NCERT Solutions For Class 9 Maths Chapter 10 Circles (Ex 10.6) Exercise 10.6

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It is advisable for students to make use of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, to study and prepare for the Senior Secondary Examination of Class 9. Students may avail themselves of the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, from the Extramarks website and mobile application. Students of Class 9 must understand the significance of Class 9 in the senior secondary academic years. The Class 9 Maths Chapter 10 Exercise 10.6 is one of the most important significant exercises that have been covered in the syllabus of the NCERT textbook of Mathematics for Class 9. Students are suggested to regularly and actively practice the Ex 10.6 Class 9 NCERT Solutions to be able to understand each step in detail. They can refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, to help themselves with the process of solving exercise problems. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, are accessible to students of Class 9 from the Extramarks website. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, can prove to be of great assistance to students in their preparation. It is highly recommended that students use the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, for their preparation for the Senior Secondary Examination of Mathematics.

NCERT Solutions For Class 9 Maths Chapter 10 Circles (Ex 10.6) Exercise 10.6

Extramarks is a well-known educational website. Students can access a variety of learning resources and study materials from the Extramarks website. Students’ academic requirements are fulfilled by Extramarks by providing them with the necessary learning resources. The Extramarks website or mobile application offers the necessary study materials to students of all classes. A variety of educational materials are offered by Extramarks, including NCERT solutions, sample papers, past years’ papers, online live sessions, revision notes, mock tests, etc. Students of Class 9 may access the required study material for their preparation for the Mathematics examination from the Extramarks website. Students may refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. Extramarks also provides revision notes of Exercise 10.6 Class 9 Maths. It is encouraged for students to use the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. Due to intermittent internet access or poor internet connections, students may occasionally be unable to access online study materials. Keeping this in mind, Extramarks offers downloadable educational tools on its website and mobile application. Students in Class 9 who might experience difficulty with their internet connection or who would prefer to refer to downloaded materials can also get the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6 in a downloadable PDF format. The downloaded PDF files are available for printing for students who prefer to learn from hardcopy study materials. One of the most trustworthy and comprehensive informational resources for Class 9 students to use during their preparation is the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, are available for download in PDF format from the Extramarks website if needed. Students may also download other learning resources from the Extramarks website. PDF files of revision notes, sample papers, practice papers, mock tests, and NCERT solutions are also available on the Extramarks website that students can download. It is essential that students pay close attention to the NCERT exercise and the problems to prepare themselves efficiently for the senior secondary examination. This is important since the senior secondary examinations are based on the NCERT curriculum.

Access NCERT Solutions Class 9 Mathematics Chapter 10 – Circles

To achieve the highest possible scores in the Senior Secondary Examination, students must make it a point to thoroughly and accurately review the NCERT curriculum. The importance of doing well in the senior secondary examination must be understood by students. The grades that students receive on the senior secondary examination help determine the extent to which they will advance academically. Students can use the NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.6 to help them prepare for the senior secondary examination. In order to help students understand the study material better, the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, have been extensively elaborated. Class 9 Maths Ex 10.6 is one of the most important exercises from the NCERT Mathematics Chapter 9 Circles syllabus. Considering circles can be a challenging topic, it is crucial that students comprehend the ideas discussed in this chapter and effectively apply them to problems and their solutions. For better preparation, students can review the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, and practice answering the questions that are part of Exercise 10.6 Class 9 of the NCERT Mathematics textbook. Students who are taking the Senior Secondary Mathematics Examination are required to study for the examination in advance. Starting the senior secondary exam preparation process early will help students prepare more effectively. It could help increase the likelihood of doing well in the concerned examination. Practice makes perfect. Students are suggested to be consistent with their practice of NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6 to make sure that their performance in the senior secondary examination is good. Performing well in the senior secondary examination is important, considering the fact that students’ grades in Class 9 make a huge difference in their performance in higher classes. For this very reason, students must make it a priority to go through the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6 on a regular basis. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, can be obtained by students for their preparation for the senior secondary examination from the Extramarks website and mobile application.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6

