A sector is the part of a circle enclosed by two radii and the corresponding arc. A segment is the part of a circle enclosed by a chord and its corresponding arc.
Circles become exam-heavy when a diagram hides a sector, segment, quadrant, or swept region inside a real object. Important Questions Class 10 Maths Chapter 11 help students practise arc length, sector area, segment area, clock hands, horse grazing, umbrella ribs, brooch design, car wipers, and lighthouse warning zones. The 2026 NCERT chapter Areas Related to Circles focuses on sectors and segments, using θ in degrees and radius r for formula-based questions.
Key Takeaways
- Sector Area: Area of sector = θ/360 × πr².
- Arc Length: Length of arc = θ/360 × 2πr.
- Segment Area: Area of segment = Area of sector − Area of corresponding triangle.
- Major Region: Major sector or major segment = Area of circle − corresponding minor region.
Important Questions Class 10 Maths Chapter 11 Structure 2026
| Concept |
Formula |
Key Variables |
| Arc length |
l = θ/360 × 2πr |
θ = central angle, r = radius |
| Sector area |
A = θ/360 × πr² |
θ = central angle, r = radius |
| Segment area |
Segment = sector − triangle |
Sector angle, chord, triangle area |
Areas Related to Circles Class 10 Chapter Overview
The chapter uses only the parts of a circle needed for sector and segment calculations. A minor sector has angle θ, while the matching major sector has angle 360° − θ.
Students also meet real-life circular regions. Clock hands, wipers, umbrellas, brooches, grazing fields, and lighthouse beams all create sector-shaped areas.
Class 10 Maths Chapter 11 Important Questions On Sector And Segment
A sector comes from two radii and an arc. A segment comes from a chord and its arc.
The same circle can have both minor and major sectors. Unless stated otherwise, “sector” and “segment” mean the minor part in NCERT.
Q1. What Is A Sector Of A Circle?
A sector is the part of a circular region enclosed by two radii and the corresponding arc.
The angle between the two radii is called the angle of the sector. If the angle is small, the region is the minor sector.
Example: A pizza slice with two straight sides and one curved side represents a sector.
Q2. What Is A Segment Of A Circle?
A segment is the part of a circular region enclosed by a chord and the corresponding arc.
A chord divides a circle into two segments. The smaller part is called the minor segment.
Example: A chord AB and minor arc AB enclose the minor segment APB.
Q3. Difference Between Sector And Segment Of Circle Class 10
A sector uses two radii, while a segment uses one chord and one arc.
A sector contains the centre if it is the minor sector with angle below 180°. A segment need not contain the centre.
Example: OAPB is a sector, while APB is a segment.
Q4. What Are Minor Sector And Major Sector?
A minor sector is the smaller sector, while a major sector is the remaining larger sector.
If the minor sector angle is θ, the major sector angle is 360° − θ. Both together form the complete circle.
Example: If θ = 120°, major sector angle = 240°.
Q5. What Are Minor Segment And Major Segment?
A minor segment is the smaller region cut by a chord, while a major segment is the larger region.
Both segments together form the entire circular region. The major segment area equals circle area minus minor segment area.
Formula:
Major segment area = πr² − Minor segment area
Area Of Sector Class 10 Questions With Answers
Sector area depends on the fraction θ/360 of the full circle. A 90° sector is one-fourth of the circle.
Always check whether the question asks for a minor sector, major sector, or quadrant.
Q6. What Is The Area Of Sector Formula Class 10?
Area of sector = θ/360 × πr².
Here, r is the radius and θ is the central angle in degrees. The formula comes from the unitary method.
Formula:
Area of sector = θ/360 × πr²
Q7. Find The Area Of A Sector With Radius 6 cm And Angle 60°.
The area of the sector is 66/7 cm².
- Given Data:
r = 6 cm
θ = 60°
π = 22/7
- Formula Used:
Area of sector = θ/360 × πr²
- Calculation:
Area = 60/360 × 22/7 × 6²
Area = 1/6 × 22/7 × 36
Area = 132/14 = 66/7
Final Result: 66/7 cm²
Q8. Find The Area Of A Sector With Radius 4 cm And Angle 30°. Use π = 3.14.
The area of the sector is 4.19 cm² approximately.
