NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.3) Exercise 11.3

Choosing between numerous books and notes can be challenging, especially while studying for examinations. It is essential to choose high-quality books to ensure that all exam-related topics are covered and presented in an understandable manner. Most teachers agree that NCERT Books are among the best books for CBSE -affiliated institutions. For a number of exams, like the IAS, UPSC, and others, NCERT texts are also crucial. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other resources are suggested for students to use in order to comprehend these books fully.

The NCERT books are significant for the following reasons:

  • Simply Written

The NCERT books are written in simple words so that everyone may understand them. Being used by so many students as exam prep materials, these publications are easy to understand. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other resources available on the Extramarks website can be used by students to comprehend and prepare the texts in NCERT books.

  • Detailed Explanations of the Subjects

The NCERT textbooks’ basics and fundamentals sections cover every subject on the syllabus. It also goes into considerable length on a number of key topics. Students will therefore be totally ready for their exams once they have thoroughly covered all of the themes from the NCERT textbooks. Practicing with the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other tools available on the Extramarks website to ensure that they have a firm understanding of the NCERT textbook material.Students are encouraged to read the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other helpful resources on the Extramarks website for a thorough understanding and clarification of their doubts.Exams require students to retain all of the material covered.

  • Authored by Professionals

Experts in their respective domains write the NCERT Books. Subject-matter experts have performed extensive research on each issue before submitting it. The specialists at Extramarks produce resources like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 that make the topics from the NCERT textbooks simpler to comprehend.

  • Ease of Access.

The NCERT textbooks are available for students to read and download by visiting the official NCERT website, where they can select any book they want or need to read  online. After finishing the NCERT books, students can use the Extramarks website to access the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other resources.Can be Used as Preparation Material for Various Exams

NCERT books are recommended by the CBSE for their exams. NCERT books are also available for students attempting the UPSC and IAS Exams. If NCERT Books are well read, many exams can be easily cleared. With the aid of tools like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other materials, students may thoroughly study with the help of NCERT books.

  • Extremely Important During Exam Preparation

One of the main benefits of reading NCERT books is that the bulk of exam questions are either directly taken from the books or are remarkably similar to ones that are already therein the book. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other similar materials can assist students in becoming thorough readers of the NCERT books.

The National Council of Educational Research and Training (NCERT) has been given the responsibility of creating and providing elementary and secondary school students’ textbooks. The best thing that can be said while discussing the significance of the NCERT Books for CBSE board exam preparation is that they put a strong emphasis on the basics to aid students in understanding the essential ideas. Tools such as the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 can also assist students in achieving that. NCERT texts are comprehensive and successful, and CBSE would scarcely expect students to read anything else. Students should use the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other similar resources to further benefit from the NCERT books.

The Department of Education in Science and Mathematics (DESM) and NCERT created some excellent problems called “Exemplar Problems” in Science and Mathematics to improve student’s learning abilities and test their comprehension, analytical thinking, and problem-solving abilities. These issues are crucial for exams, but they also aim to enhance classroom instruction. By completing the exercises in the books for self-evaluation and learning the fundamental strategies for resolving difficult problems, students can benefit themselves. Students can efficiently prepare for the exams and class tests with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3.

To give students relevant and accurate information, specialists conduct an in-depth study on every topic before compiling the NCERT books. These publications are made to provide every student, regardless of intellectual level, with relevant information and expertise in straightforward language. Each subject is covered in great detail in these books. Thus, if the NCERT books are carefully read, they can dispel all of the students’ questions and provide them with a thorough comprehension of difficult themes and disciplines. The practice for the same can be done with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and more.

The Indian government established NCERT in 1961. The organisation was created through the merger of at least seven government-run educational institutions. To harmonise India’s  educational system, NCERT was established. The primary goal was to produce NCERT books of the highest calibre that could be used nationwide to create a uniform educational system.

The NCERT books are used in CBSE-affiliated schools in India. The purpose of using these books is to create uniform books that students across India can read.The goal was to solve the issue of students reading various books and adhering to various curricula while enrolled in the same standard. The best way to make the most of the NCERT books is to use the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other similar tools available on the Extramarks website.

