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By Ankit Gupta
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Updated on 22 Jul 2026, 12:07 IST
NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles 2026-27 provide clear and easy answers to all the questions given in the latest NCERT textbook. This chapter helps students understand how to calculate the area and perimeter of different parts of a circle, such as sectors, segments, and combined figures. These concepts are useful not only for board exams but also in real-life situations involving circular objects, designs, wheels, parks, and tracks.
The chapter begins with a quick revision of important ideas like the radius, diameter, circumference, and area of a circle. It then explains how to find the area of a sector using the angle at the centre. Students also learn to calculate the length of an arc and the area of a segment. Some questions include figures made by joining circles with squares, triangles, or other shapes. The solutions show how to break these figures into smaller parts and solve them step by step.
Infinity Learn’s NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles are written in simple language so that students can easily understand every method. Each answer follows the correct formula, calculation, and final unit. The solutions also help students avoid common mistakes, such as using the wrong radius, forgetting units, or confusing the area of a sector with the area of a segment.
By practising these solutions, students can improve their problem-solving skills, revise important formulas, and prepare confidently for the 2026-27 exams. The chapter becomes easier when students understand the diagrams and practise different types of questions regularly. With the support of Infinity Learn, learners can study at their own pace, clear their doubts, and build a strong foundation in geometry. These NCERT solutions are a useful study resource for homework, classroom learning, revision, and exam preparation.
NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles 2026-27 help students understand important concepts such as the area and circumference of a circle, the length of an arc, and the area of sectors and segments. The solutions explain every question from the NCERT textbook in simple language and follow a clear step-by-step method.
Question 1: Find the area of a sector of a circle whose radius is 6 cm and whose central angle is 60°. Use π = 22/7.
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Solution:
Given:
The area of a sector is calculated using:
Area of sector = (θ/360°) × πr2
Substituting the given values:
Area of sector = (60°/360°) × (22/7) × 62
= (1/6) × (22/7) × 36
= (22 × 6)/7
= 132/7 cm2
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Final Answer: The area of the sector is 132/7 cm2, or approximately 18.86 cm2.
Question 2: Find the area of a quadrant of a circle whose circumference is 22 cm. Use π = 22/7.
Solution:
Given:
Let the radius of the circle be r cm.
The circumference of a circle is:
2πr = 22
Substituting π = 22/7:
2 × (22/7) × r = 22
r = 22 ÷ (44/7)
r = 22 × (7/44)
r = 7/2 cm
A quadrant is one-fourth of a complete circle because it forms an angle of 90° at the centre.
Area of quadrant = (1/4) × πr2
Area of quadrant = (1/4) × (22/7) × (7/2)2
= (1/4) × (22/7) × (49/4)
= 1078/112
= 77/8 cm2
Final Answer: The area of the quadrant is 77/8 cm2, or 9.625 cm2.
Question 3: The minute hand of a clock is 14 cm long. Find the area swept by the minute hand in 5 minutes. Use π = 22/7.
Solution:
Given:
In 60 minutes, the minute hand completes one full revolution of 360°.
Therefore, the angle covered in 5 minutes is:
Angle = (360°/60) × 5
= 6° × 5
= 30°
The required region is a sector of radius 14 cm and angle 30°.
Area swept = (θ/360°) × πr2
Area swept = (30°/360°) × (22/7) × 142
= (1/12) × (22/7) × 196
= (1/12) × 616
= 154/3 cm2
Final Answer: The area swept by the minute hand in 5 minutes is 154/3 cm2, or approximately 51.33 cm2.
Question 4: A chord of a circle of radius 10 cm forms a right angle at the centre. Find the area of:
Use π = 3.14.
Solution:
Given:
Part (i): Area of the Minor Segment
The minor segment is obtained by subtracting the area of the triangle formed by the two radii and the chord from the area of the 90° sector.
Step 1: Find the area of the minor sector.
Area of minor sector = (90°/360°) × πr2
= (1/4) × 3.14 × 102
= (1/4) × 3.14 × 100
= 78.5 cm2
Step 2: Find the area of the triangle.
The two radii are perpendicular because the included angle is 90°. Therefore, the triangle is a right-angled triangle.
