---
title: "NCERT Solutions Class 10 Maths Ch 11 Areas Related to Circles"
url: https://www.swavid.com/maths/class/10/chapter/areas-related-to-circles/ncert-solutions
dateModified: 2026-10-07T15:54:56+00:00
---

# NCERT Solutions Class 10 Maths Ch 11 Areas Related to Circles

This chapter's questions cover calculating areas and perimeters related to circles, including sectors, segments, arcs, and practical geometric applications. It also tests understanding of composite figures involving circles and polygons.

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## EXERCISE 11.1

### Question 1

*3 marks · Short answer*

Find the area of a sector of a circle with radius $6\text{ cm}$ if angle of the sector is $60^\circ$.

**Solution**

1. Radius $r = 6\text{ cm}$ and angle of the sector $\theta = 60^\circ$.
2. Area of the sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 6 \times 6\text{ cm}^2$
3. Area of the sector = $\frac{1}{6} \times \frac{22}{7} \times 36\text{ cm}^2 = \frac{132}{7}\text{ cm}^2$

**Answer:** $\frac{132}{7}\text{ cm}^2$

> Common mistake: Using the formula for arc length instead of area.

### Question 2

*3 marks · Short answer*

Find the area of a quadrant of a circle whose circumference is $22\text{ cm}$.

**Solution**

1. Circumference $2\pi r = 22\text{ cm}$, which gives radius $r = \frac{22}{2\pi} = \frac{22}{2 \times \frac{22}{7}} = \frac{7}{2}\text{ cm}$.
2. A quadrant of a circle is a sector with angle $\theta = 90^\circ$.
3. Area of the quadrant = $\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8}\text{ cm}^2$

**Answer:** $\frac{77}{8}\text{ cm}^2$

> Common mistake: Forgetting that a quadrant has an angle of $90^\circ$.

### Question 3

*3 marks · Short answer*

The length of the minute hand of a clock is $14\text{ cm}$. Find the area swept by the minute hand in $5\text{ minutes}$.

**Solution**

1. The minute hand describes an angle of $360^\circ$ in $60\text{ minutes}$, so in $5\text{ minutes}$ it turns through $\frac{360^\circ}{60} \times 5 = 30^\circ$.
2. The length of the minute hand is the radius $r = 14\text{ cm}$.
3. Area swept = $\frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{154}{3}\text{ cm}^2$

**Answer:** $\frac{154}{3}\text{ cm}^2$

> Common mistake: Incorrectly calculating the angle swept in 5 minutes as $5^\circ$.

### Question 4

*3 marks · Short answer*

A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use $\pi = 3.14$)

**Part (i)**

1. Given $r = 10\text{ cm}$ and $\theta = 90^\circ$. Area of minor segment = Area of sector - Area of right triangle.
2. Area of sector = $\frac{90^\circ}{360^\circ} \times 3.14 \times 10^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5\text{ cm}^2$.
3. Area of right triangle = $\frac{1}{2} \times 10 \times 10 = 50\text{ cm}^2$.
4. Area of minor segment = $78.5 - 50 = 28.5\text{ cm}^2$.

Answer (i): $28.5\text{ cm}^2$

**Part (ii)**

1. Angle of major sector = $360^\circ - 90^\circ = 270^\circ$.
2. Area of major sector = $\frac{270^\circ}{360^\circ} \times 3.14 \times 10^2 = \frac{3}{4} \times 3.14 \times 100 = 235.5\text{ cm}^2$.

Answer (ii): $235.5\text{ cm}^2$

**Answer:** Minor segment area is $28.5\text{ cm}^2$ and major sector area is $235.5\text{ cm}^2$.

> Common mistake: Using wrong angle for the major sector or incorrect formula for triangle area.

### Question 5

*3 marks · Short answer*

In a circle of radius $21\text{ cm}$, an arc subtends an angle of $60^\circ$ 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

**Part (i)**

1. Radius $r = 21\text{ cm}$ and $\theta = 60^\circ$.
2. Length of the arc = $\frac{\theta}{360^\circ} \times 2\pi r = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 2 \times 22 \times 3 = 22\text{ cm}$

Answer (i): $22\text{ cm}$

**Part (ii)**

1. Area of the sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times 22 \times 3 \times 21 = 231\text{ cm}^2$

Answer (ii): $231\text{ cm}^2$

**Part (iii)**

1. Area of triangle $\text{OAB}$ for an equilateral triangle of side $21\text{ cm}$ is $\frac{\sqrt{3}}{4} \times 21^2 = \frac{441\sqrt{3}}{4}\text{ cm}^2$.
2. Area of the segment = Area of sector $-$ Area of triangle $\text{OAB} = \left(231 - \frac{441\sqrt{3}}{4}\right)\text{ cm}^2$

Answer (iii): $\left(231 - \frac{441\sqrt{3}}{4}\right)\text{ cm}^2$

**Answer:** (i) $22\text{ cm}$, (ii) $231\text{ cm}^2$, (iii) $\left(231 - \frac{441\sqrt{3}}{4}\right)\text{ cm}^2$

> Common mistake: Using the wrong formula for the area of triangle when $\theta = 60^\circ$.

