Chapter 11: Areas Related to Circles
This chapter uses the circumference and area of a circle to find the area of a sector and the area of a segment of a circle. In the rationalised syllabus, segment problems are restricted to central angles of 60°, 90° and 120° only.
1. Circle: Basic Formulas
For a circle of radius r:
- Circumference = 2πr (also called perimeter of the circle)
- Area = πr2
- Area of a semicircle = (1/2)πr2; perimeter of a semicircle = πr + 2r
- Area of a quadrant = (1/4)πr2
- Area between two concentric circles (a ring) of radii R and r (R > r) = π(R2 − r2)
Use π = 22/7 or π = 3.14 exactly as stated in the question.
2. Sector of a Circle
- Sector: The portion of the circular region enclosed by two radii and the corresponding arc. The smaller portion is the minor sector and the larger one the major sector.
- Angle of the sector: The angle θ subtended at the centre by the arc of the sector.
- The sum of the angles of the minor and major sectors is 360°.
For a sector of angle θ (in degrees) of a circle of radius r:
- Area of the sector = (θ/360) × πr2
- Length of the arc (l) = (θ/360) × 2πr
- Perimeter of the sector = 2r + l = 2r + (θ/360) × 2πr
- Useful link: Area of sector = (1/2) × l × r (half of arc length times radius)
- Area of the major sector = πr2 − area of the minor sector = ((360 − θ)/360) × πr2
Reasoning behind the formulas: The whole circle is a sector of angle 360° with area πr2. By the unitary method, a sector of angle 1° has area πr2/360, so a sector of angle θ has area (θ/360)πr2. The arc-length formula follows in the same way from the circumference 2πr.
3. Segment of a Circle
- Segment: The portion of the circular region enclosed between a chord and the corresponding arc — minor segment (with the minor arc) and major segment (with the major arc).
Area of a segment = Area of the corresponding sector − Area of the corresponding triangle
= (θ/360) × πr2 − Area of ΔOAB (O the centre, AB the chord)
Area of ΔOAB (two sides r, included angle θ) for the allowed angles:
- θ = 60°: triangle is equilateral of side r ⇒ area = (√3/4) r2
- θ = 90°: right isosceles triangle ⇒ area = (1/2) r2
- θ = 120°: area = (√3/4) r2 (base = r√3, height = r/2)
Area of the major segment = πr2 − area of the minor segment.
4. Clock (Minute-Hand) Problems
- The minute hand sweeps 360° in 60 minutes ⇒ 6° per minute.
- The hour hand sweeps 360° in 12 hours ⇒ 30° per hour (i.e. 0.5° per minute).
- Area swept by the minute hand (length r) in t minutes = (6t/360) × πr2.
5. Wheel / Revolution Problems
Distance covered by a wheel in one complete revolution = its circumference = 2πr.
Number of revolutions = Total distance ÷ Circumference.
6. Common Board Question Patterns and Stepwise Methods
- Sector area / arc length: Identify r and θ, substitute directly. If the perimeter of a sector is given, first find the arc: l = perimeter − 2r, then use area = (1/2) l r.
- Segment area (60°, 90°, 120° only): Step 1 — area of sector (θ/360)πr2. Step 2 — area of the triangle using the table above. Step 3 — subtract. For the major segment, subtract the minor segment from πr2.
- Minute-hand sweep: Convert minutes to degrees (×6), then apply the sector-area formula.
- Ring / circular path: Use π(R2 − r2) = π(R + r)(R − r) — factorising simplifies the arithmetic.
- Grazing problems: An animal tied at a corner of a field by a rope of length r grazes a sector; at a corner of a rectangular or square field the angle is 90°, so the grazed area is a quadrant (1/4)πr2.
- Equal-parts problems (umbrella ribs, pizza, brooch): n equal sectors ⇒ each sector has angle 360°/n.
7. Values to Memorise
| θ | Fraction of circle | Triangle area in segment formula |
|---|---|---|
| 60° | 1/6 | (√3/4) r2 |
| 90° | 1/4 | (1/2) r2 |
| 120° | 1/3 | (√3/4) r2 |
Also remember √2 ≈ 1.41 and √3 ≈ 1.73 (use the value given in the question if stated).
Exam tips: (1) Read carefully whether π = 22/7 or 3.14 is to be used — using the wrong value loses accuracy marks. (2) Keep answers with units: cm, cm2 as appropriate — arc length is a length, sector/segment are areas. (3) In segment questions, never forget to subtract the triangle; writing the formula "Area of segment = Area of sector − Area of triangle" first earns method marks. (4) When the diameter is given, halve it before substituting — the single most common careless error in this chapter.
Deleted in the rationalised syllabus (2026-27) — NOT to be studied
- Problems on the area of a segment for central angles other than 60°, 90° and 120° (segment questions are restricted to these three angles only).
- Areas of combinations of plane figures (e.g. shaded-region problems combining circles with squares, triangles, etc.) — the entire section has been deleted from the rationalised NCERT.