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Class 10 Maths Case Study Questions Chapter 12 Areas Related to Circles

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Case study Questions in the Class 10 Mathematics Chapter 12  are very important to solve for your exam. Class 10 Maths Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 12  Areas Related to Circles

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Areas Related to Circles Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 12 Areas Related to Circles

Case Study/Passage-Based Questions

Question 1:

Answer: (b) 729 cm2

(ii) Area of rectangle left for car parking is

Answer: (c) 81 cm2

(iii) Radius of semi-circle is

Answer: (a) 6.75 cm

(iv) Area of a semi-circle is

Answer: (d) 71.59 cm2

(v) Find the area of the shaded region

Answer: (b) 666.82 cm2

Question 2:

A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.

Design A:  Brooch A is made with silver wire in the form of a circle with a diameter of 28mm. The wire is used for making 4 diameters which divides the circle into 8 equal parts.

Design B:  Brooch b is made of two colors – Gold and silver. The outer part is made of Gold. The circumference of the silver part is 44mm and the gold part is 3mm wide everywhere.

Refer to Design A

1. The total length of silver wire required is

Answer: b) 200 mm

2. The area of each sector of the brooch is

Answer: c) 77 mm2

Refer to Design B

3. The circumference of outer part (golden) is

a) 48.49 mm

c) 72.50 mm

d) 62.86 mm

Answer: d) 62.86 mm

4. The difference of areas of golden and silver parts is

a) 18  π

b) 44  π

c) 51  π

d) 64  π

Answer: c) 51 π

5. A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm?

Answer: c) 4

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 12 Areas Related to Circles with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Areas Related to Circles Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Class 10 Maths Case Study Questions for Chapter 10 - Circles (Published By CBSE)

Check cbse class 10 maths case study questions for chapter 10 - circles. these questions have been published by the board for class 10 mathematics..

CBSE: Case study questions for CBSE Class 10 Maths Chapter 10 - Circles are provided here which students can practice to get familiarised with the new format of questions. These questions have been published by CBSE itself. Answers to all the questions have been provided for the convenience of students. Case study questions are helpful for the preparation of the Class 10 Maths Exam 2021-2022.

Case Study Questions for Class 10 Maths Chapter 10 - Circles

CASE STUDY 1:

A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.

After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride . She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.

1. In the given figure find ∠ROQ

Answer: c) 150

2. Find ∠RQP

Answer: a) 75

3. Find ∠RSQ

Answer: b) 75

4. Find ∠ORP

Answer: a) 90

CASE STUDY 2:

Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff . The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.

1. Find the length of AD

Answer: a) 7

2. Find the Length of BE

Answer: b) 5

3. Find the length of CF

Answer: d) 3

4. If radius of the circle is 4cm, Find the area of ∆OAB

Answer: c) 24

5. Find area of ∆ABC

Answer: b) 60

Also Check:

Tips to Solve Case Study Based Questions Accurately

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CBSE Case Study Questions for Class 10 Maths Area Related to Circles Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Area Related to Circles  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

MCQ Set 1 -

Mcq set 2 -, checkout our case study questions for other chapters.

• Chapter 10: Circles Case Study Questions
• Chapter 11: Construction Case Study Questions
• Chapter 13: Surface Area and Volumes Case Study Questions
• Chapter 14: Statistics Case Study Questions

How should I study for my upcoming exams?

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Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

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Practice Assertion Reason & Case Study Based Questions

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Chapter 11 Class 10 Areas related to Circles

Click on any of the links below to start learning from Teachoo ...

Updated for new NCERT - 2023-2024 Boards.

NCERT Solutions of all exercise questions and examples of Chapter 11 Class 10 Areas related to Circle. Answers to all questions are available with video free at teachoo.

In this chapter, we will

• Revise our concepts about Area and Perimeter of Circle , and do some questions
• Then, we will see what arc , sector and segment of a circle is
• We will learn the formula for length of an arc
• and Area of sector
• Then, using Area of sector and Area of triangle formulas, we find Area of Segment
• We do some questions where different figures are combined, and we need to find their area and perimeter

Click on an exercise link below to learn from the NCERT Book way.

Or you can also learn from Teachoo way, click on any concept below Concept Wise to get started.

In concept wise, the chapter is divided into concepts. First the concept is taught, and then we solve the questions related to the concept. All questions are ordered easy to difficult. That means, easiest question is in the beginning and the more difficult question is at the end.

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Case Study on Areas Related to Circles Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Areas Related to Circles Class 10 Maths can use this page to download the PDF file. 

The case study questions on Areas Related to Circles are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Areas Related to Circles case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Areas Related to Circles Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Areas Related to Circles, therefore, they prepared a set of solutions along with the case study questions on Areas Related to Circles.

The case study on Areas Related to Circles Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Areas Related to Circles case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Areas Related to Circles Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Areas Related to Circles case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Areas Related to Circles Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Areas Related to Circles case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Areas Related to Circles Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Areas Related to Circles as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Areas Related to Circles?

Students can choose their own way to answer Case Study on Areas Related to Circles Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Areas Related to Circles Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Areas Related to Circles questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Areas Related to Circles Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Areas Related to Circles?

 A few essential things to know to solve Case Study Questions on Class 10 Areas Related to Circles are -

• Basic Formulas of Areas Related to Circles: One of the most important things to know to solve Case Study Questions on Class 10 Areas Related to Circles is to learn about the basic formulas or revise them before solving the case-based questions on Areas Related to Circles.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Areas Related to Circles case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Areas Related to Circles Class 10 Maths?

Use Selfstudys.com to find Case Study on Areas Related to Circles Class 10 Maths. For ease, here is a step-wise procedure to download the Areas Related to Circles Case Study for Class 10 Maths in PDF for free of cost.

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CBSE Class 10 Maths: Case Study Questions of Chapter 12 Areas Related to Circles PDF Download

Case study Questions in the Class 10 Mathematics Chapter 12  are very important to solve for your exam. Class 10 Maths Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based   questions for Class 10 Maths Chapter 12  Areas Related to Circles

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Areas Related to Circles Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 12 Areas Related to Circles

Case Study/Passage-Based Questions

Question 1:

Answer: (b) 729 cm2

(ii) Area of rectangle left for car parking is

Answer: (c) 81 cm2

(iii) Radius of semi-circle is

Answer: (a) 6.75 cm

(iv) Area of a semi-circle is

Answer: (d) 71.59 cm2

(v) Find the area of the shaded region

Answer: (b) 666.82 cm2

Question 2:

A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.

Design A:  Brooch A is made with silver wire in the form of a circle with a diameter of 28mm. The wire is used for making 4 diameters which divides the circle into 8 equal parts.

Design B:  Brooch b is made of two colors – Gold and silver. The outer part is made of Gold. The circumference of the silver part is 44mm and the gold part is 3mm wide everywhere.

Refer to Design A

1. The total length of silver wire required is

Answer: b) 200 mm

2. The area of each sector of the brooch is

Answer: c) 77 mm2

Refer to Design B

3. The circumference of outer part (golden) is

a) 48.49 mm

c) 72.50 mm

d) 62.86 mm

Answer: d) 62.86 mm

4. The difference of areas of golden and silver parts is

a) 18  π

b) 44  π

c) 51  π

d) 64  π

Answer: c) 51 π

5. A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm?

Answer: c) 4

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 12 Areas Related to Circles with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Areas Related to Circles Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Class 10 Maths Chapter 11 Case Based Questions - Area Related to Circles

Study case - 1.

Q2: The area of the rectangular field is (a) 420m 2 (b) 430m 2 (c) 440m 2 (d) 450m 2 Ans: (b) Explanation:  Volume of the earth taken out Area of the rectangular field = 30 × 15 = 450m 2

Q4: The area of the remaining field is (a) 402.3m 2 (b) 405m 2 (c) 410m 2 (d) 411.5m 2 Ans:  (d) Explanation:  Area of the remaining field = Area of rectangular field - area of top of pit = 450 − 38.5 − 411.5m 2 . Q5: Find the level rise in the field. (a) 0.5m (b) 3m (c) 1.12 m (d) 2.12 m Ans:  (c) Explanation: The rise in the level of field = 462/4115 = 112m.

Study Case - 2

Q5: Find the area swept by the hour hand between11 a.m.and 5 p.m. (a) 56.568 cm 2 (b) 62 cm 2 (c) 70 cm 2 (d) 72 cm 2 Ans: (a) Explanation:  Number of hours from 11 a.m. to 5 p.m. = 6 Area swept by hour hand in 1 hour = 9.428cm 2 ∴ Area swept by hour hand in 6 hours = 9.428 × 6 = 56.568cm 2

Study Case - 3

Q5: Cost of painting of the circular regions with centre A, B, C and D at Rs.5.00 per sq. cm. (a) Rs.12320.00 (b) Rs.3080.00 (c) Rs.6160.00 (d) Rs.1540.00 Ans:  (b) Explanation:  Radius of each circle =7cm ∴ Area of one circle

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Case Based Questions: Areas Related to Circles Free PDF Download

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Important Questions for CBSE Class 10 Maths Chapter 12 - Areas Related to Circles

• Class 10 Important Question
• Chapter 12: Areas Related To Circles

Crucial Practice Problems for CBSE Class 10 Maths Chapter 12: Areas Related to Circles

Class 10 Maths Chapter 12 Important Questions Areas Related to Circles are given here based on the latest pattern of CBSE for 2024-2025. Students who are preparing for the board exams can practice Areas Related to Circles Important Questions to score full marks for the questions from this chapter. Along with the Important Questions for Class 10 Maths Chapter 12 related to Circles, we have also provided Extra Questions of Chapter 12 Class 10 Maths. Students can refer to the solutions whenever they get stuck while solving a problem. Also, in the end, some practice questions are provided for students to boost their preparation for the exam.

Areas Related to Circles chapter contains many formulas and concepts thus, it is important from the examination perspective, since most of the questions and objective-based questions will come in the exam from this chapter. Students can also refer to Chapter 12 Maths Class 10 Important Questions while preparing for their exam. NCERT Solution is also available at Vedantu to help the students in scoring full marks in the board exam.

Area related to circles includes the area of a circle, segment, sector, angle and length of a circle are provided in this pdf. In this chapter, we will discuss the concepts of the perimeter (also known as circumference) and area of a circle and apply this knowledge in finding the areas of two special parts of a circular region (or a circle) known as sector and segment. You can download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.

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Study Important Questions for Class 10 Maths Chapter 12- Area Related to Circles

Very Short Answer Questions (1 Mark)

Unless stated otherwise, take \[\pi =\dfrac{22}{7}\].

1. The radii of two circles are $19$ cm and $9$ cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Ans: Let us suppose R, radius of the circle whose circumference is equal to the circumference of the two circles with radius $19$ cm and $9$ cm respectively.

Thus, $2\pi R=2\pi \left( 19 \right)+2\pi \left( 9 \right)$

$\Rightarrow R=19+9$

$\Rightarrow R=28$ cm

$\therefore$ Radius of the circle, $R=28$ cm.

2.   The circumference of a circular field is $528$ cm. Then its radius is

a.   $42$ cm

b.   $84$ cm

c.   $72$ cm

d.   $56$ cm

Ans: b. $84$ cm

Circumference of a circle is given by, $2\pi R$.

$\therefore 2\pi R=528$

$\Rightarrow R=84$ cm

3. The circumference of a circle exceeds its diameter by $180$ cm. Then its radius is

a.   $32$ cm

b.   $36$ cm

c.   $40$ cm

d.   $42$ cm

Ans: d. $42$ cm

Let the diameter of the circle be $D$.

$\Rightarrow D=2R$.

$\therefore 2\pi R=D+180$

$\Rightarrow 2\pi R=2R+180$

$\Rightarrow R=42$ cm

4. Area of the sector of angle ${{60}^{\circ }}$ of a circle with radius $10$ cm is

a.   $52\dfrac{5}{21}\text{ }c{{m}^{2}}$

b.   $52\dfrac{8}{21}\text{ }c{{m}^{2}}$

c.   $52\dfrac{4}{21}\text{ }c{{m}^{2}}$

d.   None of these

Ans: b. $52\dfrac{8}{21}\text{ }c{{m}^{2}}$

Area of the sector of a circle $=\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

$\Rightarrow$ Area $=\left( \dfrac{{{60}^{\circ }}}{{{360}^{\circ }}} \right)\times \pi {{\left( 10 \right)}^{2}}$

$\Rightarrow$ Area $=52\dfrac{8}{21}\text{ }c{{m}^{2}}$.

5.   Area of a sector of angle ${{P}^{\circ }}$ of a circle of radius R is

a.   $\dfrac{{{P}^{\circ }}}{{{180}^{\circ }}}\times 2\pi R$

b.   $\dfrac{{{P}^{\circ }}}{{{180}^{\circ }}}\times \pi {{R}^{2}}$

c.   $\dfrac{{{P}^{\circ }}}{{{360}^{\circ }}}\times 2\pi R$

d.   $\dfrac{{{P}^{\circ }}}{{{720}^{\circ }}}\times 2\pi {{R}^{2}}$

Ans: d. $\dfrac{{{P}^{\circ }}}{{{720}^{\circ }}}\times 2\pi {{R}^{2}}$

$\Rightarrow$ Area $=\left( \dfrac{{{P}^{\circ }}}{{{360}^{\circ }}} \right)\times \pi {{\left( R \right)}^{2}}$

$\Rightarrow$ Area $=\dfrac{{{P}^{\circ }}}{{{720}^{\circ }}}\times 2\pi {{R}^{2}}$.

