1
The part bounded by two radii and an arc is —
💡 Hint:
The edges are radii.
Explanation:
It is a sector.
Class 10 · Maths · Chapter 11 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Areas Related to Circles
Areas Related to Circles
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1. Read — Sector and segment · arc · area · Exercise 11.1 · summary, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
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4. The memory figure shows a minor sector in a circle, with radius r, angle θ and the arc strip.
In NCERT 2026-27 this is Chapter 11; in Bihar’s older book the same Areas Related to Circles is Chapter 12. Progress stays in this browser.
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- 1 Sector and segment
- 2 Area of a sector
- 3 Length of an arc
- 4 Segment = sector − triangle
- 5 Quadrant and a corner pasture
- 6 Wire, umbrella, wipers, light
- 7 Table designs and option (D)
- Chapter winner — every lesson at mastery ★
🎬 Shorts
This chapter has no Short on the channel yet, so ten empty cards are not shown. The notes, diagrams and checks are complete.
📄 PDF download — notes + answer key
Sector and segment
त्रिज्यखंड और वृत्तखंड · Opening of NCERT 11.1
Two names, two boundariesNotes
Two radii and the arc between them bound a sector. The angle at the centre is the angle of the sector.
The part between a chord and an arc is a segment. Unless something else is written, both names mean the minor part. The angle of the major sector is 360° − θ.
Count the boundaries on the figureActivity
In OAPB, OA and OB are radii and AB is an arc, so it is a sector. In APB, AB is a chord and the other edge is an arc, so it is a segment. Both sit in one figure. Change the name and the formula changes.
Worked exampleExample
Question: The angle at the centre is 60°. Write the angles of the minor and major sectors.
Formula: Major angle = 360° − minor angle.
Substitution: 360 − 60 = 300.
Answer: The minor angle is 60° and the major angle is 300°.
10-second revision
- Sector = two radii + arc
- Segment = chord + arc
- Assume the minor part if nothing is written
Board tip · BSEBBoard tip
In a BSEB answer write the edges of the figure first, then the name.
Board tip · CBSEBoard tip
On CBSE do not give the minor and the major the same angle.
Check your understandingall correct = mastery ★
2
Unless something else is written, sector means the major sector.
Explanation:
False — the book takes the minor part.
3
The angle of the major sector is 360° − ______.
💡 Hint:
Subtract the minor angle.
Explanation:
θ.
4
The part between a chord and an arc is —
💡 Hint:
The edge that is a chord.
Explanation:
It is a segment.
5
What is the angle of the major sector beside a minor sector of 60°?
Model answer (board format)
360° − 60° = 300°.
Check your answer:
Area of a sector
त्रिज्यखंड का क्षेत्रफल · NCERT 11.1 · formula · Exercise 11.1 question 1
The whole circle is a 360° sectorNotes
The area of a circle is πr². An angle of 1° leaves the piece πr²/360.
A sector of angle θ° = (θ/360) × πr². Question 1 of Exercise 11.1 asks this. If nothing is written, π = 22/7.
The fraction first, then πActivity
Simplify θ/360. 60/360 = 1/6. Then multiply by πr². Forgetting the square of r is the usual slip. Write the unit cm² or m², not the unit of the angle.
Worked exampleExample
Question: Radius 6 cm, angle 60°. Find the area of the sector. π = 22/7.
Formula: Area = (θ/360) × πr².
Substitution: (60/360) × (22/7) × 36 = (1/6) × (22/7) × 36 = 132/7.
Answer: 132/7 cm².
10-second revision
- Area = (θ/360) π r²
- A 60° piece is 1/6
- The unit is square centimetre
Board tip · BSEBBoard tip
In a BSEB answer show the formula line and 60/360 = 1/6 apart.
Board tip · CBSEBoard tip
On CBSE keep the π the question gives. Do not mix 22/7 and 3.14 in one answer.
Check your understandingall correct = mastery ★
1
The area of a sector of radius 6 cm and angle 60° is —
💡 Hint:
(1/6) × (22/7) × 36.
