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      Class 9 Maths

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      • Class 09
      • Class 9 Maths
      CoursesClass 09MathsClass 9 Maths
      • 1.Number System
        14
        • Lecture1.1
          Numbers and its Classification, Rational Number and its Representation on the Number Line 50 min
        • Lecture1.2
          Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number 01 hour
        • Lecture1.3
          Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers 46 min
        • Lecture1.4
          Representing Irrational Number on the Number Line 41 min
        • Lecture1.5
          Rationalisation 54 min
        • Lecture1.6
          Introduction to Exponents, Laws of Exponents, BODMAS Rule 41 min
        • Lecture1.7
          Integral Exponents of Real Numbers 57 min
        • Lecture1.8
          Integral Exponents of Real Numbers and Solving Equation for x 01 hour
        • Lecture1.9
          Rational Exponents of Real Numbers 52 min
        • Lecture1.10
          Examples Based on Rational Exponents of Real Numbers 30 min
        • Lecture1.11
          Chapter Notes – Number System
        • Lecture1.12
          NCERT Solutions – Number System Exercise 1.1 – 1.6
        • Lecture1.13
          R D Sharma Solutions Number System
        • Lecture1.14
          Revision Notes – Number System
      • 2.Polynomials
        10
        • Lecture2.1
          Introduction to Polynomials and Some Basics Questions 58 min
        • Lecture2.2
          Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation 01 hour
        • Lecture2.3
          Division of Polynomials and Division Algorithm and Remainder Theorem 51 min
        • Lecture2.4
          Factorising by Spliting Method 01 hour
        • Lecture2.5
          Factorisation By Different Methods and Factor Theorem 27 min
        • Lecture2.6
          Factorisation of Polynomial by Using Algebric Identities 01 hour
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.5
        • Lecture2.9
          R S Aggarwal Solutions Polynomials
        • Lecture2.10
          Revision Notes Polynomials
      • 3.Coordinate Geometry
        8
        • Lecture3.1
          Representation of Position on Number Line, Finding the location by Using the Axis 01 hour
        • Lecture3.2
          Quadrant, Coordinates of Point on Axis, Distance from Axis 01 hour
        • Lecture3.3
          Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis 57 min
        • Lecture3.4
          Chapter Notes – Coordinate Geometry
        • Lecture3.5
          NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
        • Lecture3.6
          R D Sharma Solutions Coordinate Geometry
        • Lecture3.7
          R S Aggarwal Solutions Coordinate Geometry
        • Lecture3.8
          Revision Notes Coordinate Geometry
      • 4.Linear Equations
        8
        • Lecture4.1
          Introduction of Linear Equation in Two Variables and Graphical Representation 01 hour
        • Lecture4.2
          Example to find the value of Constant and Graph of Linear Equation in Two Variables 57 min
        • Lecture4.3
          Equations of lines Parallel to the x-axis and y-axis 46 min
        • Lecture4.4
          Chapter Notes – Linear Equations
        • Lecture4.5
          NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
        • Lecture4.6
          R D Sharma Solutions Linear Equations
        • Lecture4.7
          R S Aggarwal Solutions Linear Equations
        • Lecture4.8
          Revision Notes Linear Equations
      • 5.Euclid's Geometry
        7
        • Lecture5.1
          Euclid’s Definitions and Euclid’s Axioms 32 min
        • Lecture5.2
          Eulcid’s Postulates and Playfair’s Axiom 37 min
        • Lecture5.3
          Chapter Notes – Euclid’s Geometry
        • Lecture5.4
          NCERT Solutions – Euclid’s Geometry Exercise 5.1 5.2
        • Lecture5.5
          R D Sharma Solutions Euclid’s Geometry
        • Lecture5.6
          R S Aggarwal Solutions Euclid’s Geometry
        • Lecture5.7
          Revision Notes Euclid’s Geometry
      • 6.Lines and Angles
        10
        • Lecture6.1
          Introduction to Lines and Angles, Types of Angle and Axioms 01 hour
        • Lecture6.2
          Vertically opposite Angles and Angle bisector 46 min
        • Lecture6.3
          Angle made by a transversal with two Lines and Parallel Lines 01 hour
        • Lecture6.4
          Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle 52 min
        • Lecture6.5
          Angle Sum Property of Triangle Including Interior and Exterior Angles 01 hour
        • Lecture6.6
          Chapter Notes – Lines and Angles
        • Lecture6.7
          NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
        • Lecture6.8
          R D Sharma Solutions Lines and Angles
        • Lecture6.9
          R S Aggarwal Solutions Lines and Angles
        • Lecture6.10
          Revision Notes Lines and Angles
      • 7.Triangles
        11
        • Lecture7.1
          Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence 42 min
        • Lecture7.2
          Questions Based on SAS Criterion 01 hour
        • Lecture7.3
          ASA Criterion of Congruence and Its based Questions 42 min
        • Lecture7.4
          Questions Based on AAS and SSS Criterion 29 min
        • Lecture7.5
          RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion 01 hour
        • Lecture7.6
          Inequalities of Triangle 42 min
        • Lecture7.7
          Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions 48 min
        • Lecture7.8
          Chapter Notes – Triangles
        • Lecture7.9
          NCERT Solutions – Triangles Exercise 7.1 – 7.5
        • Lecture7.10
          R D Sharma Solutions Triangles
        • Lecture7.11
          Revision Notes Triangles
      • 8.Quadrilaterals
        13
        • Lecture8.1
          Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite 01 hour
        • Lecture8.2
          Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3; 01 hour
        • Lecture8.3
          Theorem 4; Theorem 5; 24 min
        • Lecture8.4
          Conditions for a Quadrilateral to be a parallelogram- Theorem 6 01 hour
        • Lecture8.5
          Theorem 8 01 hour
        • Lecture8.6
          Theorem 9 51 min
        • Lecture8.7
          Theorem 9(Mid-point Theorem); 01 hour
