Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Interactive Worksheets
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Chapter Wise Worksheets
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Solved Problems and solutions
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Chapter 11: AREAS RELATED TO CIRCLES
Areas Related to Circles
Sector · Segment · Arc Length · Real-World Applications
| ◦ Sector Area | ▸ Segment Area | ∿ Arc Length | ☷ Applications |
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01
Key Definitions
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02
Core Formulae
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10+
Worked Examples
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A–D
Practice Sets
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8pt
Exam Quick-Check
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Introduction |
A circle encloses a region, and within that circular region we can identify several special sub-regions based on how we divide the circle — using radii, chords, or arcs. This chapter develops precise formulae for measuring the areas of these sub-regions and the lengths of the arcs that bound them.
These ideas have rich real-world uses: calculating the area swept by a clock hand, the grazing area available to a tethered animal, the region lit by a lighthouse beam, the surface cleaned by a windscreen wiper, and the material needed for decorative designs cut in circular shapes.
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Key Definitions with Visual Diagrams |
| Def 1 — Sector
A sector is the portion of a circular region enclosed between two radii and the corresponding arc. The angle formed at the centre between the two radii is called the angle of the sector (θ). Minor sector: the smaller region (angle θ). Major sector: the larger region (angle 360° − θ). |
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| Def 2 — Segment
A segment is the portion of a circular region enclosed between a chord and the corresponding arc. It does not include the centre. Minor segment: smaller region (lies on the side of the chord away from the centre). Major segment: larger region (contains the centre). Key relation: Area of segment = Area of sector − Area of triangle OAB |
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| Term | Definition | Angle / Condition |
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| Minor Sector | Smaller sector between two radii and minor arc | Angle = θ (where θ < 180°) |
| Major Sector | Larger sector — remainder of circle after minor sector | Angle = 360° − θ |
| Minor Segment | Region between chord and minor arc (smaller part) | = Minor Sector − △OAB |
| Major Segment | Region between chord and major arc (larger part) | = πr² − Minor Segment |
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Core Formulae — Memorise These |
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① Arc Length of a Sector
Arc Length = (θ / 360) × 2πr
Where r = radius of circle, θ = central angle in degrees. The arc is the fraction θ/360 of the full circumference 2πr. For the full circle (θ = 360°), arc length = 2πr as expected. |
② Area of a Sector
Sector Area = (θ / 360) × πr²
The sector is the fraction θ/360 of the full circle area πr². For θ = 90° (quadrant), area = πr²/4. For θ = 180° (semicircle), area = πr²/2. |
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③ Area of a Segment
Segment Area = Sector Area − Area of △OAB
= (θ/360) × πr² − (1/2) × r² × sin θ. The triangle OAB has two sides equal to r (radii) and included angle θ, so its area = ½r² sin θ. |
④ Major Sector and Major Segment
Major Sector = πr² − Minor Sector
Major Segment = πr² − Minor Segment For the major sector, use angle (360° − θ) in the formula directly: Area = ((360°−θ)/360) × πr². |
💡 Derivation — Why (θ/360)?
A full circle (360°) has area πr² and circumference 2πr. By the Unitary Method: area for 1° = πr²/360. For θ degrees: area = (θ/360) × πr². Similarly, arc length for 1° = 2πr/360, so for θ degrees: arc = (θ/360) × 2πr. This is the Unitary Method approach — the most natural way to derive both formulae.
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Worked Examples |
▶ Example 1 — Area of Sector with r = 4 cm, θ = 30°; and its Major Sector (π = 3.14)
Given:
A circle with radius r = 4 cm and sector angle θ = 30°. Find the area of the minor sector and the area of the corresponding major sector.
Solution:
Minor sector area = (θ/360) × πr²
= (30/360) × 3.14 × 4 × 4
= (1/12) × 3.14 × 16 = 50.24/12 ≈ 4.19 cm²
Major sector area = πr² − minor sector = (3.14 × 16) − 4.19 = 50.24 − 4.19 ≈ 46.05 cm²
Alternative: Major sector = ((360−30)/360) × 3.14 × 16 = (330/360) × 50.24 ≈ 46.05 cm²
▶ Example 2 — Area of Segment AYB: r = 21 cm, ∠AOB = 120° (π = 22/7)
Given:
Circle of radius 21 cm. Chord AB subtends 120° at centre O. Find the area of the minor segment AYB.
Solution:
Step 1 — Sector area: = (120/360) × (22/7) × 21 × 21 = (1/3) × (22/7) × 441 = (1/3) × 1386 = 462 cm²
Step 2 — Area of △OAB: Draw OM ⊥ AB. OA = OB = 21 cm, so △OAB is isosceles and OM bisects AB. ∠AOM = 60°.