The following were NCERT’s primary goals when it was founded:

  1. a) Conducting research in the field of education and putting innovative concepts into practise that could assist students receive a better education.
  2. b) Create the National Curriculum Framework, which will comprise the most recent course materials for all grade levels, NCERT textbooks for all grade levels, a variety of teaching aids and tools, films and videos for students, and any other pertinent study materials.
  3. c) Giving teachers instruction in the newest teaching techniques that will assist them educating students.
  4. d) To collaborate with the CBSE and other state boards to standardise the curriculum and raise academic standards.

Since then, NCERT has created NCERT books for all topics and grade levels in accordance with the published syllabi. On the basis of suggestions from the government, teachers, and students, the texts are occasionally amended. NCERT books are essential, and students should make the most of them by using NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other similar resources.

Students who are preparing for CBSE school examinations must go through the NCERT textbooks, as they are the most crucial ones. Students are responsible for making sure that all of the content in the books is fully understood. The topics in the books have been thoughtfully and beautifully articulated in a way that is self-explanatory and simple to learn. Students can further take help from the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 if they face issues while solving questions.

These books also include exercises and questions at the beginning and end of each chapter. Students must solve all the questions from the NCERT Textbooks. This can be done with the help of NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3.

Students must complete learning tools such as the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3 to ensure that they understand the concepts.After working through NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and example solutions, students are better able to understand the concepts in the chapters.They benefit from this by receiving high results on their exams. Students must go through the NCERT the Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other tools to prepare well for the exams.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.3) Exercise 11.3

Students’ test preparation can be made incredibly simple with the accessibility of materials like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. Students may easily understand a variety of difficult questions by using resources like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3, and they can effectively prepare all the questions in the exercise with the help of NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 .

There are many resources available for this specific chapter, and the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 is one of them. Students can find a range of materials for Chapter 11 of Class 7 Mathematics on the Extramarks website, including the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.

Students should make use of resources provided by Extramarks experts to prevent situations where they can lose interest in a subject. The NCERT Solutions For Class 7 Maths Chapter 11 Exercise 11.3 is just one of the several resources that are available to students with ease on Extramarks’ website. Students can successfully finish questions from Class 7 Maths Ch 11 Ex 11.3 with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. Students can find various supplementary materials along with the NCERT Class 7 Maths Chapter 11 Exercise 11.3 that are easily accessible on the Extramarks website. Students can practise the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 for Maths Class 7 Chapter 11 Exercise 11.3 to help them fully understand each question.

Access Ncert Solution for Class 7 Chapter 11 – Perimeter and Area 

Students’ test preparation can be made very simple by the availability of resources like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. Tools like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 can help students grasp all the questions with ease and prepare for all the questions from the Class 7th Maths Chapter 11 Exercise 11.3 effectively.

One of the various resources for the Maths Classes 7 Chapter 11 Exercise 11.3 is the Class 7 Maths Chapter 11 Exercise 11.3 Solutions. On the Extramarks website, students can find a variety of resources, including the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3.

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 are created by qualified mathematicians. Students can find materials for lower elementary and middle school that were produced using the same techniques for the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 on the Extramarks website. The experts suggest using resources like NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 to ensure that students fully comprehend all that is taught in class and what is expected of them to write in their exams.Students are unable to comprehend the curricula for middle and senior secondary schools if they lack a good understanding of  what is taught earlier in elementary school. Utilising tools like NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and others found on the Extramarks website increases the likelihood that students will understand the material. With the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 students may practise the questions and learn the right procedures for drawing conclusions. The step-by-step solutions in the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 are created with the correct approach to completing the questions in mind as well as the NCERT standards.

The Class 7th Maths Exercise 11.3 is one of the sub topics of Geometry. Geometry is a discipline of Mathematics that largely deals with the shapes and sizes of objects, their relative positions, and the qualities of space (from the Ancient Greek: geo- “earth,” -metron “measuring”). The Greek mathematician Euclid, commonly referred to as the “Father of Geometry,” used numerous postulates and theorems. Students are advised to use tools such as the NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.3 for Class 7 Math Chapter 11.3 to better understand the geometry subtopics.