Explanation:
132/7 cm².
2
In Exercise 11.1, π = 22/7 is used when nothing else is written.
Explanation:
True — the first line of the exercise says this.
3
The area of a sector is (θ/360) × ______.
💡 Hint:
A piece of the area of the circle.
Explanation:
πr².
4
Find the area of a sector of radius 7 cm and angle 90°. π = 22/7.
Model answer (board format)
Formula (θ/360)πr². (90/360) × (22/7) × 49 = (1/4) × 22 × 7 = 77/2. The answer is 77/2 cm².
Check your answer:
Length of an arc
चाप की लंबाई · First line of the summary · question 5 (i)
The same fraction of the circumferenceNotes
The circumference is 2πr. An angle θ° cuts the same fraction on the arc as it cuts on the area.
Arc = (θ/360) × 2πr. This is a length, not a square unit. It is the first line of the summary. Question 5 of the exercise asks the arc first, then the sector, then the segment.
Keep the length unit and the area unit apartActivity
Use cm or m on an arc. Use cm² or m² on an area. Drop the 2 in 2πr and the answer is half. In question 3 the minute hand turns 360° in 60 minutes, so 5 minutes = 30°.
Worked exampleExample
Question: Radius 21 cm, angle 60°. Find the arc length. π = 22/7.
Formula: Arc = (θ/360) × 2πr.
Substitution: (60/360) × 2 × (22/7) × 21 = (1/6) × 2 × 22 × 3 = 22.
Answer: 22 cm.
| Ask for | Formula | Unit |
|---|---|---|
| Arc | (θ/360) × 2πr | cm |
| Sector | (θ/360) × πr² | cm² |
| Circumference | 2πr | cm |
| Circle | πr² | cm² |
10-second revision
- Arc = (θ/360) × 2πr
- The unit is cm, not cm²
- 5 minutes = 30°
Board tip · BSEBBoard tip
In a BSEB answer write the three parts of question 5 in one order: arc, sector, segment.
Board tip · CBSEBoard tip
On CBSE do not swap 2πr and πr². One is a length and the other is an area.
Check your understandingall correct = mastery ★
1
The length of an arc of radius 21 cm and angle 60° is —
💡 Hint:
(1/6) × 2 × (22/7) × 21.
Explanation:
22 cm.
2
Arc length is written in square units.
Explanation:
False — it is a length.
3
The minute hand turns through ______ degrees in 5 minutes.
💡 Hint:
6 degrees each minute.
Explanation:
30°.
4
The angle swept by a 14 cm minute hand in 5 minutes is —
💡 Hint:
A full turn is 60 minutes.
Explanation:
30°. The area then comes from (30/360)πr².
5
The arc formula is (θ/360) × 2πr.
Explanation:
True — the first line of the summary.
6
How much area does a minute hand of radius 14 cm sweep in 5 minutes? π = 22/7.
Model answer (board format)
5 minutes = 30°. Formula (θ/360)πr². (30/360) × (22/7) × 196 = (1/12) × 22 × 28 = 154/3. The answer is 154/3 cm².
Check your answer:
Segment = sector − triangle
वृत्तखंड = त्रिज्यखंड − त्रिभुज · Third line of the summary · questions 4, 5, 6, 7
Cut the triangle awayNotes
The minor segment is what remains after triangle OAB is removed from the sector.
Segment = sector − triangle. At 90° the triangle has area (1/2)r². At 60° both radii and the chord are equal, so the triangle is equilateral and the area is (√3/4)r². Questions 4, 6 and 7 follow this path.
Pick the triangle from the angleActivity
If the angle is 90°, then (1/2)×10×10 = 50 cm² when r = 10 cm. If it is 60°, take the value of √3 that the question gives, often 1.73. If the sector says π = 3.14, do not bring 22/7 into that question.
Worked exampleExample
Question: r = 10 cm, angle 90°, π = 3.14. Find the minor segment.
