, so r = 7 cm.
  3. 3.A quadrant is a 90° sector.
  4. 4.Area = (90/360) × (22/7) × 7².
  5. 5.Area = 38.5 cm².
Guided Example 3: Arc and Sector

Problem
A radius of 21 cm subtends a 60° sector. Find the arc length and sector area. Take π = 22/7.

  1. 1.Arc length = (60/360) × 2 × (22/7) × 21 = 22 cm.
  2. 2.Sector area = (60/360) × (22/7) × 21² = 231 cm².
Guided Example 4: Minor Segment of a 90° Sector

Problem
A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the minor segment. Take π = 3.14.

  1. 1.Sector area = (90/360) × 3.14 × 10² = 78.5 cm².
  2. 2.The two radii form a right triangle with perpendicular sides 10 cm and 10 cm.
  3. 3.Triangle area = 1/2 × 10 × 10 = 50 cm².
  4. 4.Minor segment area = 78.5 − 50 = 28.5 cm².
Guided Example 5: Area Cleaned by Two Wipers

Problem
Two non-overlapping wipers each have length 25 cm and sweep through 115°. Find the total area cleaned in one sweep. Take π = 3.14.

  1. 1.Each wiper cleans a sector of radius 25 cm and angle 115°.
  2. 2.Area cleaned by one wiper = (115/360) × 3.14 × 25².
  3. 3.Area cleaned by one wiper ≈ 627.26 cm².
  4. 4.The wipers do not overlap, so total area ≈ 2 × 627.26 = 1254.52 cm².

Quiz

Quick check

A 90° sector is what fraction of a circle?

Quick check

Which expression gives the length of an arc subtending θ° in a circle of radius r?

Quick check

To obtain a minor segment area, what must be subtracted from the corresponding sector?

Quick check

If the minor sector angle is 70°, the major sector angle is:

Quick check

Which unit is appropriate for the area of a sector when radius is measured in centimetres?

Before You Finish the Chapter

Check that you can identify sectors and segments, calculate the fraction θ/360 correctly, separate arc-length units from area units, and find the triangle area needed in segment questions.

Practice Problems

Practice Questions
  1. Find the area of a sector of radius 9 cm and central angle 40°.
  2. A quadrant has circumference of the full circle equal to 66 cm. Find the quadrant area.
  3. A minute hand 14 cm long moves for 10 minutes. Find the area swept.
  4. A chord in a circle of radius 8 cm subtends 90° at the centre. Find the minor segment area.
  5. An arc in a circle of radius 14 cm subtends 45° at the centre. Find its length and the area of its sector.
  6. A chord of a circle of radius 15 cm subtends 60° at the centre. Find the corresponding minor and major segment areas.
  7. A chord of a circle of radius 12 cm subtends 120° at the centre. Find the corresponding minor segment area.
  8. A goat is tied at one corner of a square field with a 6 m rope. Find the area it can graze if the rope stays within the field.
  9. The rope in the previous question is increased from 6 m to 9 m. Find the increase in grazing area.
  10. A circular brooch of diameter 28 mm is divided into 8 equal sectors using diameters. Find the area of each sector.
  11. A flat circular umbrella of radius 42 cm has 8 equally spaced ribs. Find the area between two consecutive ribs.
  12. Two non-overlapping wipers, each 24 cm long, sweep an angle of 120°. Find the total area cleaned.
  13. A lighthouse warns ships over a sector of 72° up to a distance of 20 km. Find the area covered by the warning light.
  14. A circular design of radius 21 cm is divided into 6 equal sectors. Find the area of one sector.
  15. Choose the correct expression for the area of a sector of angle p° and radius R: (A) pπR²/180 (B) pπR²/360 (C) pπR/360 (D) 2pπR²/360.

Key Takeaways

Key Takeaways

• Arc length = (θ/360°) × 2πr. • Sector area = (θ/360°) × πr². • Segment area = sector area − corresponding triangle area. • Major regions can be found by subtracting minor regions from the full circle. • Most applications become straightforward once the correct central angle is identified.