NCERT Solutions for Class 9 Maths Chapter 10 Circles (Ex 10.5) Exercise 10.5

Class 9 is an important phase in a student’s career as the topics covered in Class 9 help in learning topics in Class 10. Students must focus on each subject of Class 9 to score well in the examinations. All the NCERT textbooks should be read carefully as they are the primary source for learning subjects in the CBSE Board curriculum. The NCERT textbooks contain detailed explanations of the topics and are very essential for concept-building in students. The exercises given in NCERT textbooks are very essential for preparing students for the examinations. All the questions are based on the topics and can appear in the question paper of Mathematics. Extramarks provides solutions to the exercises given in the NCERT textbooks which students can easily access in PDF format.

Mathematics plays a crucial role in building analytical thinking ability and reasoning skills among students. Mathematics equips students with the ability to perform complex modelling of scientific phenomena in their careers. Students are advised to practice as many questions as possible in Mathematics to score well in the examinations. All the topics covered in Class 9 Mathematics are helpful in building a foundation for topics to be learned in higher classes. Students are supposed to access the right kind of study materials to score well in the final examination of Mathematics. They will find all the required study materials on the Extramarks website and mobile application. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 are also available for students in Class 9.

The central theme of Chapter 10 is Circles. It is one of the significant chapters from the viewpoint of the final examination. Students are supposed to practice questions according to the syllabus to prepare correctly for the Mathematics examination. The syllabus of Mathematics can be obtained from the Extramarks website and mobile application. It is also helpful in saving time and effort for students. Class 9 students will be able to practice only those topics that are required from an examination standpoint. There are six exercises in Chapter 10, Exercise 10.6 is optional. Students should try to solve all the questions in the exercises after getting well versed in the theories of Chapter 10. Solving questions will enable students to prepare effectively for the examinations.

The question paper of Mathematics needs to be solved within a fixed period of time. Students should develop a habit of solving problems quickly so that they can answer questions within the required time in the examination. Regular practice is very necessary for developing time management skills. Class 9 students can learn better time management skills by practising questions with the help of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5. All the solutions include thorough explanations that will enable students to easily understand the steps used in solving questions. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 are very necessary for learning to give solutions in logically correct steps with clear explanations. There are many ways of solving questions in Mathematics. Students are encouraged to learn the most effective method of solving a question that is time-effective and can help them earn higher marks in the examination. Class 9 students can learn to practice questions with the best methods by taking the help of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5. Students will be able to solve questions in the examination with a positive mindset. It is necessary for students to go through all the exercises of each chapter in Class 9 Mathematics to prepare effectively for the Mathematics examination. If they come across any difficulties in solving exercises, they can access trustworthy solutions from Extramarks.

It is recommended that students practice sample papers and past years’ papers of Mathematics. Past years’ papers are beneficial for knowing the paper pattern and mark distribution of Mathematics. Students are encouraged to solve them frequently. Students will get familiar with the types of problems that appear in the final examination of Mathematics when they go through the past years’ papers of Mathematics. Past years’ papers of Class 9 Mathematics are accessible on Extramarks in PDF format. Students can also access revision notes for Mathematics on Extramarks. They are helpful in getting a quick review of topics before examinations. Formulae, definitions, theorems, etc. can be revised with the help of the revision notes.

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NCERT Solutions Class 9 Maths Chapter 10 Exercise 10.5

Class 9 students must access the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 from the Extramarks website and mobile application in PDF format. All the solutions can be referred to in offline mode as well. It is necessary for students to practice the exercises in Chapter 10 regularly as it is one of the important chapters from the Mathematics syllabus. All the theoretical parts of topics in Chapter 10 should be read carefully by students. Theories, definitions, formulae, etc. should be revised again and again for better understanding and practising exercises of Chapter 10. It is important for students to go through the solved examples given in Chapter 10 to understand the variety of questions that can be framed from the topics. Solved examples are helpful in creating a foundation for practising exercises in Chapter 10. Students are required to revise theorems of the Circles chapter as they are very essential for practising questions from the chapter. All the derivations need to be revised as well because they are crucial for understanding the implementation of formulae to solve questions. Class 9 students can take the help of the Extramarks learning platform to score well in the Mathematics examination. Teachers of Extramarks will guide students to practice questions in Mathematics. All the teachers at Extramarks are very qualified and have years of experience in teaching.