- Given Data:
r = 4 cm
θ = 30°
π = 3.14
- Formula Used:
Area of sector = θ/360 × πr²
- Calculation:
Area = 30/360 × 3.14 × 4²
Area = 1/12 × 3.14 × 16
Area = 4.186...
Final Result: 4.19 cm²
Q9. Find The Area Of The Major Sector If Radius Is 4 cm And Minor Sector Angle Is 30°.
The area of the major sector is 46.05 cm² approximately.
- Given Data:
r = 4 cm
Minor sector area = 4.19 cm²
π = 3.14
- Circle area:
πr² = 3.14 × 4²
= 3.14 × 16
= 50.24 cm²
- Major sector area:
50.24 − 4.19 = 46.05 cm²
Final Result: 46.05 cm²
Q10. Find The Area Of A Quadrant Whose Circumference Is 22 cm.
The area of the quadrant is 38.5 cm².
- Given Data:
Circumference = 22 cm
π = 22/7
- Use circumference:
2πr = 22
2 × 22/7 × r = 22
- Solve for radius:
r = 22 × 7 / 44
r = 3.5 cm
- Area of quadrant:
Area = 1/4 × πr²
Area = 1/4 × 22/7 × 3.5 × 3.5
Area = 38.5 cm²
Final Result: 38.5 cm²
Length Of Arc Class 10 Questions
Arc length is the curved boundary of a sector. It uses the same fraction θ/360, but applies it to circumference.
Students often mix arc length with sector area. Arc length has unit cm, while sector area has cm².
Q11. What Is The Length Of Arc Formula Class 10?
Length of arc = θ/360 × 2πr.
The formula gives the curved length of the sector. It uses circumference instead of area.
Formula:
Arc length = θ/360 × 2πr
Q12. In A Circle Of Radius 21 cm, An Arc Subtends 60° At The Centre. Find The Arc Length.
The length of the arc is 22 cm.
- Given Data:
r = 21 cm
θ = 60°
π = 22/7
- Formula Used:
Arc length = θ/360 × 2πr
- Calculation:
Arc length = 60/360 × 2 × 22/7 × 21
Arc length = 1/6 × 132
Arc length = 22 cm
Final Result: 22 cm
Q13. Find The Area Of The Sector Formed By The Same Arc Of Radius 21 cm And Angle 60°.
The area of the sector is 231 cm².
- Given Data:
r = 21 cm
θ = 60°
π = 22/7
- Formula Used:
Area of sector = θ/360 × πr²
- Calculation:
Area = 60/360 × 22/7 × 21²
Area = 1/6 × 22/7 × 441
Area = 231 cm²
Final Result: 231 cm²
Q14. Why Does Arc Length Use 2πr But Sector Area Uses πr²?
Arc length uses circumference, while sector area uses the area of the circle.
An arc is a curved line, so it takes a fraction of 2πr. A sector is a region, so it takes a fraction of πr².
Example: A 60° arc is 1/6 of circumference.
Area Of Segment Class 10 Important Questions
A segment is found by subtracting a triangle from a sector. The triangle is formed by two radii and the chord.
For 60°, 90°, and 120° questions, students usually need special triangle area methods.
Q15. What Is The Area Of Segment Formula Class 10?
Area of segment = Area of sector − Area of corresponding triangle.
The corresponding triangle is formed by the two radii and the chord. In ∆OAB, OA and OB are radii.
Formula:
Area of segment APB = Area of sector OAPB − Area of ∆OAB
Q16. A Chord Of A Circle Of Radius 10 cm Subtends 90° At The Centre. Find The Minor Segment Area. Use π = 3.14.
The minor segment area is 28.5 cm².
- Given Data:
r = 10 cm
θ = 90°
π = 3.14
- Area of sector:
Sector = 90/360 × πr²
Sector = 1/4 × 3.14 × 100
Sector = 78.5 cm²
- Area of triangle:
∆OAB is right-angled at O.
Area = 1/2 × 10 × 10 = 50 cm²
- Segment area:
Segment = 78.5 − 50
Segment = 28.5 cm²
Final Result: 28.5 cm²
Q17. For The Same Circle, Find The Major Sector Area.
The major sector area is 235.5 cm².