6. If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius $R$, then

a.   \[R_1+R_2=R\]

b.   \[R_1+R_2 > R\]

c.   \[R_1+R_2 < R\]

Ans: a. $R_1+R_2=R$

Thus, according to the given condition, $2\pi R_1+2\pi R_2=2\pi R$

$\Rightarrow R_1+R_2=R$.

7. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

a.   $22:7$

b.   $14:11$

c.   $7:22$

d.   $11:14$

Ans. b. $14:11$

The perimeter of a circle is given by, $2\pi R$.

The perimeter of a square is given by, $4a$.

Given, $2\pi R=4a$.

Ratio of area of circle to area of square is $\dfrac{\pi {{R}^{2}}}{{{a}^{2}}}$

$\Rightarrow \dfrac{\pi {{R}^{2}}}{{{a}^{2}}}=\dfrac{4\pi {{R}^{2}}}{{{\pi }^{2}}{{R}^{2}}}$

Hence, the ratio of their areas is, $14:11$.

8. The circumference of a circular field is $154$ m. Then its radius is

a.   $7$ m

b.   $14$ m

c.   $7.5$ m

d.   $28$ cm

Ans. Circumference of a circle is given by, $2\pi R$.

$\therefore 2\pi R=154$

$\Rightarrow R=24.5\text{ }m$.

9. The area of a circle is $394.24\text{ c}{{\text{m}}^{2}}$. Then the radius of the circle is

a.   $11.4$ cm

b.   $11.3$ cm

c.   $11.2$ cm

d.   $11.1$ cm

Ans. c. $11.2$ cm

The area of a circle is given by, $\pi {{R}^{2}}$.

$\therefore \pi {{R}^{2}}=394.24$

$\Rightarrow R=11.2$ cm.

10. If the perimeter and area of circle are numerically equal, then the radius of the circle is

a.   $2$ units

b.   $\pi$ units

c.   $4$ units

d.   $7$ units

Ans. a. $2$ units

Perimeter / Circumference of a circle is given by, $2\pi R$.

Area of a circle is given by, $\pi {{R}^{2}}$.

Given, $2\pi R=\pi {{R}^{2}}$.

$\Rightarrow R=2$ units.

11. The radius of a circle is $\dfrac{7}{\sqrt{\pi }}$ cm, then the area of the circle is

a.   $154\text{ c}{{\text{m}}^{2}}$

b.   $\dfrac{49}{\pi }\text{ c}{{\text{m}}^{2}}$

c.   $\text{22 c}{{\text{m}}^{2}}$

d.   $\text{49 c}{{\text{m}}^{2}}$

Ans. d. $49c{{m}^{2}}$

$\therefore $ Area $=\pi \left( \dfrac{49}{\pi } \right)$

$\Rightarrow $ Area $=49c{{m}^{2}}$.

12. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii $24$ cm and $7$ cm is

a.   $31$ cm

b.   $25$ cm

c.   $62$ cm

d.   $50$ cm

Ans. d. $25$ cm

Let $D$ ($D=2R$) be the diameter of the circle whose area is equal to the sum of the areas of the two circles of radii $24$ cm and $7$ cm.

$\therefore \pi {{R}^{2}}=\pi {{\left( 24 \right)}^{2}}+\pi {{\left( 7 \right)}^{2}}$

$\Rightarrow {{R}^{2}}=576+49$

$\Rightarrow R=25$ cm

13. The circumference of a circle is $528$ cm. Then its area is

a.   $22,176\text{ }c{{m}^{2}}$

b.   $22,576\text{ }c{{m}^{2}}$

c.   $23,176\text{ }c{{m}^{2}}$

d.   $24,576\text{ }c{{m}^{2}}$

Ans. a. $22,176\text{ }c{{m}^{2}}$

$\therefore 2\pi R=528$ cm

Thus, area $=\pi {{\left( 84 \right)}^{2}}$

$\Rightarrow $Area $=22,176\text{ }c{{m}^{2}}$

Very Short Answer Questions (2 Marks)

1. The radii of two circles are $8$ cm and $6$ cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Ans: Area of a circle is given by, $\pi {{R}^{2}}$.

Let $R$ be the radius of the circle whose area is equal to the sum of the areas of the two circles of radii $8$ cm and $6$ cm, respectively.

Thus, we have: $\pi {{R}^{2}}=\pi {{\left( 8 \right)}^{2}}+\pi {{\left( 6 \right)}^{2}}$

$\Rightarrow {{R}^{2}}=64+36$

$\Rightarrow {{R}^{2}}=100$

$\Rightarrow R=10$ cm.

2. Figure depicts an archery target marked with its five scoring area from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is $21$ cm and each of the other bands is $10.5$ cm wide. Find the area of the five scoring regions.

Ans: Given, diameter of the region representing Gold score as $21$ cm and the width of other bands as $10.5$ cm.

Thus, radius of the Gold scoring region, $R=\dfrac{21}{2}$cm.

Area of the Gold scoring region, $Agold=\pi {{\left( \dfrac{21}{2} \right)}^{2}}$.

$\Rightarrow Agold=346.5\text{ }c{{m}^{2}}$

Area of scoring region = Area of previous circle – Area of next circle.

$\therefore $ Area of Red scoring region, $Ared=\pi {{\left( \dfrac{21}{2}+10.5 \right)}^{2}}-\pi {{\left( \dfrac{21}{2} \right)}^{2}}$

$\Rightarrow Ared=1386-346.5$

$\Rightarrow Ared=1039.5\text{ }c{{m}^{2}}$

Area of Blue scoring region, $Ablue=\pi {{\left( \dfrac{21}{2}+10.5+10.5 \right)}^{2}}-\left( 1039.5+346.5 \right)$

$\Rightarrow Ablue=3118.5-1386$

$\Rightarrow Ablue=1732.5\text{ }c{{m}^{2}}$

Area of Black scoring region, $Ablack=\pi {{\left( \dfrac{21}{2}+10.5+10.5+10.5 \right)}^{2}}-\left( 1039.5+346.5+1732.5 \right)$

$\Rightarrow Ablack=5544-3118.5$

$\Rightarrow Ablack=2425.5\text{ }c{{m}^{2}}$

Area of White scoring region, $Awhite=\pi {{\left( \dfrac{21}{2}+10.5+10.5+10.5 \right)}^{2}}-\left( 2425.5+1039.5+346.5+1732.5 \right)$

$\Rightarrow Awhite=8662.5-5544$

$\Rightarrow Awhite=3118.5\text{ }c{{m}^{2}}$

3. The wheels of a car are of diameter $80$ cm each. How many complete revolutions does each wheel make in $10$ minutes when the car is travelling at a speed of $66$ km per hour?

Ans: Given, diameter of car wheel $=80$ cm.

$\therefore $ Radius of the car wheel $=40$ cm.

Distance covered by the wheel in one revolution is equal to the circumference of the wheel.

$\therefore 2\pi R=2\times \dfrac{22}{7}\times 40$

Thus, distance covered by the wheel in one revolution is equal to $\dfrac{1760}{7}$ cm.

Given, the distance covered in $1$ hour $=66km=6600000cm$

Hence, distance covered by the wheel in $10$ minutes $=\dfrac{6600000}{60}\times 10=1100000cm$.

Number of revolutions \[=\dfrac{\text{Total distance}}{\text{Distance covered in one revolution}}\]

Number of revolutions $=\dfrac{1100000\times 7}{1760}$

Hence, number of revolutions $=4375$.

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

Ans: Given, Angle of the sector is ${{60}^{\circ }}$.

Radius, $r=6$ cm

$\Rightarrow $ Area $=\left( \dfrac{{{60}^{\circ }}}{{{360}^{\circ }}} \right)\times \pi {{\left( 6 \right)}^{2}}$

$\Rightarrow $ Area $=\dfrac{132}{7}\text{ }c{{m}^{2}}$.

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

Ans: Given, the circumference of the circle as $22$ cm.

We know, the circumference of a circle is given by, $2\pi R$.

Therefore, $2\pi R=22$

$\Rightarrow R=3.5$ cm

For the quadrant of a circle, $\theta ={{90}^{0}}$

 Area of quadrant is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

$\Rightarrow \dfrac{{{90}^{0}}}{{{360}^{0}}}\times \dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}=\dfrac{77}{8}c{{m}^{2}}$.

$\therefore $ Area of the quadrant of the circle $=9.625\text{ }c{{m}^{2}}$.

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

Ans: Given, length of the minute hand (radius) , $r=14$ cm and from the given data we interpret the angle in the clock, $\theta ={{30}^{{}^\circ }}$.

Area swept by the minutes hand in $5$ minutes, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

$\Rightarrow A=\dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 14\times 14=\dfrac{154}{3}c{{m}^{2}}$

$\Rightarrow A=51.33\text{ c}{{\text{m}}^{2}}$.

7. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:  (Use $\pi =3.14$)

i. minor segment

Ans: Given, Radius, \[r~=10\text{ }cm\] and the angle, $\theta ={{90}^{{}^\circ }}$.

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

\[\Rightarrow A=~\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 10\times 10~\text{ }=\text{ }78.5\text{ }c{{m}^{2}}~\]

Area of $\Delta OAB$, = $\dfrac{1}{2}$ Base $\times $ Height.

\[\dfrac{1}{2}bh=~\dfrac{1}{2}\times 10\times 10~~\]

\[\Rightarrow 50\text{ }c{{m}^{2}}\].

Area of minor segment can be obtained as,

Area of minor segment $=$  Area of minor sector $-$  Area of $\vartriangle OAB$ .

\[\Rightarrow A=78.550~\]

\[\Rightarrow A=28.5\text{ }c{{m}^{2}}\]

ii. Major Segment

Ans: For major sector, radius = $10$ cm and $\theta ={{360}^{{}^\circ }}-{{90}^{{}^\circ }}={{270}^{{}^\circ }}$

$\therefore $Area of major sector \[A=~\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}~\]

\[\Rightarrow ~A=\dfrac{{{270}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \text{3}\text{.14}\times \text{10}\times \text{10}\]

\[\Rightarrow \text{A}=\text{ }235.5\text{ }c{{m}^{2}}~\].

8. A horse is tied to a peg at one corner of a square shape grass field of side $15$ m by means of a $5$ m long rope (see figure). Find:

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

Ans: Area of a sector of a circle is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

Area of the part of the field where the horse can graze is in the form of a quadrant with radius, $r=5m$.

Thus, the area is given as, \[A=\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \text{3}\text{.14}\times \text{5}\times \text{5}\]

\[\Rightarrow ~A=19.625\text{ }{{m}^{2}}\].

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

Ans: Area of quadrant with $10$m rope, the radius would be, $r=10m$. 

Thus, the grazing area for $10$ m long rope is given by, \[~A=\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 10\times 10~\].

\[\Rightarrow A=78.5\text{ }{{m}^{2}}\].

$\therefore $The increase in grazing area can be calculated by subtracting area of $5$m rope from the area of $10$m rope.

Hence, Increase in area \[=78.519.625=58.875\text{ }{{m}^{2}}\]

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 figure. Find: 

i. The total length of the silver wire required. 

Ans: Given, diameter of silver wire circle as $35$ mm 

Thus, radius is $\dfrac{35}{2}$ mm.

Circumference of a circle is given by, \[~2\pi r\].

Therefore, the circumference of the circle made by silver wire is,

\[~\Rightarrow 2\times \dfrac{22}{7}~\times \dfrac{35}{2}~=110\] mm   ……(1)

As the wire is used in making $5$ diameters,

$\therefore $ length of $5$  diameters, \[L=35\times 5\]

\[\Rightarrow L=175\] mm   ……(2) 

Total length of the silver wire required would be the sum of circumference of the circle and the length of $5$ diameters.

Hence, adding equation (1) and (2), we get 

\[\Rightarrow 110\text{ }+\text{ }175\text{ }=\text{ }285\text{ }\] mm

ii. The area of each sector of the brooch. 

Angle of each sector,$\theta =\dfrac{{{360}^{{}^\circ }}}{10}={{36}^{{}^\circ }}$.

$\therefore $The area of each sector of the brooch is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

\[\Rightarrow A=\dfrac{{{36}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times \dfrac{35}{2}\times \dfrac{35}{2}~\]

\[\Rightarrow A=~\text{ }\dfrac{385}{4}m{{m}^{2}}~\]

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

Ans: Let umbrella be a flat circle with radius,\[r=\text{ }45\text{ }cm\].

Given, the umbrella has $8$ ribs, thus the angle $\theta $ would be, $\theta =\dfrac{{{360}^{{}^\circ }}}{8}={{45}^{{}^\circ }}$

Area between two consecutive ribs of the umbrella can be obtained using the formula, \[A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}\].

\[\Rightarrow A=\dfrac{{{45}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 45\times 45~\]

\[\Rightarrow A=~\dfrac{22275}{28}\text{ }c{{m}^{2}}\]

$\Rightarrow A=795.54\text{ }c{{m}^{2}}$.

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

Ans: Given, a car has two wipers with a blade length of $25$ cm sweeping an angle of ${{115}^{\circ }}$.

Let us suppose radius, \[~r~=\text{ }25\text{ }cm\] and

Angle, $\theta ={{115}^{{}^\circ }}$.

The two wipers do not overlap.

$\therefore $ The total area cleaned at each sweep of the blade will be,  $2\times \left( \dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}} \right)$

$\Rightarrow 2\times \left( \dfrac{{{115}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 25\times 25 \right)=\dfrac{158125}{126}c{{m}^{2}}$

Hence, area cleaned at each sweep is $1254.96\text{ }c{{m}^{2}}$.