The domain of Mathematics known as Geometry connects the concepts relating to areas, Volumes, Patterns, Angles, and Distances. Geometry is the umbrella term for all concepts that are visually and spatially related. There are three different types of geometry:

  • Euclidean
  • Hyperbolic
  • Elliptical

Elliptical Geometry –

To comprehend the basics of Geometry, students study Euclidean geometry. Axioms and theorems are the foundation of Euclidean geometry, which is the study of solid and plane figures. Points and Lines, Euclid’s Axioms and Postulates, the Geometrical Proof, and Euclid’s Fifth Postulate are some of the basic ideas of Euclidean geometry. The five Euclidean Geometry postulates are used to define geometrical figures.

  • Any point to any other point is connected by a section of a straight line.
  • The length of a straight line is infinite in both directions.
  • Any specified location can serve as the circle’s centre and any length can serve as the radius.
  • Congruent angles are all just angles.
  • Any two straight lines that are equally distant from one another at two places are eternally parallel.

This may seem complex to the students but the more they use tools similar to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 that are targeted towards topics they need to understand, the better their understanding of that topic becomes.No-Euclidean Geometry-

The two non-Euclidean geometries are Spherical Geometry and Hyperbolic Geometry. In contrast to Euclidean Geometry, non-Euclidean Geometry has different hypotheses about the Nature of Parallel Lines and Angles in flat space.

  • The study of Sphere-Based Plane Geometry is known as Spherical Geometry. The shortest distance between two places along a line is how it is defined. The great circle is the name given to this Arc-Shaped Line on a Sphere. The triangle’s total angles are larger than 180 degrees.
  • A Curved Surface is referred to as Hyperbolic Geometry. Topology is where this Geometry is used. The planar triangle has a total of the angles that is less than 180 degrees, depending on the interior curvature of the curved surface.

These are themes that may not seem necessary or useful but are closely related to real life, and hence students are expected to understand and practise them. They can do so with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 .

Plane Geometry –

The study of Geometry in a plane is a component of Euclidean Geometry. The plane is a two-dimensional surface that can go on forever in both directions. Every aspect of Geometry and graph theory uses Planes. Points, Lines, and Angles are equivalent to the fundamental elements of Planes in Geometry. The fundamental geometric unit with no dimensions is the point. The Collinear Points are those that are on the same line. A line is a single-dimensional object that denotes a collection of points extending in two opposing directions; it is defined as the place where two planes join. There are no ends to a Line. A Line, Line Segment, and ray may all be distinguished easily. Lines can be Perpendicular or Parallel. Lines may or may not cross.

This is the type of Geometry that is taught initially to lower classes and Extramarks has tools that are created for those particular classes as well, along with the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3.

Solid Geometry-

Geometric solids have a three-dimensional character. Length, width, and height are the three dimensions that are taken into account. Solid shapes come in many varieties, including Cylinders, Cubes, Spheres, Cones, Cuboids, Prisms, Pyramids, and so on. These shapes take up space. They have vertices, faces, and edges that define them. In Euclidean space, the Polyhedrons and the five platonic solids have intriguing characteristics.

This is the type that’s very common for middle school students. They can practise this theme with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other similar tools.

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 can be crucial in helping students prepare for exams effectively, especially in a subject  like Mathematics where marks are awarded for each individual stepThe Extramarks website makes it simpler to locate the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. These solutions are simple to understand as well. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 are trustworthy resources as they are developed by Extramarks’ Mathematics specialists. These specialists are aware of the needs of different students, Mathematics experts make an effort to produce materials like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 that are simple for everyone to understand and profit from.

Students can always access the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other materials of a similar nature on the Extramarks website, regardless of their location or the time. Students can set up their planning conveniently because of this.

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 are designed to assist students in working through and practising Chapter 11 questions. Every question in these exercises has detailed solutions that include step-by-step directions for students. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and all the other NCERT solutions available on the Extramarks website can enhance students’ practise sessions and give them more confidence when preparing for exams.

The greatest method to comprehend any chapter of Mathematics is to practise as many problems as possible. Students can easily complete these solutions by going through the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. Students’ chances of mastering subjects that are typically termed as complex and difficult are increased as they practise using resources like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 are advised to be thoroughly read through and used frequently while practising for exams. Use of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 should not be delayed until after exams. These learning tools should be used for both homework and other assignments, such as tests in class.

