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# NCERT Solutions for Class 10 Maths Chapter 11 - Areas Related to Circles Exercise 11.1

Last updated date: 10th Sep 2024
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## NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 - FREE PDF Download

NCERT Solutions for Class 10 Areas Related to Circles Exercise 11.1 focuses on the fascinating world of circles, exploring the various concepts and formulas related to their areas. This chapter is crucial as it builds on the basic properties of circles and extends them to practical applications involving the calculation of areas. Ex 11.1 Class 10 focuses on helping students understand how to calculate the area of a circle, the area of a sector, and the area of a segment. By solving these problems, students will gain a deeper understanding of the relationships between different parts of a circle and how to apply these concepts to real-world situations. Mastery of these topics is essential for scoring well in exams and developing a strong foundation in geometry.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 11 Exercise 11.1 Class 10 | Vedantu
3. Formulas Used in Class 10th Exercise 11.1
4. Access PDF for Maths NCERT Chapter 11 Areas Related to Circles Exercise 11.1 Class 10
5. CBSE Class 10 Maths Chapter 11 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 11 Exercise 11.1 Class 10 | Vedantu

• This exercise covers the basic concepts of calculating the area and perimeter (circumference) of circles. It forms the foundation for understanding more complex problems related to circle geometry.

• Key formulas for the area (𝜋𝑟^2) and circumference (2𝜋𝑟) of circles are introduced.

• Problems in this exercise include finding the area of sectors and segments of a circle using the central angle and radius of the circle.

• The exercise includes word problems that relate to real-life situations, such as finding the area of circular gardens, paths, and other round objects.

## Formulas Used in Class 10th Exercise 11.1

1. Area of a Circle:

• A = πr²

1. Area of a Sector:

• Area of Sector = (θ/360)πr²

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1. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is ${60^ \circ }$.$\pi =\dfrac{22}{7}$

Ans:

Given that,

Radius of the circle = $r = 6cm$

Angle made by the sector with the center, $\theta = {60^ \circ }$

Let OACB be a sector of the circle making ${60^ \circ }$ angle at center O of the circle.

We know that area of sector of angle, $= \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$

Thus, Area of sector OACB $= \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times {(6)^2}$

$= \dfrac{1}{6} \times \dfrac{{22}}{7} \times 6 \times 6$

$= \dfrac{{132}}{7}c{m^2}$

Therefore, the area of the sector of the circle making 60° at the center of the circle is $\dfrac{{132}}{7}c{m^2}$.

2. Find the area of a quadrant of a circle whose circumference is 22 cm. $\text{Use } \pi =\dfrac{22}{7}$

Ans:

Given that,

Circumference = 22 cm

Let the radius of the circle be $r$.

According to the given condition,

$2\pi r = 22$

$\Rightarrow r = \dfrac{{22}}{{2\pi }}$

$= \dfrac{{11}}{\pi }$

We know that, quadrant of a circle subtends ${90^ \circ }$ angle at the center of the circle.

Area of such quadrant of the circle $= \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {r^2}$

$= \dfrac{1}{4} \times \pi \times {\left( {\dfrac{{11}}{\pi }} \right)^2}$

$= \dfrac{{121}}{{4\pi }}$

$= \dfrac{{77}}{8}c{m^2}$

Hence, the area of a quadrant of a circle whose circumference is 22 cm is $= \dfrac{{77}}{8}c{m^2}$.

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

$\text{Use } \pi =\dfrac{22}{7}$

Ans:

Given that,

Radius of clock or circle = $r$ = 14 cm.

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates ${360^ \circ }$.

Thus, in 5 minutes, minute hand will rotate $= \dfrac{{{{360}^ \circ }}}{{{{60}^ \circ }}} \times 5$

$= {30^ \circ }$

Now,

the area swept by the minute hand in 5 minutes = the area of a sector of ${30^ \circ }$ in a circle of 14 cm radius.