One of the most crucial chapters in the NCERT Mathematics curriculum for Class 9 is Chapter 10 entitled “Circles.” The topic of Circles may be rather complex. To master this chapter, students must not only comprehend the principles introduced in the chapter but also have a firm understanding of how to put those concepts to practical use. To better comprehend the practical application required in Chapter 10 Circles, students are urged to practise the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. The mathematical topics covered in the Class 9 Mathematics curriculum must be studied and understood by students. It is crucial for students to grow their all-around knowledge of mathematical principles. This significantly affects how students in Class 9 perform in the Senior Secondary Examination of Mathematics. Students can benefit greatly from the practical application required in Chapter 10 Circles, students are urged to practice the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6 as they get prepared for the Senior Secondary Mathematics Examination. To adequately understand the technical and application-based ideas given in the Chapter Circles, it is advised that students regularly practice the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. Students may use the Extramarks website or mobile application to acquire the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. Students of Class 9 can join a doubt-clearing session on the Extramarks website to clear up any queries they may have about the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. The Extramarks website offers study materials for all boards and classes. Since Extramarks’ study materials have been designed with the CBSE board examination in mind, students taking the CBSE board examination can refer to them while they prepare for their exam. The NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, which are available on the Extramarks website, are strongly advised for Class 9 students studying for the Senior Secondary Examination of Mathematics. In order to access the necessary pedagogical material, students must first register on the Extramarks website. Once they have registered on the Extramarks website, students may access whatever study material they need, including mock tests, NCERT solutions, sample papers, practice papers, past years’ papers, revision notes, and online live sessions.

NCERT Solutions for Class 9

When studying for the senior secondary examination, students should consider the weightage of  marks assigned to each topic taught in the concerned syllabus. The topic weightage for the senior secondary examination must be taken into account during preparation since it may help students irankn ranking the importance of each topic. In the Senior Secondary Examination of Mathematics, the concepts addressed in Mathematics Exercise 10.6 of Class 9 are crucial. When preparing for the senior secondary examination, the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. In order to be adequately prepared for the final exam, students must make an effort to comprehend the topics taught in the concerned curriculum. Students must adhere to a schedule that gives each topic covered in the curriculum a significant amount of time in order to adequately study for the senior secondary examination. For their preparation, students can refer to the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. Experts in the relevant field of study have prepared the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, offered by Extramarks. For the benefit of students, the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, have been meticulously broken down into steps. For better performance in the examination, students are encouraged to begin their preparation for the senior secondary examination at the beginning of the respective academic year.

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CBSE Study Materials for Class 9

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CBSE Study Materials

Experts have compiled the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6, based on the most recent Central Board of Secondary Education syllabus. For the preparation of the senior secondary examinations given by the Central Board of Secondary Education, the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6 are strongly advised. Additionally, Extramarks offers students downloadable revision materials so students may keep going over the topics they have previously thoroughly studied. Students are urged to review the revision notes for the concerned curriculum after thoroughly covering the topics in Class 9 Chapter 10 named “Circles” and working through the NCERT Solutions For Class 9 Maths Chapter 10 Exercise 10.6. Students must put their all into their preparation for the Senior Secondary Mathematics Examination. One of the most significant exams students take throughout their academic lives is the senior secondary examination for Class 9.

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Q.1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Ans

Given:Two circles with centres P and Q are intersecting at A and B.To prove:∠PAQ=∠PBQ        Proof:  In ΔAPQ and ΔBPC                 AP=BP         [Radii of same circle.]                   AQ=BQ        [Radii of same circle.]                    PQ=PQ        [Common]              ΔAPC≅ΔBPC    [By S.S.S.]              ∠PAQ=∠PBQ   [By C.P.C.T.]Thus, centres P and Q subtend equal angle at A and B.

Q.2 Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Ans

Let OP=x m and OQ=( 6−x )m. In right Δ AOP, by Pythagoras Theorem, OA 2 = AP 2 + OP 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@82E8@

OA2=(AB2)2+(x)2     [Since, perpendicular from centrebisect the chord.] OA2=(52)2+x2 OA2=254+x2      ...(i)In ΔCQO, ∠Q=90°So, by Pythagoras theorem, we have OC2= CQ2+ OQ2       =(CD2)2+(6−x)2       =(112)2+(6−x)2 OC2=1214+36−12x+x2 ...(ii)Since,  OA2=OC2     [Radius of same circle.]⇒    254+x2=1214+36−12x+x2⇒              254=1214+36−12x⇒            12x=1214+36−254               =964+36               =24+36⇒            x=6012=5Substituting value of x in equation(i), we get OA2=254+52            =254+25        OA=1254            =552            =5×2.242            =5.6 m  (approx)Thus, the radius of circle is 5.6 m (approx).

