Areas Related to Circles takes two facts you already know, the circumference 2πr and the area πr², and uses the unitary method to find the length of an arc, the area of a sector and the area of a segment. It separates minor and major sectors and segments, and applies them to clock hands, grazing animals, umbrellas, wipers, brooches and lighthouses. The chapter is short and formula-driven, so accuracy decides the marks: the right radius, the right angle, the right value of π, and the triangle step in every segment question.
Key Concepts
1. Circle Basics You Need Before This Chapter
Every formula in this chapter is built on two results about a full circle of radius r:
- Circumference (perimeter) of a circle = 2πr, which is also πd, where d = 2r is the diameter.
- Area of a circle (circular region or disc) = πr2.
The number π is the same for every circle: it is the ratio of the circumference to the diameter. It is irrational, so in calculations you use an approximation. The question tells you which one: π = 22/7 or π = 3.14. In the NCERT exercise the instruction is “Unless stated otherwise, use π = 22/7”, and some questions ask for π = 3.14 instead.
| Radius | Value of π used | Circumference 2πr | Area πr2 |
|---|---|---|---|
| 7 cm | 22/7 | 2 × 22/7 × 7 = 44 cm | 22/7 × 49 = 154 cm2 |
| 14 cm | 22/7 | 88 cm | 616 cm2 |
| 21 cm | 22/7 | 132 cm | 1386 cm2 |
| 3.5 cm | 22/7 | 22 cm | 38.5 cm2 |
| 10 cm | 3.14 | 62.8 cm | 314 cm2 |
Tip: π = 22/7 makes the working short when the radius is a multiple of 7 (7, 14, 21, 28, 35) or of 3.5, because the 7 cancels. Always use the value the question asks for.
Lengths are measured in cm, m or km. Areas are in square units: cm2, m2, km2. Always write the unit with the answer.
Finding r from the circumference. If the circumference is 22 cm, then 2 × 22/7 × r = 22, so r = 22 × 7 / 44 = 3.5 cm. Many questions give the circumference or the diameter first, so always convert to the radius before using a formula.
2. Sector and Segment: Minor and Major
Take a circle with centre O and two points A and B on it.
Sector. The portion of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle. Join OA and OB. The region bounded by radius OA, the arc from A to B, and radius OB is a sector. The angle ∠AOB at the centre is called the angle of the sector.
The two radii split the disc into two sectors. The smaller one, OAPB (where P is a point on the shorter arc), is the minor sector. The larger one, OAQB (where Q is on the longer arc), is the major sector. Since the angles at the centre make a full turn:
Angle of the major sector = 360° − ∠AOB
Segment. The portion of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. Draw chord AB. The region between chord AB and the shorter arc APB is the minor segment. The region between chord AB and the longer arc AQB is the major segment.
Remark from NCERT: when we write “segment” and “sector”, we mean the minor segment and the minor sector, unless stated otherwise.
| Feature | Sector | Segment |
|---|---|---|
| Boundary | Two radii and an arc | One chord and an arc |
| Does it use the centre? | Yes, its corner is at the centre O | No, it is cut off by the chord alone |
| Everyday picture | A slice of pizza cut from the centre | The piece left when a straight cut is made across a roti away from the centre |
| Minor + major | Minor sector + major sector = whole circle | Minor segment + major segment = whole circle |
Special sectors. A sector of angle 90° is a quadrant (one quarter of the circle). A sector of angle 180° is a semicircle (half the circle). When the chord is a diameter, the two segments are equal and each is a semicircle.
Link between the two. The minor sector OAPB is made of the minor segment APB plus the triangle OAB. This single picture gives the segment formula in Concept 5.
3. Length of an Arc
Let OAPB be a sector of a circle with centre O and radius r, and let the degree measure of ∠AOB be θ. NCERT uses the unitary method:
- The whole circle can be thought of as a sector of angle 360°. Its arc is the whole circumference, 2πr.
- So an angle of 1° at the centre gives an arc of length 2πr/360.
- So an angle of θ° gives an arc of length (2πr/360) × θ.
Length of an arc of a sector of angle θ = (θ/360) × 2πr
The fraction θ/360 tells you what part of the full circle you have. A 90° arc is 90/360 = 1/4 of the circumference, a 60° arc is 1/6 of it, and a 120° arc is 1/3 of it.
Illustration 1. Radius 21 cm, angle 60°. Arc = (60/360) × 2 × 22/7 × 21 = (1/6) × 132 = 22 cm.
Illustration 2. Radius 14 cm, angle 90°. Arc = (1/4) × 2 × 22/7 × 14 = (1/4) × 88 = 22 cm.
Finding the angle from the arc. Rearranging, θ = (arc length ÷ 2πr) × 360. For an arc of 44 cm in a circle of radius 21 cm: 2πr = 132 cm, so θ = (44/132) × 360 = (1/3) × 360 = 120°.
Perimeter of a sector. The boundary of a sector is two radii plus the arc, so
Perimeter of a sector = 2r + (θ/360) × 2πr
For a quadrant of radius 7 cm, arc = (1/4) × 44 = 11 cm and perimeter = 7 + 7 + 11 = 25 cm. Students often give only the arc (11 cm) when the question asks for the perimeter.
How the arc changes. Arc length is directly proportional to θ when r is fixed, and directly proportional to r when θ is fixed. Double the radius and the arc doubles. Double the angle and the arc doubles.
4. Area of a Sector
The same unitary method works for area. NCERT’s steps:
- When the degree measure of the angle at the centre is 360, the area of the sector is πr2 (the whole disc).
- When the degree measure of the angle at the centre is 1, the area of the sector is πr2/360.
- When the degree measure of the angle at the centre is θ, the area of the sector is (πr2/360) × θ.
Area of the sector of angle θ = (θ/360) × πr2, where r is the radius of the circle and θ is the angle of the sector in degrees.