Area of sector of angle $\theta = \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$

Thus, Area of sector of ${30^ \circ } = \dfrac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times 14 \times 14$

$= \dfrac{{11 \times 14}}{3}$

$= \dfrac{{154}}{3}c{m^2}$

Therefore, the area swept by the minute hand in 5 minutes is $\dfrac{{154}}{3}c{m^2}$.

4. A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding:  $[\text{Use }\pi =3.14]$

Ans:

Given that,

Radius of the circle $= r = 10cm$

Angle subtended by the cord = angle for minor sector$= {90^ \circ }$

Angle for major sector $= {360^ \circ } - {90^ \circ } = {270^ \circ }$

(i) Minor segment

Ans: It is evident from the figure that,

Area of minor segment ACBA = Area of minor sector OACB − Area of ΔOAB

Thus,

Area of minor sector OACB $= \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$ $= \dfrac{1}{4} \times 3.14 \times {(10)^2}$ $= 78.5c{m^2}$

Area of ΔOAB $= \dfrac{1}{2} \times OA \times OB = \dfrac{1}{2} \times {(10)^2} = 50c{m^2}$

Area of minor segment ACBA $= 78.5 - 50 = 28.5c{m^2}$

Hence, area of minor segment is $28.5c{m^2}$

(ii) Major sector

Ans: It is evident from the figure that,

Area of major sector OADB $= \dfrac{{{{270}^ \circ }}}{{{{360}^ \circ }}} = \dfrac{3}{4} \times 3.14 \times {(10)^2} = 235.5c{m^2}$.

Hence, the area of the major sector is $235.5c{m^2}$.

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.

$\pi =\dfrac{22}{7}$. Find:

Ans:

Given that,

Radius of circle = $r =$ 21 cm

Angle subtended by the given arc = $\theta = {60^ \circ }$

(i) The length of the arc

Ans: We know that, Length of an arc of a sector of angle $\theta = \dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi r$

Thus, Length of arc ACB $= \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \dfrac{{22}}{7} \times 21$

$= 22cm$

Hence, the length of the arc of a given circle is $22cm$.

(ii) Area of the sector formed by the arc

Ans: We know that, Area of sector OACB $= \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= 231c{m^2}$

Hence, the area of the sector formed by the arc of the given circle is $231c{m^2}$.

(iii) Area of the segment formed by the corresponding chord

Ans: In $OAB$,

As radius $OA = OB$

$\Rightarrow \angle OAB = \angle OBA$

$\angle OAB + \angle AOB + \angle OBA = {180^ \circ }$

$2\angle OAB + {60^ \circ } = {180^ \circ }$

$\angle OAB = {60^ \circ }$

Therefore, $OAB$ is an equilateral triangle.

Now, area of $OAB$ $= \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}$

$= \dfrac{{\sqrt 3 }}{4} \times {\left( r \right)^2}$

$= \dfrac{{\sqrt 3 }}{4} \times {\left( {21} \right)^2}$

$= \dfrac{{441\sqrt 3 }}{4}c{m^2}$

We know that, Area of segment ACB = Area of sector OACB − Area of $\Delta AOB$

$= \left( {231 - \dfrac{{441\sqrt 3 }}{4}} \right)c{m^2}$.

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the areas of the corresponding minor and major segments of the circle.

$\text{Use}\pi =\dfrac{22}{7}, \sqrt{3}=1.73$

Ans:

Given that,

Radius of circle = $r =$ 15 cm

Angle subtended by chord $= \theta = {60^ \circ }$

Area of circle $= \pi {r^2}$ $= 3.14{\left( {15} \right)^2}$

$= 706.5c{m^2}$

Area of sector OPRQ $= \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \dfrac{1}{6} \times 3.14{(15)^2} = 117.75c{m^2}$

Now, for the area of major and minor segments,

In $OPQ$,

Since, $OP = OQ$

$\Rightarrow \angle OPQ = \angle OQP$

$\angle OPQ = {60^ \circ }$

Thus, $OPQ$ is an equilateral triangle.

Area of $OPQ$ $= \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}$

$= \dfrac{{\sqrt 3 }}{4} \times {(r)^2}$

$= \dfrac{{225\sqrt 3 }}{4}$ $= 97.3125c{m^2}$.

