Class 10 — Mathematics (NCERT)

Areas Related to Circles: Sector and Segment

Class 10

✓By the end of this lesson students will be able to define a sector and a segment of a circle.
✓By the end of this lesson students will be able to calculate the area of a sector of a circle.
✓By the end of this lesson students will be able to calculate the length of an arc of a circle.
✓By the end of this lesson students will be able to calculate the area of a segment of a circle.
✓By the end of this lesson students will be able to solve problems involving areas of sectors and segments in various contexts.

Key concepts

Sector of a Circle

A sector of a circle is the region enclosed by two radii and the corresponding arc of the circle. It is like a 'slice of pizza'. A minor sector is the smaller region, and a major sector is the larger region.

Area of a sector with angle θ (in degrees) = (θ / 360°) × πr²

Length of an Arc

The length of an arc is the measure of the curved boundary of a sector. It is a part of the circumference of the circle.

Length of an arc with angle θ (in degrees) = (θ / 360°) × 2πr

Segment of a Circle

A segment of a circle is the region enclosed by a chord and its corresponding arc. A minor segment is the smaller region, and a major segment is the larger region.

Area of a minor segment = Area of the corresponding minor sector - Area of the triangle formed by the two radii and the chord.\nArea of a major segment = Area of the circle - Area of the minor segment.\nArea of triangle with two sides 'r' and included angle 'θ' = (1/2)r²sinθ

Key facts to remember

1Area of a sector with angle θ (in degrees) = (θ / 360°) × πr²
2Length of an arc with angle θ (in degrees) = (θ / 360°) × 2πr
3Area of a minor segment = Area of the corresponding minor sector - Area of the triangle formed by the two radii and the chord.
4Area of a major segment = Area of the circle - Area of the minor segment.
5Area of a triangle with two sides 'r' and included angle 'θ' = (1/2)r²sinθ.
6Area of a circle = πr².

Worked examples

Example 1

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 90°. Also, find the length of the corresponding arc. (Use π = 22/7)

IGiven: Radius (r) = 7 cm, Angle of sector (θ) = 90°.

IIArea of sector = (θ / 360°) × πr²

III= (90° / 360°) × (22/7) × (7)²

IV= (1/4) × (22/7) × 49

V= (1/4) × 22 × 7

VI= (1/4) × 154

VII= 38.5 cm²

VIIILength of arc = (θ / 360°) × 2πr

9= (90° / 360°) × 2 × (22/7) × 7

10= (1/4) × 2 × 22

11= (1/4) × 44

12= 11 cm

Answer

The area of the sector is 38.5 cm² and the length of the arc is 11 cm.

Remember to use the specified value of π.

Example 2

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use π = 3.14)

IGiven: Radius (r) = 10 cm, Angle subtended at the centre (θ) = 90°.

IIArea of minor sector = (θ / 360°) × πr²

III= (90° / 360°) × 3.14 × (10)²

IV= (1/4) × 3.14 × 100

V= (1/4) × 314

VI= 78.5 cm²

VIIThe triangle formed by the radii and the chord is a right-angled triangle (ΔOAB).

VIIIArea of ΔOAB = (1/2) × base × height = (1/2) × OA × OB

9= (1/2) × 10 × 10

10= (1/2) × 100

11= 50 cm²

12Area of minor segment = Area of minor sector - Area of ΔOAB

13= 78.5 - 50

14= 28.5 cm²

Answer

The area of the corresponding minor segment is 28.5 cm².

For a right-angled triangle formed by two radii, the area is simply (1/2)r².

Example 3

A chord of a circle of radius 14 cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment. (Use π = 22/7 and √3 = 1.73)

IGiven: Radius (r) = 14 cm, Angle subtended at the centre (θ) = 120°.

IIArea of minor sector = (θ / 360°) × πr²

III= (120° / 360°) × (22/7) × (14)²

IV= (1/3) × (22/7) × 196

V= (1/3) × 22 × 28

VI= 616/3 cm² ≈ 205.33 cm²

VIIArea of the triangle formed by the radii and the chord (ΔOAB) = (1/2)r²sinθ

VIII= (1/2) × (14)² × sin(120°)

9= (1/2) × 196 × (√3 / 2)

10= 98 × (1.73 / 2)

11= 49 × 1.73

12= 84.77 cm²

13Area of minor segment = Area of minor sector - Area of ΔOAB

14= (616/3) - 84.77

15= 205.33 - 84.77

16= 120.56 cm² (approx.)

Answer

The area of the corresponding minor segment is approximately 120.56 cm².

For angles like 120°, remember that sin(120°) = sin(180°-60°) = sin(60°) = √3/2.

Common mistakes

✗Confusing the formulas for the area of a sector and the length of an arc.
✗Incorrectly calculating the area of the triangle for the segment, especially for angles other than 90°.
✗Forgetting to subtract the area of the triangle when finding the area of a segment.
✗Using an incorrect value of π or √3 when a specific value is provided in the question.
✗Calculation errors, particularly when dealing with fractions or decimals.

Exam tips

★Always draw a neat diagram to visualise the problem. This helps in identifying the given parts and what needs to be found.
★Carefully read the question to determine whether you need to find the area of a minor sector/segment or a major sector/segment.
★Write down all given values (radius, angle, value of π) before starting the calculations.
★Show all steps clearly, especially the calculation of the area of the triangle for segments, as this carries marks.
★Double-check your calculations and ensure that the final answer includes appropriate units (e.g., cm², cm).

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