Illustration 1. Radius 14 cm, angle 90°. Area = (1/4) × 22/7 × 196 = (1/4) × 616 = 154 cm2.
Illustration 2. Radius 7 cm, angle 45°. Area = (45/360) × 154 = (1/8) × 154 = 19.25 cm2.
Major sector. Two ways, and both give the same answer:
- Area of the major sector = πr2 − area of the minor sector.
- Area of the major sector = ((360 − θ)/360) × πr2, using the angle of the major sector directly.
For radius 14 cm and θ = 90°: major sector = 616 − 154 = 462 cm2, or (270/360) × 616 = (3/4) × 616 = 462 cm2.
A shortcut linking area and arc. Write the area as (1/2) × r × (θ/360) × 2πr. The part (θ/360) × 2πr is the arc length l. So
Area of a sector = (1/2) × l × r
This looks like the area of a triangle with base l and height r, which is a good way to remember it. Check: radius 14 cm, arc 22 cm gives (1/2) × 22 × 14 = 154 cm2, the same as Illustration 1. Use it when the arc length is given and the angle is not.
How the area changes. For a fixed radius, sector area is proportional to θ. For a fixed angle, sector area is proportional to r2: double the radius and the area becomes four times as large. This is why the grazing area jumps so much when a rope is doubled (Example 10 below).
5. Area of a Segment
Let AB be a chord of a circle with centre O and radius r, and let ∠AOB = θ. The minor sector OAPB is made up of the minor segment APB and the triangle OAB. Taking the triangle away leaves the segment:
Area of the segment APB = area of the sector OAPB − area of ΔOAB = (θ/360) × πr2 − area of ΔOAB
And for the other side of the chord:
Area of the major segment AQB = πr2 − area of the minor segment APB
The sector part is routine. The real work is the area of triangle OAB. Triangle OAB is always isosceles because OA = OB = r. Here is how to handle the angles that appear in the NCERT exercises.
Case θ = 90°. The triangle is right-angled at O, with both legs equal to r. Area = (1/2) × r × r = r2/2. For r = 10 cm the triangle is 50 cm2.
Case θ = 60°. OA = OB and the angle between them is 60°, so the base angles are (180° − 60°)/2 = 60° each. The triangle is equilateral with side r. Area = (√3/4) r2. For r = 21 cm, the triangle is 441√3/4 cm2.
Case θ = 120° (the NCERT method). Draw OM perpendicular to AB. In the right triangles OMA and OMB, the hypotenuses OA = OB and the side OM is common, so ΔAMO ≅ ΔBMO by RHS congruence. So M is the mid-point of AB and ∠AOM = ∠BOM = 60°. In right triangle OMA:
- OM/OA = cos 60° = 1/2, so OM = r/2.
- AM/OA = sin 60° = √3/2, so AM = (√3/2) r, and AB = 2 AM = √3 r.
- Area of ΔOAB = (1/2) × AB × OM = (1/2) × √3 r × r/2 = (√3/4) r2.
Notice that the triangle for 120° has the same area as the equilateral triangle for 60°. The shapes are different, the areas are equal.
Case θ = 180°. The chord is a diameter and there is no triangle. Each segment is a semicircle of area πr2/2.
| θ | Sector area | Area of ΔOAB | Minor segment |
|---|---|---|---|
| 60° | πr2/6 | (√3/4) r2 | r2(π/6 − √3/4) |
| 90° | πr2/4 | r2/2 | r2(π/4 − 1/2) |
| 120° | πr2/3 | (√3/4) r2 | r2(π/3 − √3/4) |
Illustration. Radius 14 cm, θ = 90°, π = 22/7. Sector = 154 cm2, triangle = (1/2) × 14 × 14 = 98 cm2. Minor segment = 154 − 98 = 56 cm2. Major segment = 616 − 56 = 560 cm2.
Exact or approximate? When the question gives no value of √3, leave the answer in surd form, as NCERT does in Example 2: (462 − 441√3/4) cm2. When it says “use √3 = 1.73”, give a decimal answer.
6. Real-Life Problems on Sectors and Segments
The NCERT exercise is full of everyday situations. Each one hides a sector: your job is to find its radius and its angle, then decide whether the question wants an arc, a sector area or a segment area.
| Situation | Radius is | Angle is | Usually asked |
|---|---|---|---|
| Minute hand of a clock | Length of the hand | 6° per minute (360° in 60 minutes) | Area swept |
| Hour hand of a clock | Length of the hand | 30° per hour, that is 0.5° per minute | Area swept |
| Animal tied at the corner of a square or rectangular field | Length of the rope | 90° (the corner angle), so a quadrant | Grazing area |
| Umbrella with n equally spaced ribs | Radius of the umbrella | 360°/n | Area between two ribs |
| Brooch or wheel with k diameters | Half the diameter of the circle | k diameters make 2k equal sectors, each 360°/2k | Wire length, area of each sector |
| Car wiper | Length of the blade | Angle swept | Area cleaned (double it for two non-overlapping wipers) |
| Lighthouse beam | Distance the light reaches | Angle of the beam | Area of sea covered |
Method for any word problem:
- Read the radius. If a diameter or a circumference is given, convert it to r first.
- Read or work out the central angle θ in degrees.
- Decide what is asked: a length (arc or perimeter), a sector area, or a segment area.
- Write the formula, substitute, and simplify with the value of π the question asks for.
- Write the unit: cm, m or km for length; cm2, m2 or km2 for area.
Clock illustration. The minute hand turns 360° in 60 minutes, so in 5 minutes it turns 5 × 6° = 30°. In 20 minutes it turns 120°. The hour hand turns 360° in 12 hours, so in 1 hour it turns 30°.
Grazing illustration. A goat is tied with a 7 m rope to a corner of a square field of side 20 m. The field’s corner is a right angle, so the goat grazes a quadrant of radius 7 m: (1/4) × 22/7 × 49 = 38.5 m2. This works because the rope (7 m) is shorter than the side of the field (20 m), so the quadrant stays inside the field.
