# NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles (Ex 12.2)

An essential study tool for Class 10 students is the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. Students can better grasp the kinds of questions that will be asked in the CBSE Class 10 Mathematics board exams by using these NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. Additionally, offering answers to all circle-related problems aids students in effectively preparing for the CBSE board exams.

Students can ace their board examinations by studying for them in advance with the help of the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2, which are accessible on Extramarks. Students can access these NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 and get all of their conceptual questions answered.

Finding the areas of two particular ‘sections’ of a circular region known as a sector and segment requires knowledge of the perimeter (circumference) and area of a circle, which is a concept covered in detail in NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2.

The unit of Mensuration includes the chapter titled “Areas Related to Circles,” which accounts for a total of 10 marks in the board exams. At least one question from this chapter is asked in the board exams.

A list of solutions to exercises provided by Extramarks included in the chapter, apart from the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 is provided below.

Worksheet 12.2 NCERT Solutions(14 Solved Questions)

Worksheet 12.3 NCERT Solutions (16 Solved Questions)

The topics of Sections of Circles, their measurements, and Areas of Planar Figures are covered in NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. Each question’s solution has been created by experienced Extramarks mentors in accordance with the guidelines of the CBSE syllabus (2022–23).

Along with the concepts embedded in NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2, Chapter 12 consists of other significant exercises, such as-

• Exercise 12.1 Introduction
• Exercise 12.2 Perimeter and Area of a Circle
• Exercise 12.3 Areas of Sector and Segment of a Circle
• Exercise 12.4 Areas of Combinations of Plane Figures
• Exercise 12.5 Summary

The NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 by Extramarks are helpful for examination preparation for the following reasons:

• Students can improve their understanding of ideas connected to circles by using NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. Diagrams are used to illustrate questions, which makes learning more engaging and thorough.
• The language utilised in NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 is simple and clear.
• Using the methods included in the NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2, students can learn to solve problems step by step.
• The NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 would aid pupils in tackling difficult issues at their own speed.

Additionally, students can refer to the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 for acquiring an understanding of various concepts and topics. Skilled instructors created these NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 available on Extramarks with an emphasis on providing clarity on important ideas and problem-solving techniques.

Other study tools offered by Extramarks can also be used by students to help them understand the key ideas of the chapters.

The answers to NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 provide students with an overview of the question paper format, covering a range of questions including multiple-choice questions, short and long answer-type questions, and recurring questions. The more questions students correctly answer, the more certain they get that they will succeed.

From the perspective of an examination, NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 are of paramount importance.  These solutions include an introduction to the area ofa circle, its perimeter and area, its sector and segment areas, its area as a combination of plane figures, and finally, an explanation of the chapter’s summary .

The NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 includes solutions to the fourteen questions which constitute Exercise 12.2 of the chapter titled Areas Related to Circles. Cumulatively, the chapter contains 3 exercises. The first exercise consists of five questions, the second exercise consists of fourteen, and the final exercise consists of sixteen questions based on the perimeter and area of a circle, the areas of its sector and segment, and the areas of various combinations of planar figures.

The subject-matter specialists of Extramarks have developed remarkably precise and accurate NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. Extramarks provides NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 to help students with their preparation as they will encounter CBSE standard questions. The Class 10 board exams require a great deal of work and dedication to pass with flying colours. Students can enhance their learning and preparation strategies with the use of the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2, which are available on the Extramarks website and mobile application. On Extramarks, students can now obtain the Class 10 Science NCERT Solutions as well.

The NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 based on Areas Related to Circles are a crucial study guide that all students enrolled in Class 10 must acquire. Students can better comprehend the subjects presented in the chapter with the aid of these NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. Delivering NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 based on Area Related to Circles is done primarily to aid students in getting ready for their academic and competitive examinations.

Students may ace their Class 10 board examinations by studying for them in advance with the help of the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 available on Extramarks. Students may refer to the NCERT Solutions for Class 10 Chapter 12 Exercise 12.2 based on Areas Related to Circle to get answers to their questions about the chapter’s more difficult concepts. These NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 have been created by proficient instructors of Extramarks based on the most recent CBSE Board syllabus to help students in Class 10 prepare for their examinations.