Q.3 The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the center?

Ans

Let AB and CD are two parallel chords of 6 cm and 8 cm respectively. The distance of AB is 4 cm from center. Let distance of CD from center O is x cm.

In right Δ AOP, by Pythagoras Theorem, OA2= AP2+ OP2 OA2=(AB2)2+(4)2     [Since, perpendicular from centrebisect the chord.] OA2=(62)2+42 OA2=9+16        OA=25        OA=5 cm.In ΔCQO, ∠Q=90°So, by Pythagoras theorem, we have OC2= CQ2+ OQ2 (5)2=(CD2)2+x2     [∵OC=OA=5 cm]    25=(82)2+x2  x2=25−16          x=9=3Thus, the distance of chord CD from centre O is 3 cm.

Q.4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Ans

Given: AD=CE, ∠ABC is in exterior of circle with centre O.To prove:∠ABC=12(∠DOE−∠AOC)Construction: Join AC, OA, OD, OC and OE.Proof:  In ΔAOD and ΔCOE,             OA=OC        [Radius of the same circle]             OD=OE        [Radius of the same circle]             AD=CE        [Given]           ΔAOD≅ΔCOE   [By S.S.S.]∴    ∠OAD=∠OCE  [By C.P.C.T.]and ∠ODA=∠OEC  [By C.P.C.T.]But  ∠OAD=∠ODA   [Opposite angles of equal sides of atriangle are equal.]Then, ∠OAD=∠ODA=∠OCE=∠OEC=x  (let)In ΔOAC, OA=OC∴         ∠OCA=∠OAC       [Angle opposite to equal sides are equalin a triangle.]                 =y(let)and     ∠AOC=180°−2y   [Angle sum property in a triangle]In ΔODE, OD=OE∴         ∠OED=∠ODE      [Angle opposite to equal sides are equalin a triangle.]                =z (let)and     ∠DOE=180°−2z  [Angle sum property in a triangle]Now,       ∠DOE−∠AOC=(180°−2z)−(180°−2y)                            =2(y−z)12(∠DOE−∠AOC)=y−z    ...(i)                        ∠BAC=180°−(x+y)     [Linear pair of angles]and                 ∠BCA=180°−(x+y)     [Linear pair of angles]In ΔBAC, ∠ABC+∠BAC+∠BCA=180°                              ∠ABC=180°−∠BAC−∠BCA                                      =180°−{180°−(x+y)}−{180°−(x+y)}                                 ∠B=2(x+y)−180°or              2(x+y)−∠B=180°    ...(ii)In ΔBDE, ∠BDE+∠BED+∠DBE=180°                              ∠DBE=180°−∠BDE−∠BED                                    ∠B=2(x+y)−∠B−(x+z)−(x+z)                                                [From equation(ii)]                      2∠B=2(x+y)−2(x+z)                         ∠B=(y−z)    ...(iii)From equation(i) and equation(iii), we have            ∠ABC=12(∠DOE−∠AOC)  Hence proved.

Q.5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Ans

Given: Let ABCD be a rhombus and circle is drawn with           diameter AB.To prove: Circle passes through point O.Proof: In circle with diameter AB and angle in semicircle is right angle.So, ∠AOB=90°Since, diagonals of a rhombus bisect at 90°. So,                  ∠AOB=90°⇒Point O, lies on circumference. Thus, circle passes throughthe point of intersection of diagonals of rhombus.

Q.6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Ans

Given:Let ABCD be a parallelogram and circle passing through vertex A, B and C intersects at E on side CD.To prove:AD=AEProof:Since, ABCD is a parallelogram. So, ∠B=∠D     ...(i)       [Opposite angles of parallelogram areequal.]Since,ABCE is a cyclic quadrilateral, so      ∠B+∠AEC=180°  …(ii)     [Sum of opposite angles in a cylicquadrilateral is 180°.]   ∠AED +∠AEC =180°  ...(iii)       [Linear pair of angles]From equation (ii) and equation(iii), we have     ∠B+∠AEC=∠AED +∠AEC⇒                 ∠B=∠AED   ...(iv)From equation(i) and equation(iv), we have            ∠D=∠AED⇒       ∠ADE=∠AED   [∵∠D=∠ADE]So,              AE=AD       [Opposite sides of equal angles are equalin ΔADE.]                              Hence proved.