OM = OA × cos 60° = 21 × (1/2) = 21/2 cm. AM = OA × sin 60° = 21 × (√3/2) = 21√3/2 cm.
AB = 2 × AM = 21√3 cm.
Area of △OAB = (1/2) × AB × OM = (1/2) × 21√3 × (21/2) = (441√3)/4 cm²
Step 3 — Segment area: = 462 − (441√3)/4 = (21/4)(88 − 21√3) cm²
Using √3 ≈ 1.732: (441 × 1.732)/4 ≈ 190.95 → segment ≈ 462 − 190.95 ≈ 271.05 cm²
▶ Example 3 — Area of Sector: r = 6 cm, θ = 60° (π = 22/7)
Solution:
Area of sector = (θ/360) × πr²
= (60/360) × (22/7) × 6 × 6 = (1/6) × (22/7) × 36 = (22 × 36)/(7 × 6) = 792/42 = 132/7 ≈ 18.86 cm²
▶ Example 4 — Area of Quadrant of a Circle with Circumference 22 cm
Solution:
Circumference = 2πr = 22 → r = 22/(2π) = 22/(2 × 22/7) = 22 × 7/44 = 7/2 = 3.5 cm
A quadrant is a sector with θ = 90°.
Quadrant area = (90/360) × πr² = (1/4) × (22/7) × (3.5)² = (1/4) × (22/7) × 12.25 = (1/4) × 38.5 = 9.625 cm²
▶ Example 5 — Area Swept by Minute Hand (length 14 cm) in 5 Minutes
Concept:
The minute hand completes 360° in 60 minutes, so in 5 minutes it sweeps through 360° × (5/60) = 30°. The hand traces a sector of radius 14 cm and angle 30°.
Solution:
Swept area = (30/360) × (22/7) × 14 × 14
= (1/12) × (22/7) × 196 = (22 × 196)/(12 × 7) = 4312/84 = 154/3 ≈ 51.33 cm²
▶ Example 6 — Minor Segment and Major Sector: r = 10 cm, Chord Subtends 90° (π = 3.14)
Solution:
r = 10 cm, θ = 90°. Minor sector area = (90/360) × 3.14 × 100 = (1/4) × 314 = 78.5 cm²
Triangle OAB (right-angled at O since θ = 90°): Area = (1/2) × 10 × 10 = 50 cm²
Minor segment = 78.5 − 50 = 28.5 cm²
Major sector = πr² − minor sector = 314 − 78.5 = 235.5 cm²
▶ Example 7 — Arc Length, Sector Area and Segment Area: r = 21 cm, θ = 60°
Solution:
(i) Arc length = (θ/360) × 2πr = (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm
(ii) Sector area = (60/360) × (22/7) × 441 = (1/6) × 1386 = 231 cm²
(iii) Segment area = Sector − △OAB. For θ = 60°: △OAB is equilateral (OA = OB = AB = r = 21 cm).
Area of equilateral △ = (√3/4) × 21² = (√3/4) × 441 = (441√3)/4 ≈ (441 × 1.732)/4 ≈ 190.95 cm²
Segment area ≈ 231 − 190.95 ≈ 40.05 cm²
▶ Example 8 — Horse Grazing Problem: Square Field, Rope = 5 m then 10 m (π = 3.14)
Context:
A horse is tied at the corner of a square field (side 15 m) with a rope. The horse can graze in a sector. The angle at the corner of a square = 90°.
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Solution:
(i) Rope = 5 m: Grazing area = (90/360) × 3.14 × 5² = (1/4) × 3.14 × 25 = 19.625 m²
(ii) Rope = 10 m: Grazing area = (90/360) × 3.14 × 10² = (1/4) × 3.14 × 100 = 78.5 m²
Increase = 78.5 − 19.625 = 58.875 m²
▶ Example 9 — Windscreen Wiper: Length 25 cm, Sweeping 115° (Two Wipers)
Solution:
Each wiper blade: r = 25 cm, θ = 115°.
Area swept by one wiper = (115/360) × (22/7) × 25² = (115/360) × (22/7) × 625
= (115 × 22 × 625)/(360 × 7) = 1581250/2520 = 158125/252 ≈ 627.48 cm²
Total area for two non-overlapping wipers = 2 × 627.48 ≈ 1254.96 cm²
▶ Example 10 — Lighthouse Warning Area: θ = 80°, Distance = 16.5 km (π = 3.14)
Solution:
The lighthouse illuminates a sector of angle 80° over a radius of 16.5 km.
Sea area = (80/360) × 3.14 × (16.5)²
= (2/9) × 3.14 × 272.25 = (2/9) × 854.865 = 1709.73/9 ≈ 189.97 km²
▶ Example 11 — Umbrella with 8 Equal Ribs: r = 45 cm, Area Between Two Ribs
Solution:
8 equally spaced ribs divide the circle into 8 equal sectors. Each sector angle = 360°/8 = 45°.