Several important topics are covered in the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 offered by Extramarks. Students will learn the following 5 concepts in this chapter:

• The circle’s circumference can be calculated using the formula 2r.
• The area of a circle can be calculated using the formula r2.
• A sector arc length is given as /360° (2r) for a circle with radius r and an angle of degree measure.
• The formula for the area of a sector of a circle with radius r and degree measure is /360° (r2).
• Area of Triangle – Area of Corresponding Sector is how one calculates the area of a circular section.

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles (Ex 12.2) Exercise 12.2

Students can learn more about the concepts of sector and segment by working through  NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. A segment is the portion of the circular region enclosed between a chord and the corresponding arc, whereas a sector is the piece of the circular region enclosed between two radii and the corresponding arc. Students must use a variety of sums to determine the area of a segment or sector in this exercise. The level of difficulty gradually rises, allowing learners to run into a variety of issues.

There are 14 questions in all, ranging in complexity from easy to difficult, in the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. Examinees may be asked to use a formula directly for some numbers, while while others may be required to demonstratetheir mathematical comprehension abilities by applying concepts to seek appropriate answers.

The solutions to the challenging questions in these NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 combine several different concepts. For instance, a question might require students to use both the properties of circles and triangle congruence to determine the area of a segment. It’s also important to keep in mind that a sector or section can be broken down into major and smaller components. Thus, when attempting to solve such questions, it is crucial to keep these ideas in mind.

Below is a set of formulas that are extremely crucial in order to solve the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2:

• The area of the angle’s sector is equal to /360° (r2). To arrive at this equation, simply treat the entire circular portion as a 360° sector. In order to arrive at the necessary formula, students could next use the unitary approach and the area of a circle. In other words, the area is r2 if the angle at its centre has a measure of 360°. As a result, the area will be determined by the method above when the angle’s degree value is.
• Angle sector length in degrees is equal to /360° (r2). Students could use the same steps as before to discover this formula. This indicates that in order to arrive at this formula, one must use both the unitary technique and the circle’s circumference, which is given by 2r.

Downloadable PDF versions of the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 are available on the Extramarks website and mobile application. For the Class 1o board exams, these NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 are quite beneficial. The majority of the questions in Class 10 board exams are taken from NCERT textbooks, which CBSE recommends. On the Extramarks website and mobile application, students can download NCERT Solutions for every chapter as well as the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2.

## Access NCERT Solutions for Class 10 Mathematics Chapter 12 – Areas related to circles

With the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2, students can review the topic of Circles, learn to determine the number of revolutions an automobile wheel will make using its dimensions, etc., as a part of the chapter titled Areas Related to Circles. Students can also practise determining the sector’s area in a given problem with circles.

Additionally, students can practise using the solutions provided in NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 questions which require them to determine the area of a semicircle or the area of a certain design. Students can use various kinds of digital learning resources catering to the curriculum of Class 10 Mathematics. These include interactive videos, sample papers, past years’ papers, practise tests, and more from Extramarks to improve their mathematical abilities for board examination  preparation.

Areas Related to the Circle Class 10 Chapter 12 is a continuation of the previously studied chapter “Circles” from the Class 9 mathematics curriculum.There are numerous questions that incorporate principles from Class 9 in NCERT Chapter 12 Mathematics for Class 10. There are a total of 3 exercises and 35 questions in Chapter 12 of the NCERT Mathematics textbook for Class 10. The right answer to each question is provided in the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2. To ensure that every student gets the most out of the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2, qualified teachers have especially created simple-to-understand NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2.

Here are a few points to note with regards to the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2:

• This chapter, which is a component of the Mensuration unit, is worth 10 marks in the Class 10 board exams.
• It is anticipated that at least one question from these NCERT Solutions for Class 12 Exercise 12.2 may be asked in the board examinations based on the pattern of the examination as analysed from past years’ papers.
• In cases where the question is directly extracted from the exercise questions within the NCERT textbook, students won’t lose any points if they use the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 for regular practice.