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

Ans: Given, a lighthouse spreads a red light over a sector of angle ${{80}^{\circ }}$ to a distance of $16.5$ km.

Let us suppose radius, \[r~=16.5\text{ }km\] and

Angle, $\theta ={{80}^{{}^\circ }}$.

$\therefore $The area of sea over which the ships are warned is, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$\[\Rightarrow A=\dfrac{{{80}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 16.5\times 16.5~\]

\[\Rightarrow A=189.97\text{ }k{{m}^{2}}~\]

13.Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are $7$ cm and $14$cm respectively and $\angle $AOC = ${{40}^{\circ }}$.

Ans: Area of shaded region $ABDC$ can be determined by subtracting area of sector $OBD$ from the area of sector $OAC$, i.e.,

Area of shaded region $ABDC=$ Area of sector $OAC-$ Area of sector $OBD$

Area of sector is given by, \[A=~\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}~\].

Thus, Area of sector $OAC$$=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(14)}^{2}}$ 

Area of sector $OBD$$=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(7)}^{2}}$  

$\therefore $ Area of shaded region $ABDC=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(14)}^{2}}-\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(7)}^{2}}$

$\Rightarrow Area=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\left[ {{(14)}^{2}}-{{(7)}^{2}} \right]=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 7\times 21$

$\Rightarrow $ Area $=\dfrac{154}{3}~c{{m}^{2}}$.

Hence, area of the shaded region is $51.33\text{ }c{{m}^{2}}$.

14. Find the area of the shaded region in figure, if ABCD is a square of side $14$ cm and APD and BPC are semicircles. 

Ans: From the given figure,

Area of shaded region = Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC) 

$\Rightarrow $Area of the shaded region $=14\times 14-\left[ \dfrac{1}{2}\times \dfrac{22}{7}{{\left( \dfrac{14}{2} \right)}^{2}}+\dfrac{1}{2}\times \dfrac{22}{7}{{\left( \dfrac{14}{2} \right)}^{2}} \right]$

$\Rightarrow $Area of the shaded region $=196-\dfrac{22}{7}\times 7\times 7$

$\therefore $ Area of the shaded region is $42c{{m}^{2}}$.

15. Find the area of the shaded region in figure, where a circular arc of radius $6$ cm has been drawn with vertex O of an equilateral triangle OAB of side $12$ cm as centre. 

Ans: Area of the shaded region = Area of circle + Area of equilateral triangle $OAB$ - Area of the region common to the circle and the triangle.

Area of equilateral triangle is given by, $\dfrac{\sqrt{3}}{4}{{\left( a \right)}^{2}}$ ($a$ is the side)

$\Rightarrow $Area of the shaded region $=\pi {{(6)}^{2}}+\dfrac{\sqrt{3}}{4}{{(12)}^{2}}-\dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \pi {{(6)}^{2}}$

$\Rightarrow $Area of the shaded region $=30\pi +36\sqrt{3}=\left( \dfrac{660}{7}+36\sqrt{3} \right)\text{c}{{\text{m}}^{2}}$

Hence, the area of the shaded region is $156.63\text{ }c{{m}^{2}}$.

16. From each corner of a square of side $4$ cm a quadrant of a circle of radius $1$ cm is cut and also a circle of diameter of $2$ cm is cut as shown in figure. Find the area of the remaining portion of the figure:

Ans: Given, a square of side $4$ cm. Four quadrants of a circle of radius $1$ cm are cut from each corners of the square.

Also, a circle of diameter $2$ cm is cut.

$\therefore $ Area of the remaining portion = Area of square – ($4\times $ Area of quadrant of radius $1$ cm + Area of circle of diameter $2$ cm).

Area of a square is given as ${{\left( side \right)}^{2}}$.

Similarly, area of quadrant is given by $\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

Area of the remaining portion \[=4\times 4-\left[ 4\times \dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(1)}^{2}}+\dfrac{22}{7}\times {{\left( \dfrac{2}{2} \right)}^{2}} \right]\]

$\Rightarrow $ Area of the remaining portion \[=16-2\times \dfrac{22}{7}=\dfrac{68}{7}\text{ c}{{\text{m}}^{2}}\].

Hence, area of the remaining portion is $9.71\text{ }c{{m}^{2}}$.

17. In a circular table cover of radius $32$cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region). 

Ans: Given, circular table cover of radius, $r=32cm$ and an equilateral triangle $ABC$ is drawn in the middle.

Area of the design can be obtained by,

Area of design = Area of circle – Area of the equilateral triangle $ABC$.

Area of equilateral triangle is given by, $\dfrac{\sqrt{3}}{4}{{\left( a \right)}^{2}}$ ($a$ is the side).

Thus, area of design $=\pi {{\left( 32 \right)}^{2}}-\dfrac{\sqrt{3}}{4}{{\left( a \right)}^{2}}$   ……(1)

$\because $G is the centroid of the equilateral triangle. 

$\therefore $radius of the circumscribed circle =  $\dfrac{2}{3}h$ cm 

So, we have $32=\dfrac{2}{3}h$

$\Rightarrow h=48$ cm

From the figure of equilateral triangle, ${{a}^{2}}={{h}^{2}}+{{\left( \dfrac{a}{2} \right)}^{2}}$.

$\Rightarrow \dfrac{3{{a}^{2}}}{4}={{h}^{2}}$

$\Rightarrow {{a}^{2}}=\dfrac{4\times {{\left( 48 \right)}^{2}}}{3}$

$\Rightarrow a=\sqrt{3072}$

Substituting for $a$ in equation (1), we get:

Area of design $=\pi {{\left( 32 \right)}^{2}}-\dfrac{\sqrt{3}}{4}{{\left( \sqrt{3072} \right)}^{2}}$

$\Rightarrow \dfrac{22}{7}\times 1024-768\sqrt{3}=\left( \dfrac{22528}{7}-768\sqrt{3} \right)c{{m}^{2}}$

Thus, area of the design is $1888.08\text{ }c{{m}^{2}}$.

18. In figure ABCD is a square of side $14$cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region. 

Ans: Given, a square of side $14$ cm. With the corners of the square as centres four circles are drawn.

Thus, area of the shaded region = Area of the square – ($4\times $ Area of sector)

Area of a circle is given by, $\pi {{R}^{2}}$ and area of sector is given by, $\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$Area of the shaded region $=14\times 14-4\times \dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{\left( \dfrac{14}{2} \right)}^{2}}=196-\dfrac{22}{7}\times 7\times 7$

$Hence, area of the shaded region is $42\text{ }c{{m}^{2}}$.

19. The given figure depicts a racing track whose left and right ends are semi-circular. The difference between the two inner parallel line segments is \[60\text{ m}\] and they are each \[106\text{ }m\] long. If the track is \[10m\] wide, find: 

i. The distance around the track along its inner edge, 

Ans: Given that, difference between the two inner parallel line segments is \[60m\] and they are each \[106m\] long and the track is \[10m\] wide.

Distance around the track along its inner edge 

$\Rightarrow 106+106+2\times \left[ \dfrac{{{180}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 2\times \dfrac{22}{7}\times \left( \dfrac{60}{2} \right) \right]=212+60\times \dfrac{22}{7}$

$\Rightarrow 212+\dfrac{1320}{7}=\dfrac{2804}{7}m$

ii. The area of the track. 

Ans: Area of track can be calculated as,

$ \Rightarrow 106\times 10+106\times 10+2\times \left[ \dfrac{1}{2}\times \dfrac{22}{7}{{(30+10)}^{2}}-\dfrac{1}{2}\times \dfrac{22}{7}{{(30)}^{2}} \right] $

$ =1060+1060+\dfrac{22}{7}\left[ {{(40)}^{2}}-{{(30)}^{2}} \right] $

$\Rightarrow 2120+\dfrac{22}{7}\times 700=4320{{m}^{2}}$

20. In figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = $7$ cm, find the area of the shaded region. 

Ans: Given, radius of the bigger circle, OA or $R=7$ cm.

Diameter of the smaller circle is the radius of the bigger circle.

Thus, Radius of the smaller circle, $r=\dfrac{7}{2}$ cm.

Area of the circle is given by, $\pi {{R}^{2}}$ and

Area of the semicircle is given by, $\dfrac{\pi {{r}^{2}}}{2}$.

$\Delta ACB$ is made up of two right- angled triangles and we know, area of a right-angled triangle is given as  $\dfrac{1}{2}\times base\times height$.

Area of the shaded region can be obtained as:

Area of shaded region = Area of smaller circle + Area of semicircle $ACB\text{ }-$ Area of $\Delta ACB$.

Area of shaded region $=\dfrac{22}{7}\times {{\left( \dfrac{7}{2} \right)}^{2}}+\dfrac{1}{2}\times \dfrac{22}{7}\times {{(7)}^{2}}-\left( \dfrac{7\times 7}{2}+\dfrac{7\times 7}{2} \right)=\dfrac{133}{2}$

Thus, area of the shaded region is $66.5\text{ }c{{m}^{2}}$.

21. The area of an equilateral triangle ABC is $17320.5\text{ }c{{m}^{2}}$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use $\pi =3.14$ and $\sqrt{3}=1.73205$)

Ans: Given, area of the equilateral triangle $ABC=17320.5\text{ }c{{m}^{2}}$.

Also, radius of each circle is half the length of the side of the equilateral triangle.

Area of the equilateral triangle is given by, $\dfrac{\sqrt{3}}{4}\times {{\left( side \right)}^{2}}$.

Let side of the equilateral triangle $ABC$ be $a$.

$\therefore $$\dfrac{\sqrt{3}}{4}\times {{a}^{2}}=17320.5$

$\Rightarrow {{a}^{2}}=40000$

$\Rightarrow a=200$  cm

$\therefore $ radius of circle, $r=100$ cm

Area of shaded region = Area of the equilateral triangle $ABC\text{ }-$ $3\times $ Area of the sector.

We know, area of the sector of a circle $=\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

Area of shaded region $=17320.5-3\times \left( \dfrac{{{60}^{\circ }}}{{{360}^{\circ }}} \right)\times 3.14\times {{\left( 100 \right)}^{2}}$

\[\Rightarrow 17320.5-15700=1620.5cm{}^\text{2}\]

Hence, area of the shaded region is $1620.5\text{ }c{{m}^{2}}$.

22. On a square handkerchief, nine circular designs each of radius $7$cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Ans: Given, radius of circular design is $7$ cm.

Observing the figure, the side of the square is $6\times 7=42$ cm.

The handkerchief consists of $9$ circular designs.

Thus, area of the shaded region = Area of square $ABCD\text{ }-\text{ 9}\times $ Area of circular design.

Area of shaded region $=\left( 42\times 42 \right)-9\times \pi \times {{\left( 7 \right)}^{2}}$

$\Rightarrow 1764-1386=378\text{ }c{{m}^{2}}$

Hence, area of the shaded region is $378\text{ }c{{m}^{2}}$.

23. In figure, OACB is a quadrant of a circle with centre O and radius $3.5$cm. If OD = $2$ cm, find the area of the: 

i. quadrant OACB 

Ans: Given radius, $r=3.5\text{ }cm$ and

Area of the quadrant of a circle $=\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

$\therefore $ Area of the quadrant $OACB=\dfrac{{{90}^{\circ }}}{{{360}^{\circ }}}\times \dfrac{22}{7}\times {{\left( 3.5 \right)}^{2}}$

\[\Rightarrow ~\dfrac{1}{4}\times \dfrac{22}{7}\times \dfrac{35}{10}\times \dfrac{35}{10}=~\dfrac{77}{8}c{{m}^{2}}~\]

Area of the quadrant $OACB$ is $9.625\text{ }c{{m}^{2}}$.

ii. shaded region 

Area of the shaded region = Area of the quadrant $OACB\text{ }-$ Area of the triangle $BOD$.

Area of the shaded region =$9.625-\dfrac{1}{2}\times 3.5\times 2=6.125\text{ }c{{m}^{2}}$

Hence, area of the shaded region is $6.125\text{ }c{{m}^{2}}$.

24.In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = $20$  cm, find the area of the shaded region. (Use $\pi =3.14$)

Ans: Given, a square $OABC$ inscribed in a quadrant $OPBQ$ and

$OA=20$ cm.

$\therefore OA=AB=BC=OC$.

Taking the $\Delta OAB$ into consideration and applying Pythagoras Theorem,

$O{{B}^{2}}=O{{A}^{2}}+A{{B}^{2}}$

$\Rightarrow O{{B}^{2}}={{\left( 20 \right)}^{2}}+{{\left( 20 \right)}^{2}}$

$\Rightarrow OB=\sqrt{800}$

$\Rightarrow OB=20\sqrt{2}$ cm

Area of the shaded region = Area of quadrant $OPBQ$ - Area of square $OABC$.

We know, area of the quadrant of a circle $=\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

$\therefore $ Area of shaded region $=\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14{{\left( 20\sqrt{2} \right)}^{2}}-20\times 20$

$\Rightarrow \dfrac{1}{4}\times 3.14\times 800-400=228\text{ }c{{m}^{2}}$

Hence, area of the shaded region is $228\text{ }c{{m}^{2}}$.

25. AB and CD respectively are the arcs of two concentric circles of radii $21$ cm and $7$ cm and centre O (see figure). If $\angle AOB={{30}^{\circ }}$. Find the area of the shaded region.

Ans: Given, two arcs $AB$ and $CD$ of two concentric circles with radii $21$ cm and $7$ cm respectively.