Access NCERT Solutions for Class 9 Mathematics Chapter 10 – Circles

Students can access the most reliable and authentic NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 from the Extramarks learning platform. Practising exercises with the assistance of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 is helpful in boosting the confidence of students for the upcoming examination.

NCERT Solution Class 9 Maths of Chapter 10 All Exercises

Solving questions is very crucial in a subject like Mathematics and it should be done regularly. Study materials such as the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 are helpful in practising exercises of Class 9 Mathematics. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 can be accessed in PDF format from the Extramarks website and mobile application. Students can refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 in offline mode as well after downloading them.

Exercise 10.5 (Ex 10.5) Class 9 Maths Chapter 10 NCERT Solutions 

Exercise 10.5 is based on the Angle Subtended by an Arc of a Circle and Cyclic Quadrilaterals topics. It is critical to go through the topics and understand their theories in detail before beginning to solve Exercise 10.5. Students should rely on the NCERT textbook of Mathematics to practice for the examination. It contains a large number of questions that are necessary for exam preparation.  All the theories, formulae, derivations, theorems, etc. given in the NCERT textbook should be revised from time to time for clear understanding. If students are facing any difficulty even after reading from the textbook, they are advised to take help from Extramarks. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 are very crucial for solving Exercise 10.5 questions. They are accessible on the Extramarks learning platform in PDF format.

Exercise 10.5 Class 9 Maths Solutions

The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 are given in simple language for easy understanding. Students can rely on the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 to get authentic solutions for Exercise 10.5. It is advisable for students to focus on the diagrams in Class 9 Maths Chapter 10 Exercise 10.5; and they are very essential for solving questions related to the topics in the chapter. The Ex 10.5 Class 9 NCERT Solutions are beneficial for learning to solve questions with proper diagrams.

Question 1:

Students having difficulties in solving question 1 of Exercise 10.5 can take the help of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5.

Question 2:

To solve question 2, students must revise the topic given before Exercise 10.5 Class 9 Maths. Question 2 is important from the examination perspective. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 will help Class 9 students solve question 2.

Question 3:

In question 3, students are required to find the angle OPR. It is important to understand the given figure in the question, and by using the right theorem, this question can be easily solved. Students are supposed to refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 to solve question 3.

Question 4:

Question 4 is one of the important questions that students must practice to do well in the Mathematics examination. It is important to make use of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 while practising questions of Exercise 10.5.

Question 5:

To solve question 5, students are required to revise theorems preceding the Class 9 Maths Ex 10.5. All theorems must be well revised. Students can also take help from the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 to solve question 5.

Question 6:

Question 6 is based on the Cyclic Quadrilaterals topic. Students having any difficulty in solving question 6 can make use of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5. Students are advised to refer to these solutions constantly for effective preparation.

Question 7:

To find proper solutions to question 7, students are advised to make use of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5. All the solutions are easy to understand and students can utilise them efficiently.

Question 8:

Question 8 is based on the topic Angle Subtended by an Arc of a Circle topic. It is one of the critical questions of Exercise 10.5. Question 8 and other questions of Exercise 10.5 can be practised easily with the assistance of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5.

Question 9:

Students having difficulties in solving question 9 are advised to get help from the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5.

Question 10:

Students must go through the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 to find a solution for question 10 of Exercise 10.5. It is necessary for students to practice all the questions of Exercise 10.5 to score well in the Mathematics examination.