Students can also find individual exercise solutions in the Area Related to Circle chapter.Namely, the following:

• Area Related to Circle Exercise 12.1
• Area Related to Circle Exercise 12.2
• Area Related to Circle Exercise 12.3

Here is a list of all the other Class 10 Mathematics chapters that Extramarks provides NCERT Solutions for:

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• 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

Tips to efficiently utilise the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2:

• Students are advised to be familiar with the fundamentals of circles, squares, rectangles, triangles, etc. before beginning this chapter.
• They are encouraged to read and remember several formulas dealing with sectors and segments from NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2.To grasp the strategy for answering the questions, they can read through a few examples from the textbook and practise questions with the assistance of NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2.
• Following the discussion of the aforementioned issues, it is crucial to put efforts into action by working through the practise problems.
• To find the right response or learn how to tackle a certain question, one can use the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 while practising.

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.2

Students can access the PDF version of the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 by downloading it from the Extramarks website and mobile application. Questions and answers incorporated in NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 will help students review the exercise efficiently and earn higher grades. The NCERT Solutions to all exercises are available to students who register themselves on the Extramarks website and mobile application. If students have access to Class 10 Science NCERT Solutions, Mathematics NCERT Solutions, and other subject-specific solutions, studying academic disciplines like Science, Mathematics, and English would be much easier. Students can also access the NCERT Solutions for other classes offered by Extramarks.  A list of all the classes that Extramarks provides NCERT Solutions for has been mentioned below:

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

Below is a list of common terminologies that students must be familiar with when dealing with the NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2:

• The area of a circle bounded between any two of its radii and the arc next to them is known as the sector of a circle.
• A sector’s starting point will always be in the circle’s centre.
• Students must understand how to calculate the area of a circle’s sector.
• The measurement of the space included within the sector’s perimeter is known as its area.

## NCERT Solutions for Class 10 Mathematics Chapter 12 Exercises

The Class 10 Mathematics Chapter 12 consists of 3 exercises in total. There are just 5 questions in the first exercise (Ex 12.1). A total of 14 questions are included in the second exercise (Ex 12.2) whereas the third exercise has all 16 questions.

Exercise 12.1 Chapter 12:

All chapter 12 questions in the NCERT Solutions are given in-depth solutions. Students can get all of the NCERT Solutions to Mathematics Chapter 12 Exercise 12.1 in one convenient location on the Extramarks learning portal. These solutions are created by subject experts in accordance with NCERT guidelines. Chapter 12 of the Mathematics textbook for grade 10: Area of Circles Exercise 12.1 has questions and solutions that enable students to review the entire curriculum and get better grades. Students must practise the NCERT Solution in order to do well on the CBSE Class 10 examination. As a result, students will have no trouble comprehending the questions and how to approach them.

Exercise 12.2 Chapter 12:

One of the most important years in every student’s academic career is Class 10. Mathematics, Science, Social Science, and other subjects require special focus. Students who want to succeed in the Class 10 Mathematics board exams must thoroughly study the NCERT book. The NCERT Solutions for Exercise 12.2 on Areas Related to Circles can be availed through the Extramarks e-learning portal. To perform at their highest level during the class 10 board exams, students must complete all of the questions in chapter Exercise 12.2 for Class 10.

Students can easily achieve good scores in examinations by using the NCERT Solutions for Areas Related to Circles Exercise 12.2. Students can discover answers to the questions they are working on by downloading the NCERT Solutions for Areas Related to Circles Exercise 12.2 PDF.

Exercise 12.3 Chapter 12:

Students can use the NCERT Exercise 12.3 Class 10 Mathematics Solutions to help them with the 16 exercise questions from the chapter Areas Related to Circles. Students should practise using the NCERT Solutions for Class 10 Mathematics Chapter 12 exercise 12.3 in order to gain a firm grasp of all the conceptual concepts and perform well in the class 10 board exams.

Extramarks’ NCERT Class 10 Mathematics solutions are meant to deliver maximum conceptual clarity to students while boosting their problem-solving skills. The Exercise 12.3 Class 10 Mathematics NCERT solutions are additionally accessible via the Extramarks website and mobile application in PDF format.

