CBSE X Maths — Areas Related to Circles: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit VI: Mensuration

Areas Related to Circles

A sector is a “pizza slice” of a circle; a segment is the region cut off by a chord. Both are found using the central angle as a fraction of the full circle.

1. Sector and Arc Length

Area of sector = (θ/360°) × πr²

Arc length = (θ/360°) × 2πr

Worked example — area of a sector Find the area of a sector of radius 14 cm and central angle 90°.
Area = (90/360) × (22/7) × 196 = (1/4) × 616 = 154 cm²

Worked example — arc length Find the length of an arc of a circle of radius 21 cm subtending 60° at the centre.
Length = (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm

2. Area of a Segment

Area of segment = Area of sector − Area of triangle

(This syllabus restricts segment problems to central angles of 60°, 90°, or 120°.)

Worked example — area of a segment Find the area of the segment of a circle with radius 21 cm and central angle 60°.
Sector area = (60/360) × (22/7) × 441 = 231 cm². Since the angle is 60°, the triangle formed is equilateral, with area = (√3/4)r² = (√3/4)(441) ≈ 190.95 cm².
Segment area = 231 − 190.95 ≈ 40.05 cm²

Worked example — a real-world sector Find the area swept by the minute hand of a clock, of length 14 cm, in 10 minutes.
In 10 minutes, the minute hand sweeps (10/60) × 360° = 60°.
Area = (60/360) × (22/7) × 196 = 102.67 cm²

Common mistakes to watch for

Forgetting that segment area needs the triangle area subtracted — a common error is stopping at the sector area.
Using the wrong triangle-area formula for non-60° segments — only at 60° is the triangle equilateral; at 90° use ½×base×height (a right triangle with two sides = r).
Mixing up arc length (a length) with sector area (an area) — check your units.

Key Vocabulary Recap

Sector — the region between two radii and the arc between them

Segment — the region between a chord and the arc it cuts off

Syllabus Reference

This note covers CBSE Class X Mathematics Unit VI, Areas Related to Circles: area of sectors and segments, and problems on areas and perimeter/circumference, with segment problems restricted to central angles of 60°, 90°, and 120°.

Practice questions

CBSE Class X Mathematics · Unit VI: Areas Related to Circles — Practice Question Bank (no answers)

Areas Related to Circles — Practice

Covers: sector area, arc length, segment area, and circumference. 10 questions. (Use π = 22/7 unless stated otherwise.)

Find the area of a sector of a circle with radius 14 cm and central angle 90°.

Find the length of an arc of a circle of radius 21 cm subtending an angle of 60° at the centre.

Find the area of a circle of radius 7 cm.

A sector of a circle of radius 10 cm has an area of 78.5 cm² (use π=3.14). Find the central angle.

Find the area of the segment of a circle with radius 21 cm and central angle 60°.

Find the area of a quadrant of a circle of radius 7 cm.

The circumference of a circle is 44 cm. Find its area.

Find the area swept by the minute hand of a clock, of length 14 cm, in 10 minutes.

Find the perimeter of a sector with radius 5.6 cm and arc length 8 cm.

Q10

Find the area of a sector of a circle of radius 6 cm whose central angle is 60° (use π=3.14).

More Practice

Q11

Find the area of a sector of radius 21 cm and central angle 120°.

Q12

Find the area of the segment of a circle of radius 14 cm, with central angle 90°.

Q13

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

Q14

Find the area of a ring (annulus) whose outer and inner radii are 14 cm and 7 cm respectively.

Q15

The wheel of a vehicle has a diameter of 84 cm. Find how many revolutions it makes to travel 792 m.

Q16

Find the radius of a circle in which a central angle of 60° describes an arc of length 22 cm.

Q17

Find the area of a sector whose arc length is 22 cm and radius is 14 cm, using Area = ½ × arc × radius.

Q18

A horse is tied to a peg at one corner of a square-shaped grass field by a rope of length 7 m. Find the area of the field the horse can graze (assume a quarter-circle sector).

Q19

Find the area of a sector with radius 3.5 cm and arc length 2.2 cm.

Q20

Two circles of radii 19 cm and 9 cm touch each other externally. Find the radius of a circle whose circumference equals the sum of the circumferences of the two circles.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit VI: Areas Related to Circles — Solutions

Areas Related to Circles — Practice: Solutions

Sector r=14, angle=90°.

Area = (θ/360)πr²

(90/360)(22/7)(196)

✓154 cm²

Arc, r=21, angle=60°.

Length = (θ/360)×2πr

(60/360)(2)(22/7)(21)

✓22 cm

Area of circle r=7.

Area = πr²

(22/7)(49)

✓154 cm²

Area 78.5, r=10, π=3.14. Find angle.

θ = Area×360/(πr²)

78.5×360/(3.14×100)

✓90°

Segment r=21, angle=60°.

Sector area

(60/360)(22/7)(441)=231 cm²

Triangle area (equilateral, since angle=60°): (√3/4)r²

(√3/4)(441)≈190.95 cm²

✓Segment area ≈ 231 − 190.95 ≈ 40.05 cm²

Quadrant r=7.

Quadrant = sector with 90°

(90/360)(22/7)(49)

✓38.5 cm²

Circumference 44. Find area.

2πr=44 → r=44/(2×22/7)

r=7 cm

Area=πr²

(22/7)(49)

✓154 cm²

Minute hand 14cm, 10 minutes.

Angle swept in 10 min = (10/60)×360°=60°

Area=(60/360)(22/7)(196)

✓≈102.67 cm²

Perimeter, r=5.6, arc=8.

Perimeter = 2r + arc length

2(5.6)+8

✓19.2 cm

Q10

Sector r=6, angle=60°, π=3.14.

Area = (θ/360)πr²

(60/360)(3.14)(36)

✓18.84 cm²

More Practice — Solutions

Q11

Sector r=21, angle=120°.

Area=(θ/360)πr²

(120/360)(22/7)(441)

✓462 cm²

Q12

Segment r=14, angle=90°.

Sector area

(90/360)(22/7)(196)=154 cm²

Triangle area (right-angled, two sides=r): ½×14×14

98 cm²

✓Segment = 154−98 = 56 cm²

Q13

Chord r=10, right angle, π=3.14.

Sector area

(90/360)(3.14)(100)=78.5 cm²

Triangle area: ½×10×10

50 cm²

✓Segment = 78.5−50 = 28.5 cm²

Q14

Ring, outer14, inner7.

Area=π(R²−r²)

(22/7)(196−49)

✓462 cm²

Q15

Wheel diameter84cm, travel792m.

Circumference=πd=(22/7)(84)=264cm=2.64m

Revolutions=792÷2.64

✓300 revolutions

Q16

Angle60°, arc22cm. Find radius.

Arc=(θ/360)2πr → r=arc×360/(θ×2π)

22×21/44

✓r = 10.5 cm

Q17

Sector area from arc22,r=14.

Area=½×arc×r

½(22)(14)

✓154 cm²

Q18

Horse tied by 7m rope at corner.

Quarter circle: Area=(90/360)πr²

(1/4)(22/7)(49)

✓38.5 m²

Q19

Sector r=3.5, arc=2.2.

Area=½×arc×r

½(2.2)(3.5)

✓3.85 cm²

Q20

Radii19,9. Find equivalent-circumference radius.

2πR = 2π(19)+2π(9) = 2π(28)

R=28

✓R = 28 cm

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