Chapter 11 · CBSE Class 10 Maths
Areas Related to Circles
Two ways to learn — a crisp formula reference, or a story that makes sector/segment areas feel obvious.
Three scenes that build the ideas from scratch, step by step.
How this works
Three scenes, one thread: a full circle is just a 360° pie, so any smaller slice gets exactly its fair fraction of the area or arc length — and a segment is simply that slice with its triangular "crust" cut away. Read the story, then tap "The Maths" for the formal result.
Slicing the Pie
Cut a circular pizza with two straight cuts from the centre. The wedge-shaped piece — bounded by two radii and the arc between them — is a sector. The smaller piece is the minor sector; the leftover, bigger piece is the major sector.
The whole pizza is really just one giant sector spanning the full 360°. What fraction of the whole pizza does a 30° slice represent?
Full circle (360°) has area πr²
A θ° sector is exactly the fraction θ/360 of that same circle
Sector area = (θ/360) × πr²
The identical logic applies to the curved edge length — the arc. The full circle's circumference is 2πr, so a θ° sector's arc is exactly θ/360 of that.
The Slice Minus the Triangle
Now make just ONE straight cut across the pizza (a chord) instead of two cuts from the centre. The piece cut off — bounded by the chord and its arc, no radii involved — is a segment, not a sector.
A segment looks like it should be smaller than the sector with the same arc — how much smaller, exactly?
Picture the sector (with its two radii) first, then notice the segment is just that sector with the triangular piece (formed by the two radii and the chord) sliced away.
Area of segment = Area of sector − Area of the triangle (2 radii + chord)
A 30° sector of radius 4 cm has area ≈4.19 cm² (using π=3.14). The corresponding major sector, since it's everything else, is the full circle's area minus that: 50.24 − 4.19 ≈ 46.05 cm².
Trigonometry Sneaks Back In
A circle of radius 21 cm has a chord subtending 120° at the centre. Finding this segment's area means finding the enclosed triangle's area first — but 120° isn't a simple right angle, so the triangle isn't as easy as (1/2)×base×height at first glance.
Drop a perpendicular from the centre O to the chord. Since OA = OB (both radii), what does that perpendicular do to the triangle?
OA = OB, OM ⊥ AB → △AMO ≅ △BMO (RHS congruence — same idea as Chapter 10's tangent proof)
So M bisects AB, and ∠AOM = ∠BOM = 60° (half of 120°)
Now each half is a simple right triangle, and basic trigonometry (Chapter 8) gives both the half-base and the height directly.
OM/OA = cos 60° → OM = 21 × (1/2) = 21/2
AM/OA = sin 60° → AM = 21 × (√3/2)
AB = 2×AM = 21√3; Triangle area = (1/2)×AB×OM = 441√3/4
The Big Picture
Every idea in this chapter branches from one fraction: θ/360 of a full circle. A sector takes that fraction of the whole area or circumference directly; a segment takes the sector and removes its enclosed triangle.
Sector area?
(θ/360) × πr²
Arc length?
(θ/360) × 2πr
Segment area?
sector area − triangle area
Major sector/segment?
πr² − minor sector/segment
Triangle area for angle θ?
(1/2) × r² × sin θ
Want to go deeper?
Our tutors cover the full CBSE Class 10 Maths syllabus live, 1-on-1. If Chapter 11 still feels shaky after this, one session usually fixes it.
Book a Free Trial Class