### Question 6

*3 marks · Short answer*

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

**Solution**

1. Given $r = 15\text{ cm}$, $\theta = 60^\circ$, $\pi = 3.14$, and $\sqrt{3} = 1.73$.
2. Area of minor sector = $\frac{60^\circ}{360^\circ} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = 117.75\text{ cm}^2$.
3. Area of equilateral triangle $OAB$ = $\frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73 \times 225}{4} = 97.3125\text{ cm}^2$.
4. Area of minor segment = $117.75 - 97.3125 = 20.4375\text{ cm}^2$.
5. Area of major segment = Area of circle - Area of minor segment = $(3.14 \times 15^2) - 20.4375 = 706.5 - 20.4375 = 686.0625\text{ cm}^2$.

**Answer:** Minor segment area is $20.44\text{ cm}^2$ and major segment area is $686.06\text{ cm}^2$.

> Common mistake: Subtracting minor segment from total area incorrectly or using wrong decimal values.

### Question 7

*3 marks · Short answer*

A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

**Solution**

1. Area of the corresponding sector = $\frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times 3.14 \times 12 \times 12 \text{ cm}^2 = 150.72 \text{ cm}^2$.
2. For area of triangle, draw perpendicular from centre to the chord, giving two $30^\circ-60^\circ-90^\circ$ triangles with height $12 \cos 60^\circ = 6 \text{ cm}$ and base $2 \times (12 \sin 60^\circ) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ cm}$.
3. Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3} = 36 \times 1.73 = 62.28 \text{ cm}^2$.
4. Area of the segment = $\text{Area of sector} - \text{Area of triangle} = 150.72 - 62.28 = 88.44 \text{ cm}^2$.

**Answer:** $88.44 \text{ cm}^2$

> Common mistake: Forgetting to multiply by 2 for the base of the triangle or using incorrect trigonometric values for $120^\circ$.

### Question 8

*3 marks · Short answer*

A horse is tied to a peg at one corner of a square shaped grass field of side $15\text{ m}$ by means of a $5\text{ m}$ 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 $10\text{ m}$ long instead of $5\text{ m}$. (Use $\pi = 3.14$)

**Part (i)**

1. The area of the field in which the horse can graze is the area of a sector of angle $90^\circ$ and radius $5 \text{ m}$ inside the square field.
2. Area = $\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 5 \times 5 \text{ m}^2$.
3. Area = $\frac{78.5}{4} = 19.625 \text{ m}^2$.

Answer (i): $19.625 \text{ m}^2$

**Part (ii)**

1. If the rope is $10 \text{ m}$ long, the new grazing area is a sector of angle $90^\circ$ and radius $10 \text{ m}$.
2. New Area = $\frac{90^\circ}{360^\circ} \times 3.14 \times 10 \times 10 = \frac{1}{4} \times 314 = 78.5 \text{ m}^2$.
3. Increase in grazing area = $78.5 - 19.625 = 58.875 \text{ m}^2$.

Answer (ii): $58.875 \text{ m}^2$

**Answer:** $58.875 \text{ m}^2$

> Common mistake: Taking the angle as something other than $90^\circ$ for a square corner.

### Question 9

*3 marks · Short answer*

A brooch is made with silver wire in the form of a circle with diameter $35\text{ mm}$. 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.

**Part (i)**

1. Total length of silver wire required is equal to the circumference of the brooch plus the length of 5 diameters.
2. Circumference = $\pi d = \frac{22}{7} \times 35 = 110 \text{ mm}$.
3. Length of 5 diameters = $5 \times 35 = 175 \text{ mm}$.
4. Total length = $110 + 175 = 285 \text{ mm}$.

Answer (i): $285 \text{ mm}$

**Part (ii)**

1. The circle is divided into 10 equal sectors, so the angle of each sector is $\frac{360^\circ}{10} = 36^\circ$.
2. Radius $r = \frac{35}{2} \text{ mm}$.
3. Area of each sector = $\frac{36^\circ}{360^\circ} \times \pi r^2 = \frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = \frac{385}{4} = 96.25 \text{ mm}^2$.

Answer (ii): $96.25 \text{ mm}^2$

**Answer:** $285 \text{ mm}$ and $962.5 \text{ mm}^2$

> Common mistake: Including only the circumference or only the diameters in the total length of wire.

### Question 10

*3 marks · Short answer*

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

**Solution**

1. An umbrella has 8 ribs, dividing the flat circular region into 8 equal sectors.
2. Angle of each sector $\theta = \frac{360^\circ}{8} = 45^\circ$.
3. Radius of the circular umbrella $r = 45 \text{ cm}$.
4. Area between the two consecutive ribs = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 45 \times 45$.
5. Area = $\frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22275}{28} = 795.54 \text{ cm}^2$.

**Answer:** $\frac{22275}{28} \text{ cm}^2$ (or $795.54 \text{ cm}^2$)

> Common mistake: Dividing by 360 instead of 8 to find the angle between consecutive ribs.