Formula and Theorem Sheet
| Result | Formula | Notes |
|---|---|---|
| Circumference of a circle | 2πr = πd | d = 2r |
| Area of a circle | πr2 | Area of the whole disc |
| Angle of major sector | 360° − θ | θ = angle of the minor sector |
| Length of an arc | l = (θ/360) × 2πr | θ in degrees |
| Angle from arc length | θ = (l/2πr) × 360 | Rearranged arc formula |
| Perimeter of a sector | 2r + l | Two radii plus the arc |
| Area of a sector | (θ/360) × πr2 | Unitary method from πr2 for 360° |
| Area of a sector in terms of arc | (1/2) × l × r | No angle needed |
| Area of a major sector | πr2 − minor sector = ((360 − θ)/360) × πr2 | Both routes give the same answer |
| Area of a minor segment | (θ/360) × πr2 − area of ΔOAB | Sector minus triangle |
| Area of a major segment | πr2 − minor segment | Whole circle minus the small piece |
| ΔOAB when θ = 90° | r2/2 | Right-angled isosceles triangle |
| ΔOAB when θ = 60° | (√3/4) r2 | Equilateral triangle of side r |
| ΔOAB when θ = 120° | (√3/4) r2 | OM = r/2, AB = √3 r |
| Clock hands | Minute hand 6°/min; hour hand 30°/h | Converts time into the angle θ |
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Important Definitions
- Circle: the set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).
- Circular region (disc): the circle together with all the points inside it. Its area is πr2.
- Arc: a part of the circle between two points on it. The shorter one is the minor arc and the longer one is the major arc.
- Chord: a line segment joining two points on a circle. A chord through the centre is a diameter.
- Sector: the portion (or part) of the circular region enclosed by two radii and the corresponding arc.
- Angle of the sector: the angle ∠AOB formed at the centre by the two radii of the sector.
- Minor sector: the smaller of the two sectors made by two radii; its angle is θ.
- Major sector: the larger of the two sectors made by two radii; its angle is 360° − θ.
- Segment: the portion (or part) of the circular region enclosed between a chord and the corresponding arc.
- Minor segment: the smaller region cut off by a chord, on the side of the minor arc.
- Major segment: the larger region cut off by a chord, on the side of the major arc.
- Quadrant: a sector of angle 90°, one quarter of the circle.
- Semicircle: a sector of angle 180°, half the circle, cut off by a diameter.
- Length of an arc: the distance along the circle from one end of the arc to the other, equal to (θ/360) × 2πr.
Solved Examples (NCERT-Based)
Example 1: Sector and major sector (NCERT Example 1)
Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14).
Solution:
Area of the sector = (θ/360) × πr2 = (30/360) × 3.14 × 4 × 4 cm2 = 12.56/3 cm2 = 4.19 cm2 (approx.)
Area of the major sector = πr2 − area of the minor sector = (3.14 × 16 − 4.19) cm2 = (50.24 − 4.19) cm2 = 46.05 cm2 = 46.1 cm2 (approx.)
Check with the other route: ((360 − 30)/360) × 3.14 × 16 = (330/360) × 50.24 = 46.05 cm2. Both agree.
Example 2: Segment with a 120° angle (NCERT Example 2)
Find the area of the segment AYB if the radius of the circle is 21 cm and ∠AOB = 120°. (Use π = 22/7)
Solution:
Area of segment AYB = area of sector OAYB − area of ΔOAB.
Area of sector OAYB = (120/360) × 22/7 × 21 × 21 cm2 = (1/3) × 1386 cm2 = 462 cm2.
For ΔOAB, draw OM ⊥ AB. Since OA = OB and OM is common, ΔAMO ≅ ΔBMO (RHS). So M is the mid-point of AB and ∠AOM = ∠BOM = 60°.
In ΔOMA: OM/OA = cos 60° = 1/2, so OM = 21/2 cm. AM/OA = sin 60° = √3/2, so AM = 21√3/2 cm and AB = 2 AM = 21√3 cm.
Area of ΔOAB = (1/2) × AB × OM = (1/2) × 21√3 × 21/2 cm2 = 441√3/4 cm2.
Area of segment AYB = (462 − 441√3/4) cm2 = (21/4)(88 − 21√3) cm2.
(Check: (21/4) × 88 = 462 and (21/4) × 21√3 = 441√3/4.)
Example 3: Area of a sector (NCERT Exercise 11.1, Q1)
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution: Area = (60/360) × 22/7 × 6 × 6 = (1/6) × 22/7 × 36 = 6 × 22/7 = 132/7 cm2 = 18 6/7 cm2 ≈ 18.86 cm2.
Example 4: Quadrant from the circumference (NCERT Exercise 11.1, Q2)
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution: 2πr = 22, so 2 × 22/7 × r = 22, giving r = 7/2 = 3.5 cm.
Area of the quadrant = (1/4) × πr2 = (1/4) × 22/7 × 7/2 × 7/2 = (1/4) × 77/2 = 77/8 cm2 = 9.625 cm2.
Example 5: Minute hand of a clock (NCERT Exercise 11.1, Q3)
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution: The minute hand turns 360° in 60 minutes, so 6° per minute. In 5 minutes it turns θ = 30°. The radius is the length of the hand, r = 14 cm.
Area swept = (30/360) × 22/7 × 14 × 14 = (1/12) × 616 = 154/3 cm2 = 51 1/3 cm2 ≈ 51.33 cm2.
Example 6: Chord subtending a right angle (NCERT Exercise 11.1, Q4)
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)
Solution:
Area of the minor sector = (90/360) × 3.14 × 100 = 78.5 cm2.
ΔOAB is right-angled at O with OA = OB = 10 cm, so its area = (1/2) × 10 × 10 = 50 cm2.
(i) Minor segment = 78.5 − 50 = 28.5 cm2.
(ii) Major sector = πr2 − minor sector = 314 − 78.5 = 235.5 cm2.