Now,

Area of minor segment PRQP = Area of sector OPRQ − Area of $OPQ$

$= 117.75 - 97.3125$

$= 20.4375c{m^2}$

Area of major segment PSQP = Area of circle − Area of minor segment PRQP

$= 706.5 - 20.4375$

$= 686.0625c{m^2}$

Therefore, the areas of the corresponding minor and major segments of the circle are $20.4375c{m^2}$ and $686.0625c{m^2}$ respectively.

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle. $\text{Use}\pi =\dfrac{22}{7}\text{ and }\sqrt{3}=1.73$

Ans:

Draw a perpendicular OV on chord ST bisecting the chord ST such that SV = VT

Now, values of OV and ST are to be found.

Therefore,

In $OVS$,

$\cos {60^ \circ } = \dfrac{{OV}}{{OS}}$

$\Rightarrow \dfrac{{OV}}{{12}} = \dfrac{1}{2}$

$\Rightarrow OV = 6cm$

Also, $\dfrac{{SV}}{{SO}} = \sin {60^ \circ }$

$\Rightarrow \dfrac{{SV}}{{12}} = \dfrac{{\sqrt 3 }}{2}$

$\Rightarrow SV = 6\sqrt 3$

Now, $ST = 2SV$

$= 2 \times 6\sqrt 3 = 12\sqrt 3 cm$

Area of $OST = \dfrac{1}{2} \times ST \times OV$

$= \dfrac{1}{2} \times 12\sqrt 3 \times 6$

$= 62.28c{m^2}$

Area of sector OSUT $= \dfrac{{{{120}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(12)^2}$

$= 150.42c{m^2}$

Area of segment SUTS = Area of sector OSUT − Area of $OVS$

$= 150.72 - 62.28$

$= 88.44c{m^2}$

Hence, the area of the corresponding segment of the circle is $88.44c{m^2}$.

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 the given figure). $\text{Use }\pi =3.14$

Find

Ans:

From the above figure, it is clear that the horse can graze a sector of ${90^ \circ }$ in a circle of 5 m radius.

Hence,

$\theta \text{ }\!\!~\!\!\text{ }={{90}^{{}^\circ }}$

$r=5m$

(i) The area of that part of the field in which the horse can graze.

Ans: It is evident from the figure,

Area that can be grazed by horse = Area of sector OACB

$= \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}$

$= \dfrac{1}{4} \times 3.14 \times {(5)^2}$ $= 19.625{m^2}$

(ii) The increase in the grazing area if the rope were 10 m long instead of 5 m.

Ans: It is evident from the figure,

Area that can be grazed by the horse when length of rope is 10 m long

$= \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {(10)^2}$

$= 78.5{m^2}$

Therefore, the increase in grazing area for horse $= (78.5 - 19.625){m^2} = 58.875{m^2}$.

9. A brooch is made with silver wire in the form of a circle with a diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. $\pi =\dfrac{22}{7}$

Find:

Ans:

Given that,

Radius of the circle $= r$ $= \dfrac{{diameter}}{2}$$= \dfrac{{35}}{2}mm$

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° (i.e., 360°/10=36°) at the center of the circle.

(i) The total length of the silver wire required.

Ans:

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Circumference of brooch $= 2\pi r$

$= 2 \times \dfrac{{22}}{7} \times \left( {\dfrac{{35}}{2}} \right)$

$= 110mm$

Length of wire required $= 110 + \left( {5 \times 35} \right)$

$= 285mm$

Therefore, The total length of the silver wire required is $285mm$.

(ii) The area of each sector of the brooch.

Ans:

Area of each sector $= \dfrac{{{{36}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \dfrac{1}{{10}} \times \dfrac{{22}}{7} \times {\left( {\dfrac{{35}}{2}} \right)^2}$

$= \dfrac{{385}}{4}m{m^2}$

Hence, The area of each sector of the brooch is $\dfrac{{385}}{4}m{m^2}$.

10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. $\pi =\dfrac{22}{7}$

Ans:

Given that,

Radius of the umbrella $= r$$= 45cm$

There are 8 ribs in an umbrella.