The ease of access to the internet has made it much simpler than before for students to consult other sorts of study materials in addition to their course textbooks. The value of using reliable information sources and training tools cannot be overstated. The ability for students to practise at their convenience is greatly facilitated by materials like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 are readily available on the Extramarks website, students may effortlessly practise and feel assured about their test preparations.

The Math Class 7th Exercise 11.3 is about Perimeter and area. The whole length of any closed shape boundary is known as its Perimeter. Students might have a sizable square-shaped farm, for instance. Students now decide to fence their farm to protect it from stray animals. Finding the total length of the farm’s boundary is as simple as multiplying the length of one side of the farm by 4. There are a lot of situations like this when students can be applying the Perimeter-Finding Notion without even realising it. The use of NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 can help students, better understand this concept. These are a few basics that are part of the topic taught under Chapter 11 of Class 7 Mathematics. These can all be better understood with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 .

The whole distance encircling a shape is referred to as its Perimeter. It is the length of any two-dimensional geometric shapes boundary or outline. Depending on the size, the perimeter of several figures can be the same. Consider a triangle built of an L-length wire, for instance. If all the sides are the same length, the same wire can be used to create a square. Students are advised to make use of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 so that they do not get overwhelmed by this sub-topic of Geometry and get good practise with it.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3 

All vocations in the world depend on Mathematics. The application of Mathematics and numbers to many facets of life is crucial. It is used to advance and explain the theories of other branches of Science, particularly those in Physics, Chemistry, Astronomy, Geography, etc. It allows intellectuals to conduct several experiments to test their theories. Extramarks provides tools like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other similar tools can help students master important subjects like Mathematics. Tools such as the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 available on the Extramarks website can help students master Mathematics and make a career in various fields.

Everyone owns a cell phone, which requires a basic understanding of abilities and numbers. One must be familiar with the functions of the phone’s numbers. With today’s technology, using a cell phone for anything from talking and chatting to Internet browsing is possible.

Mathematics also aids in the learning of data entry in many fields that only depend on Mathematics, including Engineering, Physics, Computer Programming, Accounting, and Banking. An essential tool for computing is Mathematics.

The core of computer operation is binary Mathematics. The most fundamental form of Mathematics utilised in Computer Science to represent each number in the machine is a binary number. Students can make the most of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other tools available on the Extramarks website to better understand these concepts that have their basis in Mathematics. Algebra, Statistics, and Calculus I, and Calculus II are some of the mathematical procedures frequently utilised in computer programming. The use of tools created by the experts at Extramarks that are targeted towards Mathematics, for example the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 can help students understand the real life uses of Mathematics

Technical innovation: Mathematics is crucial to the creation of contemporary technological tools like the Internet, which is a necessity of daily life and makes it easier for people to communicate, find information, and navigate.

Mathematics is frequently employed in animation design. When working with moving and changing objects, this enables the animator to extrapolate features of geometric figures and discover the unknown from a straightforward set of equations. The animation designer demonstrates how the designs are animated, transformed, and zoomed in and out using Linear Algebra. Students interested in these fields should have a strong foundation in mathematics, which they can achieve by using the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other similar tools available on the Extramarks website.

Mathematics is a key element in all engineering disciplines and is extensively employed in Architecture. Mathematics is used by architects to determine the square footage of rooms and structures as well as the measurements of the ground area and the space needed for additional areas like parking, plumbing, and other areas. A lot of students go into this field after completing their schooling. Students are advised to develop a healthy relationship with Mathematics in order to have a successful career in this field. They can do so by using resources such as the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and many other tools available on the Extramarks website.

An essential component of the sports world is Mathematics. It significantly affects how effectively sports are played. A person’s cognitive and decision-making abilities are enhanced by Mathematics. These abilities are crucial for a sportsperson because they enable him to choose the best course of action for his squad. The player can choose the path and angle the ball will take to reach the goal with the aid of Engineering and Trigonometry. A person who struggles with Mathematics may find it difficult to estimate situations accurately and make wise choices.

This is a field that many students find very closely related to Mathematics. Students should try to look at such real-life implementations of Mathematics in order to develop a liking for the subject and not fear it. They can improve their performance in the subject by using the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other tools created by Extramarks’ mathematics experts for students.