Example 7: Arc, sector and segment together (NCERT Exercise 11.1, Q5)
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord.
Solution:
(i) Arc = (60/360) × 2 × 22/7 × 21 = (1/6) × 132 = 22 cm.
(ii) Sector = (60/360) × 22/7 × 21 × 21 = (1/6) × 1386 = 231 cm2.
(iii) With θ = 60° and OA = OB, ΔOAB is equilateral with side 21 cm. Its area = (√3/4) × 212 = 441√3/4 cm2. Segment = (231 − 441√3/4) cm2.
Example 8: Minor and major segments with 60° (NCERT Exercise 11.1, Q6)
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
Solution:
Sector = (60/360) × 3.14 × 225 = 706.5/6 = 117.75 cm2.
ΔOAB is equilateral with side 15 cm: area = (1.73/4) × 225 = 389.25/4 = 97.3125 cm2.
Minor segment = 117.75 − 97.3125 = 20.4375 cm2.
Area of the circle = 3.14 × 225 = 706.5 cm2. Major segment = 706.5 − 20.4375 = 686.0625 cm2.
Example 9: Segment with 120° (NCERT Exercise 11.1, Q7)
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Solution:
Sector = (120/360) × 3.14 × 144 = 452.16/3 = 150.72 cm2.
Draw OM ⊥ AB. As in NCERT Example 2, ∠AOM = 60°, OM = 12 cos 60° = 6 cm, AM = 12 sin 60° = 6√3 cm, so AB = 12√3 cm.
Area of ΔOAB = (1/2) × 12√3 × 6 = 36√3 = 36 × 1.73 = 62.28 cm2.
Segment = 150.72 − 62.28 = 88.44 cm2.
Example 10: The grazing horse (NCERT Exercise 11.1, Q8)
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
Solution: The corner of a square is 90°, and both rope lengths are less than 15 m, so the horse grazes a quadrant whose radius is the rope length.
(i) Area = (1/4) × 3.14 × 5 × 5 = 78.5/4 = 19.625 m2.
(ii) With a 10 m rope: (1/4) × 3.14 × 100 = 78.5 m2. Increase = 78.5 − 19.625 = 58.875 m2.
Doubling the rope made the area four times as large (19.625 × 4 = 78.5), because area depends on r2.
Example 11: The silver brooch (NCERT Exercise 11.1, Q9)
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
Solution: r = 35/2 = 17.5 mm.
(i) Circumference = πd = 22/7 × 35 = 110 mm. Five diameters = 5 × 35 = 175 mm. Total wire = 110 + 175 = 285 mm.
(ii) Ten equal sectors, so each angle = 360°/10 = 36°. Area of each = (36/360) × 22/7 × 17.5 × 17.5 = (1/10) × 962.5 = 96.25 mm2 (= 385/4 mm2).
Example 12: Umbrella ribs (NCERT Exercise 11.1, Q10)
An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution: 8 equally spaced ribs make 8 equal sectors, each of angle 360°/8 = 45°.
Area = (45/360) × 22/7 × 45 × 45 = (1/8) × 22/7 × 2025 = 44550/56 = 22275/28 cm2 ≈ 795.54 cm2.
Example 13: Car wipers (NCERT Exercise 11.1, Q11)
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution: Each blade sweeps a sector of radius 25 cm and angle 115°. The wipers do not overlap, so total area = 2 × one sector.
Total = 2 × (115/360) × 22/7 × 25 × 25 = (115/180) × 22/7 × 625 = (23/36) × 13750/7 = 316250/252 = 158125/126 cm2 ≈ 1254.96 cm2.
Example 14: Lighthouse warning zone (NCERT Exercise 11.1, Q12)
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution: r = 16.5 km, θ = 80°. r2 = 272.25.
Area = (80/360) × 3.14 × 272.25 = (2/9) × 854.865 = 1709.73/9 = 189.97 km2 (approx.).
Competency-Based Questions (with answers)
1. Case-based: The pizza party
A round pizza of radius 14 cm is cut from the centre into 8 equal slices. (Use π = 22/7)
(a) What is the angle of each slice at the centre? 360°/8 = 45°.
(b) Find the area of one slice. (45/360) × 22/7 × 196 = (1/8) × 616 = 77 cm2.
(c) Find the length of crust on one slice and the perimeter of the slice. Crust (arc) = (1/8) × 2 × 22/7 × 14 = (1/8) × 88 = 11 cm. Perimeter = 14 + 14 + 11 = 39 cm.
(d) Three friends eat 5 slices. What area of pizza is left? 3 slices remain: 3 × 77 = 231 cm2. The 3 slices form one sector of angle 3 × 45° = 135°, and (135/360) × 616 = 231 cm2 agrees.
2. Case-based: The garden sprinkler
A lawn sprinkler fixed at a point throws water up to 21 m and turns back and forth through an angle of 120°. (Use π = 22/7)
(a) What shape is the watered region? A sector of radius 21 m and angle 120°.
(b) Find the area watered. (120/360) × 22/7 × 441 = (1/3) × 1386 = 462 m2.
(c) Find the length of the curved edge of the watered region. (1/3) × 2 × 22/7 × 21 = (1/3) × 132 = 44 m.
(d) If the gardener wants to water 693 m2 with the same reach, what angle is needed? (θ/360) × 1386 = 693, so θ/360 = 1/2 and θ = 180°.
3. Case-based: The swinging pendulum
The bob of a pendulum 35 cm long swings through an angle of 72° from one extreme position to the other. (Use π = 22/7)
(a) Find the length of the path of the bob in one swing. Arc = (72/360) × 2 × 22/7 × 35 = (1/5) × 220 = 44 cm.
(b) Find the area swept by the string in one swing. Using (1/2) × l × r = (1/2) × 44 × 35 = 770 cm2. Check: (1/5) × 22/7 × 1225 = (1/5) × 3850 = 770 cm2.