Area of triangle = (1/2) × Base × Height
= (1/2) × 10 × 10
= 50 cm2
Step 3: Subtract the triangle from the sector.
Area of minor segment = 78.5 − 50
= 28.5 cm2
Answer for Part (i): The area of the minor segment is 28.5 cm2.
Part (ii): Area of the Major Sector
The angle of the major sector is:
360° − 90° = 270°
Area of major sector = (270°/360°) × πr2
= (3/4) × 3.14 × 102
= (3/4) × 314
= 235.5 cm2
Answer for Part (ii): The area of the major sector is 235.5 cm2.
Question 5: In a circle of radius 21 cm, an arc forms an angle of 60° at the centre. Use π = 22/7 and find:
Solution:
Given:
Part (i): Length of the Arc
The length of an arc is:
Arc length = (θ/360°) × 2πr
Arc length = (60°/360°) × 2 × (22/7) × 21
= (1/6) × 2 × 22 × 3
= (1/6) × 132
= 22 cm
Answer for Part (i): The length of the arc is 22 cm.
Part (ii): Area of the Sector
Area of sector = (θ/360°) × πr2
= (60°/360°) × (22/7) × 212
= (1/6) × (22/7) × 441
= (1/6) × 22 × 63
= 231 cm2
Answer for Part (ii): The area of the sector is 231 cm2.
Part (iii): Area of the Segment
The two sides joining the centre to the endpoints of the chord are radii. Therefore, they are equal.
The angle between these radii is 60°. The remaining two angles of the triangle are equal and have a total of:
180° − 60° = 120°
Each of the two equal angles is:
120° ÷ 2 = 60°
Thus, all three angles are 60°, so the triangle is equilateral. Each side of the triangle is 21 cm.
Area of an equilateral triangle = (√3/4) × Side2
Area of triangle = (√3/4) × 212
= (√3/4) × 441
= (441√3)/4 cm2
Area of segment = Area of sector − Area of triangle
Area of segment = 231 − (441√3)/4 cm2
Using √3 ≈ 1.732:
Area of triangle ≈ (441 × 1.732)/4
≈ 190.95 cm2
Area of segment ≈ 231 − 190.95
≈ 40.05 cm2
Answer for Part (iii): The exact area of the segment is 231 − (441√3)/4 cm2, which is approximately 40.05 cm2.
Question 6: A chord of a circle of radius 15 cm forms an angle of 60° at the centre. Find the areas of the corresponding minor and major segments. Use π = 22/7 and √3 = 1.73.
Solution:
Given:
Step 1: Find the area of the complete circle.
Area of circle = πr2
= (22/7) × 152
= (22/7) × 225
= 4950/7 cm2
≈ 707.14 cm2
Step 2: Find the area of the 60° sector.
Area of sector = (60°/360°) × πr2
= (1/6) × (22/7) × 225
= 825/7 cm2
≈ 117.86 cm2
Step 3: Find the area of the triangle formed by the chord and the two radii.
Since the two radii are equal and the angle between them is 60°, the other two angles are also 60°. Therefore, the triangle is equilateral with side 15 cm.
Area of equilateral triangle = (√3/4) × Side2
= (1.73/4) × 152
= (1.73/4) × 225
= 97.3125 cm2
Area of the Minor Segment
Area of minor segment = Area of sector − Area of triangle
= 117.8571 − 97.3125
= 20.5446 cm2
Therefore, the area of the minor segment is approximately 20.54 cm2.
Area of the Major Segment
Area of major segment = Area of circle − Area of minor segment
= 707.1429 − 20.5446
= 686.5983 cm2
Therefore, the area of the major segment is approximately 686.60 cm2.
| Segment | Area |
| Minor segment | Approximately 20.54 cm2 |
| Major segment | Approximately 686.60 cm2 |
Question 7: A chord of a circle of radius 12 cm forms an angle of 120° at the centre. Find the area of the corresponding segment. Use π = 22/7 and √3 = 1.73.
Solution:
Given:
The required segment is the region between the chord and the corresponding minor arc. Its area is:
Area of segment = Area of 120° sector − Area of triangle
Step 1: Find the area of the 120° sector.