- Given Data:
r = 10 cm
Minor sector angle = 90°
π = 3.14
- Circle area:
πr² = 3.14 × 10²
= 314 cm²
- Minor sector area:
78.5 cm²
- Major sector area:
314 − 78.5 = 235.5 cm²
Final Result: 235.5 cm²
Q18. A Chord Of Radius 21 cm Subtends 60° At The Centre. Find The Segment Area.
The segment area is 231 − 441√3/4 cm².
- Given Data:
r = 21 cm
θ = 60°
π = 22/7
- Sector area:
Sector = 60/360 × 22/7 × 21²
Sector = 231 cm²
- Triangle type:
∆OAB has OA = OB = 21 cm and ∠AOB = 60°.
So ∆OAB is equilateral.
- Triangle area:
Area of ∆OAB = √3/4 × side²
= √3/4 × 21²
= 441√3/4 cm²
- Segment area:
Segment = 231 − 441√3/4
Final Result: 231 − 441√3/4 cm²
Q19. A Chord Of Radius 15 cm Subtends 60° At The Centre. Find The Minor Segment Area. Use π = 3.14 And √3 = 1.73.
The minor segment area is 20.44 cm² approximately.
- Given Data:
r = 15 cm
θ = 60°
π = 3.14
√3 = 1.73
- Sector area:
Sector = 60/360 × 3.14 × 15²
Sector = 1/6 × 3.14 × 225
Sector = 117.75 cm²
- Triangle area:
∆OAB is equilateral.
Area = √3/4 × 15²
Area = 1.73/4 × 225
Area = 97.31 cm²
- Segment area:
Segment = 117.75 − 97.31
Segment = 20.44 cm²
Final Result: 20.44 cm²
Q20. A Chord Of Radius 15 cm Subtends 60° At The Centre. Find The Major Segment Area.
The major segment area is 686.06 cm² approximately.
- Given Data:
r = 15 cm
Minor segment = 20.44 cm²
π = 3.14
- Circle area:
πr² = 3.14 × 15²
= 3.14 × 225
= 706.5 cm²
- Major segment area:
Major segment = Circle area − Minor segment
= 706.5 − 20.44
= 686.06 cm²
Final Result: 686.06 cm²
Q21. A Chord Of Radius 12 cm Subtends 120° At The Centre. Find The Segment Area. Use π = 3.14 And √3 = 1.73.
The segment area is 88.44 cm² approximately.
- Given Data:
r = 12 cm
θ = 120°
π = 3.14
√3 = 1.73
- Sector area:
Sector = 120/360 × 3.14 × 12²
Sector = 1/3 × 3.14 × 144
Sector = 150.72 cm²
- Triangle area:
Area of ∆OAB = 1/2 × r² × sin 120°
sin 120° = √3/2
Area = 1/2 × 144 × √3/2
Area = 36√3 = 62.28 cm²
- Segment area:
Segment = 150.72 − 62.28
Segment = 88.44 cm²
Final Result: 88.44 cm²
Area Swept By Minute Hand Questions
A clock hand sweeps a sector as it moves. The angle depends on time.
In 60 minutes, the minute hand covers 360°. In 5 minutes, it covers 30°.
Q22. The Length Of The Minute Hand Is 14 cm. Find The Area Swept In 5 Minutes.
The area swept by the minute hand is 154/3 cm².
- Given Data:
r = 14 cm
Time = 5 minutes
π = 22/7
- Angle swept:
60 minutes = 360°
5 minutes = 5/60 × 360° = 30°
- Formula Used:
Area of sector = θ/360 × πr²
- Calculation:
Area = 30/360 × 22/7 × 14²
Area = 1/12 × 22/7 × 196
Area = 154/3 cm²
Final Result: 154/3 cm²
Q23. What Angle Does A Minute Hand Sweep In 20 Minutes?
A minute hand sweeps 120° in 20 minutes.
- Full circle in 60 minutes:
360°
- Angle in 1 minute:
360°/60 = 6°
- Angle in 20 minutes:
20 × 6° = 120°
Final Result: 120°
Q24. A Minute Hand Of Length 21 cm Moves For 10 Minutes. Find The Area Swept.
The area swept is 231 cm².