The angle between two consecutive ribs is subtending $\dfrac{{{{360}^ \circ }}}{8} = {45^ \circ }$ at the center of the assumed flat circle.

Area between two consecutive ribs of the assumed circle $= \dfrac{{{{45}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \dfrac{1}{8} \times \dfrac{{22}}{7} \times {\left( {45} \right)^2}$

$= \dfrac{{22275}}{{28}}c{m^2}$

Hence, the area between the two consecutive ribs of the umbrella is $\dfrac{{22275}}{{28}}c{m^2}$.

11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of ${115}^{\circ}$. Find the total area cleaned at ${115^ \circ }$ each sweep of the blades. $\pi =\dfrac{22}{7}$

Ans:

Given that,

Each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.

Area of sector $= \dfrac{{{{115}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {\left( {25} \right)^2}$

$= \dfrac{{158125}}{{252}}c{m^2}$

Area swept by 2 blades $= 2 \times \dfrac{{158125}}{{252}}$

$= \dfrac{{158125}}{{126}}c{m^2}$.

Therefore, the total area cleaned at each sweep of the blades is $\dfrac{{158125}}{{126}}c{m^2}$.

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

Ans:

Given that,

The lighthouse spreads light across a sector (represented by the shaded part in the figure) of ${80^ \circ }$ in a circle of 16.5 km radius.

Area of sector OACB $= \dfrac{{{{80}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \dfrac{2}{9} \times 3.14 \times {(16.5)^2}$

$= 189.97k{m^2}$

Hence, the area of the sea over which the ships are warned is $189.97k{m^2}$.

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

Ans:

Given in the figure,

The designs are segments of the circle.

Radius of the circle is 28cm.

Consider segment APB and chord AB is a side of the hexagon.

Each chord will substitute at $\dfrac{{{{360}^ \circ }}}{6} = {60^ \circ }$ at the center of the circle.

In $OAB$,

Since, $OA = OB$

$\Rightarrow \angle OAB + \angle OBA + \angle AOB = {180^ \circ }$

$2\angle OAB = {180^ \circ } - {60^ \circ } = {120^ \circ }$

$\angle OAB = {60^ \circ }$

Therefore, $OAB$ is an equilateral triangle.

Area of $OAB$ $= \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}$

$= \dfrac{{\sqrt 3 }}{4} \times {\left( {28} \right)^2}$

$= 333.2c{m^2}$

Area of sector OAPB $= \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \dfrac{1}{6} \times \dfrac{{22}}{7} \times {(28)^2}$

$= \dfrac{{1232}}{3}c{m^2}$

Area of segment APBA = Area of sector OAPB − Area of ∆OAB

$= \left( {\dfrac{{1232}}{3} - 333.2} \right)c{m^2}$

Therefore, area of designs $= 6 \times \left( {\dfrac{{1232}}{3} - 333.2} \right)c{m^2}$

$= 464.8c{m^2}$

Now, given that Cost of making 1 $c{m^2}$ designs = Rs 0.35

Cost of making 464.76 $c{m^2}$ designs $= 464.8 \times 0.35 =$162.68

Therefore, the cost of making such designs is Rs 162.68.

14. Tick the correct answer in the following: Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is $(A)\dfrac{P}{{180}} \times 2\pi R$ $(B)\dfrac{P}{{180}} \times 2\pi {R^2}$ $(C)\dfrac{P}{{180}} \times \pi R$ $(D)\dfrac{P}{{720}} \times 2\pi {R^2}$

Ans:

We know that area of sector of angle $\theta = \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {R^2}$

So, Area of sector of angle $P = \dfrac{P}{{{{360}^ \circ }}}\left( {\pi {R^2}} \right)$

$= \left( {\dfrac{P}{{{{720}^ \circ }}}} \right)\left( {2\pi {R^2}} \right)$

Hence, (D) is the correct answer.