Also, $\angle AOB={{30}^{\circ }}$

Area of the shaded region = Area of arc $AB$ - Area of arc $CD$.

$\therefore $ Area of shaded region $=\left( \dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}} \right)\times \dfrac{22}{7}\times {{\left( 21 \right)}^{2}}-\left( \dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}} \right)\times \dfrac{22}{7}\times {{\left( 7 \right)}^{2}}$

$\Rightarrow \left( \dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}} \right)\times \dfrac{22}{7}\times \left[ {{\left( 21 \right)}^{2}}-{{\left( 7 \right)}^{2}} \right]=\dfrac{308}{3}\text{ }c{{m}^{2}}$

Hence, area of the shaded region is $\text{102}\text{.67 }c{{m}^{2}}$.

26. In figure, ABC is a quadrant of a circle of radius $14$ cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Ans: Given, a quadrant $ABC$ of a circle of radius $14\text{ }cm$.

Let us take, $\Delta ABC$ and apply Pythagoras Theorem.

$\therefore A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}$

$\Rightarrow {{\left( 14 \right)}^{2}}+{{\left( 14 \right)}^{2}}=B{{C}^{2}}$

$\Rightarrow BC=\sqrt{392}$

$\Rightarrow BC=14\sqrt{2}\text{ }cm$.

$\therefore $ Radius of the semicircle $=\dfrac{14\sqrt{2}}{2}\text{=7}\sqrt{2}\text{ }cm$.

Area of the shaded region $BPCQ=$ Area of $BCQB$ - (Area of $BACP$ - Area of $\Delta ABC$).

$\therefore $ Area of shaded region \[=\dfrac{{{180}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}{{\left( 7\sqrt{2} \right)}^{2}}-\left[ \dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{\left( 14 \right)}^{2}}-\dfrac{14\times 14}{2} \right]~\]

$\Rightarrow \dfrac{1}{2}\times \dfrac{22}{7}\times 98-\left( \dfrac{1}{4}\times \dfrac{22}{7}\times 196-98 \right)=98\text{ }c{{m}^{2}}$

Hence, area of the shaded region is $\text{98 }c{{m}^{2}}$.

27.Calculate the area of the designed region in figure common between the two quadrants of circles of radius $8$ cm each. 

Ans: Given, two quadrants each of radii $8$ cm which has a design in the common region.

Let us consider the figure as:

Applying Pythagoras Theorem in the right- angled triangle $ABC$, \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[~\Rightarrow A{{C}^{2}}=\text{ }{{\left( 8 \right)}^{2}}+\text{ }{{\left( 8 \right)}^{2}}=\text{ }2{{\left( 8 \right)}^{2}}~\]

\[~\Rightarrow ~AC=\sqrt{128}=8\sqrt{2}~~cm~\]

Draw a perpendicular, \[BM\bot AC.~\]

Then \[AM=MC=~\text{ }\dfrac{1}{2}AC~\]

\[\Rightarrow ~\dfrac{1}{2}\times 8\sqrt{2}~=~4\sqrt{2}~cm~\]

From the figure the right triangle AMB, 

$(AB)^2$ = $(AM)^2 +(BM)^2$ [Pythagoras theorem] 

\[\Rightarrow {{\left( 8 \right)}^{2}}={{\left( 4\sqrt{2} \right)}^{2}}+B{{M}^{2}}\]

\[\Rightarrow B{{M}^{2}}=6432=32\]

\[\Rightarrow BM\text{ }=4\sqrt{2}~~cm\]

Area of  \[\Delta ABC\text{ }=\dfrac{1}{2}~~\times \text{ }AC\text{ }\times \text{ }BM~\]

\[\Rightarrow ~\dfrac{8\sqrt{2}\times 4\sqrt{2}}{2}~=\text{ }32\text{ }c{{m}^{2}}~\]

Area of half shaded region = Area of quadrant $ABC$ - Area of $\Delta ABC$.

$\therefore $Area of half shaded region $=\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{\left( 8 \right)}^{2}}-32$

\[\Rightarrow ~16\times \dfrac{22}{7}-32~=~\dfrac{128}{7}\text{ }c{{m}^{2}}~\]

Thus, area of designed region can be obtained as, twice the area of half shaded region.

\[\Rightarrow 2\text{ }\times \dfrac{128}{7}~~=~\dfrac{256}{7}\text{ }c{{m}^{2}}~\]

Hence, area of the designed region is $36.57\text{ }c{{m}^{2}}$.

28. Find the circumference of a circle of diameter $14$cm. 

Ans: Given, diameter of circle, d$=14$ cm

Circumference of a circle is given by, $2\pi r$

Since, $d=14$ cm

$\Rightarrow r=\dfrac{14}{2}=7$ cm

$\therefore r=7cm$

Thus, the circumference of the circle is, $C=2\pi r$

$\Rightarrow C=2\times \dfrac{22}{7}\times 7$

$\Rightarrow C=44$ cm

The circumference of the circle is $44$ cm.

29. The diameter of a circular pond is $17.5$m. It is surrounded by a path of width $3.5$m. Find the area of the path. 

Ans: Given, diameter of circular pond, $d=17.5$ m.

$\therefore $ Radius, \[r=\dfrac{17.5}{2}=8.25\] m

It is surrounded by a path of width $3.5$ m.

$\therefore $Outer radius \[R\] is the sum of circular pond radius with the width of the surrounded path.

\[\Rightarrow R=8.75+3.5=12.25\] m

Equation for area of path $A$  is given by $\pi \left[ {{R}^{2}}-{{r}^{2}} \right]$.

$\Rightarrow A=\pi \left[ \left( R+r \right)\left( R-r \right) \right]$

$\Rightarrow A=\dfrac{22}{7}\left[ \left( 12.25+8.75 \right)\left( 12.25-8.75 \right) \right]$

$\Rightarrow A=\dfrac{22}{7}\times 21\times 3.50$

$\Rightarrow A=231~{{\text{m}}^{2}}$

Hence, area of the path is  $231~{{\text{m}}^{2}}$.

30. Find the area of the shaded region where ABCD is a square of side $14$ cm.

Ans: Given, side of square, $a=14$ cm.

Area of the shaded region can be calculated by subtracting area of $4$ circles from the area of square $ABCD$, i.e.,

Area of shaded region = Area of square $ABCD$ $-\text{ }4\times $Area of circle.

Area of the given square $ABCD$ is given by, $A=14cm\times 14cm$

$\Rightarrow A=196c{{m}^{2}}$

From the figure, we observe that the diameter of each circle is half the side of the square.

Thus, diameter of circle $=\dfrac{14}{2}=7$ cm

 $\therefore $ Radius of each circle $=\dfrac{7}{2}cm$

Area of the circle $=\pi {{r}^{2}}$  

$\Rightarrow \dfrac{22}{7}\times {{\left( \dfrac{7}{2} \right)}^{2}}c{{m}^{2}}$   

$\therefore $ Area of shaded region \[=196\text{ }c{{m}^{2}}\left( 4\times 38.5 \right)\text{ }c{{m}^{2}}~\]

$\Rightarrow \left( 196-154 \right)c{{m}^{2}}=42\text{ }c{{m}^{2}}$

Hence, area of the shaded region is \[42\text{ }c{{m}^{2}}\].

31. The radius of a circle is $20$cm. Three more concentric circles are drawn inside it in such a manner that it is divided into four parts of equal area. Find the radius of the largest of the three concentric circles. 

Ans: Given radius of the circle, $r=20$ cm

Area of a circle is given by, $A=\pi {{r}^{2}}$

Substituting for radius, $r$.

$\therefore A=\pi {{\left( 20 \right)}^{2}}$ 

$\Rightarrow A=400\pi c{{m}^{2}}$

As the circle is divided into four parts of equal area, area of each circle would be,

$\Rightarrow \dfrac{1}{4}400\pi c{{m}^{2}}=100\pi c{{m}^{2}}$

Let $R$ be the radius of the largest of the three circles and its area is $400\pi -100\pi =300\pi cm$.

Thus the radius, $R$ can be calculated as

$\Rightarrow \pi {{R}^{2}}=300\pi $

$\text{ }{{\text{R}}^{2}}=300$

$\therefore R=\sqrt{300}\text{ =}10\sqrt{3}$ cm

Radius of the largest of the three concentric circles is $10\sqrt{3}$ cm.

32. OACB is a quadrant of a circle with centre O and radius $7$ cm. If OD = $4$cm, then find area of shaded region. 

Ans: Given radius $r=7$ cm and $OD=4$ cm

Area of quadrant is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

From the figure it is clear that, angle $\theta ={{90}^{\circ }}$

$\therefore $ Area of quadrant  $OACB=\dfrac{{{90}^{\circ }}}{{{360}^{\circ }}}\pi {{\left( 7 \right)}^{2}}$

$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\times 49=\dfrac{77}{2}c{{m}^{2}}$

Area of $\Delta OAD=\dfrac{1}{2}\left( 7\times 4 \right)$

Area of shaded region = Area of quadrant $OACB$$-$ Area of $\Delta AOB$

$\Rightarrow \dfrac{77}{2}-\dfrac{1}{2}\left( 7\times 4 \right)=\dfrac{49}{2}c{{m}^{2}}$

Thus, area of the shaded region is $24.5\text{ }c{{m}^{2}}$.

33. A pendulum swings through on angle of ${{30}^{{}^\circ }}$and describes an arc $8.8$ cm in length. Find the length of pendulum. 

Ans: Given, pendulum swings at angle ${{30}^{\circ }}$ , with an arc of length $8.8$ cm.

Sketching the figure, we have:

In the below given figure let $r$ be the length of pendulum and $\angle

AOB={{30}^{{}^\circ }}$ then,$\angle AOB=\dfrac{\pi }{{{180}^{{}^\circ }}}\times {{30}^{{}^\circ }}$

$\Rightarrow \angle AOB=\dfrac{\pi }{6}$

Use $\theta =\dfrac{l}{r}$ to determine the length of the pendulum.

$\Rightarrow \dfrac{\pi }{6}=\dfrac{8.8}{r}$

 $\therefore r=\dfrac{8.8\times 6}{\pi }$

$\Rightarrow r=16.8cm$

The length of the pendulum is $16.8$ cm.

34. The cost of fencing a circular field at the rate of Rs. $24$ per metre is Rs. $5280$. The field is to be ploughed at the rate of Rs.$0.50$per $m^2$. Find the cost of ploughing the field. (Take $\pi =\dfrac{22}{7}$)

Ans: As per the question the cost of fencing a circular field at the rate of Rs$24$, per meter is Rs \[5280.\]

With Rs. $24$, the length of fencing $=1$ metre 

$\therefore $ for Rs.$5280$, the length fencing$=\dfrac{1}{24}\times 5280=220$meters

$\therefore $Perimeter i.e., circumference of the field $=220$meters 

Let $r$ be the radius of the field.

Circumference is given by, $2\pi R$ 

$\therefore 2\pi r=220$

$\Rightarrow r=\dfrac{220\times 7}{2\times 22}=35m$

Area of the field, $A=\pi {{r}^{2}}$.

$\Rightarrow \pi {{\left( 35 \right)}^{2}}=1225\pi {{m}^{2}}$

Rate = Rs.\[0.50\] per m2 

Total cost of ploughing the field

$\Rightarrow Rs.\left( 1225\pi \times 0.50 \right)=Rs.\dfrac{1225\times 22\times 1}{7\times 2}$  

$\Rightarrow Rs.\left( 175\times 11 \right)=Rs.1925$

Hence, the cost of ploughing the field is Rs.$1925$.

35. Find the difference between the area of regular hexagonal plot each of whose side $72$ m and the area of the circular swimming take in scribed in it. (Take $\pi =\dfrac{22}{7}$)

Ans: Given, side of hexagonal plot \[=72\]m

Area of equilateral triangle is given by$A=\dfrac{\sqrt{3}}{4}{{(side)}^{2}}$

$\therefore A\Delta AOB=\dfrac{\sqrt{3}}{4}{{(72)}^{2}}=1296\sqrt{3}{{m}^{2}}$

Area of hexagonal plot $=6\times $ Area of triangle $AOB$ 

$\Rightarrow 6\times 1296\sqrt{3}=13468.032~{{\text{m}}^{2}}$

Now,$O{{C}^{2}}=O{{A}^{2}}-A{{C}^{2}}$

$\Rightarrow {{(72)}^{2}}-{{\left( \dfrac{72}{2} \right)}^{2}}~=5184-1296=3888$

$\Rightarrow O{{C}^{2}}=3888$

$\Rightarrow OC=\sqrt{3888}=62.35~\text{m}$

Area of circular region, $A=\pi {{r}^{2}}$ 

$\Rightarrow \dfrac{22}{7}\times {{\left( 62.35 \right)}^{2}}=12,219.42\text{ }{{m}^{2}}$.

Then the difference is, $13468.032~{{\text{m}}^{2}}-12,219.42\text{ }{{m}^{2}}=1248.603\text{ }{{m}^{2}}$.

36. In the given figure areas have been drawn of radius $21$cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Ans: Given, areas drawn with radius $21$ cm at vertices, A,B,C and D.

Thus, required area is the area of the circle with radius $21$ cm.

$\Rightarrow A=\pi {{\left( 21 \right)}^{2}}=\dfrac{22}{7}\times 21\times 21$ 

$\Rightarrow A=1386c{{m}^{2}}$

37. A wheel has diameter \[84cm\],  find how many complete revolutions it must make to complete $792$ meters. 

Ans: Given, diameter of wheel, \[d=2r\text{ }=\text{ }84\] cm.

