CBSE Class 10 Mathematics Areas Related to Circles Notes

1. Introduction

This chapter deals with the areas of plane figures related to circles. We study the perimeter and area of a circle and its parts — sectors and segments. These concepts are essential for solving real-world problems and frequently appear in CBSE board examinations.

Key Topics in This Chapter

• Perimeter (Circumference) and Area of a Circle

• Area of a Sector of a Circle

• Area of a Segment of a Circle

• Areas of Combinations of Plane Figures

2. Basic Concepts — Circle Review

2.1 Parts of a Circle

Before diving into areas, it is important to recall the key parts of a circle and their definitions:

• Radius (r): The distance from the centre of the circle to any point on the circle.

• Diameter (d): The longest chord passing through the centre. d = 2r.

• Chord: A line segment joining any two points on the circle.

• Arc: A part of the circumference of a circle.

• Sector: The region bounded by two radii and the corresponding arc.

• Segment: The region bounded by a chord and the arc it cuts off.

2.2 Minor and Major Parts

Part	Minor	Major
Arc	Smaller arc (< semicircle)	Larger arc (> semicircle)
Sector	Region of smaller arc	Region of larger arc
Segment	Region of minor arc & chord	Region of major arc & chord

3. Perimeter and Area of a Circle

3.1 Circumference (Perimeter)

The perimeter of a circle is called its circumference. It is the total length of the boundary of the circle.

Circumference = 2πr = πd

where r = radius, d = diameter, π ≈ 3.14159 or 22/7

3.2 Area of a Circle

The area of a circle is the region enclosed within its boundary (circumference).

Area of Circle = πr²

Also expressed as: Area = (π/4) × d²

Worked Example 3.1

Problem: Find the circumference and area of a circle with radius 7 cm. (Use π = 22/7)

Solution:

Circumference = 2πr = 2 × (22/7) × 7 = 44 cm

Area = πr² = (22/7) × 7 × 7 = 154 cm²

Answer: Circumference = 44 cm, Area = 154 cm²

4. Area of a Sector of a Circle

4.1 Definition

A sector is the region between two radii of a circle and the arc between them. The angle formed by the two radii at the centre is called the angle of the sector (denoted by θ).

Important Note

When θ = 360°, the sector is the entire circle.

When θ = 180°, the sector is a semicircle.

The minor sector has a smaller angle; the major sector has a larger angle.

4.2 Formulas for Sector

Area of Sector = (θ/360°) × πr²

Length of Arc = (θ/360°) × 2πr

Perimeter of Sector = 2r + Arc Length

= 2r + (θ/360°) × 2πr

Worked Example 4.1

Problem: A sector of a circle with radius 12 cm has an angle of 60°. Find its area and arc length. (Use π = 3.14)

Solution:

Area = (θ/360) × πr² = (60/360) × 3.14 × 12 × 12

= (1/6) × 3.14 × 144 = 75.36 cm²

Arc Length = (60/360) × 2 × 3.14 × 12

= (1/6) × 75.36 = 12.56 cm

Perimeter of sector = 2r + arc = 24 + 12.56 = 36.56 cm

5. Area of a Segment of a Circle

5.1 Definition

A segment of a circle is the region between a chord and the arc it subtends. There are two types: minor segment (smaller region) and major segment (larger region).

5.2 Formulas for Segment

Area of Minor Segment = Area of Sector - Area of Triangle

Area of Minor Segment = (θ/360°) × πr² - (1/2) × r² × sin θ

Area of Major Segment = Area of Circle - Area of Minor Segment

= πr² - Area of Minor Segment

Key Formula — Area of Triangle in Sector

When the sector has angle θ and radius r:

Area of triangle (OAB) = (1/2) × r² × sin θ

For common angles:

sin 30° = 1/2 | sin 45° = √2/2 | sin 60° = √3/2 | sin 90° = 1

Worked Example 5.1

Problem: A chord of a circle of radius 10 cm subtends an angle of 90° at the centre.

Find the area of the corresponding minor segment. (Use π = 3.14)

Solution:

Area of Sector = (90/360) × 3.14 × 10² = (1/4) × 314 = 78.5 cm²

Area of Triangle = (1/2) × r² × sin 90° = (1/2) × 100 × 1 = 50 cm²

Area of Minor Segment = 78.5 - 50 = 28.5 cm²

6. Areas of Combinations of Plane Figures

Many practical problems require finding areas that are combinations of circles, sectors, segments, triangles, rectangles, and other plane figures. The approach is to break down complex shapes into simpler, known shapes.