When painting, Spatial Calculations are crucial. Calculating how much paint is required to complete a task, such as painting a wall, is necessary. In 3D art, mathematical calculations are used to decide how to create flat objects so that they appear three-dimensional from a particular vantage point. When a particular angle is taken when looking at 2D flat art, the proportions are warped to create the appearance of a 3D item. This may come as a shock to students. This is how important Mathematics is. It is even a part of fields like Art and Painting which many might conventionally consider the opposite of Mathematics.

There are even more parts of real life that are closely associated with mathematics. Understanding mathematics is always an added skill that helps in leading a better life. Students are advised to try to understand the subject at a young age.Most students get overwhelmed by the subject because of their lack of background knowledge. This is the reason why many students dislike or fear the subject of Mathematics.

Such problems can and should be handled at an early age. The students can do so with the help of the expert-made resources available on the Extramarks website. These are available for elementary classes as well. The NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3 is just one of the many educational resources for Mathematics available on the Extramarks website.

After taking help from the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3, students can refer to the FAQ section of the website for any further doubts and questions.

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 has step-by-step solutions to the questions that are asked in the 11.3 exercise of Chapter 11 in the Class 7 Mathematics book. Students can find extra questions and their solutions on the Extramarks website easily. However, the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3 only address the questions found in NCERT textbooks.

Q.1

Find the circumference of the circles with the following radius: Take π=227a 14 cm b 28 mm c 21 cm

Ans

(a) Circumference of a circle is =2πr =2× 22 7 ×14 =2×22×2 =88cm (b) Circumference of a circle is =2πr =2× 22 7 ×28 =2×22×4 =176mm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0482@

(c) Circumference of a circle is =2πr =2× 22 7 ×21 =2×22×3 =132cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A3AF@

Q.2

Find the area of the following circles, given that:a radius = 14 mm Take π=227b diameter = 49 mc radius = 5 cm

Ans

(a) Area of a circle=π r 2 = 22 7 ( 14 ) 2 =616 mm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7CF8@

(b) Radius= Diameter 2 = 49 2 =24.5mm Area of a circle=π r 2 = 22 7 ( 24.5 ) 2 =1886.5 m 2 (c) Area of a circle=π r 2 = 22 7 ( 5 ) 2 = 550 7 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DCD5@

Q.3 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =22/7)

Ans

Since, circumference=2πr So, we get MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbofatjabbMgaPjabb6gaUjabbogaJjabbwgaLjabbYcaSiabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbwha1jabb2gaTjabbAgaMjabbwgaLjabbkhaYjabbwgaLjabb6gaUjabbogaJjabbwgaLjabg2da9iabbkdaYiabec8aWjabbkhaYbqaaiabbofatjabb+gaVjabbYcaSiabbccaGiabbEha3jabbwgaLjabbccaGiabbEgaNjabbwgaLjabbsha0baaaa@6CE5@

154=2× 22 7 ×r r= 154×7 2×22 = 49 2 m=4.5m Now, Area=π r 2 = 22 7 × 49 2 × 49 2 =1886.5 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A695@

Q.4

A gardener wants to fence a circular garden of diameter21m. Find the length of the rope he needs to purchase, if hemakes 2 rounds of fence. Also find the costs of the rope, if it cost ₹ 4 per meter. Take π=227

Ans

Given, d=21 m So, r= 21 2 m Circumference=2πr =2× 22 7 × 21 2 =66m Length of the rope required for fencing=2×66 m =132 m Cost of 1 m rope =₹4 So, cost of 132 m rope=₹ 4×132 =₹528MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2405@

Q.5 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Ans

Outer radius of circular sheet=4 cm Inner radius of circular sheet=3 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@99EC@

Remaining area=( 3.14×3×3 ) =50.2428.26 =21.98 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@884A@

Q.6

Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15.Takeπ= 3.14

Ans

Circumference=2πr =2×3.14× d 2 =2×3.14 1.5 2 =4.71m Cost of 1 m of lace=₹ 15 Therefore cost of 4.71 m of lace=₹15×4.71 =₹70.65MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EFD4@

Q.7 Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Ans

Radius=5 cm Length of curved part=πr = 22 7 ×5 =15.71 cm Total permieter=Length of the curved part+Length of diameter =15.71+10 =25.71 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@045B@

Q.8

Find the cost of polishing a circular tabletop of diameter1.6 m, if the rate of polishing is Rs 15/m2.Takeπ= 3.14

Ans

Diameter=1.6 m Radius= 1.6 2 =0.8 m Area=3.14×0.8×0.8 =2 .0096 m 2 Cost for polishing 1 m2 area=₹15 Cost for polishing 2.0096 m2 area=₹15×2.0096 =₹30.14 So, it will cost ₹30.14 for polishing such circular table.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3FF3@

Q.9

Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more.