Distance covered in one revolution = circumference

Circumference of the circle is given by $2\pi r$.

Distance covered in one revolution $=2\pi r=\pi \left( 2r \right)=\dfrac{22}{7}\times 84$

$\Rightarrow 22\times 12=264$ cm

Thus, for distance covered\[264cm\], number of revolutions$=1$

$\therefore $For distance covered $792$m/ \[79200\] cm, number of revolutions$=\dfrac{1}{264}\times 79200=300$ .

Number of revolutions to complete $792$ m is $300$.

38. The given figure is a sector of a circle of radius \[10.5cm\]. Find the perimeter of the sector. (Take $\pi =\dfrac{22}{7}$ ) 

Ans: Given radius, $r=10.5$ cm

We know that circumference, i.e., perimeter of a sector of angle ${{P}^{{}^\circ }}$ of a circle with radius $R$ is given by, $\dfrac{{{P}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 2\pi R+2R$.

$\therefore $Required perimeter can be calculated as,

$\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{2\times 22}{7}\times \left( 10.5 \right)+2\left( 10.5 \right)=\dfrac{1}{6}\times \dfrac{44}{7}\times \dfrac{21}{2}+21$ 

Thus, the perimeter of a sector is $32$ cm.

39. A car had 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 each sweep of the blades. 

Ans: Given, radius of each wiper\[=\text{ }25\] cm and sweeping angle, $\theta ={{115}^{{}^\circ }}$.

Since, area of sector is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

Total area cleaned at each sweep of the blades $=2\left( \dfrac{115}{360}\times \dfrac{22}{7}\times 25\times 25 \right)$

$\Rightarrow \dfrac{230\times 22\times 5\times 25}{72\times 7}=\dfrac{230\times 11\times 125}{36\times 7}$

$\Rightarrow \dfrac{115\times 11\times 125}{18\times 7}=1254.96~\text{c}{{\text{m}}^{2}}$

Thus, the total area cleaned at each sweep of the blades is $1254.96~\text{c}{{\text{m}}^{2}}$

40. In the given figure arcs have been drawn with radii \[14\text{ }cm\] each and with centres P, Q and R. Find the area of the shaped region. 

Ans: Given, arc with radius $14$ cm each.

Area of sector is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

From the figure area of shaded region is,$A=3\times \dfrac{{{60}^{\circ }}}{{{360}^{{}^\circ }}}\times \pi {{\left( 14 \right)}^{2}}=\dfrac{22}{7}\times \dfrac{{{\left( 14 \right)}^{2}}}{2}$

$\Rightarrow A=11\times 2\times 14=308c{{m}^{2}}$

Thus, area of shaded region is $308\text{ }c{{m}^{2}}$.

41. The radii of two circles are \[19cm\] and \[9cm\] respectively. Find the radius of the circle which has its circumference equal to the sum of the circumference of the two circles.

Ans: Given, radii of two circles as $19$ cm and $9$ cm.

Circumference of the circle is given by \[C=2\pi r\].

C1 = circumference of the 1st circle

\[\Rightarrow 2\pi \left( 19 \right)=\text{ }38\pi \text{ }cm~\]

C2 = circumference of the 2nd circle

\[\Rightarrow 2\pi \left( 19 \right)=18\pi \text{ }cm~\]

Circumference of the required circle \[=\] sum of circumferences of two circles

\[\Rightarrow 2\pi r=38\pi +18\pi \text{ }cm~\]

\[\Rightarrow 2\pi r=56\pi \]

\[\Rightarrow r=28cm\]

Thus, the radius of the required circle is $28$ cm.

42. A car travels \[0.99\,km\] distance in which each wheel makes \[450\] complete revolutions. Find the radius of its wheel 

Ans: Given, distance travelled by a wheel in 450 complete revolutions \[=0.99\] km \[=\text{ }990m\]

Distance travelled in one revolution $=\dfrac{990}{450}=\dfrac{11}{5}$.

Let $r$ be the radius of the wheel.

Circumference of the circular field is given by, $2\pi r$.

$\therefore 2\pi r=\dfrac{11}{5}$

$\Rightarrow 2\times \dfrac{22}{7}r=\dfrac{11}{5}$

$\Rightarrow r=\dfrac{7\times 100}{20}=35m$.

43. A sector is cut from a circle of diameter \[21cm\]. If the angle of the sector is \[\mathbf{15}{{\mathbf{0}}^{{}^\circ }}\] find its area. 

Ans: Given, diameter, \[d=\text{ }21cm~\]

Thus, radius can be obtained as,  \[r=~\text{ }\dfrac{21}{2}cm~\]

Angle of sector $={{150}^{{}^\circ }}$ 

Area of the sector is given by,  $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$.

Hence, area  of the sector is, $A=\dfrac{5}{12}\times \dfrac{22}{7}\times \dfrac{21}{2}\times \dfrac{21}{2}=\dfrac{5\times 11\times 21}{4\times 2}$ 

$\Rightarrow A=144.38\text{ }c{{m}^{2}}$.

44. In the given figure AOBCA represent a quadrant of area \[~9.625c{{m}^{2}}\].  Calculate the area of the shaded portion. 

Ans: Given, area of quadrant $9.625\text{ }c{{m}^{2}}$.

From the figure, $OD=2$ cm and $OA=3.5$ cm.

Hence, the area of the shaded region is $6.125\text{ }c{{m}^{2}}$.

Short Answer Questions (3 Marks)

1. Find the area of the shaded region if \[PQ=24cm,\text{ }PR=7cm\] and O is the centre of the circle. 

Ans: Given \[PQ=24\text{ }cm\] and \[PR=7\text{ }cm\]

Since QR is a diameter passing through the centre O of the circle 

Angle of semi-circle, $\angle RPQ={{90}^{{}^\circ }}$

$\therefore Q{{R}^{2}}=P{{R}^{2}}+P{{Q}^{2}}$

$\Rightarrow {{7}^{2}}+{{24}^{2}}=625={{25}^{2}}$

Diameter of the circle \[=\text{ }25cm~\]

Thus radius, $r=\dfrac{25}{2}cm$

Area of the semi-circle$=\dfrac{1}{2}\pi {{r}^{2}}$

$\therefore $Area $=\dfrac{1}{2}\times \dfrac{22}{7}\times \dfrac{25}{2}\times \dfrac{25}{2}=\dfrac{11\times 625}{28}c{{m}^{2}}$

Also, area of  $\Delta PQR=\dfrac{1}{2}PR\times PQ$

$\Rightarrow \dfrac{1}{2}\times 7\times 24=84c{{m}^{2}}$

From the given figure,

 Area of the shaded region = Area of the semi-circle with O as centre and OQ as radius – area of $\Delta PQR$

Hence, area of the shaded region can be calculated as,

Area of the shaded region $=\dfrac{11\times 625}{28}-84=\dfrac{11\times 625-84\times 28}{28}$

$\Rightarrow \dfrac{6875-2352}{28}=\dfrac{4523}{28}$

Area of the shaded region  $=161.54 c{{m}^{2}}$.

2. The perimeter of a sector of a circle of radius\[~5.7\text{ }m\] is \[27.2\text{ }m\].Calculate:

(i). The length of arc of the sector in cm.

Ans: Given radius, $r=5.7$ m and perimeter $=27.2$ m. 

Then \[OA=OB=5.7m~\]

$OA+OB+arcAB=27.2m$

$\Rightarrow 5.7m+5.7m+arcAB=27.2$ m

$\Rightarrow arcAB=27.2-11.4=15.8$ m

The length of the arc of the sector is $15.8$ m.

ii. The area of the sector in cm2 correct to the nearest $cm^2$ 

Ans: Area of sector OACB$=\left( \dfrac{1}{2}\times radius\times arc \right)$

$\Rightarrow \left( \dfrac{1}{2}\times 5.7\times 15.8 \right)=45.03c{{m}^{2}}$

Area of sector correct to nearest $c{{m}^{2}}=45c{{m}^{2}}$

3. Find the area of the shaded region in the given figure where ABCD is a square of side \[10cm\]  and semi-circles are drawn with each side of the square as diameter. $\left[ \pi =3.14 \right]$

Ans: Given, side of square ABCD $=10\text{ cm}$                                                     

Area of square ABCD $={{\left( side \right)}^{2}}={{\left( 10 \right)}^{2}}=100\text{ }c{{m}^{2}}$                           

Given that semicircle is drawn with side of square as diameter.   

Diameter of semicircle $=10\text{ }cm$.                                                                  

Radius of semicircle $=5\text{ }cm$                                                                      

Area of semi-circle $=\dfrac{1}{2}\pi {{r}^{2}}$                                                                       

Area $=\dfrac{1}{2}\times 3.14\times {{\left( 5 \right)}^{2}}=39.25\text{ }c{{m}^{2}}$                                                                 

Area of 4 semi-circles – Area of shaded region = Area of square ABCD  

$\therefore $Area of shaded region = Area of 4 semi-circles - Area of square ABCD 

Area of shaded region $=4\times 39.25-100=57c{{m}^{2}}$

4. In the given figure $\Delta ABC$  is an equilateral triangle inscribed in a circle of radius \[4\text{ }cm\] and centre O. Show that the area of the shaded region is $\dfrac{4}{3}\left( 4\pi -3\sqrt{3} \right)c{{m}^{2}}$.

Ans: Given, $\Delta ABC$ as an equilateral triangle inscribed in circle with radius, $r=4$ cm.

In  $\Delta OBD$, Let $BD=a,$$OB=4cm$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{a}{4}$

$\Rightarrow a=\dfrac{4\sqrt{3}}{2}$

$\Rightarrow a=2\sqrt{3}~\text{cm}$

$\Rightarrow BC=2a=2\times 2\sqrt{3}=4\sqrt{3}~\text{cm}$

$\dfrac{OD}{4}=\cos {{60}^{{}^\circ }}\Rightarrow OD=4\times \dfrac{1}{2}=2~\text{cm}$

Area of shaded region = Area of sector OBPC – Area of $\Delta OBC$

$\Rightarrow \dfrac{120}{{{360}^{{}^\circ }}}\times \pi \times {{4}^{2}}-\dfrac{1}{2}\times 4\sqrt{3}\times 2=\dfrac{4}{3}[4\pi -3\sqrt{3}]\text{c}{{\text{m}}^{2}}$

5. The radii of two circles are \[8cm\]and \[6cm\]respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles. 

Ans: Given, radii of two circles, $r1=8\text{ cm}$ and $r2=6\text{ cm}$.

Area of a circle is given by $\pi {{r}^{2}}$

Let $A1$ = Area of the first circle

$\Rightarrow A1=\pi {{\left( 8 \right)}^{2}}=64\pi c{{m}^{2}}$ 

$A2=$Area of the second circle

$\Rightarrow A2=\pi {{\left( 6 \right)}^{2}}=36\pi c{{m}^{2}}$  

Total area $=A1+A2$,

$\Rightarrow 64\pi +36\pi =100\pi c{{m}^{2}}$  

Let R be the radius of the circle with area \[{{A}_{1}}+\text{ }{{A}_{2}}~\],

$\therefore \pi {{R}^{2}}=100\pi $

$\Rightarrow R=10cm$

Hence, required radius \[=\text{ }10cm~\]

6. A chord of a circle of radius \[10cm\] subtends a right angle at the centre. Find the area of the corresponding: (Use $\pi =3.14$) 

i. minor sector

Ans: Given, radius of circle, $r=10$ cm subtending an angle ${{90}^{\circ }}$ at the centre. $\therefore \angle AOB={{90}^{\circ }}$.

Area of minor sector $=\dfrac{\theta }{360}\pi {{r}^{2}}$

$\Rightarrow \dfrac{90}{360}\times 3.14\times {{\left( 10 \right)}^{2}}=\dfrac{1}{4}\left( 3.14 \right)\left( 100 \right)$  

$\Rightarrow \dfrac{314}{4}=78.5c{{m}^{2}}$

Area of minor sector is $78.5c{{m}^{2}}$.

ii. major sector 

Ans: Area of circle $=\pi {{r}^{2}}=\pi {{\left( 10 \right)}^{2}}$

Area of major sector = Area of circle – Area of minor sector 

$\Rightarrow \pi {{\left( 10 \right)}^{2}}-\dfrac{90}{360}\pi {{\left( 10 \right)}^{2}}=\left( 3.14 \right)\left( 100 \right)-\dfrac{1}{4}\left( 3.14 \right)\left( 100 \right)$

$\Rightarrow 314-78.50=235.5c{{m}^{2}}$

Area of major sector is \[235.5\text{ }c{{m}^{2}}\].

iii. minor segment

Ans:   Area of minor segment = Area of minor sector OAB – Area of $\Delta OAB$

$\because $Area of $\Delta OAB=\dfrac{1}{2}\left( OA \right)\left( OB \right)\sin \angle AOB$

$=\dfrac{1}{2}\left( OA \right)\left( OB \right)\left( \because \angle AOB={{90}^{{}^\circ }} \right)$

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

$\Rightarrow \dfrac{1}{4}\left( 3.14 \right)\left( 100 \right)-50=25\left( 3.14 \right)-50$

$\Rightarrow 78.50-50=28.5c{{m}^{2}}$

Area of minor segment is \[28.5\text{ }c{{m}^{2}}\].

iv. major segment 

Ans: Area of major segment = Area of the circle – Area of minor segment

$\Rightarrow \pi {{\left( 10 \right)}^{2}}-25.5=100\left( 3.14 \right)-28.5$

$\Rightarrow 314-18.5=285.5c{{m}^{2}}$

Area of major segment is \[285.5\text{ }c{{m}^{2}}\].