Concepts like the area and perimeter of polygons, circles, segments, and sectors are covered in Class 10 Maths Chapter 12 Exercise 12.2. People often see circular shapes in daily life, including bangles, brooches, bicycle wheels, etc. They need to be able to precisely determine the area and perimeter of these values in order to use them effectively. Remarkably efficient and experienced mentors of Mathematics have created the NCERT Solutions for Area Related to Circle Class 10 Exercise 12.2 based on Class 10 Maths Ex 12.2 to provide students with a comprehensive learning experience. Extramarks has provided solutions to Ex. 12.2 of the NCERT class 10 math textbook for a more thorough understanding.

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

Ans

$\begin{array}{l}\text{It is given that radius of the circle is 6 cm and angle of the}\\ \text{sector is 60°.}\\ \text{Area of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \therefore \text{Area of a sector of angle}60\mathrm{°}=\frac{60\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{6}^{2}=\frac{132}{7}{\text{cm}}^{2}\end{array}$

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

Ans

$\begin{array}{l}\text{Let the radius of the circle be}\mathrm{r}.\\ \text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Circumference of the circle}=\text{22 cm}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{\pi r}=\text{22 cm}\\ \text{or}\mathrm{r}=\frac{\text{22}}{2}×\frac{7}{22}=\frac{7}{2}\\ \text{Quadrant of a circle subtends an angle of 90° at the centre.}\\ \therefore \text{Area of a quadrant of the given circle}=\frac{90\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(\frac{7}{2}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{77}{8}{\text{cm}}^{2}\end{array}$

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

Ans

$The length of minute hand,r=14 cm In 60 minutes, a minute hand subtends an angle of 360° at the centre. So, in 5 minutes a minute hand subtends an angle of 360°×5 60° =30° ∴Area swept by the minute hand in 5 minutes = 30° 360° × 22 7 × ( 14 ) 2 = 154 3 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@420E@$

Q.4 A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)

Ans

$\begin{array}{l}\text{Let AB be the chord of the circle subtending 90° angle at}\\ \text{centre O of the circle.}\\ \left(\left(\mathrm{i}\right)\text{) Area of minor sector OACB}=\frac{90°}{360°}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}×3.14×10×10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=78.5{\text{cm}}^{2}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{1}{2}×\text{OA}×\text{OB}=\frac{1}{2}×10×10=50{\text{cm}}^{2}\\ \text{Area of minor segment ACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Area of minor sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=78.5{\text{cm}}^{2}-50{\text{cm}}^{2}=28.5{\text{cm}}^{2}\\ \left(\left(\mathrm{ii}\right)\right)\text{Area of major sector OADB}=\frac{360°-90°}{360°}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{270°}{360°}×3.14×10×10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=235.5{\text{cm}}^{2}\end{array}$

Q.5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

1. the length of the arc
2. area of the sector formed by the arc
3. area of the segment formed by the corresponding chord

Ans

$\begin{array}{l}\text{Radius of the given circle}=\mathrm{r}=21\text{cm}\\ \text{Angle subtended by the given arc}=60\mathrm{°}\\ \text{Length of an arc of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}×2\mathrm{\pi r}\\ \therefore \text{Length of arc ACB}=\frac{60\mathrm{°}}{360\mathrm{°}}×2×\frac{22}{7}×21\text{\hspace{0.17em}}=22\text{cm}\\ \text{Area of sector OACB}=\frac{60\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×21×21=231{\text{cm}}^{2}\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\angle \text{OBA [As OA}=\text{OB]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\angle \text{OAB}+60\mathrm{°}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\frac{180\mathrm{°}-60\mathrm{°}}{2}=\frac{120\mathrm{°}}{2}=60\mathrm{°}\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}×{\left(\text{side}\right)}^{2}=\frac{\sqrt{3}}{4}×{\left(\text{21}\right)}^{2}=\frac{441\sqrt{3}}{4}{\text{cm}}^{2}\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=\left(231-\frac{441\sqrt{3}}{4}\right){\text{cm}}^{2}\end{array}$

