Class 10 Mathematics Chapter 11 Revision Summary Strictly NCERT

1. Chapter at a glance

  • A sector is the portion of the circular region enclosed by two radii and the corresponding arc; a segment is the portion enclosed between a chord and the corresponding arc.
  • Minor sector/segment corresponds to the smaller part; major sector/segment corresponds to the larger part (angle of major sector = 360° – θ).
  • Unless stated otherwise, “sector” and “segment” refer to the minor sector and minor segment.
  • Area of a sector with central angle θ° is obtained from the area of the full circle (πr²) by the unitary method.
  • Length of the arc corresponding to a sector is likewise obtained from the circumference (2πr) by the unitary method.
  • Area of a segment equals area of the sector minus area of the isosceles triangle formed by the two radii and the chord.
  • Area of major sector = πr² – area of minor sector; area of major segment = πr² – area of minor segment.

2. Definitions, theorems and results

  • Sector of a circle: portion of the circular region enclosed by two radii and the corresponding arc. Angle of the sector is the central angle ∠AOB = θ°.
  • Segment of a circle: portion of the circular region enclosed between a chord and the corresponding arc.
  • Minor sector/segment: the smaller of the two parts formed; major sector/segment: the larger part.
  • Remark (NCERT): When we write “segment” and “sector” we mean the minor segment and minor sector respectively, unless stated otherwise.
  • No formal theorems, lemmas or axioms are stated that require proof in the Board examination. All results are derived directly by the unitary method from the known area of the circle (πr²) and circumference (2πr).

3. Formula sheet

Formula Meaning of symbols
Arc length = (θ/360) × 2πr θ = central angle in degrees, r = radius
Area of sector = (θ/360) × πr² θ = central angle in degrees, r = radius
Area of segment = [(θ/360) × πr²] – area of ΔOAB θ = central angle, r = radius; ΔOAB is the triangle formed by the two radii and the chord
Area of major sector = πr² – minor sector area
Area of major segment = πr² – minor segment area

4. Solved-example patterns

  • Pattern A (Sector area only): Given r and θ, compute (θ/360)πr²; for major sector subtract from πr² or use (360–θ)/360 × πr².
  • Pattern B (Arc length + sector): First find arc length = (θ/360)2πr, then sector area = (θ/360)πr²; both use the same ratio.
  • Pattern C (Segment area): Compute sector area, then find area of ΔOAB (drop perpendicular from centre to chord, use congruence or sin/cos 30°/60° values) and subtract.
  • Pattern D (Major/minor pair): Calculate minor quantity and subtract from πr² to obtain the major counterpart.
  • Pattern E (Composite figures – grazing, umbrella, etc.): Identify the relevant sector(s) of given radius and angle; add or subtract areas as required by the geometry of the figure.

5. Common mistakes and exam pitfalls

  • Using the major-sector formula when the question asks for the minor sector (or vice versa).
  • Forgetting to subtract the area of ΔOAB when finding segment area.
  • Omitting units (cm², m²) or mixing degree and radian measures.
  • Taking θ as the reflex angle instead of the smaller central angle given in the figure.
  • Using an incorrect value of π or √3 without checking the question’s instruction.
  • Neglecting to verify that the triangle area calculation uses the correct height (perpendicular from centre to chord) or the midpoint property from congruence.

A study aid reviewed by GFIS faculty — always verify with your textbook and teacher.