4. Source-based: The unitary method
NCERT derives the area of a sector like this: “When degree measure of the angle at the centre is 360, area of the sector = πr2. So, when the degree measure of the angle at the centre is 1, area of the sector = πr2/360.”
(a) For a circle of radius 6 cm, what is the area of the sector for 1°? (Use π = 3.14) 3.14 × 36/360 = 113.04/360 = 0.314 cm2.
(b) Using (a), find the area of a sector of angle 50° in the same circle. 50 × 0.314 = 15.7 cm2.
(c) Apply the same reasoning to arc length. A 360° angle gives the whole circumference 2πr, so 1° gives 2πr/360 and θ° gives (θ/360) × 2πr.
5. Assertion-Reason
Assertion (A): The area of a sector of radius 7 cm whose arc is 11 cm long is 38.5 cm2.
Reason (R): The area of a sector with arc length l and radius r is (1/2) × l × r.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a) (1/2) × 11 × 7 = 38.5 cm2, and R is exactly the rule used.
6. Assertion-Reason
Assertion (A): If the radius of a circle is doubled, the area of a sector of the same angle is also doubled.
Reason (R): For a fixed angle, the area of a sector is proportional to the square of the radius.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (d) Area = (θ/360)πr2, so doubling r multiplies the area by 4. A is false and R is true.
7. Assertion-Reason
Assertion (A): A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segment is 28.5 cm2 (use π = 3.14).
Reason (R): Area of a segment = area of the corresponding sector + area of the corresponding triangle.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (c) 78.5 − 50 = 28.5 cm2, so A is true. The triangle is subtracted, so R is false.
8. Error analysis: The wrong triangle
A student finds the area of the minor segment of a circle of radius 14 cm cut off by a chord subtending 60° at the centre. The working is: sector = (1/6) × 22/7 × 196 = 308/3 cm2; triangle = (1/2) × 14 × 14 = 98 cm2; segment = 308/3 − 98 = 14/3 cm2.
(a) Find the error. The formula (1/2) × r × r works only when the angle at O is 90°. With 60°, ΔOAB is equilateral with side 14 cm.
(b) Correct the answer (use √3 = 1.73). Triangle = (√3/4) × 196 = 49√3 = 49 × 1.73 = 84.77 cm2. Sector = 308/3 ≈ 102.67 cm2. Segment ≈ 102.67 − 84.77 = 17.90 cm2. Exact form: (308/3 − 49√3) cm2.
9. Competency MCQ: Which expression is the area of a sector?
The area of a sector of angle p (in degrees) of a circle with radius R is:
(A) (p/180) × 2πR (B) (p/180) × πR2 (C) (p/360) × 2πR (D) (p/720) × 2πR2
Answer: (D) (p/720) × 2πR2 = (p/360) × πR2, the sector formula. Option (C) is the arc length, (A) is twice the arc length and (B) is twice the sector area.
Important Questions for Board Exams
1-Mark Questions
Q1. Find the area of a quadrant of a circle of radius 7 cm. (Use π = 22/7)
(1/4) × 22/7 × 49 = 38.5 cm2.
Q2. An arc of a circle of radius 21 cm subtends 60° at the centre. Find its length.
(1/6) × 2 × 22/7 × 21 = 22 cm.
Q3. The angle of a minor sector is 75°. What is the angle of the corresponding major sector?
360° − 75° = 285°.
Q4. The area of a sector is one-fifth of the area of its circle. Find the angle of the sector.
θ/360 = 1/5, so θ = 72°.
Q5. Find the perimeter of a semicircular protractor of radius 7 cm. (Use π = 22/7)
Curved part πr = 22 cm, plus the diameter 14 cm. Perimeter = 36 cm.
2-Mark Questions
Q6. The minute hand of a clock is 21 cm long. Find the area swept by it in 20 minutes.
Angle = 20 × 6° = 120°. Area = (120/360) × 22/7 × 441 = (1/3) × 1386 = 462 cm2.
Q7. The area of a sector of angle 40° is 352/7 cm2. Find the radius. (Use π = 22/7)
(40/360) × 22/7 × r2 = 352/7 ⇒ (1/9) × 22 × r2 = 352 ⇒ r2 = 352 × 9/22 = 144 ⇒ r = 12 cm.
Q8. An arc of length 44 cm lies on a circle of radius 21 cm. Find the angle it subtends at the centre and the area of the sector.
Circumference = 132 cm, so θ = (44/132) × 360 = 120°. Area = (1/2) × 44 × 21 = 462 cm2.
Q9. Find the perimeter of a quadrant of a circle of radius 14 cm.
Arc = (1/4) × 88 = 22 cm. Perimeter = 14 + 14 + 22 = 50 cm.
3-Mark Questions
Q10. A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the areas of the minor and major segments. (Use π = 22/7)
Sector = (1/4) × 616 = 154 cm2. Triangle = (1/2) × 14 × 14 = 98 cm2. Minor segment = 154 − 98 = 56 cm2. Circle = 616 cm2, so major segment = 616 − 56 = 560 cm2.
Q11. A cow is tied with a 14 m rope at one corner of a rectangular field 20 m long and 16 m wide. Find the area of the field the cow can graze and the area it cannot graze. (Use π = 22/7)
The corner is 90° and 14 m is less than both sides, so the grazed region is a quadrant: (1/4) × 22/7 × 196 = 154 m2. Field = 20 × 16 = 320 m2. Ungrazed = 320 − 154 = 166 m2.
Q12. A chord of a circle of radius 12 cm subtends 60° at the centre. Find the area of the minor segment. (Use π = 3.14, √3 = 1.73)
Sector = (1/6) × 3.14 × 144 = 75.36 cm2. Equilateral triangle = (1.73/4) × 144 = 62.28 cm2. Segment = 75.36 − 62.28 = 13.08 cm2.
