Grade 10 core mathematics Measurements and Geometry – Area of a Part of a Circle (12 lessons) Notes
Core Mathematics — 2.0 Measurements and Geometry
Subtopic 2.6: Area of a Part of a Circle (12 lessons) — target age: 15
This unit (12 lessons) develops students' ability to compute and apply areas of parts of circles: annulus (ring), sector, annular sector, segment, and the common region (lens) of two intersecting circles. Emphasis is on understanding, deriving formulas, solving real-life problems and using metric units appropriate to Kenya (mm, cm, m).
- Identify and outline the sub-sub-strands:
- Area of an annulus
- Area of a sector of a circle
- Area of an annular sector
- Area of a segment of a circle
- Area of the common region (lens) between two intersecting circles
- Applications of areas of parts of a circle
- Determine the area of an annulus in different situations.
- Compute the area of a sector in real-life situations.
- Find the area of an annular sector in different situations.
- Work out the area of a segment of a circle (using sector minus triangle).
- Determine the area of the common region between two intersecting circles (lens).
- Apply these areas to practical Kenyan contexts (e.g., ring-shaped garden beds, circular windows, pizza slices, manhole covers).
- Explore uses of these areas in projects and investigations.
- Area of a circle: A = π r²
- Area of annulus (ring) with outer radius R and inner radius r: A = π (R² − r²)
- Sector area with central angle θ (degrees): A = (θ/360) · π r². In radians: A = ½ r² θ.
- Annular sector (between two concentric radii R and r, angle θ): A = (θ/360) · π (R² − r²)
- Segment (central angle θ in degrees): Area(segment) = Area(sector) − Area(triangle)
For an isosceles triangle with sides r,r and included angle θ: triangle area = ½ r² sin θ (θ in radians or convert sin for degrees). - Common area (lens) of two circles radius r and R with centers distance d:
A = r² cos⁻¹((d² + r² − R²)/(2 d r)) + R² cos⁻¹((d² + R² − r²)/(2 d R)) − ½ √{(−d+r+R)(d+r−R)(d−r+R)(d+r+R)}
(Use calculators; derive from sum of two circular segments.) - Use π ≈ 3.142 or 22/7 where appropriate; indicate which approximation is used.
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Lesson 1 — Review of circle basics (1 lesson)
- Objectives: recall radius, diameter, circumference, area of circle, units.
- Activity: measure lids and compute areas; check units (cm²).
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Lesson 2 — Annulus (1 lesson)
- Outline annulus and derive A = π(R² − r²).
- Classwork: ring-shaped flowerbed examples (outer R, inner r).
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Lesson 3 — Sector — definition and formula (1 lesson)
- Derive sector area from fraction of circle: A = (θ/360)πr².
- Group activity: slices of pizza, water tank sector.
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Lesson 4 — Annular sector (1 lesson)
- Combine annulus and sector ideas: A = (θ/360)π(R² − r²).
- Practical: ring-shaped road markings or circular garden beds.
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Lesson 5 — Segment — concept and formula (1 lesson)
- Segment = sector − triangle. Show triangle area = ½ r² sin θ (θ in radians) or derive using ½ab sin C.
- Example: area of a shallow pond cap cut by a straight path.
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Lesson 6 — Segment problems (1 lesson)
- Practice: given r and θ (degrees), compute segment area. Emphasize calculator use for sin and cos and unit practice.
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Lesson 7 — Intersecting circles: sketch and reasoning (1 lesson)
- Show two circles overlap; area of overlap = sum of two circular segments.
- Derive expressions using cos⁻¹ and triangle area (general formula provided).
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Lesson 8 — Intersecting circles: worked examples (1 lesson)
- Examples with equal radii (simpler) and unequal radii; use calculators; check special cases (no overlap, one inside another).
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Lesson 9 — Applications: real life contexts (1 lesson)
- Problems: cover for a circular tank with co-axial hole (annulus), pizza slices, circular beams cutouts, traffic roundabout markings.
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Lesson 10 — Mixed problem solving (1 lesson)
- Higher-order problems combining several parts (e.g., annular sector cut by chord forming segments).