Formula: Segment = (θ/360)πr² − (1/2)r².
Substitution: (90/360) × 3.14 × 100 − 50 = 78.5 − 50 = 28.5.
Answer: 28.5 cm².
10-second revision
- Subtract, do not add
- At 90° the triangle is (1/2)r²
- At 60° the triangle is (√3/4)r²
Board tip · BSEBBoard tip
In a BSEB answer find the sector and the triangle on two separate lines, then subtract.
Board tip · CBSEBoard tip
On CBSE do not replace √3 = 1.73 and π = 3.14 with values from outside the question.
Check your understandingall correct = mastery ★
1
For r = 10 cm and angle 90°, the minor segment is —
💡 Hint:
78.5 − 50.
Explanation:
28.5 cm². π = 3.14.
2
Segment = sector + triangle.
Explanation:
False — the triangle is subtracted.
3
At 60° the equilateral triangle has area (______/4) r².
💡 Hint:
Square root of three.
Explanation:
(√3/4) r².
4
In question 4 the direct way to the major sector is —
💡 Hint:
The whole circle.
Explanation:
Subtract the minor sector from πr².
5
r = 10 cm, angle 90°, π = 3.14. Find the major sector.
Model answer (board format)
Circle = 3.14 × 100 = 314 cm². Minor sector = 78.5 cm². Major = 314 − 78.5 = 235.5 cm².
Check your answer:
Quadrant and a corner pasture
चतुर्थांश और कोने का चरागाह · Exercise 11.1 questions 2 and 8
90° is a quarter of the circleNotes
A quadrant has angle 90°, so its area is (1/4)πr². Find the radius first from the circumference 2πr.
If the circumference is 22 cm and π = 22/7, then r = 7/2 cm. That is question 2. In question 8 the rope is tied at a corner of a square, so the horse grazes a quarter circle.
If the rope grows, subtract the two quartersActivity
If the rope is 5 m, the area is (1/4)π × 25. If the rope is 10 m, it is (1/4)π × 100. The increase = larger − smaller. The side of the square is 15 m, and both 5 m and 10 m are shorter than the side, so the pasture stays a quarter circle.
Worked exampleExample
Question: Area of a quadrant of a circle whose circumference is 22 cm. π = 22/7.
Formula: 2πr = 22, then area = (1/4)πr².
Substitution: 2 × (22/7) × r = 22, so r = 7/2. Area = (1/4) × (22/7) × (49/4) = 77/8.
Answer: 77/8 cm².
10-second revision
- Quadrant = (1/4)πr²
- Circumference 22 cm gives r = 3.5 cm
- A rope at a corner cuts a quarter circle
Board tip · BSEBBoard tip
In a BSEB answer find r first, then the area. Do not jump to 77/8.
Board tip · CBSEBoard tip
On CBSE, if the increase is asked, keep both areas and subtract.
Check your understandingall correct = mastery ★
1
The radius of a circle of circumference 22 cm is —
💡 Hint:
2 × (22/7) × r = 22.
Explanation:
r = 7/2 cm = 3.5 cm.
2
A rope tied at the corner of a square lets the animal graze a semicircle.
Explanation:
False — the corner is 90°, so a quarter circle.
3
The area of that quadrant is ______ cm².
💡 Hint:
(1/4)πr² and r = 7/2.
Explanation:
77/8 cm².
4
A 5 m rope is at a corner. π = 3.14. Find the grazing area.
Model answer (board format)
A quarter circle. Formula (1/4)πr². (1/4) × 3.14 × 25 = 19.625. The answer is 19.625 m².
Check your answer:
Wire, umbrella, wipers, light
तार, छाता, वाइपर, प्रकाश · Exercise 11.1 questions 9 to 12
Add or share equal sectorsNotes
In question 9 the five diameters join the length of wire: circumference + 5 × diameter. Ten equal sectors mean one area is a tenth of the circle.