6.1 General Strategy

1. Identify the composite figure and break it into simple shapes.

2. Calculate the area of each simple shape separately.

3. Add or subtract areas as required by the problem.

4. Always check units — keep everything in the same unit.

6.2 Common Combination Types

Combination Type	Approach	Formula Used
Circle + Square	Square with circle inside/outside	Area of square ± πr²
Sector + Triangle	Sector minus triangle = segment	Sector area − (1/2)r² sin θ
Ring (Annulus)	Outer circle − inner circle	π(R² − r²)
Horse-track / Running track	Rectangle + two semicircles	lb + πr²
Flower / Petal design	Sum of sectors or segments	n × (sector area)

Worked Example 6.1 — Ring (Annulus)

Problem: Two concentric circles have radii 21 cm and 14 cm. Find the area of the ring.

(Use π = 22/7)

Solution:

Area of outer circle = π × 21² = (22/7) × 441 = 1386 cm²

Area of inner circle = π × 14² = (22/7) × 196 = 616 cm²

Area of ring = 1386 − 616 = 770 cm²

Worked Example 6.2 — Horse-track / Running Track

Problem: A running track has two parallel straight sections of 100 m each, connected by

semicircular ends of radius 35 m. Find the total area of the track. (Use π = 22/7)

Solution:

Area of rectangular region = 100 × (2 × 35) = 100 × 70 = 7000 m²

Area of two semicircles = πr² = (22/7) × 35² = (22/7) × 1225 = 3850 m²

Total area = 7000 + 3850 = 10,850 m²

7. Formula Summary Table

The following table summarizes all key formulas from this chapter for quick revision:

Shape / Concept	Formula	Variables
Circumference of Circle	C = 2πr = πd	r = radius, d = diameter
Area of Circle	A = πr²	r = radius
Length of Arc (Sector)	l = (θ/360°) × 2πr	θ = angle in degrees
Area of Sector	A = (θ/360°) × πr²	θ = angle in degrees
Perimeter of Sector	P = 2r + (θ/360°) × 2πr	r = radius, θ = angle
Area of Minor Segment	A = (θ/360°)πr² − (1/2)r² sin θ	θ = angle in degrees
Area of Major Segment	A = πr² − Area of Minor Seg.
Area of Ring (Annulus)	A = π(R² − r²)	R = outer, r = inner radius
Area of Semicircle	A = πr²/2	r = radius
Perimeter of Semicircle	P = πr + 2r	r = radius

8. Key Points to Remember

• Always use π = 22/7 when the radius is a multiple of 7, and π = 3.14 otherwise (unless specified).

• The angle θ in all sector/segment formulas must be in degrees, not radians.

• A semicircle is a special sector where θ = 180°.

• A quadrant is a sector where θ = 90°, so its area = (1/4)πr².

• Area of minor segment = Area of sector − Area of triangle (not the other way).

• Area of major segment = Total circle area − Area of minor segment.

• For combination problems, always draw a diagram before solving.

• The perimeter of a segment = chord length + arc length (not sector perimeter).

9. Practice Questions

Test your understanding with the following questions, categorized by marks and difficulty level as per the CBSE board exam pattern.

9.1 — 1 Mark Questions (VSA)

5. What is the area of a quadrant of a circle whose circumference is 44 cm? (Use π = 22/7)

6. If the radius of a circle is doubled, what is the ratio of areas of the new circle to the original?

7. The arc length of a sector of a circle with radius 6 cm is 5π cm. Find the angle of the sector.

8. A sector has an area of 38.5 cm² with radius 7 cm. Find the angle of the sector. (π = 22/7)

9. What is the perimeter of a semicircle of radius 14 cm? (Use π = 22/7)

9.2 — 3 Mark Questions (SA)

10. Find the area of the shaded region where ABCD is a square of side 14 cm and four semicircles are drawn taking each side as diameter. (Use π = 22/7)

11. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor and major segments. (Use π = 3.14, √3 = 1.73)

12. Two circles touch externally. The sum of their areas is 130π cm² and the distance between their centres is 14 cm. Find the radii of the circles.

13. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find: (i) total length of silver wire required; (ii) area of each sector.

14. From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut. Find the area of the remaining portion.

9.3 — 5 Mark Questions (LA)

15. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of that part of the field in which the horse can graze; (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

16. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) length of the arc; (ii) area of the sector formed by the arc; (iii) area of the segment formed by the corresponding chord. (Use π = 22/7)

17. ABCD is a flower bed. If OA = 21 m and OC = 14 m, find the area of the flower bed. (ABCD is a quadrant, with O at centre)

18. A round table cover has six equal designs as shown. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm². (Use √3 = 1.7)

19. On a square handkerchief, nine circular designs, each of radius 7 cm, are made. Find the area of the remaining portion of the handkerchief. (Use π = 22/7)

10. Sector vs. Segment — Quick Comparison

Property	Sector	Segment
Bounded by	Two radii + one arc	One chord + one arc
Has vertex at centre?	Yes	No
Area formula	(θ/360°) × πr²	(θ/360°)πr² − (1/2)r² sin θ
Perimeter	2r + arc length	Chord + arc length
Special case (θ=180°)	Semicircle	Semicircular region without triangle
Used in combinations?	Frequently	Yes, especially petal/leaf shapes