Q13. The minute hand of a clock is 7 cm long. Find the area swept by it between 9:00 a.m. and 9:35 a.m. (Use π = 22/7)
35 minutes = 35 × 6° = 210°. Area = (210/360) × 22/7 × 49 = (7/12) × 154 = 1078/12 = 539/6 cm2 ≈ 89.83 cm2.
5-Mark Questions
Q14. In a circle of radius 21 cm, a chord AB subtends an angle of 120° at the centre O. Find (i) the length of the minor arc AB, (ii) the area of the minor sector, (iii) the area of the minor segment, (iv) the perimeter of the minor segment. (Use π = 22/7, √3 = 1.73)
(i) Arc = (1/3) × 132 = 44 cm.
(ii) Sector = (1/3) × 1386 = 462 cm2.
(iii) Draw OM ⊥ AB. ∠AOM = 60°, OM = 21 × 1/2 = 10.5 cm, AM = 21 × √3/2, so AB = 21√3 cm. Triangle = (1/2) × 21√3 × 10.5 = 441√3/4 = 441 × 1.73/4 = 190.73 cm2 (approx.). Segment = 462 − 190.73 = 271.27 cm2 (approx.).
(iv) The segment is bounded by the arc and the chord. AB = 21 × 1.73 = 36.33 cm. Perimeter = 44 + 36.33 = 80.33 cm.
Q15. ABCD is a square of side 14 cm. With each corner A, B, C and D as centre, a quadrant of radius 7 cm is drawn inside the square. Find the area of the part of the square not covered by the quadrants. (Use π = 22/7)
Each quadrant has radius 7 cm and angle 90°, the angle of the square’s corner. Since 7 + 7 = 14, neighbouring quadrants just touch at the mid-points of the sides and do not overlap.
Area of one quadrant = (1/4) × 22/7 × 49 = 38.5 cm2. Four quadrants = 154 cm2 (the same as one full circle of radius 7 cm).
Area of the square = 14 × 14 = 196 cm2. Uncovered area = 196 − 154 = 42 cm2.
Q16. ABC is an equilateral triangle of side 12 cm. With each vertex as centre, a sector of radius 6 cm is drawn inside the triangle. Find the area of the triangle not covered by the three sectors. (Use π = 3.14, √3 = 1.73)
Each angle of an equilateral triangle is 60°, so each sector has angle 60° and radius 6 cm. Since 6 + 6 = 12, the sectors meet at the mid-points of the sides without overlapping.
Total angle of the three sectors = 3 × 60° = 180°, so together they equal a semicircle of radius 6 cm: (1/2) × 3.14 × 36 = 56.52 cm2.
Area of the triangle = (√3/4) × 144 = 36√3 = 36 × 1.73 = 62.28 cm2.
Uncovered area = 62.28 − 56.52 = 5.76 cm2.
Common Mistakes and Examiner Tips
- Using the diameter as the radius. In the brooch question the diameter is 35 mm, so r = 17.5 mm. Fix: write “r = …” as your first line in every answer.
- Mixing up arc and area formulas. Arc uses 2πr; area uses πr2. Fix: check the unit. A length comes out in cm, an area in cm2.
- Giving the arc when the perimeter is asked. The perimeter of a sector is 2r + arc. Fix: underline the word “perimeter” and add both radii.
- Adding the triangle in a segment question. Segment = sector − triangle. Fix: picture the sector as a slice, then cut off the triangular part near the centre.
- Using (1/2) × r × r for every triangle OAB. That works only for 90°. Fix: for 60° use the equilateral triangle (√3/4)r2; for 120° drop the perpendicular OM, as in NCERT Example 2.
- Wrong angle for clock hands. The minute hand turns 6° per minute; the hour hand turns 30° per hour. Fix: convert time to degrees before touching the formula.
- Forgetting the second wiper. Two non-overlapping wipers clean twice the area of one. Fix: read “two” and “do not overlap” as “multiply by 2”.
- Using the wrong value of π. Fix: use the value in the question. If none is given in an NCERT-style question, use 22/7, as the exercise instructs.
- Rounding too early. Rounding the sector area and then subtracting a rounded triangle area can shift the final digit. Fix: keep full decimals until the last step, or keep the answer exact in surd form when no value of √3 is given.
- Finding the major segment as πr2 minus the minor sector. Fix: major segment = πr2 − minor segment; major sector = πr2 − minor sector. Match the words.
- Wrong unit or no unit. Fix: km for the lighthouse radius gives km2 for the area; mm for the brooch gives mm2.
- Using the wrong number of sectors. 8 ribs make 8 sectors, but 5 diameters make 10 sectors. Fix: count the regions the lines actually create.
Quick Revision Points
- Circumference of a circle = 2πr; area = πr2.
- A sector is bounded by two radii and an arc; a segment by a chord and an arc.
- ∠AOB is the angle of the sector; the major sector has angle 360° − ∠AOB.
- “Sector” and “segment” on their own mean the minor ones.
- Length of an arc = (θ/360) × 2πr.
- Area of a sector = (θ/360) × πr2.
- Area of a sector = (1/2) × arc length × radius.
- Perimeter of a sector = 2r + arc length.
- Major sector = πr2 − minor sector = ((360 − θ)/360) × πr2.
- Area of a segment = area of sector − area of ΔOAB.
- Major segment = πr2 − minor segment.
- ΔOAB for 90°: r2/2. For 60°: equilateral, (√3/4)r2. For 120°: (√3/4)r2 using OM = r/2 and AB = √3 r.
- A quadrant is a 90° sector; a semicircle is a 180° sector.
- Minute hand: 6° per minute. Hour hand: 30° per hour.
- A rope tied at the corner of a square or rectangle gives a quadrant of radius equal to the rope.
- n equally spaced ribs make n sectors of 360°/n; k diameters make 2k sectors.
- For a fixed angle, doubling the radius doubles the arc and makes the area four times as large.
- Use π = 22/7 unless the question says 3.14.
- Leave √3 in the answer unless a decimal value is given.