7. In Akshita’s house, there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 140 cm and the difference of their circumference is\[88cm\], find the diameter of the circular top and bottom. 

Ans: Given, sum of radii of circular top and bottom \[=\text{ }140cm~\]

Let radius of top \[=r\text{ }cm~\]

$\therefore $ Radius of bottom$=\left( 140-r \right)cm$

Circumference of top$=2\pi r$ cm 

Circumference of bottom$=2\pi \left( 140-r \right)cm$

Difference of circumference $=\left[ 2\pi r-2\pi \left( 140-r \right) \right]cm$  

By the given conditions in the problem, 

$2\pi r-2\pi \left( 140-r \right)=88$

$\Rightarrow 2\pi \left[ r-140+r \right]=88$

$\Rightarrow 2r-140=\dfrac{88}{2\left( \dfrac{22}{7} \right)}$

$\Rightarrow 2r=140+14-154$

$\Rightarrow r=\dfrac{154}{2}$

$\therefore $Radius of top is obtained as \[77cm~\]

Diameter of top can be calculated as,

$\Rightarrow 2\times 77=154cm$

Radius of bottom can be obtained as,

 \[\Rightarrow 140\text{ }\text{ }r=140\text{ }\text{ }77\]

$\therefore $ Diameter of bottom $2\times 63=126cm~$.

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

Ans: Given, radius of circle,$r=15\text{ }cm$ subtending an angle ${{60}^{\circ }}$ at centre.

$\therefore \angle AOB={{60}^{\circ }}$.

Area of a sector of a circle is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

Area of minor segment $=$ Area of minor sector OAB $-$ area of $\Delta OAB$ $\Rightarrow \dfrac{60}{360}\pi {{(15)}^{2}}-\dfrac{\sqrt{3}}{4}{{(15)}^{2}}=\dfrac{1}{6}(3.14)(225)-\dfrac{(1.73)}{4}(225)$

$\Rightarrow 117.75-97.3125=20.4375~\text{c}{{\text{m}}^{2}}$

Area of major segment = Area of the circle – area of minor segment

$\Rightarrow \pi {{(15)}^{2}}-20.4375=(3.14)(225)-20.4375$

$\Rightarrow 706.5-20.4375=686.0625~\text{c}{{\text{m}}^{2}}$

Area of minor segment is $20.4375~\text{c}{{\text{m}}^{2}}$ and area of major segment is $686.0625\,~\text{c}{{\text{m}}^{2}}$.

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

Ans: Given, radius of cover, $r=28\text{ }cm$. Table has six equal designs and the cost of making the design is Rs. $0.35$ per $c{{m}^{2}}$.

Area of equilateral triangle is given by,$A=\dfrac{\sqrt{3}}{4}{{a}^{2}}$ .

From the figure area of one design $=\dfrac{60}{360}\times \pi {{\left( 28 \right)}^{2}}$– Area of equilateral $\Delta OAB$

$\Rightarrow \dfrac{\pi }{6}\times {{\left( 28 \right)}^{2}}-\dfrac{\sqrt{3}}{4}{{\left( 28 \right)}^{2}}={{\left( 28 \right)}^{2}}\left( \dfrac{\pi }{6}-\dfrac{\sqrt{3}}{4} \right)$

$\Rightarrow {{\left( 28 \right)}^{2}}\left( \dfrac{22}{7\times 6}-\dfrac{1.7}{4} \right)$ 

Total cost can be calculated as follows,

$\Rightarrow 6\times {{(28)}^{2}}\left( \dfrac{22}{42}-\dfrac{17}{40} \right)\times 0.35\times \dfrac{440-357}{21\times 40}$

$\Rightarrow 21\times {{(28)}^{2}}\left( \dfrac{11}{21}-\dfrac{17}{40} \right)=\dfrac{21\times {{(28)}^{2}}}{100}\times \dfrac{440-357}{21\times 40}$

\[\Rightarrow \dfrac{28\times 28\times 83}{40}=\dfrac{7\times 28\times 83}{1000}\]

$\Rightarrow Rs.16.268$

The total cost of making the design is $Rs.16.268$.

10. ABCD is a flower bed. If \[OA\text{ }=\text{ }21m\] and \[DC\text{ }=\text{ }14m\]. Find the area of the bed.

Ans:  Given,

\[OA\text{ }=\text{ }R\text{ }=\text{ }21m\]

 \[OC\text{ }=\text{ }r\text{ }=\text{ }14m~\]

Area of the flower bed (i.e., shaded portion) $=$ area of quadrant of a circle of radius $R$ of the quadrant of a circle of radius $r$

$\Rightarrow \dfrac{1}{4}\pi {{R}^{2}}-\dfrac{1}{4}\pi {{r}^{2}}=\dfrac{\pi }{4}\left( {{R}^{2}}-{{r}^{2}} \right)$

$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\left[ {{(21)}^{2}}-{{(14)}^{2}} \right]{{m}^{2}}$

$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\times (21\times 14)(21-14){{m}^{2}}$

$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\times 35\times 7~{{\text{m}}^{2}}$

$\Rightarrow 192.5~{{\text{m}}^{2}}$

Hence, area of the flower bed is $192.5~{{\text{m}}^{2}}$.

11. 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. 

Ans: Given radius, \[r=\text{ }21\text{ }cm\] and angle $\theta ={{60}^{{}^\circ }}$.

Length of arc \[=\dfrac{\theta }{360{}^\circ }\times 2\pi r~\]

\[\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 2\times \dfrac{22}{7}\times 21~~=\text{ }22\text{ }cm~\]

ii. area of the sector formed by the arc. 

Ans: Area of the sector \[=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}\]

\[~\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 21\times 21~~=\text{ }231\text{ }c{{m}^{2}}\]

iii. area of the segment formed by the corresponding chord 

Ans: Area of segment formed by corresponding chord \[=~\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}-~Area\text{ }of~\text{ }\Delta OAB~\]

$\Rightarrow $Area of segment $=231-$  Area of $\Delta OAB$   …(1)

In right triangles $OMA$ and $OMB$, 

\[OA\text{ }=\text{ }OB\] [Radii of same circle] 

\[OM\text{ }=\text{ }OM\] [Common] 

\[\therefore \Delta OMA\cong \Delta OMB\] [RHS congruency] 

\[~\therefore AM\text{ }=\text{ }BM\] [By CPCT] 

Here $M$ is the midpoint.

\[\Rightarrow ~~\dfrac{1}{2}AB\text{ }and~\text{ }\angle AOM\text{ }=~\text{ }\angle BOM\]

\[\Rightarrow \dfrac{1}{2}\angle AOB\text{ }=\dfrac{1}{2}\times {{60}^{{}^\circ }}={{30}^{{}^\circ }}\]

In right angled \[\Delta OMA\],

$\cos {{30}^{{}^\circ }}=\dfrac{\text{OM}}{\text{OA}}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\text{OM}}{21}$

$\Rightarrow \text{OM}=\dfrac{21\sqrt{3}}{2}~cm$

Also $\sin {{30}^{{}^\circ }}=\dfrac{\text{AM}}{\text{OA}}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\text{AM}}{21}$

$\Rightarrow 2AM=21cm$

$\Rightarrow AB=21cm$

 $\therefore $ Area of $\Delta AOB=\dfrac{1}{2}\times AB\times OM$

\[\Rightarrow \dfrac{1}{2}\times 21\times \dfrac{21\sqrt{3}}{2}=\dfrac{441\sqrt{3}}{4}\]cm2

Using eq. $\left( 1 \right)$,

Area of segment formed by corresponding chord \[=~\left( 231-\dfrac{441\sqrt{3}}{4} \right)\text{ }c{{m}^{2}}~\].

12. A chord of a circle of radius \[15\text{ }cm\]subtends an angle of ${{60}^{{}^\circ }}$ at the centre. Find the area of the corresponding segment of the circle.  (Use $\pi =3.14$ and $\sqrt{3}=1.73$)

Ans: Given radius,\[~r~=\text{ }15\text{ }cm\] and angle $\theta ={{60}^{{}^\circ }}$

Area of minor sector can be obtained from  $\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$ 

\[\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 15\times 15=117.75c{{m}^{2}}\]

For, Area of $\Delta AOB$, 

Draw $OM$, where $OM\bot AB$. 

 In right angled \[\Delta OMA\],

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\text{OM}}{15}$

$\Rightarrow \text{OM}=\dfrac{15\sqrt{3}}{2}~cm$

Also, $\sin {{30}^{{}^\circ }}=\dfrac{\text{AM}}{\text{OA}}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\text{AM}}{15}$

$\Rightarrow 2AM=15cm$

$\Rightarrow AB=15cm$

$\therefore $ Area of $\Delta AOB=\dfrac{1}{2}\times AB\times OM$

$\Rightarrow \dfrac{1}{2}\times 15\times \dfrac{15\sqrt{3}}{2}=\dfrac{225\sqrt{3}}{3}$

$\Rightarrow \dfrac{225\times 1.73}{4}=97.3125c{{m}^{2}}$

$\therefore $Area of minor segment = Area of minor sector – Area of  $\Delta AOB$

\[\Rightarrow 117.75\text{ }\text{ }97.3125\text{ }=\text{ }20.4375\text{ }c{{m}^{2}}~\]

And, Area of major segment  $=\pi {{r}^{2}}-$ Area of minor segment 

\[\Rightarrow 706.5\text{ }\text{ }20.4375\text{ }=\text{ }686.0625\text{ }c{{m}^{2}}~\]

13. A chord of a circle of radius 12 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$\text{ }\sqrt{3}=1.73$ )

Ans: Given radius, \[r~~=\text{ }12cm\] and angle $\theta ={{120}^{{}^\circ }}$.

Area of corresponding sector can be obtained by using the formula $\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

\[\Rightarrow \dfrac{{{120}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 12\times 12=\text{ }150.72\text{ }c{{m}^{2}}\]

\[\Rightarrow \dfrac{1}{2}\angle AOB\text{ }=\dfrac{1}{2}\times {{120}^{{}^\circ }}={{60}^{{}^\circ }}\]

$\cos {{60}^{{}^\circ }}=\dfrac{\text{OM}}{\text{OA}}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\text{OM}}{12}$

$\Rightarrow \text{OM}=6~\text{cm}$

Also $\text{ }\sin {{60}^{{}^\circ }}=\dfrac{\text{AM}}{\text{OA}}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\text{AM}}{12}$

$\Rightarrow 2AM=12\sqrt{3}cm$

$\Rightarrow AB=12\sqrt{3}cm$

$\Rightarrow \dfrac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}$

$\Rightarrow 36\times 1.73=62.28c{{m}^{2}}$

Area of corresponding segment = Area of corresponding sector – Area of $\Delta AOB$

\[\Rightarrow 150.72\text{ }\text{ }62.28\text{ }=\text{ }88.44\text{ }c{{m}^{2}}~\]

14. Find the area of the shaded region  in  figure,  if \[PQ\text{ }=\text{ }24\text{ }cm,\text{ }PR\text{ }=\text{ }7\text{ }cm\] and $O$is  the centre of the circle. 

Long Answer Questions (4 Marks)

1. Three horses are tethered with \[7m\] long ropes at the three corners of a triangular field having sides \[20m,\text{ }34m\] and  \[42\text{ }m\] . Find the area of the plot which can be grazed by the horses. Also, find the area of the plot which remains ungrazed. 

Ans: Given, length of ropes at the three corners, $l=33$ m and the sides of the triangle as $20$ m, $34$ m and $42$ m.

From the figure let, 

$\angle A=\theta {{1}^{{}^\circ }}$, $\angle B=\theta {{2}^{{}^\circ }}$ and $\angle C=\theta {{3}^{{}^\circ }}$

Area which can be grazed by three horses $=$ Area of sector with central angle $\theta {{1}^{{}^\circ }}$ and radius \[7\text{ }m\]$+$ Area of sector with central angle $\theta {{2}^{{}^\circ }}$ and radius \[7\text{ }cm\]$+$ Area of sector with central angle $\theta {{3}^{{}^\circ }}$ and radius \[7\text{ }cm~\]

$\Rightarrow \dfrac{\pi {{r}^{2}}\theta {{1}^{{}^\circ }}}{360}+\dfrac{\pi {{r}^{2}}\theta {{2}^{{}^\circ }}}{{{360}^{{}^\circ }}}+\dfrac{\pi {{r}^{2}}\theta {{3}^{{}^\circ }}}{{{360}^{{}^\circ }}}$

$\Rightarrow \dfrac{\pi {{r}^{2}}}{{{360}^{{}^\circ }}}\left( \theta {{1}^{{}^\circ }}+\theta {{2}^{{}^\circ }}+\theta {{3}^{{}^\circ }} \right)$

$\Rightarrow \dfrac{\pi {{r}^{2}}}{{{360}^{{}^\circ }}}\times {{180}^{\circ }}$

( $\because $Sum of three angles of a  $\Delta ={{180}^{{}^\circ }}$)

$\Rightarrow \dfrac{22}{7}\times \dfrac{7\times 7\times {{180}^{{}^\circ }}}{{{360}^{{}^\circ }}}=77{{m}^{2}}$ 

Sides of plot \[ABC\] are given in the problem as,

 \[a=\text{ }20m~\], \[b\text{ }=\text{ }34m\] and \[c=42m\]

$\because $Semi perimeter,$s=\dfrac{20+34+42}{2}=48m$ 

$\therefore $Area of triangular plot \[=\] Area of $\Delta ABC$

$\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=\sqrt{48\times 28\times 14\times 6}=336{{m}^{2}}$

Area grazed by the horses \[=\text{ }77{{m}^{2}}~\]

$\therefore $Ungrazed area can be calculated by subtracting area grazed by the horses from the area of the triangular plot.

\[\Rightarrow \] Ungrazed area \[=\left( 33677 \right)~=\text{ }259\text{ }{{m}^{2}}~\]

2. In the given figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side $56$ m.  If the corner of each circular flower bed is the point of enter section O of the diagonals of the square lawn. Find the sum of the areas of the lawn and the flower beds. 

