NCERT Solutions for Class 10 Maths Chapter 11: Areas Related to Circles (NCERT 2026–27)

These Class 10 Maths Chapter 11 solutions cover Areas Related to Circles from the NCERT textbook (Reprint 2026–27). Every question of Exercise 11.1 is solved step by step, with full working for areas of sectors and segments, arc lengths and combination figures, so you can revise quickly and score full marks.

Class: 10 Subject: Mathematics Chapter: 11 Name: Areas Related to Circles Exercises: Exercise 11.1 (14 questions) Session: 2026–27
On this page

Chapter 11 Overview

Chapter 11 of Class 10 Maths, Areas Related to Circles, builds on the area of a circle (πr2) and its circumference (2πr) to find the areas of two important parts of a circular region — the sector (the region between two radii and the arc) and the segment (the region between a chord and the arc). Using the unitary method you derive a single formula for the area of a sector of any angle, the length of its arc, and then obtain the area of a segment as “sector minus triangle.” The chapter applies these ideas to real situations — clock hands, grazing animals, wiper blades, lighthouses and decorative designs. The Class 10 Maths Chapter 11 solutions below solve every question of Exercise 11.1 step by step.

Key Concepts & Definitions

Sector of a circle: the part of the circular region enclosed by two radii and the arc between them. The angle between the two radii is the angle of the sector (θ).

Minor & major sector: the smaller region is the minor sector; the larger is the major sector. The angle of the major sector is 360° − θ.

Segment of a circle: the part of the circular region enclosed between a chord and its corresponding arc. The smaller is the minor segment, the larger the major segment.

Arc: a part of the circle’s boundary (circumference) cut off by the two radii of a sector.

Quadrant: a sector whose angle is 90°, i.e. one-fourth of the circle.

Key idea: Area of segment = Area of corresponding sector − Area of the triangle formed by the two radii and the chord.

Important Formulas (Chapter 11)

Area of a circle: πr2  •  Circumference: 2πr.

Area of a sector of angle θ (in degrees): (θ/360) × πr2.

Length of an arc of a sector of angle θ: (θ/360) × 2πr.

Area of a segment = (θ/360) × πr2 − area of the triangle made by the two radii and the chord.

Area of an equilateral triangle of side a (used when θ = 60°): (√3/4)a2.

Area of a triangle with two sides r and included angle θ: (1/2)r2 sinθ — useful for the θ = 120° segment.

Major part = whole − minor part: major sector = πr2 − minor sector; major segment = πr2 − minor segment.

Exercise 11.1 Solutions

Unless stated otherwise, use π = 22/7. Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

SOLUTION Area of sector = (θ/360) × πr2 with r = 6 cm and θ = 60°. = (60/360) × (22/7) × 6 × 6 = (1/6) × (22/7) × 36. = (22 × 6)/7 = 132/7 = 18.86 cm2 (approx.), i.e. 132/7 cm2.

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

SOLUTION Circumference 2πr = 22 ⇒ r = 22/(2π) = 22 × 7 / (2 × 22) = 7/2 = 3.5 cm. A quadrant has angle 90°, so its area = (90/360) × πr2 = (1/4) × (22/7) × (7/2)2. = (1/4) × (22/7) × (49/4) = (22 × 49)/(4 × 7 × 4) = 1078/112 = 9.625 cm2, i.e. 77/8 cm2.

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

SOLUTION In 60 minutes the minute hand turns 360°, so in 5 minutes it turns (5/60) × 360° = 30°. The hand sweeps a sector of radius r = 14 cm and angle 30°. Area = (30/360) × (22/7) × 14 × 14 = (1/12) × (22/7) × 196. = (22 × 196)/(12 × 7) = 4312/84 = 154/3 = 51.33 cm2 (approx.).

4. 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 Here r = 10 cm and θ = 90° (the chord subtends a right angle). (i) Minor segment = area of minor sector − area of ▵OAB. Minor sector = (90/360) × 3.14 × 102 = (1/4) × 314 = 78.5 cm2. The triangle is right-angled at O with both legs equal to r, so area = (1/2) × 10 × 10 = 50 cm2. Minor segment = 78.5 − 50 = 28.5 cm2. (ii) Major sector has angle 360° − 90° = 270°. = (270/360) × 3.14 × 100 = (3/4) × 314 = 235.5 cm2.

5. 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 Here r = 21 cm and θ = 60°. (i) Arc length = (θ/360) × 2πr = (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm. (ii) Area of sector = (60/360) × (22/7) × 21 × 21 = (1/6) × (22 × 63) = (1/6) × 1386 = 231 cm2. (iii) Area of segment = area of sector − area of ▵OAB. With OA = OB = 21 and θ = 60°, ▵OAB is equilateral of side 21. Area of triangle = (√3/4) × 212 = (√3/4) × 441 = (441√3)/4 cm2. ∴ Area of segment = (231 − (441√3)/4) cm2 = 21/4 (44 − 21√3) cm2 ≈ 40.05 cm2.