Weightage in Board Exams
The marks for this chapter are set in the current CBSE Class 10 Maths course structure, so check that document for the exact figure. The chapter is assessed in a predictable way because the whole chapter rests on three formulas: arc length, sector area and segment area.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Choosing the correct sector formula; angle of a major sector; the effect of doubling the radius; area from arc length |
| Short answers | Area swept by a clock hand; area of a quadrant from its circumference; radius or angle from a given sector area |
| Longer answers | Minor and major segments for 60°, 90° or 120°, including the triangle OAB step; grazing, umbrella, wiper and brooch problems |
| Case-based | A real object, such as a pizza, sprinkler, pendulum or lighthouse, split into parts: angle, arc, sector area, then a reverse question |
Most errors in this chapter are small slips: the diameter taken as the radius, the triangle added instead of subtracted, the wrong value of π, or a missing unit. Work every NCERT Exercise 11.1 question and both solved examples first, then practise segment questions until the triangle step is automatic.
Class 10 Maths · Chapter 11 – swipe through all 10 cards to understand the whole chapter.
Two radii and an arc
A sector is the part of a circular region enclosed by two radii and the corresponding arc.
On its own, the word sector means the minor sector.
- ∠AOB at the centre is the angle of the sector
- Minor sector + major sector = whole circle
- Quadrant = 90° sector, semicircle = 180° sector
A chord and an arc
A segment is the part of a circular region enclosed between a chord and the corresponding arc.
A sector always has its corner at the centre; a minor segment never contains the centre.
- Minor segment lies on the side of the shorter arc
- Major segment lies on the side of the longer arc
- A diameter cuts the circle into two equal semicircles
Part of the circumference
By the unitary method, 360° gives the whole circumference 2πr, so θ degrees give θ/360 of it.
r = 21 cm, θ = 60°: l = (1/6) × 132 = 22 cm
- Angle from arc: θ = (l/2πr) × 360
- Perimeter of a sector = 2r + l
- Double r or double θ and the arc doubles
Area of a sector
The whole disc is a 360° sector of area πr2, so a sector of angle θ has θ/360 of that area.
r = 14 cm, θ = 90°: A = (1/4) × 616 = 154 cm2
- With arc length l: A = (1/2) × l × r
- Double the radius and the area becomes 4 times
- Area is in square units: cm2, m2, km2
Major sector and major segment
The larger part is always the whole circle minus the smaller part.
Major segment = πr2 − minor segment
- r = 4 cm, θ = 30°, π = 3.14: minor sector ≈ 4.19 cm2
- Major sector = 50.24 − 4.19 ≈ 46.05 cm2
- Subtract a segment for a segment, a sector for a sector
Sector minus triangle
The minor sector is the minor segment plus triangle OAB, so remove the triangle.
r = 14 cm, θ = 90°: 154 − 98 = 56 cm2
- Triangle OAB is isosceles: OA = OB = r
- Never add the triangle
- Leave √3 in the answer unless a value is given
The triangle step
The area of triangle OAB depends on the angle at the centre.
For 120°, draw OM ⊥ AB: OM = r/2, AB = √3 r
- 60°: the triangle is equilateral with side r
- 120°: ΔAMO ≅ ΔBMO by RHS, so ∠AOM = 60°
- (1/2) × r × r works only for 90°
Time into angle
A clock hand sweeps a sector whose radius is the length of the hand.
14 cm minute hand, 5 min: θ = 30°, area = 154/3 cm2
- 20 minutes of the minute hand = 120°
- One hour of the minute hand = the full circle
- Convert time to degrees before using the formula
Find r and θ first
Grazing, umbrellas, wipers, brooches and lighthouses each hide a sector.
Horse on a 5 m rope at a square’s corner: (1/4) × 3.14 × 25 = 19.625 m2
- Rope at a corner of a square or rectangle: quadrant
- n equal ribs: angle 360°/n; k diameters: 2k sectors
- Two non-overlapping wipers: double one sector
Where marks are lost
The formulas are few, so most lost marks come from small slips.
Use π = 22/7 unless the question says 3.14.
- Halve the diameter before using any formula
- Perimeter of a sector includes both radii
- Write the unit: cm for length, cm2 for area
📝 Practice Areas Related to Circles - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
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Why D: Step 1 (circumference = 2πr): 2 × (22/7) × 6.3 = 44 × 0.9 = 39.6 cm. Step 2 (arc = (θ/360) × circumference): θ = 360 × 11 / 39.6 = 100°.
Why not A: 10° is a decimal-place slip, as if 6.3 were read as 63.
Why not B: a 60° arc here is 39.6/6 = 6.6 cm, not 11 cm.
Why not C: a 45° arc here is 39.6/8 = 4.95 cm, not 11 cm.
Remember: θ = 360 × arc / (2πr); check by putting θ back in.
Why D: Step 1 (area of circle, πr²): (22/7) × 7² = 154 cm². Step 2 (16 equal sectors, each 22.5°): 154/16 = 77/8 cm².
Why not A: 77/4 cm² is 154/8, the circle cut into 8 sectors, not 16.
Why not B: 77 cm² is half the circle, a semicircle.
Why not C: 154 cm² is the whole circle; it was never divided.
Remember: n equal sectors each have area πr²/n and angle 360°/n.
Why C: Step 1 (hour hand rate): 360° in 720 minutes, so 0.5° per minute. Step 2 (time elapsed): 7:00 to 8:10 is 70 minutes, so angle = 70 × 0.5 = 35°. The length 7 cm is not needed for the angle.
Why not A: (35/4)° is a quarter of the answer, from using 0.125° per minute.
Why not B: (35/2)° counts only 35 minutes, or uses 0.25° per minute.
Why not D: 70° uses 1° per minute, twice the true rate.
Remember: hour hand angle = 0.5° × minutes elapsed.