Question 11:

Question 11 requires students to prove that angle CAD is equal to angle CBD in a circle. It is important for students to refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 to be able to solve question 11.

Question 12:

The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.5 are very useful study resources for practising Exercise 10.5. Students can easily solve question 12 with their help.

NCERT Solutions for Class 9

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Q.1 In figure below, A,B and C are three points on a circle with center O such that ∠BOC = 30° and ∠AOB = 60°.
If D is a point on the circle other than the arc ABC, find ∠ADC.

Ans

Given:In circle, with centre O, ∠BOC=30° and  ∠AOB=60°.To find: ∠ADCSol:   ∵∠AOC=60°+30°                    =90°Since, the angle subtended by an arc at the centre is double theangle subtended by it at any point on the remaining part of the circle.     ∴∠AOC=2 ∠ADC⇒            90°=2 ∠ADC⇒      ∠ADC=90°2                  =45°Thus, the required angle is 45°.

Q.2 A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Ans

Given:Let O is the centre of circle and chord AB is equal to the radius of circle.To find: ∠ACB and ∠ADB.Proof: In ΔAOB, OA=AB=OB. So, ∠AOB=60°       [Angle of equilateral triangle.]             ∠ACB=12∠AOB  [By theorem]                       =12×60°                       =30°Since, ​ADBC is a cyclic quadrilateral. So, ∠ACB+∠ADB=180°⇒     30°+∠ADB=180°⇒          ∠ADB=180°−30°                      =150°

Q.3 In the figure below, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Ans

Given:In circle with centre O, ∠PQR=100°.To find:∠OPRSolution: Since, The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.∴ Reflex∠POR=2 ∠PQR                    =2×100°                    =200°Now,  ∠POR+Reflex∠POR=360°                   ∠POR+200°=360°                         ∠POR=360°−200°                                      =160°In ΔOPR,        OP=OR          ⇒       ∠ORP=∠OPR   [Angles opposite to equal sides are equal.]∠ORP+∠OPR+∠POR=180°   [By angle sum property.]   ∠OPR+∠OPR+160°=180°                               2 ∠OPR=180°−160°                               2 ∠OPR=20°⇒                    ∠OPR=10°Thus, the ∠OPR is 10°.

Q.4 In the figure below, ∠ABC = 69°, ∠ACB=31°, find ∠BDC.

Ans

Given: In circle, ∠ABC = 69°,∠ACB=31°.To find:∠BDCSolution: In ΔABC, ∠BAC+∠ABC+∠ACB=180°                        ∠BAC+69°+31°=180°                                          ∠BAC=180°−100°                                               =80°Since, angles in same segment of a circle are equal.                                       ∠BDC=∠BAC                                         =80°

Q.5 In the figure below A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Ans

Given:In a circle, chords AC and BD intersect at E. ∠BEC=130°           and ∠ECD=20°.To find: ∠BAC.Solution:   ∠CEB+∠CED=180°         [Linear pair of angles.]                130°+∠CED=180°                        ∠CED=180°−130°                               =50°In ΔDEC,   ∠ECD+∠CED+∠CDE=180°               20°+50°+∠CDE=180°                       ∠CDE=180°−70°                            =130°⇒                          ∠BDC=130°Since, angles in same segment of a circle are equal.                         ∠BAC=∠BDC                             =130°Thus, the required angle ∠BAC is 130°.

Q.6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Ans

Given:ABCD is a cyclic quadrilateral. ∠DBC=70°,∠BAC=30°.To find:∠BCD, ∠ECD(if AB=BC)Since, angles in the same segments of the circle are equal.                    So, ∠DAC=∠DBC                                    =70°      ∴               ∠DAB=∠DAC+∠BAC                                    =70°+30°                                    =100°Since, sum of opposite angles in a cyclic quadrilateral is 180°.Then,           ∠DAB+∠DCB=180°                 100°+∠DCB=180°                         ∠DCB=180°−100°or                      ∠BCD=80°If AB=BC

