Chapter – 11

Areas Related to Circles

In this post we have given the detailed notes of class 10 Maths Chapter 11 (Areas Related to Circles) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 11: Areas Related to Circles

Perimeter and Area of a Circle – A Recap

The distance covered by travelling once around a circle is called its perimeter, usually called its circumference. The circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter π (read as ‘pi’).

Circumference / Diameter = π

Circumference = π × Diameter = π × 2r = 2πr  (where r is the radius of the circle)

Note: The numerical value of π is generally taken as 22/7 or 3.14 (approximately), since π is an irrational number whose decimal expansion is non-terminating and non-repeating.

Area of a circle = π times the square of its radius, i.e., A = πr²

Example 1

Question: The cost of fencing a circular field at the rate of ₹24 per metre is ₹5280. The field is to be ploughed at the rate of ₹0.50 per square metre. Find the cost of ploughing the field. (Take π = 22/7)

Solution: Length of the fence (i.e. circumference) = Total cost / Rate = 5280 / 24 = 220 m

So, circumference of the field = 220 m

Let the radius of the field be r metres, then:

2πr = 220

2 × 22/7 × r = 220

r = (220 × 7) / (2 × 22) = 35 m

So, the radius of the field is 35 m.

Area of the field = πr² = 22/7 × 35 × 35 = 3850 m²

Cost of ploughing 1 m² = ₹0.50

Total cost of ploughing the field = 3850 × 0.50 = ₹1925

Example 2

Question: The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Solution: Circumference of first circle = 2π(19)

Circumference of second circle = 2π(9)

Let R be the radius of the required circle. Then:

2πR = 2π(19) + 2π(9) = 2π(19 + 9) = 2π(28)

∴ R = 28 cm

Areas of Sector and Segment of a Circle

Sector: The region enclosed by two radii and the corresponding arc is called a sector of the circle. The angle between the two radii is called the angle of the sector.

Segment: The region enclosed by a chord and the corresponding arc is called a segment of the circle.

Note: Unless stated otherwise, ‘sector’ and ‘segment’ refer to the minor sector and minor segment respectively.

Area of a Sector

Let OAPB be a sector of a circle with centre O and radius r, and let the angle of the sector (∠AOB) be θ (in degrees).

We know that the area of the complete circle (a sector with angle 360°) is πr². Using the unitary method:

When the angle at the centre is 360°, area of the sector (whole circle) = πr²

When the angle at the centre is 1°, area of the sector = πr² / 360

When the angle at the centre is θ°, area of the sector = θ/360 × πr²

Area of sector with angle θ = θ/360 × πr²

Length of an Arc

Similarly, using the fact that the length of the complete circumference (360° arc) is 2πr, the length of an arc corresponding to a sector of angle θ is:

Length of arc = θ/360 × 2πr

Area of a Segment

The area of a segment can be found by subtracting the area of the corresponding triangle from the area of the sector.

Area of segment = Area of the corresponding sector − Area of the corresponding triangle

Area of minor segment = θ/360 × πr² − Area of ΔOAB

Area of major sector = πr² − Area of minor sector, and Area of major segment = πr² − Area of minor segment.

Example 3

Question: Find the area of a sector of a circle with radius 4 cm and angle 30°. Also, find the area of the corresponding major sector. (Take π = 3.14)

Solution: Radius r = 4 cm, θ = 30°

Area of the minor sector = θ/360 × πr² = 30/360 × 3.14 × 4 × 4

= 1/12 × 3.14 × 16 = 50.24/12 = 4.19 cm² (approximately)

Area of the circle = πr² = 3.14 × 16 = 50.24 cm²

Area of major sector = Area of circle − Area of minor sector = 50.24 − 4.19 = 46.05 cm² ≈ 46.1 cm²

Example 4

Question: Find the area of the segment of a circle, given that the radius of the circle is 21 cm and the angle of the corresponding sector is 120°. (Take π = 22/7)

Solution: Radius r = 21 cm, ∠AOB = θ = 120°

Area of segment AYB = Area of sector OAYB − Area of ΔOAB  …(1)

Area of sector OAYB = 120/360 × 22/7 × 21 × 21 = 462 cm²  …(2)

To find the area of ΔOAB, draw OM ⊥ AB. Since OA = OB, ΔAMO ≅ ΔBMO (RHS congruence). So M is the midpoint of AB, and ∠AOM = ∠BOM = ½ × 120° = 60°.

Let OM = x cm. Then in right triangle OMA:

OM/OA = cos 60° = ½  ⇒  x/21 = ½  ⇒  x = 21/2 cm

AM/OA = sin 60° = √3/2  ⇒  AM = 21√3/2 cm

So AB = 2 × AM = 21√3 cm

Area of ΔOAB = ½ × AB × OM = ½ × 21√3 × 21/2 = 441√3/4 cm²  …(3)

From (1), (2) and (3):

Area of segment = {462 − 441√3/4} cm² = 21/4 (88 − 21√3) cm²

Example 5

Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of (i) the corresponding minor segment, and (ii) the corresponding major sector. (Take π = 3.14)

Solution: Radius r = 10 cm, θ = 90°

(i) Area of minor segment:

Area of minor segment = Area of sector OAPB − Area of ΔOAB

= θ/360 × πr² − ½ × r × r  (since ΔOAB is right-angled at O)

= 90/360 × 3.14 × 10 × 10 − ½ × 10 × 10

= 78.5 − 50 = 28.5 cm²

(ii) Area of major sector:

Area of major sector = πr² − Area of minor sector

= 3.14 × 100 − 78.5 = 314 − 78.5 = 235.5 cm²

Example 6

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

Solution: Radius r = 21 cm, θ = 60°

(i) Length of arc = θ/360 × 2πr = 60/360 × 2 × 22/7 × 21 = 1/6 × 132 = 22 cm

(ii) Area of sector = θ/360 × πr² = 60/360 × 22/7 × 21 × 21 = 1/6 × 1386 = 231 cm²

(iii) Area of segment = Area of sector − Area of ΔOAB

Since θ = 60° and OA = OB = r, ΔOAB is equilateral, so its area = √3/4 × r² = √3/4 × 21 × 21 = (441√3)/4 cm²

Area of segment = 231 − (441√3)/4 = 231 − 190.94 ≈ 40.06 cm²

Key Points to Remember

• Circumference of a circle = 2πr; Area of a circle = πr² (where r is the radius).
• π is an irrational number, generally taken as 22/7 or 3.14 for calculations.
• A sector is the region bounded by two radii and the corresponding arc; a segment is the region bounded by a chord and the corresponding arc.
• Area of a sector with angle θ = θ/360 × πr².
• Length of an arc with angle θ = θ/360 × 2πr.
• Area of a segment = Area of the corresponding sector − Area of the corresponding triangle.
• Area of major sector = πr² − Area of minor sector; Area of major segment = πr² − Area of minor segment.
• When the angle of the sector is 60°, the triangle formed with the two radii is equilateral, so its area = √3/4 × r².

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