Class 10 Maths Chapter 11: Areas Related to Circles

Areas Related to Circles is Chapter 11 of NCERT Class 10 Maths. It teaches you how to find the length of an arc and the areas of a sector and a segment of a circle, which are simply fractions of the full circle. These Class 10 Maths Chapter 11 notes cover all the formulas, a clear method and fully worked examples for quick revision.

What you will learn

• The circumference and area of a circle as the starting point
• The meaning of arc, sector and segment, and their minor and major parts
• Formulas for length of an arc and area of a sector using the central angle θ
• How to find the area of a segment by subtracting a triangle from a sector
• Real-life uses such as the area swept by a clock hand or a wiper

Key Concepts and Formulas

Circle Basics

Every formula in this chapter is built from the two basic results for a full circle of radius r. A full circle has a central angle of 360°.

Circumference = 2πr
Area of circle = πr²

• Use π = 22/7 when the radius is a multiple of 7, otherwise π = 3.14, unless the question says which value to use.

Arc and Sector

An arc is a part of the circumference. A sector is the region enclosed by two radii and the arc between them, like a slice of pizza. If the central angle is θ, the sector is the fraction θ/360 of the whole circle.

Length of arc = (θ/360°) × 2πr
Area of sector = (θ/360°) × πr²
Area of sector = ½ × arc length × r
Perimeter of sector = 2r + arc length
Area of major sector = πr² − area of minor sector

• Minor sector: central angle less than 180°. Major sector: the remaining part, with angle 360° − θ.

Segment of a Circle

A chord divides a circle into two regions called segments. The smaller one is the minor segment and the larger one is the major segment. The minor segment is the sector minus the triangle formed by the two radii and the chord.

Minor segment = sector − triangle
Major segment = πr² − minor segment
For θ = 90°: triangle = ½r²
  segment = πr²/4 − r²/2
For θ = 60°: triangle = (√3/4)r²
  segment = πr²/6 − (√3/4)r²

• For θ = 90° the triangle is right-angled, and for θ = 60° it is equilateral.
• In CBSE board questions on segments, the central angle is usually 60° or 90°.

Clock and Rotation Problems

Many questions describe motion that sweeps out a sector. Convert time or rotation into a central angle first.

Minute hand: 360° in 60 min = 6° per minute
Hour hand: 360° in 12 h = 30° per hour

• The length of the hand is the radius of the sector it sweeps.

Solved Examples

Example 1: Find the area of a sector of a circle of radius 14 cm with central angle 45°.

Area = (θ/360) × πr²
= (45/360) × (22/7) × 14 × 14
= (1/8) × 616
∴ Area of sector = 77 cm²

Example 2: Find the length of an arc of a circle of radius 21 cm that subtends 120° at the centre.

Arc = (θ/360) × 2πr
= (120/360) × 2 × (22/7) × 21
= (1/3) × 132
∴ Arc length = 44 cm

Example 3: The minute hand of a clock is 10.5 cm long. Find the area swept by it in 20 minutes.

θ = 20 × 6° = 120°
Area = (120/360) × (22/7) × 10.5 × 10.5
= (1/3) × 346.5
∴ Area swept = 115.5 cm²

Example 4: A chord of a circle of radius 12 cm subtends 90° at the centre. Find the areas of the minor and major segments (π = 3.14).

Sector = ¼ × 3.14 × 144 = 113.04 cm²
Triangle = ½ × 12 × 12 = 72 cm²
Minor segment = 113.04 − 72 = 41.04 cm²
Circle = 3.14 × 144 = 452.16 cm²
Major segment = 452.16 − 41.04
∴ Minor = 41.04 cm², Major = 411.12 cm²

Example 5: A chord of a circle of radius 6 cm subtends 60° at the centre. Find the area of the minor segment (π = 3.14, √3 = 1.73).

Sector = (1/6) × 3.14 × 36 = 18.84 cm²
Triangle = (1.73/4) × 36 = 15.57 cm²
Segment = 18.84 − 15.57
∴ Minor segment = 3.27 cm²

Example 6: Find the perimeter of a quadrant of a circle of radius 7 cm.

Arc = ¼ × 2 × (22/7) × 7 = 11 cm
Perimeter = r + r + arc
= 7 + 7 + 11
∴ Perimeter = 25 cm

Example 7: An arc of length 22 cm is cut from a circle of radius 14 cm. Find the central angle.

(θ/360) × 2 × (22/7) × 14 = 22
(θ/360) × 88 = 22
θ = 22 × 360 ÷ 88
∴ θ = 90°

Common Mistakes to Avoid

• Forgetting to add the two radii when finding the perimeter of a sector. The arc length alone is not the perimeter.
• Using the diameter in place of the radius. Read whether the question gives the radius or the diameter.
• Subtracting the wrong triangle for a segment. For 60° use the equilateral triangle formula, and for 90° use ½r².
• Mixing units or values of π in the same answer. Pick one value of π and keep units in cm, cm² or m² consistently.

Quick Revision

Arc length = (θ/360) × 2πr
Area of sector = (θ/360) × πr² = ½ × arc × r
Perimeter of sector = 2r + arc length
Segment = sector − triangle
Major part = πr² − minor part

• Minute hand: 6° per minute; hour hand: 30° per hour

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