Ans

Circumference=2πr=44 cm 2× 22 7 ×r=44 r=7 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabboeadjabbMgaPjabbkhaYjabbogaJjabbwha1jabb2gaTjabbAgaMjabbwgaLjabbkhaYjabbwgaLjabb6gaUjabbogaJjabbwgaLjabg2da9iabbkdaYiabec8aWjabbkhaYjabg2da9iabbsda0iabbsda0iabbccaGiabbogaJjabb2gaTbqaaiaaxMaacaaMe8UaaGjbVlabbkdaYiabgEna0oaalaaabaGaeGOmaiJaeGOmaidabaGaeG4naCdaaiabgEna0kabdkhaYjabg2da9iabisda0iabisda0aqaaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaemOCaiNaeyypa0JaeG4naCJaeeiiaaIaee4yamMaeeyBa0gaaaa@8092@

Area=π r 2 = 22 7 ×7×7 =154 cm 2 If the wire is bent into a square, then the length of each side would be= 44 4 =11 cm Area of square= ( 11 ) 2 =121 cm 2 Therefore, circle encloses more area. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1AB3@

Q.10

From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth1 cm are removed. (as shown in the adjoining figure).Find the area of the remaining sheet. Takeπ=227

Ans

Area of bigger cirle= 22 7 ×14×14 =616 cm 2 Area of 2 small circles=2×π r 2 =2× 22 7 ×3.5×3.5 =77 cm 2 Area of rectangle=Length×Breadth =3×1= 3 cm 2 Remaining area of sheet= 616 cm 2 77 cm 2 3 cm 2 = 536 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3FAA@

Q.11

A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? Take π= 3.14

Ans

Area of square-shaped sheet= ( side ) 2 = ( 6 cm 2 ) 2 =36 cm 2 Area of circle=3.14×2×2 =12.56 cm 2 Remaining area= 36 cm 2 12 .56 cm 2 =23 .44 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0452@

Q.12 The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? (Take π = 3.14)

Ans

Circumference=2πr=31.4cm 2×3.14×r=31.4 r= 31.4 2×3.14 =5 cm So, Area=3.14×5×5 =78 .50 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AB49@

Q.13 A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Ans

Radius of the flower bed= 66 2 =33m So, radius of flower bed and path together=33+4 =37m Area of flower baed and path together=3.14×37×37 =4298.66 m 2 Area of flower bed =3.14×33×33 =3419 .46 m 2 Area of path =4298 .66 m 2 3419 .46m 2 =879 .20m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6874@

Q.14

A circular flower garden has an area of 314 m2. A sprinkle rat the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden?Takeπ= 3.14

Ans

Area=π r 2 =314 m 2 3.14× r 2 =314 r 2 =100 r=10 m Yes, the sprinkle will water the whole garden. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B781@

Q.15 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Ans

Radius of outer circle=19 m Circumference=2πr =2×3.14×19 =119.32 m Radius of the inner circle=1910=9 m Circumference=2πr =2×3.14×9 =56.52 m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@18E1@

Q.16 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π=22/7)

Ans

Radius, r=28 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGsbGucqqGHbqycqqGKbazcqqGPbqAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGYbGCcqGH9aqpcqqGYaGmcqqG4aaocqqGGaaicqqGJbWycqqGTbqBaaa@5395@

Circumference=2πr =2× 22 7 ×28 =176 cm Number of rotatinos= Total distance to be covered Circumference of the wheel = 352 m 176cm = 35200 176 =200 Therefore, it will rotate 200 times. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@36A6@

Q.17 The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)

Ans

Since, Distance travelled by the tip of minute hand =Circumference of the clock =2πr =2×3.14×15 =94.2 cm Therfore, minute hand move 94.2 cm in 1 hour. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F536@

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