Ans: Given, side of square lawn $=56$ m.

Area of the square lawn $ABCD$, $A={{\left( 56 \right)}^{2}}$

 Let $OA=OB=x$

$\therefore $By Pythagoras theorem, 

$\Rightarrow {{x}^{2}}+{{x}^{2}}={{56}^{2}}$

$\Rightarrow 2{{x}^{2}}=56\times 56$

$\Rightarrow {{x}^{2}}=28\times 56$

Again, area of sector $OAB$ $=\dfrac{90}{360}\times \pi {{x}^{2}}=\dfrac{1}{4}\pi {{x}^{2}}$

Substituting for $x$, we get

$\Rightarrow $ Area of sector $OAB=\dfrac{1}{4}\times \dfrac{22}{7}\times 28\times 56{{m}^{2}}$ 

Area of  $\Delta OAB=\dfrac{1}{4}56\times 56{{m}^{2}}$

(Here, $\angle AOB={{90}^{{}^\circ }}$ since square $ABCD$ is divided into 4 right triangles)

$\therefore $ Area of flowerbed$AB=\left( \dfrac{1}{2}\times \dfrac{22}{7}\times 28\times 56-\dfrac{1}{4}56\times 56 \right){{m}^{2}}$

$\Rightarrow \dfrac{1}{4}\times 28\times 56\left( \dfrac{22}{7}-2 \right)$ 

$\Rightarrow \dfrac{7\times 56\times 8}{7}$

Area of flower bed  $AB=448{{m}^{2}}$

Similarly, area of the other flowerbed \[=\text{ }448{{m}^{2}}~\]

Total area of the lawn and the flowerbeds $=56\times 56+448+448=4032{{m}^{2}}$

3. An elastic belt is placed round the rein of a pulley of radius $5$ cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, $10$ cm from O. Find the length of the best that is in contact with the rim of the pulley. Also find the shaded area. 

Ans: Given, radius of rein of pulley, $OA=5$ cm.

Thus, $\cos \theta =\dfrac{OA}{OP}$

$\Rightarrow \cos \theta =\dfrac{5}{10}=\dfrac{1}{2}~$

$\Rightarrow \theta ={{60}^{{}^\circ }}$

$\Rightarrow \angle AOB={{120}^{{}^\circ }}$

$\therefore \operatorname{Arc}AB=\dfrac{120\times 2\times \pi \times 5}{360}~\text{cm}$

$\Rightarrow \dfrac{10\pi }{3}~\text{cm}\left[ \because l=\dfrac{\theta }{360}\times 2\pi r \right]$

Length of the belt that is in contact with the rim of the pulley can be calculated by subtracting length of arc AB from Circumference of the rim.

Length of the belt $=2\pi \times 5cm-\dfrac{10}{3}\pi cm$

Length of the belt $=\dfrac{20}{3}\pi cm$

Now, area of sector OAQB$=\dfrac{{{120}^{\circ }}}{{{360}^{\circ }}}\times \pi \times {{5}^{2}}cm=\dfrac{25}{3}\pi c{{m}^{2}}$ $\left[ \text{Since Area=}\dfrac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}} \right]$

$\left[ \because AP=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}cm \right]$

Area of quadrilateral OAPB is given by, $2(\text{Area of }\Delta OAP)=25\sqrt{3}c{{m}^{2}}$

Hence, shaded area can be obtained as,

Area of shaded region$=25\sqrt{3}-\dfrac{25}{3}\pi $

Hence, area of shaded region is $17.12\text{ }c{{m}^{2}}$.

Download Class 10 Maths Chapter 12 Important Questions PDF

Area of a circle.

The area is the region enclosed by the circumference of the circle. The area covered by one complete cycle of the radius of the circle on a two-dimensional plane is known as the area of that circle.

Area of the circle = πr2 square units.

Where r is the radius of the circle.

Area of Semicircle

To find the area of the semi-circle we need half of the area of the circle.

(Image will be uploaded soon)

The formula for the area of semicircle = \[\frac{\pi r^{2}}{2}\] square unit.

Area of a Ring

We know the area of circle πr2. Here the ring is having two radii inner and outer radius. Area of the ring i.e. the coloured part shown in the above figure is calculated by subtracting the area of the inner circle from the area of the bigger circle.

Area of the ring = πR2 - πr2 = π(R2 - r2)

Where R = radius of outer circle

r = radius of the inner circle

The Sector of a Circle

The sector of a circle is the region of a circle enclosed by an arc and two radii. The smaller area is called the minor sector and the larger area is called the major sector of the circle.

Areas of Sectors of a Circle

The area formed by an arc and the two radii joining the endpoints of the arc is known as a sector of a circle.

In the above diagram, OACB is the minor sector. OADB is the major sector.

Minor Sector

The area including arc  ∠AOB with point C is called Minor Sector. So, OACB is the minor sector. The area included in ∠AOB is the angle of the minor sector.

Area of the sector of angle \[\theta = \frac{\theta}{360} \times \pi r^{2}\]

Major Sector

The area included in ∠AOB with point D is called the Major Sector. So OADB is the major sector. The angle of the major sector is 360° – ∠AOB.

If the degree measure of the angle at the center is 360, area of the sector = πr2.

Formula to find the area of Major Sector = πr2 -  Area of the Minor Sector

Area of the circle can also be calculated by adding an area of a minor sector and area of a major sector.

Note: Area of Minor Sector + Area of Major Sector = Area of the Circle

To Calculate Length of an Arc of a Sector of Angle θ

An arc is the piece of the circumference of the circle so the length of an arc can be calculated as the θ part of the circumference.

Formula to calculate the length of an arc of a sector of angle \[\theta = \frac{\theta}{360} \times 2\pi r\].

Areas of Segments of the Circle

The area made by an arc and a chord of a circle is called the segment of the Circle.

Minor Segment

The minor segment is defined as the region bounded by the chord and the minor arc.

In the above diagram, AB is a chord of the circle. The area made by chord AB and arc X is known as the minor segment.

Formulas to calculate Area of Minor Segment = Area of Minor Sector – Area of ∆ABO

\[= \frac{\theta}{360} \times \pi r^{2} - \frac{1}{2} r^{2} sin \theta\]

Major Segment

The major segment is defined as the region bounded by the chord and the major arc.

Formulas to calculate Area of Major Segment = πr2 -  Area of Minor Segment

Note: Area of major segment + Area of minor segment = Area of the circle.

Areas of Combinations of Plane Figures

A plane figure is a geometric figure that has no thickness, it lies entirely on one plane. In this section, we will learn how to calculate the areas of some plane figures which are combinations of more than one plane figure.

To find the area of the composite area of the shapes, simply find the area of each shape and add them together. The order in which we calculate the areas does not matter and the commutative property states that it does not matter which order you add them in.

Following are the steps to find the area of combined figures:

Step I: First we divide the given combined figure into its simple geometrical shapes.

Step II: Then calculate the area of simple geometrical shapes separately.

Step III: Finally, to find the area of the combined figure we need to add or subtract these areas.

Below are some examples of combinations of plane figures:

Above diagram is a combination of one rectangle and two semi-circles. So, to find an area of the figure, first, find an area of the rectangle and then add the area of two semi-circles.

Area of the given figure = Area of the rectangle + 2 x Area of semi-circles.

Above diagram is the form of a ring.

Area of the  required region = \[\frac{3}{4} (\pi r_{1}^{2} - \pi r_{2}^{2})\]

Above diagram is a combination of circle and square. To find the area of a given figure, first, calculate the area of the circle then subtract the area of the square.

Area of the required region = πr2 - a2

Important formula to remember:

Area of circle = πr2

Length of an arc of a sector of a circle with radius r and angle with degree measure θ is \[\frac{\theta}{360} \times 2\pi r\].

Area of a sector of a circle with radius r and angle with degrees measure θ is \[\frac{\theta}{360} \times \pi r^{2}\]

Area of the segment of a circle = Area of the corresponding sector – Area of the corresponding triangle.

Conclusion:

Refer to this Important Questions For Class 10 Maths Chapter 12 for boards exam preparation. All the important concepts are available in the given PDF. The formula for the area of a circle also helps us calculate the area of circle sectors and segments as well. Download important questions of ch 12 Maths Class 10 from Vedantu site or app for more information.

Important Related Links for CBSE Class 10 Maths

Vedantu's offering of "Important Questions for CBSE Class 10 Maths Chapter 12 - Areas Related to Circles" is a valuable resource for students seeking to excel in their mathematics studies. The comprehensive collection of questions and solutions provides a deep understanding of the intricate concepts related to circles and their areas. By using this platform, students can gain confidence in tackling complex problems, enhance their problem-solving skills, and prepare effectively for their CBSE Class 10 examinations. Vedantu's commitment to quality education is evident in the well-structured and engaging content, making it an indispensable tool for students aiming to achieve academic success in mathematics.

FAQs on Important Questions for CBSE Class 10 Maths Chapter 12 - Areas Related to Circles

1. How many sums are there in class 10 Maths Chapter 12 Areas Related to Circles?

Chapter 12 of Maths for class 10 teaches about the application of the chapter ‘Circles’. There are a total of 35 sums in all the exercises combined together from the chapter. All of these problems are unique in their own way and offer excellent practice and knowledge to students. At times, however, these can get very confusing. The Important questions by Vedantu for this chapter can help students here because of their simple language structure.

2. What students will study in chapter 12 Areas Related to Circles of class 10 Maths?

In this chapter, students learn the different applications of circles in terms of their areas. Concepts like the area of circles, their circumference, and the various parts of it, segment, sector, angle, and length, everything is discussed in this chapter. Important Questions for chapter 12 is specifically designed to focus on the important areas by providing students with questions so that they can practice with them and score good results for questions from this chapter by Vedantu.

3. What are the benefits of using important questions for chapter 10 of class 12 Maths?

Important Questions provided by Vedantu are good for students to learn ‘Areas Related to Circles’. They can access the questions free of cost by visiting the link-Important Questions for class 10 maths, and they can be assured of their quality content as each question is kept up to date and follows the latest curriculum. Moreover, the easily understandable structure provides them with a good practice experience for good marks. Since these questions are from the perspective of the Board exam paper, these are excellent sources for studying.

4. State the important formulae in chapter 12 of class 10 Maths.

Chapter 12 is named 'Areas Related to Circles' and this chapter is practically an application of the previous chapter in class 10 named 'Circles'. So, this chapter has some important formulae to solve various problems: 

• Area of circle: $\pi r^2$

Area of a semicircle: $\dfrac{1}{2}\pi r^2$

Area of quadrant: $\dfrac{\pi r^2}{4}$ 

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

Area of Ring: $\pi R^2 - \pi r^2$

The perimeter of the sector of Circle: 2 Radius + $\dfrac{\theta}{360^\circ} \times 2\pi r$

Length of arc = $\dfrac{2\pi r\theta}{360^\circ}

To know more students can download the vedantu app.

5. How to understand chapter 12 of class 10 Maths?

Circles are round figures without any sides or edges. Their applications are taught to students in chapter 12 of class 10 Maths where students learn how to solve different questions related to the areas of circles, the length and circumference, and a whole lot of other things. Since this chapter has a whole lot of things to learn, students can use Important Questions prepared for this chapter as it offers them an examination patterned structure, which is very beneficial.

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Case Study Questions for Class 10 Maths Chapter 10 Circles

• Last modified on: 10 months ago
• Reading Time: 3 Minutes

Question 1:

A student draws two circles that touch each other externally at point K with centres A and B and radii 6 cm and 4 cm, respectively as shown in the figure.

Based on the above information, answer the following questions. (i) The value of PA = (a) 10 cm (b) 5 cm (c) 13 cm (d) Can’t be determined

(ii) The value of BQ= (a) 4 cm (b) 5 cm (c) 6 cm (d) 18 cm

(iii) The value of PK = (a) 13 cm (b) 15 cm (c) 16 cm (d) 18 cm

(iv) The value of QY = (a) 2 cm (b) 5 cm (c) 1 cm (d) 3 cm

(v) If two circles touch externally, then the number of common tangents can be drawn is (a) 1 (b) 2 (c) 3 (d) None of these

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Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables C hapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

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• Class 10 Maths
• Chapter 12: Areas Related Circles

Important Questions Class 10 Maths Chapter 12 Areas Related to Circles

Important questions for Class 10 Maths Chapter 12 Areas Related to Circles are given here based on the new pattern of CBSE. Students preparing for the board exams can practice these questions of Areas Related to Circles For Class 10 to score full marks for the questions from this chapter. We have also provided detailed solutions, along with the Important questions of Class 10 Maths Chapter 12 Areas Related to Circles. Students can refer to the solutions whenever they get stuck solving a problem. Also, in the end, some practice questions are provided for students to boost their preparation.

This chapter contains many formulas and concepts. Thus, it is important from the examination perspective since most of the questions, and objective-based questions will come in the exam from this chapter. Students can also refer to these important questions as a part of their exam preparation. NCERT Solutions are also available at BYJU’S to help the students in scoring full marks.