In question 10 the eight ribs share the angle 360/8 = 45°. In question 11 the two wipers do not overlap, so add both areas. One wiper has angle 115°. In question 12 the angle is 80° and the distance is 16.5 km, with π = 3.14.
What is being added, length or areaActivity
The brooch wire is a length, unit mm. Each sector is an area, mm². The area between two ribs of the umbrella is one sector. The total wiper area is 2 × (115/360)πr². The lighthouse is also a single sector.
Worked exampleExample
Question: A brooch of diameter 35 mm also uses 5 diameters. Total length of wire. π = 22/7.
Formula: Length = 2πr + 5 × (2r).
Substitution: r = 35/2. 2 × (22/7) × (35/2) + 5 × 35 = 110 + 175 = 285.
Answer: 285 mm.
10-second revision
- 5 diameters = the diameter five times in the wire
- 8 ribs give an angle of 45°
- Two wipers are added, not overlapped
Board tip · BSEBBoard tip
In a BSEB answer write the question number, then say in one word whether it is a length or an area.
Board tip · CBSEBoard tip
On CBSE do not turn 115° into 90°. If there are two blades, the factor 2 must remain.
Check your understandingall correct = mastery ★
1
The length of wire for diameter 35 mm and 5 extra diameters is —
💡 Hint:
Circumference 110, five diameters 175.
Explanation:
285 mm.
2
The angle between eight equal ribs is 45°.
Explanation:
True — 360/8 = 45.
3
In ten equal sectors, one area is ______ of the circle.
💡 Hint:
Ten equal pieces.
Explanation:
1/10.
4
Two non-overlapping wipers, each of 115°, the factor in the total area is —
💡 Hint:
Add two blades.
Explanation:
2 × (115/360) π r².
5
The lighthouse region is a major sector, because the angle is 80°.
Explanation:
False — 80° is a minor angle.
6
An umbrella has radius 45 cm and 8 equal ribs. Write the area between two ribs in terms of π.
Model answer (board format)
The angle is 45°. Formula (45/360)πr² = (1/8)π × 2025 = 2025π/8 cm².
Check your answer:
Table designs and option (D)
मेज के डिजाइन और विकल्प (D) · Exercise 11.1 questions 13 and 14 · summary 11.2
A design is a sum of segmentsNotes
Question 13 has six equal designs. Each design is a segment of angle 60°. Total design area = 6 × (one segment). Then the rate is ₹0.35 per cm². √3 = 1.7.
In question 14 the correct area is (p/360)πR². That equals (p/720) × 2πR², option (D). The summary repeats only three facts: arc, sector, segment.
Match an option to the formulaActivity
(p/180) × 2πR is a piece of half the circumference. (p/180)πR² treats the angle as double. (p/360)×2πR is an arc. Only (p/720)×2πR² is the area, because 2/720 = 1/360.
Worked exampleExample
Question: Show that (p/720) × 2πR² and (p/360)πR² are the same.
Formula: 2/720 = 1/360.
Substitution: (p/720) × 2πR² = p × (2/720) × πR² = p × (1/360) × πR².
Answer: Both are (p/360)πR². Option (D) is correct.
10-second revision
- Six designs = six segments
- Option (D) is the correct area
- The summary has three formulas
Board tip · BSEBBoard tip
In a BSEB answer write the line 2/720 = 1/360.
Board tip · CBSEBoard tip
On CBSE do not pick the arc option as an area. The unit is hidden.
Check your understandingall correct = mastery ★
1
The area of a sector of angle p and radius R is —
💡 Hint:
2/720 = 1/360.
Explanation:
Option (D).
2
(p/360) × 2πR is an area.
Explanation:
False — it is an arc length.
3
In six equal designs each central angle is ______ degrees.
💡 Hint:
360/6.
Explanation:
60°.
4
The number of formulas in summary 11.2 is —
💡 Hint:
Arc, sector, segment.
Explanation:
Three.
5
By what factor is (p/180)πR² larger than the correct area formula?
Model answer (board format)
The correct formula is (p/360)πR². (p/180) = 2 × (p/360), so it is double.