Why C: Step 1 (circumference = 2πr): 2 × (22/7) × 2.8 = 44 × 0.4 = 17.6 cm. Step 2 (arc = (θ/360) × circumference): θ = 360 × 2.2 / 17.6 = 360/8 = 45°.
Why not A: a 50° arc here is 17.6 × 50/360 ≈ 2.44 cm, not 2.2 cm.
Why not B: a 60° arc here is 17.6/6 ≈ 2.93 cm.
Why not D: 30° gives 17.6/12 ≈ 1.47 cm.
Remember: arc/circumference = 2.2/17.6 = 1/8, and one eighth of 360° is 45°.
Why A: Step 1 (arc is a fraction of the circumference): circumference = 2 × (22/7) × 21 = 132 cm. Step 2 (fraction θ/360): 60/360 = 1/6, so arc = 132/6 = 22 cm.
Why not B: 44 cm is (60/360) × 2 × 2πr, the arc doubled, as if the diameter were used in place of the radius.
Why not C: 88 cm is four times the arc, a slip in the fraction of 360°.
Why not D: 11 cm is half the arc, which drops the factor 2 in 2πr.
Remember: arc = (θ/360) × 2πr; a 60° arc is one sixth of the circumference.
Why D: Step 1 (sector area gives θ): (θ/360) × π × 36² = 54π, so θ/360 = 54/1296 = 1/24, θ = 15°. Step 2 (arc = (θ/360) × 2πr): (1/24) × 2π × 36 = 3π cm. (Shortcut: area = (1/2) × l × r, so 54π = 18l and l = 3π cm.)
Why not A: 8π cm does not come from the 1/24 fraction of the 72π cm circumference.
Why not B: 6π cm is twice the arc, from using area = l × r without the half.
Why not C: 4π cm matches neither formula; it is a division slip.
Remember: sector area = (1/2) × arc × radius, just like a triangle with base l and height r.
Why D: the angles at the centre of a full circle add up to 360°, so 5 equal slices each get 360°/5 = 72°.
Why not A: 60° is the angle for 6 equal slices.
Why not B: 90° is the angle for 4 equal slices, a quadrant.
Why not C: 45° is the angle for 8 equal slices.
Remember: n equal sectors means 360°/n each.
Why D: Step 1 (sector = (θ/360) × πr²): (72/360)πr² = πr²/5 = 40π. Step 2 (solve for r): r² = 200, so r = √200 = 10√2 units.
Why not A: 200 is r², the square root was never taken.
Why not B: 100 is half of r², a slip in both the fraction and the root.
Why not C: 20 would give a sector area of (1/5)π × 400 = 80π, twice 40π.
Remember: after dividing out π and the fraction, you have r², so take the square root last.
Why B: Step 1 (circumference = 2πr): 2π × (60/π) = 120 cm, so π cancels. Step 2 (arc = (θ/360) × circumference): (θ/360) × 120 = 20, so θ = 360 × 20/120 = 60°.
Why not A: 30° would give an arc of only 120/12 = 10 cm, half of the 20 cm wire.
Why not C: 90° would need an arc of 30 cm, a quarter of 120 cm.
Why not D: 50° gives an arc of 16.7 cm, not 20 cm; it is a guess, not a solved value.
Remember: θ = 360 × arc / circumference.
Why B: Step 1 (hour hand rate): it turns 360° in 12 hours = 720 minutes, so 0.5° per minute. Step 2 (time elapsed): 7:20 to 7:55 is 35 minutes, so angle = 35 × 0.5 = 17.5° = (35/2)°. The length 6 cm is not needed for the angle.
Why not A: (35/4)° uses 0.25° per minute, half the true rate.
Why not C: 35° uses 1° per minute, twice the true rate.
Why not D: 70° treats the 35 minutes as if each minute moved the hand 2°.
Remember: minute hand 6° per minute, hour hand 0.5° per minute.
Chapter Navigation
Previous: Circles Class 10 Notes
Next: Surface Areas and Volumes Class 10 Notes
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- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
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Frequently Asked Questions
A sector is the part of a circular region enclosed by two radii and the corresponding arc, so its corner is at the centre, like a slice of pizza. A segment is the part enclosed between a chord and the corresponding arc, so it is cut off by the chord alone and the minor segment never contains the centre. Each comes as a minor and a major part, and on their own the words mean the minor one.
For a circle of radius r and a sector of angle θ in degrees, area of sector = (θ/360) × πr². It comes from the unitary method: the whole circle is a sector of 360° with area πr², so 1° gives πr²/360 and θ degrees give θ times that. If the arc length l is known, area = (1/2) × l × r.
Length of an arc of a sector of angle θ = (θ/360) × 2πr. The fraction θ/360 is the part of the full circle, and 2πr is the full circumference. For radius 21 cm and angle 60°, the arc is (1/6) × 132 = 22 cm. The perimeter of the sector adds the two radii: 2r + arc.
Area of the minor segment = area of the corresponding sector minus area of triangle OAB. The triangle has two sides equal to r. For 90° its area is r²/2. For 60° it is equilateral with area (√3/4)r². For 120° drop a perpendicular OM to the chord: OM = r/2 and AB = √3 r, so the area is again (√3/4)r².
Subtract the minor part from the whole circle. Major sector = πr² minus minor sector, which also equals ((360 − θ)/360) × πr². Major segment = πr² minus minor segment. Match the words carefully: taking the minor sector away from the circle gives the major sector, and the minor segment gives the major segment.
The minute hand turns 360° in 60 minutes, which is 6° per minute. Convert the time to an angle, take the length of the hand as the radius, and use (θ/360) × πr². For a 14 cm hand over 5 minutes, θ = 30° and the area is (1/12) × 616 = 154/3 cm², about 51.33 cm².
Use the value the question gives. The NCERT exercise says to use π = 22/7 unless stated otherwise, and some questions ask for π = 3.14. 22/7 is quick when the radius is a multiple of 7 or 3.5. If the question gives no value for √3, leave the answer in surd form, such as (462 − 441√3/4) cm².