### Question 11

*3 marks · Short answer*

A car has two wipers which do not overlap. Each wiper has a blade of length $25\text{ cm}$ sweeping through an angle of $115^\circ$. Find the total area cleaned at each sweep of the blades.

**Solution**

1. Each wiper sweeps through an angle $\theta = 115^\circ$ with a blade length $r = 25 \text{ cm}$.
2. Area cleaned by one wiper = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{115^\circ}{360^\circ} \times \frac{22}{7} \times 25 \times 25$.
3. Since there are two wipers, the total area cleaned = $2 \times \frac{115}{360} \times \frac{22}{7} \times 625$.
4. Total area = $\frac{23 \times 11 \times 625}{126} = \frac{158125}{126} = 1254.48 \text{ cm}^2$.

**Answer:** $\frac{158125}{126} \text{ cm}^2$ (or $1254.48 \text{ cm}^2$)

> Common mistake: Forgetting to multiply by 2 for both wipers.

### Question 12

*3 marks · Short answer*

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^\circ$ to a distance of $16.5\text{ km}$. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$)

**Solution**

1. The lighthouse spreads light over a sector of angle $\theta = 80^\circ$ and radius $r = 16.5 \text{ km}$.
2. Area of the sea = $\frac{\theta}{360^\circ} \times \pi r^2$.
3. Substitution: $\frac{80^\circ}{360^\circ} \times 3.14 \times 16.5 \times 16.5$.
4. Calculation: $\frac{2}{9} \times 3.14 \times 272.25 = \frac{1709.73}{9} = 189.97 \text{ km}^2$.

**Answer:** $189.97 \text{ km}^2$

> Common mistake: Using $360$ instead of $80$ or miscalculating the decimal multiplication.

### Question 13

*3 marks · Short answer*

A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is $28\text{ cm}$, find the cost of making the designs at the rate of $\text{₹ } 0.35 \text{ per cm}^2$. (Use $\sqrt{3} = 1.7$)

**Solution**

1. The round table cover has 6 equal designs, which are segments of the circle subtending an angle of $\frac{360^\circ}{6} = 60^\circ$ at the centre.
2. Area of one design (segment) = Area of sector of angle $60^\circ$ - Area of equilateral triangle of side $28\text{ cm}$.
3. Area of 1 sector = $\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 28 \times 28 = \frac{1}{6} \times \frac{22}{7} \times 784 = \frac{1232}{3}\text{ cm}^2$.
4. Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times r^2 = \frac{1.7}{4} \times 28 \times 28 = 1.7 \times 7 \times 28 = 333.2\text{ cm}^2$.
5. Area of one design = $\frac{1232}{3} - 333.2 = 410.67 - 333.2 = 77.47\text{ cm}^2$, and total area of 6 designs = $6 \times 77.47 = 464.82\text{ cm}^2$.
6. Total cost of making the designs at the rate of $\text{₹ } 0.35\text{ per cm}^2$ = $464.82 \times 0.35 = \text{₹ } 162.68$ (approx).

**Answer:** ₹ 162.68

> Common mistake: Multiplying the area of only one segment or miscalculating the angle subtended at the centre.

### Question 14

*1 mark · MCQ*

Tick the correct answer in the following : Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is

- $\frac{p}{180} \times \pi R$
- $\frac{p}{180} \times \pi R^2$
- $\frac{p}{360} \times 2\pi R$
- $\frac{p}{720} \times 2\pi R^2$

**Solution**

1. The formula for the area of a sector of angle $\theta$ is $\frac{\theta}{360^\circ} \times \pi r^2$.
2. Replacing $\theta$ with $p$ and $r$ with $R$, we get $\frac{p}{360} \times \pi R^2$, which can also be written as $\frac{p}{720} \times 2\pi R^2$.
3. Therefore, option (D) is correct.

**Answer:** (D) $\frac{p}{720} \times 2\pi R^2$

> Common mistake: Confusing the area of a sector formula with the arc length formula.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles?

For the 2026-27 session, the chapter contains Exercise 11.1 which has 14 questions in total. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page only.

### Which topics do the questions in this chapter cover?

The questions cover concepts such as arc length, sector area, segment area, area of a quadrant, and area of a sector. They also include practical applications like the area of a sector formed by a horse grazing or the area of an umbrella's consecutive ribs.

### What types of questions are asked in Exercise 11.1?

Exercise 11.1 primarily features multiple choice questions and short answer questions. These test your understanding of calculating the area of minor segments, major sectors, and areas of sea over which a lighthouse spreads light.

### How should I approach the harder application-based questions in this chapter?

For complex problems involving combinations of plane figures, first draw a clear diagram and identify the basic geometric shapes involved. SwaVid's detailed solutions on this page break down these steps clearly to help you master the approach.

### Is the free PDF for these Class 10 Maths solutions available on SwaVid?

Yes, SwaVid provides a free PDF with comprehensive solutions for this chapter on this page only. Following these structured steps will help you present your answers neatly to score full marks in your exams.

## Related pages

- [Exercise 11.1 solutions](https://www.swavid.com/maths/class/10/chapter/areas-related-to-circles/ncert-solutions/exercise-11-1)
- [Areas Related to Circles: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/areas-related-to-circles)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student