6. 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 Here r = 15 cm and θ = 60°. Area of minor sector = (60/360) × 3.14 × 152 = (1/6) × 3.14 × 225 = (1/6) × 706.5 = 117.75 cm2. Since θ = 60°, ▵OAB is equilateral of side 15. Area = (√3/4) × 152 = (1.73/4) × 225 = (389.25)/4 = 97.3125 cm2. Minor segment = 117.75 − 97.3125 = 20.4375 cm2 ≈ 20.44 cm2. Area of circle = 3.14 × 225 = 706.5 cm2. Major segment = area of circle − minor segment = 706.5 − 20.4375 = 686.0625 cm2 ≈ 686.06 cm2.

7. 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 Here r = 12 cm and θ = 120°. Area of sector = (120/360) × 3.14 × 122 = (1/3) × 3.14 × 144 = (1/3) × 452.16 = 150.72 cm2. Area of ▵OAB with two sides r and included angle 120° = (1/2)r2 sin120° = (1/2) × 144 × (√3/2) = 36√3. = 36 × 1.73 = 62.28 cm2. ∴ Area of segment = 150.72 − 62.28 = 88.44 cm2.

8. 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 (see Fig. 11.8). 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 At a corner of the square the field opens through a right angle (90°), so the horse can graze over a quarter circle of radius equal to the rope length. (i) With rope 5 m: grazing area = (90/360) × π × 52 = (1/4) × 3.14 × 25 = (1/4) × 78.5 = 19.625 m2. (ii) With rope 10 m: grazing area = (1/4) × 3.14 × 102 = (1/4) × 314 = 78.5 m2. Increase = 78.5 − 19.625 = 58.875 m2. (10 m is less than the 15 m side, so the quarter circle still fits inside the field.)

9. 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 as shown in Fig. 11.9. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

SOLUTION Diameter = 35 mm, so radius r = 35/2 = 17.5 mm. (i) Total wire = circumference of the circle + length of 5 diameters. Circumference = 2πr = 2 × (22/7) × 17.5 = 110 mm. Five diameters = 5 × 35 = 175 mm. Total length = 110 + 175 = 285 mm. (ii) The 5 diameters divide the circle into 10 equal sectors, so each sector has angle 360°/10 = 36°. Area of each sector = (36/360) × πr2 = (1/10) × (22/7) × (17.5)2 = (1/10) × (22/7) × 306.25. = (22 × 306.25)/(70) = 6737.5/70 = 96.25 mm2 = 385/4 mm2.

10. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

SOLUTION The 8 ribs divide the flat circle into 8 equal sectors, so the angle between two consecutive ribs = 360°/8 = 45°. Area between two ribs = (45/360) × πr2 = (1/8) × (22/7) × 452 = (1/8) × (22/7) × 2025. = (22 × 2025)/(56) = 44550/56 = 22275/28 cm2 ≈ 795.54 cm2.

11. 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 r = 25 cm and angle 115°. Area cleaned by one wiper = (115/360) × (22/7) × 252 = (115/360) × (22/7) × 625. As the two wipers do not overlap, total area = 2 × that area = (230/360) × (22/7) × 625 = (23/36) × (22/7) × 625. = (23 × 22 × 625)/(36 × 7) = 316250/252 = 158125/126 cm2 ≈ 1254.96 cm2.

12. 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 The warned region is a sector of radius r = 16.5 km and angle 80°. Area = (80/360) × πr2 = (2/9) × 3.14 × (16.5)2 = (2/9) × 3.14 × 272.25. = (2/9) × 854.865 = 1709.73/9 = 189.97 km2 (approx.).

13. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2. (Use √3 = 1.7)

SOLUTION Six equal designs around the centre means each design subtends 360°/6 = 60° at the centre, with r = 28 cm. Each design is a segment (sector − triangle). Area of one sector = (60/360) × (22/7) × 282 = (1/6) × (22/7) × 784 = (1/6) × 2464 = 410.6667 cm2. Since θ = 60°, the triangle is equilateral of side 28: area = (√3/4) × 282 = (1.7/4) × 784 = 1.7 × 196 = 333.2 cm2. Area of one design (segment) = 410.6667 − 333.2 = 77.4667 cm2. Area of all 6 designs = 6 × 77.4667 = 464.8 cm2. Cost = 464.8 × ₹ 0.35 = ₹ 162.68 (approx.).

14. Tick the correct answer in the following:
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

SOLUTION Area of a sector = (angle/360) × πR2 = (p/360) × πR2. Rewrite to match the options: (p/360) × πR2 = (p/720) × 2πR2 (multiplying numerator and denominator suitably). ∴ The correct answer is (D) (p/720) × 2πR2.

Common Mistakes to Avoid

Watch out for these

  • Forgetting to convert time to an angle in clock problems — the minute hand turns 6° per minute (360° in 60 min), so 5 min = 30°.
  • Using the wrong π: read the question — some parts ask for π = 22/7 and others for π = 3.14.
  • Confusing segment with sector: a segment = sector − triangle; never subtract the triangle from a sector you have not computed.
  • For a 60° segment, remembering the triangle is equilateral (area (√3/4)a2); for other angles use (1/2)r2 sinθ.
  • Mixing up minor and major: the major sector/segment angle is 360° − θ, or use major part = whole circle − minor part.
  • Grazing & design problems: first work out the sector angle from the geometry (corner of a square = 90°; 6 designs = 60° each; 8 ribs = 45° each).

