Class 10 Maths Notes — Chapter 11: Areas Related to Circles
Circumference, area, and sector/segment computations for circles.
Detailed NCERT notes
- Circumference of a circle = 2πr; area = πr².
- For a sector of angle θ (in degrees): length of arc = (θ/360) · 2πr; area of sector = (θ/360) · πr².
- Area of a segment = area of sector − area of triangle formed by two radii and chord.
- For minor and major segments: major segment area = πr² − minor segment area.
- For angle in radians (rarely used at Class 10): arc length = rθ, sector area = ½r²θ.
- Combination figures: circle inscribed in square (r = a/2), square inscribed in circle (diagonal = 2r), semicircles on sides, etc.
- In problems mixing shapes: identify each region, compute area separately, add or subtract.
- Common π values: use 22/7 when r is a multiple of 7, and 3.14 otherwise (unless stated).
Formulas & key results
- Circle: C = 2πr, A = πr²
- Sector: arc = (θ/360)·2πr; area = (θ/360)·πr²
- Segment area = sector area − ½ r² sin θ (or minus triangle area)
Mind map
- Sector, segment, chord
- Combinations with squares/triangles
- Choice of π (22/7 vs 3.14)
Tricks & shortcuts
- For 60° sector with radius r: area = πr²/6; equilateral triangle formed if chord joins endpoints.
- Add/subtract areas region by region for compound figures.
Common mistakes to avoid
- Mixing degrees and radians.
- Using r² instead of πr² for area.
- Forgetting the triangle subtraction in a segment.
Competency-based questions & answers
- Q. Find the area of a sector with r = 6 and θ = 60°.A. (60/360)·π·36 = 6π ≈ 18.86 cm².
- Q. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the minor segment.A. Sector = (90/360)·π·100 = 25π; triangle = ½·10·10 = 50; segment = 25π − 50 ≈ 28.5 cm².
- Q. The wheel of a car has radius 35 cm. How many revolutions does it make to travel 11 km?A. Circumference = 2·22/7·35 = 220 cm; revolutions = 1100000/220 = 5000.