Area between two consecutive ribs = (45/360) × (22/7) × 45²
= (1/8) × (22/7) × 2025 = (22 × 2025)/(8 × 7) = 44550/56 = 22275/28 ≈ 795.54 cm²
▶ Example 12 — Segment Area: r = 12 cm, θ = 120° (π = 3.14, √3 = 1.73)
Solution:
r = 12 cm, θ = 120°.
Sector area = (120/360) × 3.14 × 144 = (1/3) × 452.16 = 150.72 cm²
△OAB: OM ⊥ AB, ∠AOM = 60°. OM = 12 × cos 60° = 12 × 0.5 = 6 cm. AM = 12 × sin 60° = 12 × (1.73/2) = 10.38 cm. AB = 20.76 cm.
Area of △OAB = (1/2) × 20.76 × 6 = 62.28 cm²
Segment area = 150.72 − 62.28 = 88.44 cm²
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Practice Sets A – D |
A1. Find the arc length and area of a sector with r = 7 cm and θ = 90°. (π = 22/7) [Arc = (90/360)×2×(22/7)×7 = 11 cm; Area = (90/360)×(22/7)×49 = 38.5 cm²]
A2. A sector has arc length 22 cm and radius 14 cm. Find the angle of the sector. (π = 22/7) [22 = (θ/360)×2×(22/7)×14; θ = 22×360×7/(2×22×14) = 90°]
A3. Find the area of a sector of angle 120° in a circle of radius 21 cm. (π = 22/7) [(120/360)×(22/7)×441 = (1/3)×1386 = 462 cm²]
A4. The minute hand of a watch is 6 cm long. What area does it sweep in 15 minutes? (π = 3.14) [(90/360)×3.14×36 = (1/4)×113.04 = 28.26 cm²]
B1. Find the area of the minor segment of a circle of radius 14 cm where the chord subtends 60° at the centre. (π = 22/7, △ is equilateral) [Sector = (1/6)×(22/7)×196 = 308/3 cm²; △ area = (√3/4)×196 ≈ 84.87 cm²; Segment ≈ 308/3 − 84.87 ≈ 17.8 cm²]
B2. A chord of a circle of radius 15 cm subtends 60° at the centre. Find areas of minor and major segments. (π = 3.14, √3 = 1.73) [Minor sector = (1/6)×3.14×225 = 117.75 cm²; △area = (1.73/4)×225 ≈ 97.31 cm²; Minor seg ≈ 20.44 cm²; Major seg = 3.14×225 − 20.44 ≈ 686.06 cm²]
B3. Find the area of a segment of a circle of radius 10 cm where the central angle is 90°. (π = 3.14) [Sector = (1/4)×3.14×100 = 78.5 cm²; △area = (1/2)×10×10 = 50 cm²; Segment = 28.5 cm²]
B4. The area of a sector of a circle is 77 cm² and the arc length is 22 cm. Find the radius. [Area = (1/2)×arc×r → 77 = (1/2)×22×r → r = 7 cm]
C1. A cow is tied at the corner of a rectangular field (30 m × 20 m) with a rope of 14 m. What area can the cow graze? The corner angle is 90°. (π = 22/7) [Grazing area = (90/360)×(22/7)×196 = (1/4)×616 = 154 m²]
C2. A pendulum of length 30 cm swings through an angle of 40°. Find the arc length traced by its tip and the area swept. (π = 3.14) [Arc = (40/360)×2×3.14×30 = 20.93 cm; Area = (40/360)×3.14×900 = 314.15 cm²]
C3. A sector-shaped ventilation panel has radius 1.2 m and angle 150°. Find the area of the panel. (π = 3.14) [(150/360)×3.14×1.44 = (5/12)×4.52 ≈ 1.88 m²]
C4. A fan blade of length 20 cm sweeps through 60° in one rotation. How much area does one blade sweep? (π = 22/7) [(60/360)×(22/7)×400 = (1/6)×(8800/7) = 8800/42 ≈ 209.5 cm²]
D1. A brooch has circular shape with diameter 35 mm and 5 diameters dividing it into 10 equal sectors. Find (i) total silver wire needed for the circle and 5 diameters (ii) area of each sector. (π = 22/7) [(i) Circumference = (22/7)×35 = 110 mm; 5 diameters = 5×35 = 175 mm; Total = 285 mm (ii) Each sector = (1/10)×(22/7)×(35/2)² = (1/10)×(22/7)×306.25 ≈ 96.25 mm²]
D2. A pizza of diameter 30 cm is cut into 8 equal slices. Find the arc length of one slice and the area of one slice. (π = 3.14) [θ = 45°; r = 15 cm; Arc = (45/360)×2×3.14×15 = 11.78 cm; Area = (45/360)×3.14×225 = 88.31 cm²]
D3. A round table cover (radius 28 cm) has six equal designs shaped like a segment of a circle, each spanning 60°. Find the total area of the designs. (√3 = 1.7, π = 22/7) [Each design = segment at 60°; Sector area = (1/6)×(22/7)×784 = 411.33 cm²; △ area = (1.7/4)×784 = 332.8 cm²; Segment ≈ 78.53 cm²; Total = 6 × 78.53 ≈ 471.2 cm²]
D4. The area of a sector is 2/5 of the total circle area. Find the angle of the sector. [2/5 = θ/360 → θ = 144°]
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Chapter Summary |
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∿ Arc Length Formula
L = (θ/360) × 2πr Fraction of full circumference equal to θ/360. θ must be in degrees. |
◦ Sector Area Formula