Practice MCQs & Assertion–Reason

1. The area of a sector of angle θ in a circle of radius r is:

(a) (θ/180)πr2    (b) (θ/360)πr2    (c) (θ/360)2πr    (d) (θ/720)πr2

2. The length of the arc of a sector of angle 60° in a circle of radius 21 cm (π = 22/7) is:

(a) 11 cm    (b) 22 cm    (c) 44 cm    (d) 66 cm

3. The area swept by the minute hand of length 14 cm in 5 minutes is:

(a) 154/3 cm2    (b) 154 cm2    (c) 308 cm2    (d) 77 cm2

4. The angle of the major sector when a chord subtends a right angle at the centre is:

(a) 90°    (b) 180°    (c) 270°    (d) 360°

5. Area of a quadrant of a circle of radius 3.5 cm (π = 22/7) is:

(a) 9.625 cm2    (b) 38.5 cm2    (c) 19.25 cm2    (d) 7 cm2

6. The area of a segment of a circle equals:

(a) sector + triangle    (b) sector − triangle    (c) triangle − sector    (d) circle − sector

7. If 8 ribs of an umbrella are equally spaced, the angle between two consecutive ribs is:

(a) 30°    (b) 40°    (c) 45°    (d) 60°

8. A horse tied at a corner of a square field can graze a region shaped like a:

(a) semicircle    (b) quarter circle    (c) full circle    (d) segment

9. For a 60° segment, the triangle formed by the two radii and the chord is:

(a) right-angled    (b) isosceles only    (c) equilateral    (d) scalene

10. The area of a circle of radius r that is 270° of a sector is what fraction of the whole circle?

(a) 1/4    (b) 1/2    (c) 3/4    (d) 2/3

Answer key: 1-(b), 2-(b), 3-(a), 4-(c), 5-(a), 6-(b), 7-(c), 8-(b), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The area of a sector of angle 60° in a circle of radius 6 cm is 132/7 cm2.

Reason: Area of a sector = (θ/360) × πr2.

A-R 2. Assertion: The area of a segment is found by subtracting the triangle from the corresponding sector.

Reason: A segment is the region between a chord and its arc.

A-R 3. Assertion: The minute hand of a clock sweeps an angle of 30° in 5 minutes.

Reason: The minute hand turns 360° in 60 minutes, i.e. 6° per minute.

A-R 4. Assertion: For a chord subtending 60° at the centre, the triangle formed with the two radii is equilateral.

Reason: All three sides of that triangle are equal because two are radii and the 60° angle forces the third side equal too.

A-R 5. Assertion: The area of the major sector equals πr2 minus the area of the minor sector.

Reason: The angle of the major sector is θ − 360°.

Answer key: 1-(A), 2-(B), 3-(A), 4-(A), 5-(C). (A-R 5: Assertion true, but the angle of the major sector is 360° − θ, so the Reason is false.)

Quick Revision Summary

  • Area of a circle = πr2; circumference = 2πr.
  • Area of a sector of angle θ = (θ/360) × πr2; arc length = (θ/360) × 2πr.
  • Area of a segment = area of corresponding sector − area of the triangle.
  • For θ = 60° the triangle is equilateral, area = (√3/4)a2; in general use (1/2)r2 sinθ.
  • Major part = whole circle − minor part; major sector angle = 360° − θ.
  • In applications, find the sector angle from the geometry first: corner of a square = 90°, 6 equal designs = 60° each, 8 ribs = 45° each, 5 minutes on a clock = 30°.

How to score full marks in this chapter

Always write the sector/segment formula first, then substitute neatly. Check which value of π the question demands (22/7 or 3.14) and keep that throughout. For segments, compute the sector and the triangle separately and subtract. When a figure is described in words, label the angle of the sector before plugging numbers in — most marks are lost from a wrong angle, not wrong arithmetic. Carry units (cm2, m2, km2) at every step.

Frequently Asked Questions

What is Class 10 Maths Chapter 11 about?

Chapter 11, Areas Related to Circles, teaches how to find the area of a sector and a segment of a circle and the length of an arc, using the formulas (θ/360)πr2 for the sector area and (θ/360)2πr for the arc, and “sector minus triangle” for a segment. It applies these to real-life figures like clocks, wiper blades and grazing fields.

How many exercises are there in Class 10 Maths Chapter 11?

The 2026–27 NCERT textbook has one exercise — Exercise 11.1 with 14 questions — all of which are solved step by step on this page.

What is the difference between a sector and a segment?

A sector is the region enclosed by two radii and the arc between them, while a segment is the region enclosed between a chord and its corresponding arc. The area of a segment is found by subtracting the triangle (formed by the two radii and the chord) from the area of the corresponding sector.

Are these Class 10 Maths Chapter 11 solutions free?

Yes. All solutions are free and follow the official NCERT Class 10 Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

← Chapter 10: Circles Class 10 Maths Chapter 12: Surface Areas and Volumes →
Scroll to Top