Q.6

$A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. ( Use π=3.14 and 3 =1.73 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeqabeqadiWaceGabeqabeWaaqaafeaakqaabeqaaiaahgeacaqGGaGaaC4yaiaahIgacaWHVbGaaCOCaiaahsgacaqGGaGaaC4BaiaahAgacaqGGaGaaCyyaiaabccacaWHJbGaaCyAaiaahkhacaWHJbGaaCiBaiaahwgacaqGGaGaaC4BaiaahAgacaqGGaGaaCOCaiaahggacaWHKbGaaCyAaiaahwhacaWHZbGaaeiiaiaahgdacaWH1aGaaeiiaiaahogacaWHTbGaaeiiaiaahohacaWH1bGaaCOyaiaahshacaWHLbGaaCOBaiaahsgacaWHZbGaaeiiaiaahggacaWHUbGaaeiiaiaahggacaWHUbGaaC4zaiaahYgacaWHLbGaaeiiaiaah+gacaWHMbaabaGaaCOnaiaahcdacqGHWcaScaqGGaGaaCyyaiaahshacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahogacaWHLbGaaCOBaiaahshacaWHYbGaaCyzaiaac6cacaqGGaGaaCOraiaahMgacaWHUbGaaCizaiaabccacaWH0bGaaCiAaiaahwgacaqGGaGaaCyyaiaahkhacaWHLbGaaCyyaiaahohacaqGGaGaaC4BaiaahAgacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahogacaWHVbGaaCOCaiaahkhacaWHLbGaaC4CaiaahchacaWHVbGaaCOBaiaahsgacaWHPbGaaCOBaiaahEgaaeaacaWHTbGaaCyAaiaah6gacaWHVbGaaCOCaiaabccacaWHHbGaaCOBaiaahsgacaqGGaGaaCyBaiaahggacaWHQbGaaC4BaiaahkhacaqGGaGaaC4CaiaahwgacaWHNbGaaCyBaiaahwgacaWHUbGaaCiDaiaahohacaqGGaGaaC4BaiaahAgacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahogacaWHPbGaaCOCaiaahogacaWHSbGaaCyzaiaac6cacaqGGaaabaWaaeWaaeaacaWHvbGaaC4CaiaahwgacaqGGaGaeqiWdaNaeyypa0JaaC4maiaac6cacaWHXaGaaCinaiaabccacaWHHbGaaCOBaiaahsgacaqGGaWaaOaaaeaacaWHZaaaleqaaOGaeyypa0JaaCymaiaac6cacaWH3aGaaC4maaGaayjkaiaawMcaaaaaaa@D1CC@$

Ans

$\begin{array}{l}\text{Radius of the given circle}=\mathrm{r}=15\text{cm}\\ \text{Angle subtended by the given chord}=60\mathrm{°}\\ \text{Area of sector OACB}=\frac{60\mathrm{°}}{360\mathrm{°}}×3.14×15×15=117.75{\text{cm}}^{2}\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\angle \text{OBA [As OA}=\text{OB]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\angle \text{OAB}+60\mathrm{°}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\frac{180\mathrm{°}-60\mathrm{°}}{2}=\frac{120\mathrm{°}}{2}=60\mathrm{°}\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}×{\left(\text{side}\right)}^{2}=\frac{\sqrt{3}}{4}×{\left(\text{15}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{225×1.73}{4}{\text{cm}}^{2}=97.3125{\text{cm}}^{2}\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=117.75-97.3125=20.4375{\text{cm}}^{2}\\ \text{Area of major segment ADB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Area of circle}-\text{Area of segment ACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }{\left(15\right)}^{2}-20.4375\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3.14×225-20.4375=706.5-20.4375\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=686.0625{\text{cm}}^{2}\end{array}$

Q.7

$\begin{array}{l}\text{A chord of a circle of radius 12 cm subtends an}\\ \text{angle of 120° at the centre. Find the area of the}\\ \text{corresponding segment of the circle.}\\ \left(\mathbf{Use}\text{}\mathrm{\pi }=\mathbf{3}.\mathbf{14}\text{}\mathbf{and}\text{}\sqrt{3}=\mathbf{1}.\mathbf{73}\right)\end{array}$