- Given Data:
r = 21 cm
Time = 10 minutes
π = 22/7
- Angle swept:
10 × 6° = 60°
- Sector area:
Area = 60/360 × 22/7 × 21²
Area = 1/6 × 22/7 × 441
Area = 231 cm²
Final Result: 231 cm²
Horse Grazing Problem Class 10
A horse tied to a corner of a square field grazes a quadrant. The corner acts as the centre.
When the rope length changes, the grazing area changes with r².
Q25. A Horse Is Tied At A Corner Of A Square Field Of Side 15 m With A 5 m Rope. Find The Grazing Area.
The horse can graze 19.625 m².
- Given Data:
Rope length = 5 m
Field corner angle = 90°
π = 3.14
- Shape formed:
Grazing area = quadrant of radius 5 m
- Calculation:
Area = 90/360 × πr²
Area = 1/4 × 3.14 × 5²
Area = 19.625 m²
Final Result: 19.625 m²
Q26. Find The Increase In Grazing Area If The Rope Becomes 10 m Long.
The increase in grazing area is 58.875 m².
- Area with 10 m rope:
Area = 1/4 × 3.14 × 10²
Area = 78.5 m²
- Area with 5 m rope:
Area = 19.625 m²
- Increase:
78.5 − 19.625 = 58.875 m²
Final Result: 58.875 m²
Q27. Why Is The Horse Grazing Region A Quadrant?
The grazing region is a quadrant because the horse is tied at a corner of a square.
A corner forms an angle of 90°. The rope traces a circular arc inside that 90° region.
Example: A 5 m rope forms a quadrant of radius 5 m.
Umbrella Ribs Area Problem Class 10
Umbrella ribs divide a circular umbrella into equal sectors. The central angle comes from 360° divided by the number of ribs.
The NCERT umbrella question uses 8 equally spaced ribs and radius 45 cm.
Q28. An Umbrella Has 8 Equally Spaced Ribs. Find The Area Between Two Consecutive Ribs If Radius Is 45 cm.
The area between two consecutive ribs is 795.54 cm² approximately.
- Given Data:
Number of ribs = 8
r = 45 cm
π = 22/7
- Angle between two ribs:
θ = 360°/8 = 45°
- Area between ribs:
Area = 45/360 × 22/7 × 45²
Area = 1/8 × 22/7 × 2025
Area = 44550/56
Final Result: 795.54 cm² approximately
Q29. Why Is The Area Between Two Umbrella Ribs A Sector?
The area between two umbrella ribs is a sector because two ribs act as radii.
The curved edge of the umbrella forms the arc. The two ribs and the arc enclose a sector.
Example: With 8 ribs, each sector has angle 45°.
Brooch Sector Problem Class 10
A circular brooch uses circumference and diameters together. Wire length includes the circle boundary and the internal diameters.
The same figure also divides the circle into equal sectors.
Q30. A Brooch Has Diameter 35 mm And 5 Diameters Inside. Find The Total Length Of Silver Wire.
The total wire length is 285 mm.
- Given Data:
Diameter = 35 mm
Number of internal diameters = 5
π = 22/7
- Circumference of circle:
C = πd
C = 22/7 × 35 = 110 mm
- Length of 5 diameters:
5 × 35 = 175 mm
- Total wire length:
110 + 175 = 285 mm
Final Result: 285 mm
Q31. The Same Brooch Is Divided Into 10 Equal Sectors. Find Area Of Each Sector.
The area of each sector is 96.25 mm².
- Given Data:
Diameter = 35 mm
Radius = 17.5 mm
Number of sectors = 10
π = 22/7
- Circle area:
Area = πr²
Area = 22/7 × 17.5 × 17.5
Area = 962.5 mm²
- Area of each sector:
962.5/10 = 96.25 mm²
Final Result: 96.25 mm²
Car Wiper Area Problem Class 10
A car wiper sweeps a sector with radius equal to blade length. Two non-overlapping wipers give twice the sector area.
NCERT uses two blades of length 25 cm and angle 115°.
Q32. A Car Has Two Wipers Which Do Not Overlap. Each Blade Is 25 cm And Sweeps 115°. Find The Total Area Cleaned.
The total area cleaned is 1254.96 cm² approximately.