Then,                ∠BAC=∠BCA    [Opposite angles of equal sides areequal.]                             30°=∠BCASince,​                  ∠BCD=80°Then,  ∠BCA+∠ACD=80°                30°+∠ECD=80°⇒                ∠ECD=80°−30°                           =50°

Q.7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans

Given:ABCD is a cyclic quadrilateral.AC and BD are the  diameters of the circle.To prove: ABCD is a rectangle.        Proof: Since, angle is semicircle is right angle. So, ∠ABC=90°      [AC is diameter.]  and    ∠ADC=90°      [AC is diameter.] Now, ∠BAD=90°      [BD is diameter.]  and    ∠BCD=90°      [BD is diameter.]Since, each angles of ABCD is 90°, so ABCD is a rectangle.

Q.8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Ans

Given:Let ABCD be trapezium in which AD=BC. To prove: ABCD is a cyclic quadrilateral.Construction: Draw EC∥AD and produce AB upto E.           Proof: Since, AE∥DC and AD∥EC                   So, AECD is a prallelogram. Then, AD=EC but AD=BC             ⇒          EC=BC                     And ∠CEA=∠CDA      [Opposite angles of a parallelogram are equal.]In​ ΔCEB,              ∠CBE=∠CEB       [Opposite angles of equal sidesare equal.]            ∠CBE+∠CBA=180°         [Linear pair of angles]            ∠CEB+∠CBA=180°         [∵∠CBE=∠CEB]⇒      ∠CEA+∠CBA=180°⇒      ∠CDA+∠CBA=180°        [∵∠CEA=∠CDA]∵∠CDA+∠CBA +∠ABC+∠ADC =360°  [Angle sum property in quadrilateral.]                180°+∠ABC+∠ADC =360°                          ∠ABC+∠ADC =360°−180°                             ∠ABC+∠ADC=180°Since, sum of opposite angles is 180°, so ABCD is a cyclic quadrilateral.

Q.9 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the figure below). Prove that ∠ACP=∠QCD.

Ans

Given:Intersection points of two circles are B and C. Two line segments ABD and PBQ are drawn to intersect the circles at A, D and P,Q respectivelyTo prove:∠ACP=∠QCDProof:  Since, angles in the same segments of a circle are equal. ∠ACP=∠ABP     …(i)  and     ∠QCD=∠QBD    …(ii)But        ∠ABP=∠QBD    …(iii)     [Vertical opposite angles]⇒        ∠ACP=∠QCD               [From equation(i) and equation(ii) and equation (iii)]                        Hence proved.

Q.10 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans

Given:ABC is a triangle. Two circles are drawn with          diameter AB and AC respectively.To prove: Point of intersection(D) of circles lie on BC.

Proof:  ∠ADC=90°       [Angle in semi circle is 90°.] and ∠ADB=90°       [Angle in semi circle is 90°.]           ∠ADC +∠ADC=90°+90°                             ∠BDC=180°⇒BDC is a straight line.⇒D lies on BC i.e., third side of triangle.Thus, point of intersection of both circles lie on third side of triangle.

Q.11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD=∠CBD.

Ans

Given:ΔABC and ΔDAC have equal hypotenous AC.To prove:∠CAD=∠CBDProof:In quadrilateral ABCD,              ∠B+∠D=180° So, quadrilateral ABCD is cyclic.Then, ∠CAD=∠CBD     [Angles in same segment of circle are equal.]                              Hence Proved.

Q.12 Prove that a cyclic parallelogram is a rectangle.

Ans

Given:ABCD is a cyclic parallelogram.To prove:ABCD is a rectangle.Proof:   ∠A+∠C=180°       [ABCD is a cyclic quadrilateral.]                  ∠A=∠C         [Opposite​ angles of a parallelogram.]Then,   ∠A=∠C=90°Since, a parallelogram will be a rectangle if any angle is right angle. Thus, ABCD is a rectangle.

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