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Lesson 11 — Written test / summative assessment (1 lesson)
- Test covering all sub-strands; include practical word problems and an investigative question.
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Lesson 12 — Project / consolidation and exploration (1 lesson)
- Group project: measure real objects (manhole cover, ring-shaped garden), model, compute areas, present findings and usage.
Outer radius R = 10 cm, inner radius r = 6 cm. Area = π(R² − r²) = π(100 − 36) = 64π cm² ≈ 64 × 3.142 = 201.1 cm².
Radius r = 8 m, central angle θ = 45°. Area = (45/360)π(8²) = (1/8)π(64) = 8π m² ≈ 25.13 m².
r = 10 cm, central angle θ = 60°.
Sector area = (60/360)π(100) = (1/6)π(100) = (50/3)π ≈ 52.36 cm².
Triangle area (isosceles) = ½ r² sin θ = ½ ×100 × sin 60° = 50 × (√3/2) ≈ 50 × 0.8660 = 43.30 cm².
Segment area ≈ 52.36 − 43.30 = 9.06 cm².
Two circles radius r = 5 cm, centres d = 6 cm apart. Use standard formula or sum two equal segments.
Compute cos⁻¹((d/2r)) etc. (Teacher to demonstrate calculator steps). Use formula given in summary.
- Hands-on measuring: bring lids, tyres, plates, manhole covers; measure radii and compute areas of circle and annulus.
- Group experiment: mark a circular garden bed with inner hole (annulus) and calculate area of soil to be planted — connect to agriculture class.
- Real-life problem: design a circular window with a decorative ring (annular sector) — compute glass area and frame area for budgeting.
- Use simple geometry software (GeoGebra) or a scientific calculator for segments and lens areas; show step-by-step calculator key presses for inverse cosine and square roots.
- Set short projects: measure local infrastructure items (manhole cover, water tank) and prepare a one-page report including diagrams and area calculations.
- Encourage use of units: convert mm² ↔ cm² ↔ m² where necessary; discuss appropriate unit for area in each problem.
- Find the area of an annulus with outer diameter 30 cm and inner diameter 18 cm. (Give exact and approximate answers.)
- A circular pizza radius 12 cm is cut into a sector of angle 120°. Find the area of this slice.
- An annular sector has inner radius 3 m, outer radius 6 m and central angle 90°. Find its area.
- Find the area of the segment of a circle with radius 7 cm cut off by a chord that subtends an angle of 90° at the centre.
- Two circles of radii 8 cm and 6 cm have centres 10 cm apart. Find the area of their overlap (use calculator and show steps).
Answers (brief): 1) A = π(15²−9²)=π(225−81)=144π cm² ≈ 452.39 cm². 2) (120/360)π(12²)= (1/3)π144 = 48π ≈150.8 cm². 3) (90/360)π(36−9)= (1/4)π27 = 6.75π ≈21.21 m². 4) Sector area=(90/360)π49= (1/4)π49=12.25π≈38.48 cm²; triangle area=½·49·sin90°=24.5; segment ≈13.98 cm². 5) Use formula — calculator required; teacher to provide worked solution in class.
- Formative: quick quizzes on deriving formulas and simple calculations; observe measuring activities.
- Summative (lesson 11): paper with 6–8 questions including at least one practical/contextual problem and one lens/segment problem. Allow calculators for inverse trig and square roots.
- Marking: check units, showing working (derivation of sector/segment), correct use of π and accuracy of approximations.
- Rulers, compasses, protractors, scientific calculators.
- GeoGebra (optional) for dynamic exploration.
- Examples from local context: plates, tyres, water tank lids, circular garden plots.
- Reference: KCSE-style past paper questions on circle parts (teachers to collect relevant past questions).
- Start with visual and physical activities to make abstract areas meaningful.
- Emphasise reasoning: e.g., segment = sector − triangle rather than memorisation only.
- Use simple angles (30°, 60°, 90°, 180°) for initial practice then move to arbitrary angles requiring calculators.
- Relate to other topics (trigonometry, Pythagoras) when finding chord lengths or triangle areas inside circles.