Area of sector = (120°/360°) × πr2
= (1/3) × (22/7) × 122
= (1/3) × (22/7) × 144
= 1056/7 cm2
≈ 150.86 cm2
Step 2: Find the area of the triangle.
Draw a perpendicular from the centre to the chord. This perpendicular bisects both the chord and the 120° central angle.
Therefore, each of the two right-angled triangles has an angle of:
120° ÷ 2 = 60°
In one of the right-angled triangles:
Half of the chord = 12 × sin 60°
= 12 × (√3/2)
= 6√3 cm
Therefore, the complete chord is:
Chord length = 2 × 6√3
= 12√3 cm
The perpendicular distance from the centre to the chord is:
Height = 12 × cos 60°
= 12 × (1/2)
= 6 cm
Area of triangle = (1/2) × Base × Height
= (1/2) × 12√3 × 6
= 36√3 cm2
Using √3 = 1.73:
Area of triangle = 36 × 1.73
= 62.28 cm2
Step 3: Find the area of the segment.
Area of segment = 150.8571 − 62.28
= 88.5771 cm2
Final Answer: The area of the corresponding segment is approximately 88.58 cm2.
Question 8: A horse is tied to a peg placed at one corner of a square grass field. Each side of the field is 15 m long. The horse is tied with a rope of length 5 m. Use π = 3.14 and find:
Solution:
Since the horse is tied at a corner of the square field, it can move through an angle of 90°. The grazing region is therefore a quadrant of a circle.
Part (i): Grazing Area with a 5 m Rope
Given:
Grazing area = (90°/360°) × πr2
= (1/4) × 3.14 × 52
= (1/4) × 3.14 × 25
= 19.625 m2
Answer for Part (i): The horse can graze an area of 19.625 m2.
Part (ii): Increase in Area with a 10 m Rope
The new rope is 10 m long. Since 10 m is less than the 15 m side of the field, the complete quadrant still lies inside the field.
New grazing area = (90°/360°) × π × 102
= (1/4) × 3.14 × 100
= 78.5 m2
Increase in grazing area = New area − Original area
= 78.5 − 19.625
= 58.875 m2
Answer for Part (ii): The grazing area increases by 58.875 m2.
Question 9: A brooch is made from silver wire in the shape of a circle with a diameter of 35 mm. Five additional diameters are made with the same wire, dividing the circle into 10 equal sectors. Use π = 22/7 and find:
Solution:
Given:
Part (i): Total Length of Silver Wire
The wire is used for:
Length of the circular boundary:
Circumference = πd
= (22/7) × 35
= 110 mm
Length of five diameters:
5 × 35 = 175 mm
Total wire required = 110 + 175
= 285 mm
Answer for Part (i): The total length of silver wire required is 285 mm.
Part (ii): Area of Each Sector
Since the circle is divided into 10 equal sectors, the angle of each sector is:
360° ÷ 10 = 36°
Area of each sector = (36°/360°) × πr2
= (1/10) × (22/7) × (35/2)2
= (1/10) × (22/7) × (1225/4)
= 26950/280
= 385/4 mm2
385/4 = 96.25
Answer for Part (ii): The area of each sector is 385/4 mm2, or 96.25 mm2.
Question 10: An umbrella has 8 ribs placed at equal distances from one another. Treating the open umbrella as a flat circle of radius 45 cm, find the area between two consecutive ribs. Use π = 22/7.
Solution:
Given:
The 8 ribs divide the circular umbrella into 8 equal sectors.
The angle between two consecutive ribs is:
360° ÷ 8 = 45°
The area between two consecutive ribs is the area of one 45° sector.
Area = (45°/360°) × πr2
= (1/8) × (22/7) × 452
= (1/8) × (22/7) × 2025
= 44550/56
= 22275/28 cm2
22275/28 ≈ 795.54
Final Answer: The area between two consecutive ribs is 22275/28 cm2, or approximately 795.54 cm2.
Question 11: A car has two windscreen wipers that do not overlap. Each wiper has a blade 25 cm long and moves through an angle of 115°. Find the total area cleaned in one sweep of both blades. Use π = 22/7.
Solution:
Given:
Step 1: Find the area cleaned by one wiper.