Ans

$\begin{array}{l}\text{Let AB is a chord of a circle of radius 12 cm which subtends an}\\ \text{angle of 120° at the centre}\mathrm{O}\text{.}\\ \text{Radius of the given circle}=\mathrm{r}=\text{OA}=\text{OB}=12\text{cm}\\ \text{Angle subtended by the given chord}=120\mathrm{°}\\ \text{Area of sector OACB}=\frac{120\mathrm{°}}{360\mathrm{°}}×3.14×12×12=150.72{\text{cm}}^{2}\\ \text{Let us draw a perpendicular OD on chord AB. It bisects the}\\ \text{chord AB.}\\ \therefore \text{AD}=\text{DB}\\ \text{In}\mathrm{\Delta }\text{ODA,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{OD}}{\mathrm{OA}}=\mathrm{cos}60\mathrm{°}\\ \mathrm{or}\text{}\frac{\mathrm{OD}}{12}=\frac{1}{2}\\ \mathrm{or}\text{OD}=6\text{cm}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin60°}=\frac{\mathrm{AD}}{\mathrm{OA}}=\frac{\mathrm{AD}}{12}\\ \mathrm{or}\text{}\frac{\sqrt{3}}{2}=\frac{\mathrm{AD}}{12}\\ \mathrm{or}\text{}\mathrm{AD}=6\sqrt{3}\text{cm}\\ \mathrm{Now},\\ \mathrm{AB}=2\mathrm{AD}=2×6\sqrt{3}=12\sqrt{3}\text{cm}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{1}{2}×\mathrm{AB}×\mathrm{OD}=\frac{1}{2}×12\sqrt{3}×6=62.28{\text{cm}}^{2}\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=150.72-\text{}62.28=88.44{\text{cm}}^{2}\end{array}$

Q.8

$\begin{array}{l}\text{A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m}\\ \text{long rope}\left(\text{see the following figure}\right)\text{. Find}\\ \text{}\left(\text{i}\right)\text{the area of that part of the field in which the horse can graze.}\\ \text{}\left(\text{ii}\right)\text{the increase in the grazing area if the rope were 10 m long instead of 5 m.}\left(\mathbf{Use}\text{}\mathrm{\pi }=\mathbf{3}.\mathbf{14}\right)\end{array}$

Ans

$\begin{array}{l}\left(\left(\mathrm{i}\right)\right)\text{The horse can graze a sector of 90° in a circle of radius 5 m.}\\ \text{Area that can be grazed by horse}\\ \text{}=\text{Area of sector of 90° in a circle of radius 5 m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{90°}{360°}×3.14×25=19.625{\text{m}}^{2}\\ \left(\left(\mathrm{ii}\right)\right)\text{When length of rope is 10 m,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}area that can be grazed by horse}\\ \text{}=\frac{90°}{360°}×3.14×100=78.5{\text{m}}^{2}\\ \mathrm{Increase}\text{}\mathrm{in}\text{grazing area}=\left(78.5-19.625\right){\text{m}}^{2}=58.875{\text{m}}^{2}\end{array}$

Q.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 the following figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Ans

$\begin{array}{l}\text{Total length of wire required will be the length of 5 diameters}\\ \text{and the circumference of the brooch.}\\ \text{Radius of circle}=\frac{35}{2}\text{mm}\\ \text{Circumference of brooch}=2\mathrm{\pi r}=2×\frac{22}{7}×\frac{35}{2}=110\text{mm}\\ \text{Length of wire required}=110+5×35=110+175=285\text{mm}\\ \text{From the given figure, we observe that each of 10 sectors of the}\\ \text{circle subtends 36° at the centre of the circle.}\\ \text{Therefore, area of each sector}=\frac{36\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{10}×\frac{22}{7}×{\left(\frac{35}{2}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{385}{4}{\text{mm}}^{2}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{There}\text{are 8 ribs in the given umbrella. The arc between two}\\ \text{consecutive ribs subtends}\frac{360\mathrm{°}}{8}=45\mathrm{°}\text{at the centre of the}\\ \text{assumed flat circle.}\end{array}$