Q.7 AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.

Ans

Given:In circle with centre O, AC and BD are two chords which            bisect each other.To prove:(i) AC and BD are diameters, (ii)ABCD is a rectangle.Proof:In ΔAOB and ΔCOD    AO=OC              [Given]      ∠AOC=∠BOD        [Vertical opposite angles]    BO=OD            [Given]       ΔAOB≅ΔCOD         [By SAS]∴         AB=DC            [By C.P.C.T.]           In ΔAOD and ΔBOC    AO=OC          [Given]      ∠AOD=∠BOC     [Vertical opposite angles]    DO=OB          [Given]      ΔAOD≅ΔBOC     [By SAS]∴           AD=BC          [By C.P.C.T.]Since, opposite sides of quadrilateral ABCD are equal,so ABCD is a prallelogram.Then, ∠A=∠C  …(i)   [Opposite angles of a parallelogram are equal.]But ABCD is a cyclic quadrilateral.So,   ∠A+∠C=180°…(ii)From equation(i) and (ii), we have          ∠A=90°If any angle in a parallelogram is right angle, then it is a rectangle. Here, in parallelogram ABCD,  ∠A=90°, so ABCD is a rectangle                                                Hence proved.

Q.8

Bisectors of angles A, B and C of a triangle ABC intersect itscircumcircle at D ,E and F respectively. Prove that theangles of the triangle DEF are 90°−12A, 90°−12Band 90°−12C.

Ans

Given:ABC is a triangle.AD, BE and CF are bisector of ∠A, ∠B and ∠C respectively.To prove: ∠D=90°−12A, ∠E=90°−12B,∠F=90°−12C

Proof: ∠BEF=∠BCF        [Angles in same segment of circleare equal.]                  =C2            ∠BED=∠BAD         [Angles in same segment of circleare equal.]                   =A2So,       ∠E=∠BEF+∠BED                 =C2+A2                 =12(C+A)                 =12(180°−B)                =90°−12BSimilarly,             ∠F=90°−12C and ∠D=90°−12A.Thus, ∠D=90°−12A,  ∠E=90°−12B and ∠F=90°−12C.

Q.9 Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Ans

Given: Two congruent circles intersect at A and B. PAQ                intersects circles at P and Q respectively.To prove: BP=BQ        Proof:Since, AB is common chord in two congruent triangles. So, the angles subtended by common arc at any point on the circumference of both circles will be equal. ∴∠BPA=∠BQA⇒          ∠BPQ=∠BQPIn ΔBPQ,           ∠BPQ=∠BQP⇒              BQ=BP         [Opposite sides of equal anglesare equal.]⇒              BP=BQ

Q.10 In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Ans

Given: AD is bisector of ∠A in ΔABC.To prove: AD and pependicular of BC intersect at point D on             circumcircle.Proof:In ΔBOM and ΔCOM            OB=OC           [radius of circle]           ∠OMB=∠OMC    [Each 90°]            MB=CM           [Given]     ∴     ΔBOM≅ΔCOM     [By S.A.S.]     ∴     ∠BOM=∠COM     [By C.P.C.T.]⇒           ∠BOC=2 ∠BOM   ...(i)The angle subtended by an arc at the centre is double the anglesubtended by it at any point on the remaining part of the circle.                ∠BOC=2 ∠BAC     ...(ii)Similarly , for arc BD,                 ∠BOD=2 ∠BAD    ...(iii)Since, AD is bisector of ∠A, then            ∠BAD=12∠BAC                 =12×12∠BOC    [From equation(ii)]           ∠BAD=12∠BOM         [From equation(i)]       12∠BOD=12∠BOM          [From equation(iii)]⇒    ∠BOD=∠BOM⇒perpendicular bisector of BC and bisector of ∠A intersect      circumcircle at the point D.

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