- Given Data:
Number of wipers = 2
r = 25 cm
θ = 115°
π = 22/7
- Area cleaned by one wiper:
Area = 115/360 × 22/7 × 25²
- Calculation:
Area = 115/360 × 22/7 × 625
Area = 627.48 cm² approximately
- Total area cleaned:
2 × 627.48 = 1254.96 cm²
Final Result: 1254.96 cm² approximately
Q33. Why Is A Wiper Sweep Treated As A Sector?
A wiper sweep is treated as a sector because the blade rotates about a fixed point.
The blade length acts as the radius. The angle swept by the blade becomes the central angle.
Example: A 25 cm blade sweeping 115° forms a sector of radius 25 cm.
Lighthouse Sector Problem Class 10
A lighthouse beam spreads over a sector of the sea. Its radius is the distance reached by the light.
The angle of the beam gives the sector angle.
Q34. A Lighthouse Spreads Red Light Over 80° To A Distance Of 16.5 km. Find The Area Warned. Use π = 3.14.
The warned sea area is 189.97 km² approximately.
- Given Data:
r = 16.5 km
θ = 80°
π = 3.14
- Formula Used:
Area of sector = θ/360 × πr²
- Calculation:
Area = 80/360 × 3.14 × 16.5²
Area = 2/9 × 3.14 × 272.25
Area = 189.97 km²
Final Result: 189.97 km² approximately
Q35. Why Does The Lighthouse Question Use Sector Area?
The lighthouse question uses sector area because the light spreads between two boundary rays.
The lighthouse is the centre of the sector. The distance reached by light is the radius.
Example: An 80° light beam forms a sector of angle 80°.
Round Table Cover Design Questions
The round table cover question combines sector and triangle areas. The shaded design appears in each of six equal parts.
A 60° sector is common here because six equal sectors divide 360°.
Q36. A Round Table Cover Has 6 Equal Designs And Radius 28 cm. What Is The Angle Of Each Sector?
The angle of each sector is 60°.
- Complete circle angle:
360°
- Number of equal designs:
6
- Angle of each sector:
360°/6 = 60°
Final Result: 60°
Q37. Find The Area Of One 60° Sector Of Radius 28 cm. Use π = 22/7.
The area of one sector is 410.67 cm² approximately.
- Given Data:
r = 28 cm
θ = 60°
π = 22/7
- Sector area:
Area = 60/360 × 22/7 × 28²
Area = 1/6 × 22/7 × 784
Area = 2464/6
Final Result: 410.67 cm² approximately
Q38. If A Design Uses The Segment In Each 60° Sector, How Is Its Area Found?
The design area is found by subtracting the equilateral triangle from the 60° sector.
- Sector angle:
60°
- Triangle formed by two radii:
Equilateral triangle of side r
- Segment formula:
Segment area = 60° sector area − √3/4 × r²
Example: For r = 28 cm, triangle area = √3/4 × 28².
NCERT Class 10 Maths Chapter 11 Questions On Formula Selection
Many wrong answers come from selecting the wrong circular part. Sector, segment, arc, and circumference represent different measures.
Check the unit first. Length uses cm, while area uses cm².
Q39. Which Formula Should Students Use For Area Of Sector?
Students should use Area of sector = θ/360 × πr².
This formula applies when two radii and an arc enclose the required region. The angle θ must be in degrees.
Example: For θ = 90°, the sector is one-fourth of the circle.
Q40. Which Formula Should Students Use For Area Of Segment?
Students should use Area of segment = Area of sector − Area of triangle.
The triangle is formed by the two radii and the chord. This formula applies to minor segment questions.
Example: For θ = 60°, the triangle is equilateral.
Q41. How Can Students Identify Whether A Question Needs Arc Length Or Sector Area?
Students should check whether the question asks for curved boundary or region.
Arc length is a line measure and uses cm. Sector area is a region measure and uses cm².
Example: “Length of arc” needs θ/360 × 2πr.
Q42. What Is The Correct Option For Area Of A Sector Of Angle p And Radius R?
The correct option is p/360 × πR².
- Full circle angle:
360°
- Full circle area:
πR²
- Sector area for angle p:
p/360 × πR²
Final Result: p/360 × πR²
Important Questions Class 10 Maths All Chapters