$\begin{array}{l}\text{Area between two consecutive ribs of circle}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{45\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{8}×\frac{22}{7}×{\left(45\right)}^{2}=\frac{22275}{28}{\text{cm}}^{2}\end{array}$

Q.11 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Ans

$\begin{array}{l}\mathrm{It}\text{is obvious that each blade of wiper will sweep an area of}\\ \text{a sector of 115° in a circle of 25 cm radius.}\\ \\ \text{Area of a sector of 115° in a circle of 25 cm radius}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{115\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{115\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(25\right)}^{2}=\frac{158125}{252}{\text{cm}}^{2}\\ \therefore \text{Area swept by two blades}=2×\frac{158125}{252}{\text{cm}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{158125}{126}{\text{cm}}^{2}\end{array}$

Q.12 To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Ans

$\begin{array}{l}\mathrm{It}\text{is obvious that the lighthouse spreads light across a sector}\\ \text{of 80° in a circle of 16.5 km radius.}\\ \therefore \text{Required area}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{80\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{80\mathrm{°}}{360\mathrm{°}}×3.14×{\left(16.5\right)}^{2}=189.97{\text{km}}^{2}\end{array}$

Q.13

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

Ans

$\begin{array}{l}\mathrm{It}\text{is obvious from the above figure that the designs are}\\ \text{segments of the circle.}\\ \text{Let us consider the segment APB. Chord AB is a side of}\\ \text{the hexagon. Each chord subtends}\frac{360°}{6}=60°\text{at the centre}\\ \text{of the circle.}\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \angle \text{OAB}=\angle \text{OBA [As OA}=\mathrm{OB}\right]\\ \angle \text{AOB}=60°\\ \mathrm{Also},\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180°\\ \mathrm{or}\text{2}\angle \text{OAB}=180°-60°=120°\\ \mathrm{or}\text{}\angle \text{OAB}=60°\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}×{\left(\text{side}\right)}^{2}=\frac{\sqrt{3}}{4}×{\left(\text{28}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=333.2{\text{cm}}^{2}\\ \therefore \text{Area of sector OAPB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{60°}{360°}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}×\frac{22}{7}×{\left(28\right)}^{2}=\frac{1232}{3}{\text{cm}}^{2}\\ \text{Area of segment APB}=\text{Area of sector OAPB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{1232}{3}-333.2\right){\text{cm}}^{2}\\ \therefore \text{Area of designs}=6×\left(\frac{1232}{3}-333.2\right){\text{cm}}^{2}=464.8{\text{cm}}^{2}\\ \text{Cost of making}464.8{\text{cm}}^{2}\text{design}=464.8×0.35=₹162.68\end{array}$

Q.14

$\begin{array}{l}\text{Tick the correct answer in the following:}\\ \text{Area of a sector of angle p}\left(\text{in degrees}\right)\text{of a circle}\\ \text{with radius R is}\\ \text{\hspace{0.17em}\hspace{0.17em}(A)\hspace{0.17em}\hspace{0.17em}}\frac{\text{P}}{180}×2\mathrm{\pi }\text{R \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(B)\hspace{0.17em}\hspace{0.17em}}\frac{\text{P}}{180}×\mathrm{\pi }{\text{R}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}(C)}\frac{\text{P}}{360}×2\mathrm{\pi }\text{R \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(D)}\frac{\text{P}}{720}×2\mathrm{\pi }{\text{R}}^{2}\end{array}$

Ans

$\begin{array}{l}\text{We know that area of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}×{\mathrm{\pi R}}^{2}\\ \therefore \text{Area of a sector of angle}\mathrm{P}=\frac{\mathrm{P}}{360\mathrm{°}}×{\mathrm{\pi R}}^{2}=\frac{\mathrm{P}}{360\mathrm{°}}×\frac{2}{2}{\mathrm{\pi R}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{P}}{720\mathrm{°}}×2{\mathrm{\pi R}}^{2}\\ \mathrm{Hence},\text{Option}\left(\text{D}\right)\text{is the correct answer.}\end{array}$