A = (θ/360) × πr² Fraction of full circle area equal to θ/360. Use correct value of π. |
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▸ Segment Area Formula
Segment = Sector − △OAB = (θ/360)×πr² − (1/2)r² sin θ. Triangle area for special angles: 90°→½r², 60°→(√3/4)r², 120°→(√3/4)r². |
🔄 Major vs Minor
Major = πr² − Minor Applies to both sectors and segments. Major sector angle = 360° − θ. Can compute directly by substituting (360°−θ). |
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△ Triangle Area Shortcuts
θ = 60°: Equilateral △; Area = (√3/4)r² θ = 90°: Right △; Area = r²/2 θ = 120°: △ area = (√3/4)r² (same as 60°) |
☷ Real-World Applications
Clock hands, grazing animals, lighthouse beams, windscreen wipers, pizza slices, umbrella panels, brooch designs, fan blades, irrigation sprinklers — all use sector/segment area and arc length formulae. |
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8-Point Exam Quick-CheckTop mistakes students make and how to avoid them |
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① Always use the correct value of π
Use π = 22/7 when radius is a multiple of 7. Use π = 3.14 otherwise, or when specifically instructed. Never mix them in one calculation. |
② Sector vs Segment — they are different
A sector includes the centre (pie-slice shape). A segment does NOT include the centre — it sits between a chord and its arc. Segment = Sector − Triangle. |
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③ Arc length and area both use θ/360
Arc = (θ/360) × 2πr. Area = (θ/360) × πr². The only difference is 2πr vs πr². Don’t confuse which formula applies to length vs area. |
④ θ = 60° gives an equilateral triangle
When two radii form a 60° angle, and both are equal (= r), the triangle OAB is equilateral with all sides = r. Its area = (√3/4)r². This shortcut is used repeatedly in segment problems. |
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⑤ θ = 90° gives a right-angled triangle at O
For a chord subtending 90° at centre, triangle OAB is right-angled at O. Area = (1/2) × r × r = r²/2. This is the simplest case for segment problems. |
⑥ Major = πr² − Minor (for both)
Whether it’s sectors or segments, the major one always equals the full circle area minus the minor one. Or compute directly: major sector uses angle (360°−θ). |
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⑦ Real-world angle = clock or geometry angle
Minutes to angle: 1 minute = 6°; 5 min = 30°; 15 min = 90°. Corner angle of square = 90°; equilateral triangle corner = 60°. Always identify θ before plugging in. |
⑧ Area relation: Area = (1/2) × arc × r
A useful shortcut: since arc = (θ/360) × 2πr, sector area = (1/2) × arc length × r. This lets you find area directly from arc length without re-computing θ. |
Grade 10 Maths Chapter 11 — Areas Related to Circles covers the calculation of arc length, sector area, segment area, and their applications in real-world contexts. Students learn to use the unitary method to derive the key formulae: arc length = (θ/360) × 2πr and sector area = (θ/360) × πr², where r is the radius and θ is the central angle in degrees. The segment area formula — which subtracts the triangle area from the sector area — is central to this chapter, and mastery of triangle areas for standard angles (30°, 60°, 90°, 120°) is essential. Important applications include finding the area swept by a clock’s minute hand, the grazing region for a tethered animal tied at the corner of a field, the sea area covered by a lighthouse beam, the area cleaned by car windscreen wipers, and the material needed for brooch or decorative designs. This fully original revision page includes key definitions with visual table diagrams, all core formulae with derivation, 12 fully solved examples, four graded practice sets with complete answers, a chapter summary, and an 8-point exam quick-check — perfect preparation for any assessment on areas related to circles for Class 10.
Keywords: areas related to circles class 10, sector area formula, segment area formula, arc length formula, minor segment major sector, areas of sectors and segments grade 10