## Conclusion

Class 10 Maths Chapter 11.1, "Areas Related to Circles," provides a solid foundation for understanding and calculating the areas of circles, sectors, and segments. By mastering these calculations, students gain essential skills that are applicable in both academic and real-world scenarios. Ex 11.1 Class 10 emphasizes the importance of precise formula application and enhances problem-solving abilities. Completing these problems will not only prepare students for exams but also build confidence in handling geometric concepts, paving the way for advanced studies in mathematics.

## CBSE Class 10 Maths Chapter 11 Other Study Materials

 S.No. Important Links for Chapter 11 Areas Related To Circles 1 Class 10 Chapter 11 Areas Related to Circles Important Questions 2 Class 10 Chapter 11 Areas Related to Circles Revision Notes 3 Class 10 Chapter 11 Areas Related to Circles NCERT Exemplar Solution 4 Class 10 Chapter 11 Areas Related to Circles RD Sharma Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 11 - Areas Related to Circles Exercise 11.1

1. What is the formula for the area of a circle in 11.1 Class 10 Maths NCERT Solutions?

The formula for the area of a circle is A=πr², where r is the radius of the circle.

2. What is a segment of a circle, and how is its area calculated?

A segment of a circle is covered in Class 10 Ch 11 Maths Ex 11.1 and it is the region between a chord and the corresponding arc. Its area is found by subtracting the area of the triangle formed by the chord and the radius from the area of the sector.

3. What is the relationship between the radius, diameter, and circumference of a circle?

According to Class 10th Maths Chapter 11 Exercise 11.1, The radius is half the diameter, and the circumference is the distance around the circle. The relationships are given by C=2πr and D=2r, where CC is the circumference, r is the radius, and D is the diameter.

4. Can you give an example of a real-life application of calculating the area of a sector?

One real-life application is calculating the area of a slice of pizza, which is essentially a sector of a circle. By knowing the radius of the pizza and the angle of the slice, one can determine the area of the slice using the sector area formula. This is how Ex 11.1 Class 10 NCERT Solutions helps students in real world scenarios.

5. In which questions of exercise 11.1 of 10th Maths, students face difficulty while solving?

Exercise 11.1 of Class 10 Maths deals with Circles. It is a very important and comparatively easy and interesting chapter in the NCERT syllabus from which the students can score good marks. However, a few questions in Exercise 11.1 are a bit difficult for the students to solve. These are questions 3, 7, 9, 10, and 13. One must carefully analyse and understand these five questions to be able to solve them and also refer to the NCERT Solutions for Class 10 maths for help.

6. Which questions are easy in Exercise 11.1 of 10th Maths?

Exercise 11.1 from NCERT Maths Class 10 is a vital exercise that will give the students a potential edge to score well in the board exams. In this exercise, most of the questions are quite easy to understand and solve. The problems that are easy and therefore the favourites of students are 1, 2, 4, 5, 6, 8, 11, 12 and 14. Students enjoy solving these questions.

7. Is Exercise 11.1 important?

Exercise 11.1 of Class 10 Maths Chapter 11 is an important exercise in Class 10. This exercise deals with the questions related to the area of circles. These topics form a major part of the CBSE examination and also a part of the study curriculum in higher classes. Students must be clear with this chapter to ace their examinations. If you want proper guidance, refer to NCERT Solutions for Class 10 Maths for help.

8. From where can I download the NCERT Solutions for  Exercise 11.1 of 10th Maths?

The NCERT Solutions for Exercise 11.1 of Class 10 Maths is available free of cost on the Vedantu website (vedantu.com). If you want to download CBSE NCERT Solutions for Exercise 11.1 of 10th Maths, you can follow these steps:

1. Click NCERT Solutions for Class 10 Maths.

2. The next page will contain the Exercise 11.1 Solutions PDF.

3. Click on it, and it will redirect you to the next page, where you will find the link to download.

9. How to get full marks in Class 10 Maths?

The only way to get full marks in Maths is by regular practice. Firstly, you will have to understand the concept of the chapter and then start solving the exercise questions. You can take help from Vedantu. Once you get acquainted with the topic, you can repeatedly practise the NCERT Solutions for Class 10 Maths to bring in the accuracy and eventually score full marks in the exam.