Area cleaned by one wiper = (115°/360°) × πr2
= (115/360) × (22/7) × 252
= (115/360) × (22/7) × 625
= 158125/252 cm2
Step 2: Find the area cleaned by both wipers.
Total area = 2 × (158125/252)
= 158125/126 cm2
158125/126 ≈ 1254.96
Final Answer: The total area cleaned by both wipers in one sweep is 158125/126 cm2, or approximately 1254.96 cm2.
Question 12: A lighthouse sends out red light over a sector of angle 80° up to a distance of 16.5 km to warn ships about underwater rocks. Find the area of the sea covered by the warning light. Use π = 3.14.
Solution:
Given:
The area covered by the light is the area of an 80° sector.
Area of sector = (θ/360°) × πr2
= (80°/360°) × 3.14 × 16.52
Simplifying the angle fraction:
80/360 = 2/9
Also:
16.52 = 272.25
Therefore:
Area = (2/9) × 3.14 × 272.25
= 2 × 3.14 × 30.25
= 189.97 km2
Final Answer: The lighthouse warns ships over an area of 189.97 km2.
Question 13: A circular table cover contains six equal designs. The radius of the cover is 28 cm. Find the total cost of making the designs at the rate of Rs. 0.35 per cm2. Use √3 = 1.7 and π = 22/7.
Solution:
Given:
Each design is a circular segment. The six equal chords form a regular hexagon inside the circle.
The central angle corresponding to each design is:
360° ÷ 6 = 60°
The triangle formed by two radii and one side of the hexagon has two equal sides. Since its central angle is 60°, its other two angles are also 60°. Therefore, it is an equilateral triangle with side 28 cm.
Step 1: Find the area of one 60° sector.
Area of one sector = (60°/360°) × πr2
= (1/6) × (22/7) × 282
= (1/6) × (22/7) × 784
= 1232/3 cm2
≈ 410.67 cm2
Step 2: Find the area of the equilateral triangle.
Area of triangle = (√3/4) × Side2
= (1.7/4) × 282
= (1.7/4) × 784
= 333.2 cm2
Step 3: Find the area of one design.
Area of one design = Area of sector − Area of triangle
= 1232/3 − 333.2
= 410.6667 − 333.2
= 77.4667 cm2
Step 4: Find the total area of all six designs.
Total area of designs = 6 × 77.4667
= 464.8 cm2
Step 5: Calculate the cost.
Total cost = Total area × Cost per cm2
= 464.8 × 0.35
= Rs. 162.68
| Calculation | Result |
| Area of one sector | 1232/3 cm2 |
| Area of one equilateral triangle | 333.2 cm2 |
| Area of one design | Approximately 77.47 cm2 |
| Total area of six designs | 464.8 cm2 |
| Total cost | Rs. 162.68 |
Final Answer: The cost of making all six designs is Rs. 162.68.
Question 14: Choose the correct expression for the area of a sector with angle P degrees in a circle of radius R.
Solution:
The area of a complete circle is πR2.
A sector with angle P degrees forms the fraction P/360 of the complete circle.
Therefore:
Area of sector = (P/360) × πR2
Multiply the numerator and denominator expression by 2:
(P/360) × πR2
= (P/720) × 2πR2
This matches the fourth option.
Final Answer: Option (D), (P/720) × 2πR2, is correct.
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NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles 2026-27 provide clear, step-by-step answers to the questions given in the NCERT textbook. They help students understand concepts related to circles, sectors, segments, arcs, and combined figures.
This chapter covers the circumference and area of a circle, the length of an arc, the area of a sector, and the area of a segment. It also includes questions based on figures formed by combining circles with other geometrical shapes.
The solutions explain each problem using the correct formulas and calculation steps. They help students revise important concepts, practise different types of questions, avoid common mistakes, and prepare confidently for school and board examinations.
Yes, the solutions are written in simple language and follow an easy step-by-step approach. Infinity Learn provides clear explanations that help students understand the diagrams, formulas, and methods used to solve each question.
Regular practice helps students remember formulas, improve calculation speed, and understand how to solve questions involving sectors, segments, and circular figures. It also builds confidence and improves overall performance in Class 10 Maths.