Areas Related to Circles Class 10 Notes (2026-27) — CBSE
Class 10 Maths Chapter 11 notes: circumference and area of a circle, length of an arc, and area of a sector and a segment, with all key formulas and examples.
Areas Related to Circles — Class 10 Maths Notes
Chapter Snapshot
This chapter applies the circle's circumference and area to parts of a circle: the arc, the sector (a "pizza slice"), and the segment (the region between a chord and an arc). You use these to find areas of shaded regions in combined figures.
Board relevance: a Mensuration scorer. Expect a sector/segment area question and a shaded-region problem. Choose π = 22/7 when the radius is a multiple of 7, otherwise π = 3.14.
Syllabus note (rationalised): the areas of combinations of plane figures section has been trimmed; concentrate on the circle, arc, sector, and segment formulas.
Key Concepts & Definitions
- Circle: the set of all points at a fixed distance (radius r) from a centre.
- Arc: a part of the circle (curved boundary). A minor arc is shorter, a major arc longer.
- Sector: the region bounded by two radii and the arc between them (like a slice). The minor sector has the smaller angle; the major sector the larger.
- Segment: the region bounded by a chord and the arc it cuts off. Minor segment (smaller) and major segment (larger).
Formulas
For a circle of radius r, with a sector of central angle θ (in degrees):
Quantity Formula
Circumference of circle 2πr
Area of circle πr²
Length of arc of sector (θ/360) × 2πr
Area of sector (θ/360) × πr²
Area of segment (area of sector) − (area of triangle)
Perimeter of sector arc length + 2r
- π = 22/7 (use when r is a multiple of 7) or 3.14.
- Area of the triangle in a segment (two radii + chord) = ½ r² sin θ (or ½ × base × height for standard angles like 60°, 90°).
- Major segment area = area of circle − minor segment area.
Worked Examples
Example 1 — Area & circumference: A circle has radius 21 cm. Find its area and circumference (π = 22/7).
Area = πr² = (22/7)(21²) = (22/7)(441) = 1386 cm². Circumference = 2πr = 2(22/7)(21) = 132 cm.
Example 2 — Sector area & arc length: A sector of a circle of radius 14 cm has central angle 90°. Find its area and arc length (π = 22/7).
Area = (90/360)πr² = (1/4)(22/7)(196) = 154 cm². Arc = (90/360)(2πr) = (1/4)(2)(22/7)(14) = 22 cm.
Example 3 — Segment area: Find the area of the minor segment of a circle (radius 10 cm) cut off by a chord subtending 90° at the centre (π = 3.14).
Sector area = (90/360)πr² = (1/4)(3.14)(100) = 78.5 cm². Triangle (right-angled, legs = r) = ½(10)(10) = 50 cm².
Minor segment = 78.5 − 50 = 28.5 cm².
Example 4 — Shaded region: A square of side 14 cm has a quarter circle drawn at each corner (radius 7 cm). Find the shaded area outside the quarter circles.
Four quarter-circles = one full circle of radius 7 → area = (22/7)(49) = 154 cm². Square area = 14² = 196 cm².
Shaded area = 196 − 154 = 42 cm².
Example 5 — Ring (annulus): A circular path 7 m wide surrounds a circular pond of radius 21 m. Find the area of the path (π = 22/7).
Outer radius R = 21 + 7 = 28 m, inner radius r = 21 m. Path area = πR² − πr² = π(R² − r²) = (22/7)(28² − 21²) = (22/7)(784 − 441) = (22/7)(343) = 1078 m². (For any ring, area = π(R² − r²) = π(R + r)(R − r).)
Example 6 — Angle from area: A sector of a circle of radius 6 cm has area 9π cm². Find its central angle.
(θ/360)πr² = 9π → (θ/360)(36) = 9 → θ/360 = 1/4 → θ = 90°. (Rearranging the sector-area formula lets you find the angle when the area is given.)
Example 7 — Two coloured regions: A circle of radius 10 cm is divided by a diameter; one semicircle is shaded. Its area = ½ × πr² = ½ × 3.14 × 100 = 157 cm². A semicircle is just a 180° sector, and a quadrant is a 90° sector — recognising these fractions saves time.
Important Question Patterns
1. Circle basics (2 marks): area/circumference from radius or diameter; find radius given area or circumference.
2. Sector (2–3 marks): area of a sector and arc length from the central angle; perimeter of a sector (arc + 2r); the angle from a known area.
3. Segment (3 marks): minor segment = sector − triangle; often with θ = 60°, 90°, or 120° so the triangle area is easy.
4. Shaded regions (3–4 marks): combinations — circle in a square, quarter circles at corners, two overlapping circles, a running track — add and subtract areas carefully.
5. Rotating/clock problems (2–3 marks): the angle swept by a clock hand or a fan blade → sector area/arc length.
⚡ Quick Revision
- Circle: circumference 2πr, area πr². Use π = 22/7 (r multiple of 7) or 3.14.
- Arc length = (θ/360)·2πr; sector area = (θ/360)·πr².
- Perimeter of sector = arc + 2r.
- Segment area = sector area − triangle area (triangle = ½r² sin θ).
- Major segment = circle area − minor segment.
- Shaded region: break into circles/sectors/squares and add or subtract.
- Four quarter-circles of equal radius = one full circle.
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