NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3 (Ex 10.3)
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Mathematics is an extremely pragmatic subject. It helps lay the foundation for a number of careers like Engineering, Actuarial Science, Computer Information Systems, Computer Science, Statistics, Mathematics Education, Economics, etc. It is essential that students understand the Mathematical concepts to establish a successful career in these fields. Having a good understanding of Mathematical concepts is crucial for developing many skills. Skills like logical reasoning and quantitative aptitude are developed with substantial knowledge of Mathematics. These skills are also required to perform well in competitive examinations like NEET, CUET, CAT, JEE Mains, JEE Advance, etc. Students must strive to have a deep understanding of the mathematical concepts that are taught in the NCERT syllabus for Class 9. Students may refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, to have a good grasp of the mathematical concepts covered in the Mathematics syllabus for Class 9. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 can be utilised by students of Class 9 for their preparation for the Senior Secondary Examination of Mathematics.
NCERT Solutions for Class 9 Maths Chapter 10 Circles (Ex 10.3) Exercise 10.3
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Access NCERT Solutions for Class 9 Maths Chapter10 Circles
Students of Class 9 appearing for the Senior Secondary Examination are recommended to access the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 for their preparation. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, are prepared in an elaborate manner. They have been explained in detail for students to be able to have a better and deeper understanding of the concepts and principles introduced in the Class 9 Maths Chapter 10 Exercise 10.3. It is essential that students thoroughly comprehend the topics covered in the syllabus of Maths Class 9 Chapter 10 Exercise 10.3 since it is one of the most significant topics covered in the NCERT Mathematics syllabus of Class 9. The Class 9 Chapter 10 Maths Exercise 10.3 accounts for a higher topic weightage in the Senior Secondary Examination; hence, it is advisable for students to cover the concerned topics in depth.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3
Chapter 10 Maths Class 9 Exercise 10.3 is one of the most important exercises because it covers a lot of theories and principles that are essential for an overall understanding of Chapter 10 Mathematics. It is crucial that students cover the topics introduced in this exercise seriously. To prepare for the Senior Secondary Examination, students may use the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, which can be downloaded from the Extramarks website or learning application. For many reasons, like spending less time on screens, students often find it more helpful to study from hardcopy study material instead of softcopy study material. Catering to this requirement of students, Extramarks also provides downloadable PDF files for students to access, which can later be printed by students. Students of Class 9 may also download the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, to help themselves with their preparation for the Mathematics examination.
Basic Facts about Circles
The NCERT Mathematics Chapter 10 entitled Circles gives an introduction to various applications of formulas on circles. It is one of the most important chapters. Students must understand the basic concepts and facts about circles before diving into the exercises concerned with the chapters. While solving the exercise problems in the Chapter Circles, students can take help from the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 provided by Extramarks. It is advisable for students to practice NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 repeatedly to gain more confidence with regard to the preparation of the concerned chapter.
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Extramarks provides NCERT Solutions for all classes. Students of all classes can attain the NCERT solution to all exercise problems in all subjects. Students who are appearing for the CBSE board examination are highly encouraged to refer to the NCERT solutions as the NCERT curriculum has been prepared in accordance with the CBSE requirements. For students in Class 9, it is advisable to study with the help of NCERT solutions. Students can obtain the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 from Extramarks to be well prepared for the senior secondary examination they would be appearing for. Practising the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 regularly can help students gain more clarity about the concepts that have been covered in the said exercise.
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It is advisable for students to refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 while solving the exercise problems. It is beneficial to do so for a number of reasons. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, have been designed by experts who are well versed in the concerned field. They have been prepared in accordance with the requirements of the CBSE Board. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, are helpful for the preparation of students as they are broken down into steps and explained in detail. They can be helpful to students as they are easy to comprehend. Since they are easy to comprehend, students can refer to the solutions as a primary text of reference for their preparation.
Clears Concepts
It is highly crucial that students have a clear understanding of all the concepts covered in the syllabus to be able to perform well in the senior secondary examination. Mathematical concepts can be difficult to understand and master. To be able to do both, students must be consistent with their practice of the exercise questions and answers. Students may practice solving the exercise problems while referring to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3. It is important to have a strong foundation in Mathematics to be able to score higher marks. Students need to be able to thoroughly understand all the concepts, which can be done by solving exercise problems regularly. Students can make use of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 for practising for the preparation for the Senior Secondary Examination of Mathematics.