Check your answer:
❓ Full question bank — with answers and explanations — 65 questions
Pick a type. The 35 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ120 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/12Pick one option. A wrong try brings a hint.
1
The part bounded by two radii and an arc is —
💡 Hint:
Look at the edges.
Explanation:
A sector.
2
For r = 6 cm, θ = 60°, π = 22/7, the area is —
💡 Hint:
(1/6)πr².
Explanation:
132/7 cm².
3
For r = 21 cm and θ = 60°, the arc is —
💡 Hint:
(1/6)×2πr.
Explanation:
22 cm.
4
If the circumference is 22 cm, the quadrant area is —
💡 Hint:
r = 7/2.
Explanation:
77/8 cm².
5
The minor segment equals —
💡 Hint:
Remove the triangle.
Explanation:
Sector − triangle.
6
The angle of the minute hand in 5 minutes is —
💡 Hint:
6° each minute.
Explanation:
30°.
7
A 5 m rope at a corner grazes —
💡 Hint:
The corner is 90°.
Explanation:
A quarter circle.
8
The angle between eight ribs is —
💡 Hint:
360/8.
Explanation:
45°.
9
The correct option in question 14 is —
💡 Hint:
1/360.
Explanation:
(D).
10
The brooch wire length is —
💡 Hint:
Circumference + five diameters.
Explanation:
285 mm.
11
The major sector is —
💡 Hint:
Whole minus minor.
Explanation:
πr² − minor sector.
12
At 60° the triangle area is —
💡 Hint:
Equilateral.
Explanation:
(√3/4) r².
True or false
0/81
Unless written otherwise, a sector is the minor one.
Explanation:
True — the book remark.
2
Segment = sector + triangle.
Explanation:
False — subtract.
3
The unit of an arc is cm².
Explanation:
False — cm.
4
In Exercise 11.1, π = 22/7 unless a question says 3.14.
Explanation:
True.
5
Eight ribs share an angle of 45°.
Explanation:
True — 360/8.
6
(p/360)×2πR is an area.
Explanation:
False — it is an arc.
7
If two wipers do not overlap, the area doubles.
Explanation:
True — add both 115° sweeps.
8
Major segment = πr² − minor segment.
Explanation:
True.
Fill in the blanks
0/81
Sector = (θ/360) × ______.
💡 Hint:
πr².
Explanation:
Area of the circle.
2
Arc = (θ/360) × ______.
💡 Hint:
2πr.
Explanation:
Circumference.
3
If the circumference is 22 cm, r = ______ cm.
💡 Hint:
3.5 cm.
Explanation:
2πr = 22.
4
5 minutes = ______ degrees.
💡 Hint:
30.
Explanation:
360 in 60 minutes.
5
The correct letter in question 14 is ______.
💡 Hint:
D.
Explanation:
(p/720)×2πR².
6
The angle for eight ribs is ______ degrees.
💡 Hint:
45.
Explanation:
360/8.
7
For r = 6 cm and 60°, the area is ______ cm².
💡 Hint:
132/7.
Explanation:
(1/6)×(22/7)×36.
8
The brooch wire is ______ mm long.
💡 Hint:
285.
Explanation:
110 + 175.
Match
0/21
Match the name with the formula.
Column B: A. (1/4)πr² · B. (θ/360)×2πr · C. (θ/360)πr² · D. Sector − triangle
1. Arc
2. Sector
3. Segment
4. Quadrant
Explanation:
The arc has 2πr, the sector has πr², the segment subtracts, and a quadrant is a quarter.
2
Match the question number with its job.
Column B: A. Correct option (D) · B. 60° sector, r = 6 · C. Quadrant from circumference · D. Arc, sector, segment
1. 11.1 item 1
2. 11.1 item 2
3. 11.1 item 5
4. 11.1 item 14
Explanation:
1 is the area, 2 is the quadrant, 5 has three demands, 14 is the option.