Students can also get access to Class 10 Maths Chapter 12 Areas Related Circles MCQs here.

• Area Perimeter Formula
• Important 4 Marks Question For Cbse Class 10 Maths
• Maths Formulas For Class 10

Important Questions & Answers For Class 10 Maths Chapter 12 Areas Related to Circles

Some of the important questions for areas related to circles for class 10 chapter 12 are given below. The questions include both short-answer type questions and long answer type questions along with HOTS questions.

Short Answer Type Questions

1. If the radius of a circle is 4.2 cm, compute its area and circumference.

Area of a circle = πr 2

So, area = π(4.2) 2 = 55.44 cm 2

Circumference of a circle = 2πr

So, circumference = 2π(4.2) = 26.4 cm

2. What is the area of a circle whose circumference is 44 cm?

From the question,

Or, r = 22/π

Now, area of circle = πr 2 = π × (22/π) 2

So, area of circle = (22×22)/π = 154 cm 2

3. Calculate the area of a sector of angle 60°. Given, the circle has a radius of 6 cm.

The angle of the sector = 60°

Using the formula,

The area of sector = (θ/360°)×π r 2

= (60°/360°) × π r 2 cm 2

Or, area of the sector = 6 × 22/7 cm 2 = 132/7 cm 2

4. A chord subtends an angle of 90°at the centre of a circle whose radius is 20 cm. Compute the area of the corresponding major segment of the circle.

Point to note:

Area of the sector = θ/360 × π × r 2

Base and height of the triangle formed will be = radius of the circle

Area of the minor segment = area of the sector – area of the triangle formed

Area of the major segment = area of the circle – area of the minor segment

Radius of circle = r = 20 cm and

Angle subtended = θ = 90°

Area of the sector = θ/360 × π × r 2 = 90/360 × 22/7 × 20 2

Or, area of the sector = 314.2 cm 2

Area of the triangle = ½ × base × height = ½ × 20 × 20 = 200 cm 2

Area of the minor segment = 314.2 – 200 = 114.2 cm 2

Area of the circle = π × r 2  = (22/7) × 20 2  = 1257.14

Area of the major segment = 1257.14 – 114.2 = 1142 .94 cm 2

So, the area of the corresponding major segment of the circle = 1142 .94 cm 2

5. A square is inscribed in a circle. Calculate the ratio of the area of the circle and the square.

As the square is inscribed in a circle, a diagonal of the square will be = the diameter of the circle.

Let “r” be the radius of the circle and “d” be the length of each diagonal of the square.

Length of the diagonal of a square = side (s) × √2

And, s × √2 = 2r

Or, s = √2r

We know, the area of the square = s 2

Thus, the area of the square = (√2r) 2 = 2r 2

Now, the area of the circle = π × r 2

∴ Area of the circle : area of the square = π × r 2 : 2r 2 = π : 2

So, the ratio of the area of the circle and the square is π : 2.

6. Find the area of the sector of a circle with a radius of 4cm and of angle 30°. Also, find the area of the corresponding major sector.

Radius = r = 4 cm, θ=30° Area of sector = [𝜃/360]×𝜋𝑟 2 = 30/360×3.14×(4) 2 = 1/12×3.14×4×4 = 1/3×3.14×4 = 12.56/3 cm 2 = 4.19 cm 2 Area of major sector = ((360 − θ)/360)×𝜋𝑟 2 = ((360 − 30))/360×3.14×(4) 2 = 330/360×3.14×4×4 = 11/12×3.14×4×4 = 46.05 cm 2

7. Calculate the perimeter of an equilateral triangle if it inscribes a circle whose area is 154 cm 2

Here, as the equilateral triangle is inscribed in a circle, the circle is an incircle.

Now, the radius of the incircle is given by,

r = Area of triangle/semi-perimeter

In the question, it is given that area of the incircle = 154 cm 2

So, π × r 2 = 154

Or, r = 7 cm

Now, assume the length of each arm of the equilateral triangle to be “x” cm

So, the semi-perimeter of the equilateral triangle = (3x/2) cm

And, the area of the equilateral triangle = (√3/4) × x 2

We know, r = Area of triangle/semi-perimeter

=> 7 = √3x/6

Or, x = 42/√3

Multiply both numerator and denominator by √3

So, x = 42√3/3 = 14√3 cm

Now, the perimeter of an equilateral triangle will be = 3x = 3 × 14√3 = 72.7 cm.

Long Answer Type Questions

Q.1: The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280. The field is to be ploughed at the rate of Rs. 0.50 per m 2 . Find the cost of ploughing the field (Take π = 22/7).

Length of the fence (in metres) = Total cost/Rate = 5280/24 = 220

So, the circumference of the field = 220 m

If r metres is the radius of the field, then 2πr = 220

2 × (22/7) × r = 220

r = (220 × 7)/ (2 × 22)

Hence, the radius of the field = 35 m

Area of the field = πr 2

= (22/7) × 35 × 35

= 22 × 5 × 35 m 2

= 3850 sq. m.

Cost of ploughing 1 m 2 of the field = Rs. 0.50

So, the total cost of ploughing the field = 3850 × Rs. 0.50 = Rs. 1925

Q.2: The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2πr = 80 π cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm

It is given that the distance covered by the car in 1 hr = 66km

Converting km into cm we get,

Distance covered by the car in 1hr = (66 × 10 5 ) cm

In 10 minutes, the distance covered will be = (66 × 10 5 × 10)/60 = 1100000 cm/s

∴ Distance covered by car = 11 × 10 5 cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels) = 11 × 105 /80 π = 4375.

Q.3: Find the area of the sector of a circle with a radius of 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14)

Let OAPB be the sector.

Area of the major sector = [(360 – θ)/ 360] × πr 2

=[(360 – 30)/360] × 3.14 × 4 × 4

= (330/360) × 3.14 × 16

= 46.05 cm 2

= 46.1 cm 2 (approx)

Q.4: Find the area of the segment AYB shown in the figure, if the radius of the circle is 21 cm and ∠ AOB = 120°. (Use π = 22/7).

Area of the segment AYB = Area of sector OAYB – Area of ∆ OAB …..(1)

Area of the sector OAYB = (120/360) × (22/7) × 21 × 21 = 462 cm 2 ……(2)

Draw OM ⊥ AB.

OA = OB (radius)

Therefore, by RHS congruence, ∆ AMO ≅ ∆ BMO.

M is the mid-point of AB and ∠ AOM = ∠ BOM = (1/2) × 120° = 60°

Let OM = x cm

In triangle OMA,

OM/OA = cos 60°

OM = 21/2 cm

AM/OA = sin 60°

AM/21 = √3/2

AM = 21√3/2 cm

AB = 2 AM = 2 (21√3/2) = 21√3 cm

Area of triangle OAB = (½) × AB × OM

= (½) × 21√3 × (21/2)

= (441/4)√3 cm 2 ………..(3)

From (1), (2) and (3),

Area of the segment AYB = [462 – (441/4)√3] cm 2

Q.5:  Find the area of the shaded design in the given figure, where ABCD is a square of 10 cm and semicircles are drawn with each side of the square as diameter. (Use π = 3.14).

Let us assign I, II, III and IV for the unshaded regions.

Give that, side of square ABCD = 10 cm

The sides of a square are also the diameters of semicircles.

The radius of semicircle = 10/2 = 5 cm

Now, area of the region I +III = Area of square ABCD – Area of two semicircles of radius 5 cm

= (10) 2 – 2 × (½) π ×(5) 2

= 100 – 3.14 × 25

= 100 – 78.5

= 21.5 cm 2

Area of the region II + Iv = 21.5 cm 2

Area of the shaded region = Area of square ABCD – Area of the region (I + II + III + IV)

= 100 – 2× 21.5

= 100 – 43

Q.6: A round table cover has six equal designs as shown in the 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 . (Use 3 = 1.7)

Total number of equal designs = 6

∠AOB = 360°/6 = 60°

The radius of the cover = 28 cm

Cost of making design = Rs. 0.35 per cm 2

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its area will be √3/4 × a 2

Here, a = OA

∴ Area of equilateral ΔAOB = √3/4 × 28 2 = 333.2 cm 2

Area of sector ACB = (60°/360°) × π r 2 cm 2

= 410.66 cm 2

Area of a single design = area of sector ACB – Area of ΔAOB

= 410.66 cm 2 – 333.2 cm 2 = 77.46 cm 2

∴ Area of 6 designs = 6 × 77.46 cm 2 = 464.76 cm 2

So, the total cost of making design = 464.76 cm 2 × Rs. 0.35 per cm 2

= Rs. 162.66

Q.7: In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region (pink and yellow regions together).

Radius of larger circle, R = 7 cm

Radius of smaller circle, r = 7/2 cm

Height of ΔBCA = OC = 7 cm

Base of ΔBCA = AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm 2

Area of larger circle = πR 2 = 22/7 × 7 2 = 154 cm 2

Area of larger semicircle = 154/2 cm2 = 77 cm 2

Area of smaller circle = πr 2 = 22/7 × 7/2 × 7/2 = 77/2 cm 2

Area of the shaded region = Area of the larger circle – Area of a triangle – Area of larger semicircle + Area of the smaller circle

Area of the shaded region = (154 – 49 – 77 + 77/2) cm 2

= 66.5 cm 2

Q.8:  In the figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (take π = 3.14)

We know that the angle in the semicircle is the right angle.

Thus, ∠ACB = 9° 

By Pythagoras theorem,

BC 2 + AC 2 = AB 2

BC 2 = AB 2 – AC 2

= (13) 2 – (12) 2

= 169 – 144

⇒ BC = 5 cm

From the given,

Diameter of circle = AB = 13 cm

Radius of semicircle = AB/2 = 13/2 cm

Area of the shaded region = Area of the semicircle – Area of right triangle ABC

= (1/2)πr 2 – (1/2) × BC × AC

= (1/2) × 3.14 × (13/2) × (13/2) – (1/2) × 5 × 12

= 66.33 – 30

= 36.33 cm 2

Practice Questions For Class 10 Maths Chapter 12 Areas Related to Circles

• A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5m, find the increase in the area of the grassy lawn in which the calf can graze.
• Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm.

• Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.
• If the difference between the circumference and the radius of a circle is 37 cm, then using π = 22/7, calculate the circumference (in cm) of the circle.
• The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
• A pipe of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle. [use π = 22/7]

Video Lessons on Circles

Introduction to circles.

Parts of a Circle

Area of a Circle

All about Circles

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CBSE 10th Standard Maths Subject Circles Case Study Questions 2021

By QB365 on 22 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) Number of tangents that can be drawn on a circle is

(iii) Number of tangents that can be drawn to a circle from a point not on it, is

(iv) Number of secants that can be drawn to a circle from a point on it is

(v) A line that touches a circle at only one point is called

(ii) The value of BQ =

(iii) The value of CQ =

(iv) Which of the following is correct?

(v) Radius of the pit is

(iii) The value of PK =

(iv) The value of QY =

(v) Which of the following is true?

(ii) On the basis of which of the following congruency criterion, \(\Delta \mathrm{OAP} \cong \Delta \mathrm{OBP} ?\)

(iii) If \(\angle\) AOB = 150°, then \(\angle\) APB =

(iv) If \(\angle\) APB = 40°, then \(\angle\) BAO =

(v) If \(\angle\) ABO = 45°, then which of the following is correct option?

(iv) PT is a tangent to a circle with centre 0 and diameter = 40 cm. If PT = 21 cm, then OP =

*****************************************

Cbse 10th standard maths subject circles case study questions 2021 answer keys.

(i) (a) (ii) (d) (iii) (b) (iv) (a) (v) (c)

Here in right angled triangle ABC, AB = 6 m and BC= 8 cm. \(\therefore\) By Pythagoras theorem  \(A C=\sqrt{(A B)^{2}+(B C)^{2}}\) \(=\sqrt{(8)^{2}+(6)^{2}}=\sqrt{100}=10 \mathrm{~m}\) Also, AP = x m. (i) (c): AR=AP = xm      ..(1) [Since, length of tangents drawn from an external point are equal] (ii) (b): BQ = BR = AB - AR = (6 - x) m         (Using (1)) (iii) (d): CQ = CP = AC - AP = (10 - x) m Also, CQ = BC - BQ = BC - BR = 8 - (6 - x) = 2 + x (iv) (b): Since, CQ = 10 - x = 2 + x \(\Rightarrow\) 8 = 2x \(\Rightarrow\) x = 4 \(\therefore\)   AR = AP = 4 m, BR = BQ = 2 m and CP = CQ = 6 m Also, OQ. \(\perp\) BQ and OR.l BR \(\therefore\) BROQ is a square. (v) (a): Radius of the circle, OR = BR = 2 cm

Here, AS = 5 cm, BT = 4 cm  [ \(\therefore\) Radii of circles] (i) (c): Since, radius at point of contact is perpendicular to tangent. \(\therefore\) By Pythagoras theorem, we have \(P A=\sqrt{P S^{2}+A S^{2}}=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{~cm}\) (ii) (b): Again by Pythagoras theorem, we have \(B Q=\sqrt{T Q^{2}+B T^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5 \mathrm{~cm}\) (iii) (d): PK = PA + AK = 13 + 5 = 18 cm (iv) (c): QY = BQ - BY = 5 - 4 = 1 cm (v) (c): PS 2 = PA 2 - AS 2 = PA 2 - AK 2 = (PA + AK)(PA - AK) = PK.PX [ \(\because\) AK = AX]

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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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