Boosts Confidence and Speed
Solving the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 regularly and repeatedly is highly beneficial for the preparation process of students. This can help boost the confidence of students in relation to the Senior Secondary Examination of Mathematics. Solving the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 can also help familiarise students with the syllabus, which can help them gain confidence. This also helps students reduce the time it takes them to solve problems. Regularly solving the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, can help students prioritise the questions in accordance with their importance and, in turn, manage their time during the Senior Secondary Examination of Mathematics.
Solving Unique Problems
It is crucial to the preparation process that students stay consistent with the revision of NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3. However, it is also essential that students practice solving unique problems, as this can help them enhance their understanding and application of mathematical concepts. Solving unique problems will help students challenge their capabilities in the concerned topic and help improve them. Students can access practice papers to do so. Students may access practice papers from the Extramarks website or mobile application. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, can also be acquired by students from the Extramarks website and learning application. Students’ preparation can benefit from solving practice problems and sample papers. They may experience a boost in their confidence by practising solving exercise problems and solutions as well as solving unique problems.
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NCERT Solutions for Class 9
Extramarks can be relied upon by students to provide adequate study material and educational resources. Students of Class 9 can access the study material required for all exercises in all the subjects that are part of the Class 9 NCERT curriculum. For the preparation for the examination of senior secondary classes, it is advisable that students take advantage of the learning resources available on the Extramarks website or mobile application. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, may be availed by students of Class 9 for their preparation for the Mathematics examination. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3, provided by Extramarks, can prove to be a great tool to help with the preparation for the Senior Secondary Examination of Mathematics. Students may also download the concerned study in PDF format from the Extramarks website.
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The NCERT curriculum is designed according to the requirements of the Central Board of Secondary Education. The NCERT solutions can be accessed from the Extramarks website or mobile application. The study material and learning resources that are available on the Extramarks website and mobile application have been curated based on the requirements of the CBSE guidelines. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 can be acquired by students of Class 9 for their preparation from the Extramarks website, as the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.3 have been prepared keeping in mind the CBSE Board guidelines and conditions.
Q.1 Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Ans
Different pairs of circles are given below:
There are zero common points in figure (a) and (d).
There is one common point in figure (b). There are two common points in figure (c).
Thus, maximum number of common points is 2.
Q.2 Suppose you are given a circle. Give a construction to find its centre.
Ans
Given: Let there are three point A, B and C on the circumference.
To find: Centre of circle.
Steps of construction:
(i) Join A and B, B and C.
(ii) Draw perpendicular bisector of AB and BC.
The intersection point O, of perpendicular bisector is called centre of circle.
Q.3 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Ans
$\begin{array}{l}\text{Given}:\text{Two circles with centre O and O}\u2019\text{intersect at A and B}.\\ \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} AB is common chord of both circles}.\\ \text{To prove}:\text{Centres O and O}\u2019\text{lie on the perpendicular bisector of}\\ \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} AB i}.\text{e}.,\text{OO}\u2019\text{is perpendicular bisector of AB}.\\ \text{Proof}:\text{In}\mathrm{\Delta OAO}\u2018\text{and}\mathrm{\Delta OBO}\u2018\\ \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}}\mathrm{OA}=\mathrm{OB}\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}}\left[\text{radii of same circle.}\right]\\ \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}}\mathrm{OO}\u2018=\mathrm{O}\u2018\mathrm{O}\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}}\left[\text{Common}\right]\\ \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}}\mathrm{O}\u2018\mathrm{A}=\mathrm{O}\u2018\mathrm{B}\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}}\left[\text{radii of same circle.}\right]\\ \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}}\mathrm{\Delta OAO}\u2018\cong \mathrm{\Delta OBO}\u2018\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \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 \mathrm{AOO}\u2018=\angle \mathrm{BO}\u2018\mathrm{O}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \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 \mathrm{AO}\u2018\mathrm{O}=\angle \mathrm{BOO}\u2018\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta OAP}\text{and}\mathrm{\Delta OBP}\\ \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}}\mathrm{OA}=\mathrm{OB}\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}}\left[\text{radii of same circle.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOP}=\angle \mathrm{BO}\u2018\mathrm{P}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \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}}\mathrm{OP}=\mathrm{OP}\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}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta OAP}\cong \mathrm{\Delta OBP}\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}}\left[\mathrm{By}\text{S.A.S.}\right]\end{array}$ \begin{array}{l}So,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AP=BP\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle OPA=\angle OPB\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ But\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle OPA+\angle OPB=180\xb0\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle OPA=\angle OPB=90\xb0\\ \Rightarrow \text{OP is perpendicular bisector of AB}\text{.}\\ \Rightarrow OO\u2018\text{is perpendicular bisector of AB}\text{.}\\ O\text{and O\u2019 lies on perpendicular bisector of AB}\text{. Hence proved}\text{.}\end{array}
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