Assertion–reason
0/5Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): For r = 6 cm and 60°, the area is 132/7 cm².
Reason (R): The area is (θ/360)πr² and π = 22/7.
Explanation:
Both are true and R is the basis of the calculation.
2
Assertion (A): Segment = sector + triangle.
Reason (R): The minor segment is what remains after the triangle inside the sector is removed.
Explanation:
A is false. R is true, so the operation is subtraction.
3
Assertion (A): If the circumference is 22 cm, then r = 3.5 cm.
Reason (R): The quadrant area equals the whole circumference.
Explanation:
A is true. R is false, the area is 77/8 cm².
4
Assertion (A): In question 14, (D) is correct.
Reason (R): Arc length is (θ/360)×2πr.
Explanation:
Both are true, but R does not explain option (D).
5
Assertion (A): A rope tied at a corner gives the area of a quarter circle.
Reason (R): A corner of a square is 90°, so θ/360 = 1/4.
Explanation:
Both are true and R explains A.
Coefficient practice
0/4These chemistry lines are not a result of this maths chapter. They are only practice in filling coefficients. A blank means 1.
1
This is coefficient practice and not a result of this chapter. In the water line, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice and not a result of this chapter. In the line of H2 and O2, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
3
This is coefficient practice and not a result of this chapter. In this line for filling coefficients, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
4
This is coefficient practice and not a result of this chapter. In the water line with blank places, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
Classify
0/21
Place each demand in its formula family.
Arc length
Sector
360° − θ
Segment
Explanation:
An arc is a length. A sector and a segment are areas. 360 − θ is an angle.
2
Place each question in its type.
Question 1, r = 6, 60°
Question 2, circumference 22 cm
Question 9, brooch
Question 14
Explanation:
1 is a sector, 2 is a quadrant, 9 also has a wire length, and 14 is the option.
Very short answer
0/51
Write the sector area formula.
Model answer (board format)
(θ/360) × πr².
Check your answer:
2
Write the arc formula.
Model answer (board format)
(θ/360) × 2πr.
Check your answer:
3
Write the one-line rule for a segment.
Model answer (board format)
Segment = corresponding sector − corresponding triangle.
Check your answer:
4
What is the usual value of π in Exercise 11.1?
Model answer (board format)
22/7, unless the question writes 3.14.
Check your answer:
5
Write the correct option of question 14 and its simpler form.
Model answer (board format)
(D) (p/720)×2πR², which equals (p/360)πR².
Check your answer:
Short answer
0/41
r = 14 cm, angle 30°. Find the arc and the sector. π = 22/7.
Model answer (board format)
Arc = (30/360)×2×(22/7)×14 = (1/12)×2×22×2 = 22/3 cm. Area = (30/360)×(22/7)×196 = 154/3 cm².
Check your answer:
2
r = 10 cm, 90°, π = 3.14. Find the minor segment.
Model answer (board format)
Sector = (1/4)×3.14×100 = 78.5 cm². Triangle = 50 cm². Segment = 28.5 cm².
Check your answer:
3
A corner rope grows from 5 m to 10 m. π = 3.14. Find the increase in area.
Model answer (board format)
Smaller = (1/4)×3.14×25 = 19.625 m². Larger = (1/4)×3.14×100 = 78.5 m². Increase = 58.875 m².
Check your answer:
4
One sector of a brooch of diameter 35 mm, when there are ten equal sectors. π = 22/7.
Model answer (board format)
r = 35/2 mm. One area = (1/10)×(22/7)×(35/2)² = (1/10)×(22/7)×(1225/4) = 96.25 mm².
Check your answer:
Long answer
0/31
r = 21 cm, angle 60°, π = 22/7, √3 = 1.73. Find the arc, the sector and the minor segment.
Model answer (board format)
Arc = (60/360)×2×(22/7)×21 = 22 cm. Sector = (60/360)×(22/7)×441 = 231 cm². Triangle = (1.73/4)×441 = 190.7325 cm². Segment = 231 − 190.7325 = 40.2675 cm².
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2
In one line each, what do questions 1, 2, 5, 8, 9 and 14 of Exercise 11.1 ask?
Model answer (board format)
1: area for r = 6 cm, 60°. 2: quadrant area when the circumference is 22 cm. 5: arc, sector and segment for r = 21 cm, 60°. 8: grazing area of a corner rope. 9: brooch wire and one sector. 14: the correct area option (D).
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3
r = 7 cm. Find the 90° sector and its arc. Then say how large the remaining major sector is. π = 22/7.
Model answer (board format)
Arc = (90/360)×2×(22/7)×7 = 11 cm. Sector = (1/4)×(22/7)×49 = 77/2 cm². Circle = 154 cm². Major = 154 − 38.5 = 115.5 cm².
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BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination.
1
The area of a sector of r = 7 cm and 90° is —
💡 Hint:
A quarter circle.
Explanation:
77/2 cm².
2
A warning light has angle 80°. This sector is —
💡 Hint:
80 < 180.
Explanation:
Minor.
3
The major angle, when the minor is 80°, is ______ degrees.
💡 Hint:
280.
Explanation:
360 − 80.
4
r = 16.5 km, angle 80°, π = 3.14. Find the area of the sea under warning.
Model answer (board format)
Formula (θ/360)πr². (80/360)×3.14×(16.5)² = (2/9)×3.14×272.25 = 189.97 km².
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5
A second hand is 7 cm long. What area does it sweep in 1 minute? π = 22/7.
Model answer (board format)
In 1 minute it makes a full turn, angle 360°. The area is the whole circle: (22/7)×49 = 154 cm².
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6
Assertion (A): In ten equal sectors each area is one tenth of the circle.
Reason (R): The equal angle is 36° and 36/360 = 1/10.
Explanation:
Both are true and R explains A.
CBSE-style questions
0/6These are competency-based practice questions. They are not a copy of any year’s paper.
1
A student puts πr² into the arc. For r = 21 cm and 60°, the correct arc is —
💡 Hint:
The formula with 2πr.
Explanation:
22 cm. πr² is the area.
2
Two wipers, each of length 25 cm and angle 115°, the total area is —
💡 Hint:
Two areas, r² = 625.
Explanation:
2 × (115/360) π × 625 cm².
3
Assertion (A): A segment of any angle is the sum of the sector and the triangle.
Reason (R): The summary says segment = corresponding sector − corresponding triangle.
Explanation:
A is false. R is true.
4
Assertion (A): The option (p/720)×2πR² is the correct area.
Reason (R): 2/720 = 1/360, so it becomes (p/360)πR².
Explanation:
Both are true and R explains A.
5
A student writes the major sector as 360° + θ. For θ = 60°, write the correct major angle and what it means.
Model answer (board format)
The sum is wrong. The major angle is 360 − 60 = 300°. It is the remaining part, not a second turn.
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6
In question 9, why are the wire length and one sector area in different units?
Model answer (board format)
The wire adds the circumference and the diameters, so the unit is mm. One sector is an area, so the unit is mm². They are not written in one unit.
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🏛️ Board exam corner — Bihar Board (BSEB)— CBSE
Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.
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CBSE-style · competency-based
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🧠 What you learned + equation sheet
What you learned
- A minor sector is bounded by two radii and an arc.
- A minor segment lies between a chord and an arc.
- The length of an arc is (θ/360) × 2πr.
- The area of a sector is (θ/360) × πr².
- Segment = sector − triangle. The major part is the whole circle minus the minor part.
- In Exercise 11.1, π = 22/7 when nothing else is written. Use 3.14 only where the question says so.
| What | Keep this |
|---|---|
| Sector | (θ/360) π r² |
| Arc | (θ/360) × 2πr |
| Segment | sector − triangle |
| Major sector | πr² − minor sector |
| Quadrant | 90° sector, area (1/4)πr² |
| Value of π | 22/7 unless the question says 3.14 |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.