        • Lecture8.8
          Theorem 10 27 min
        • Lecture8.9
          Chapter Notes – Quadrilaterals
        • Lecture8.10
          NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
        • Lecture8.11
          R D Sharma Solutions Quadrilaterals
        • Lecture8.12
          R S Aggarwal Solutions Quadrilaterals
        • Lecture8.13
          Revision Notes Quadrilaterals
      • 9.Area of Parallelogram
        11
        • Lecture9.1
          Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area 57 min
        • Lecture9.2
          Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle 01 hour
        • Lecture9.3
          Triangles on the same Base and between the same Parallels 56 min
        • Lecture9.4
          Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels 54 min
        • Lecture9.5
          Sums Based on Median of a Triangle 52 min
        • Lecture9.6
          Two Triangles on the same Base and equal Area lie b/w the same parallel 01 hour
        • Lecture9.7
          Summary of All Concepts and Miscellaneous Questions 56 min
        • Lecture9.8
          Chapter Notes – Area of Parallelogram
        • Lecture9.9
          NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
        • Lecture9.10
          R D Sharma Solutions Area of Parallelogram
        • Lecture9.11
          Revision Notes Area of Parallelogram
      • 10.Constructions
        7
        • Lecture10.1
          Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector 42 min
        • Lecture10.2
          Construction of Different types of Triangle 43 min
        • Lecture10.3
          Chapter Notes – Constructions
        • Lecture10.4
          NCERT Solutions – Constructions Exercise
        • Lecture10.5
          R D Sharma Solutions Constructions
        • Lecture10.6
          R S Aggarwal Solutions Constructions
        • Lecture10.7
          Revision Notes Constructions
      • 11.Circles
        11
        • Lecture11.1
          Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4 01 hour
        • Lecture11.2
          Problems Solving 48 min
        • Lecture11.3
          Problems Based on Parallel Chords, Common Chord of Circles 01 hour
        • Lecture11.4
          Problems Based on Concentric Circles, Triangle in Circle 41 min
        • Lecture11.5
          Equal Chords and Their Distance, Theorem-5, 6, 7, 8 01 hour
        • Lecture11.6
          Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10 01 hour
        • Lecture11.7
          Chapter Notes – Circles
        • Lecture11.8
          NCERT Solutions – Circles Exercise
        • Lecture11.9
          R D Sharma Solutions Circles
        • Lecture11.10
          R S Aggarwal Solutions Circles
        • Lecture11.11
          Revision Notes Circles
      • 12.Heron's Formula
        8
        • Lecture12.1
          Introduction of Heron’s Formula; Area of Different Triangles 01 hour
        • Lecture12.2
          Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2) 53 min
        • Lecture12.3
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1) 32 min
        • Lecture12.4
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2) 38 min
        • Lecture12.5
          Chapter Notes – Heron’s Formula
        • Lecture12.6
          NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
        • Lecture12.7
          R D Sharma Solutions Heron’s Formula
        • Lecture12.8
          Revision Notes Heron’s Formula
      • 13.Surface Area and Volume
        16
        • Lecture13.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere 40 min
        • Lecture13.2
          Type-1: Surface Area and Volume of Cube and Cuboid 27 min
        • Lecture13.3
          Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder 28 min
        • Lecture13.4
          Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1) 25 min
        • Lecture13.5
          Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere 27 min
        • Lecture13.6
          Type-5 Short Answer Types Questions, 12 min
        • Lecture13.7
          Chapter Notes – Surface Area and Volume of a Cuboid and Cube
        • Lecture13.8
          Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
        • Lecture13.9
          Chapter Notes – Surface Area ad Volume of A Right Circular Cone
        • Lecture13.10
          Chapter Notes – Surface Area ad Volume of A Sphere
        • Lecture13.11
          NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
        • Lecture13.12
          R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
        • Lecture13.13
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
        • Lecture13.14
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
        • Lecture13.15
          R D Sharma Solutions Surface Areas And Volumes of A Sphere
        • Lecture13.16
          Revision Notes Surface Areas And Volumes
      • 14.Statistics
        15
        • Lecture14.1
          Data; Presentation of Data; Understanding Basic terms; 01 hour
        • Lecture14.2
          Questions; 52 min
        • Lecture14.3
          Bar Graph, Histogram, Frequency Polygon; 01 hour
        • Lecture14.4
          Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency, 01 hour
        • Lecture14.5
          Examples Based on Mean of Ungrouped Data 56 min
        • Lecture14.6
          Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data) 01 hour
        • Lecture14.7
          Mean; Mean for an ungrouped freq. Dist. Table 25 min
        • Lecture14.8
          Chapter Notes – Statistics – Tabular Representation of Statistical Data
        • Lecture14.9
          Chapter Notes – Statistics – Graphical Representation of Statistical Data
        • Lecture14.10
          Chapter Notes – Statistics – Measures of Central Tendency
        • Lecture14.11
          NCERT Solutions – Statistics Exercise 14.1 – 14.4
        • Lecture14.12
          R D Sharma Solutions Tabular Representation of Statistical Data
        • Lecture14.13
          R D Sharma Solutions Graphical Representation of Statistical Data
        • Lecture14.14
          R S Aggarwal Solutions Statistics
        • Lecture14.15
          Revision Notes Statistics
      • 15.Probability
        8
        • Lecture15.1
          Activities; Understanding term Probability; Experiments; 44 min
        • Lecture15.2
          Event; Probability; Empirical Probability- Tossing a coin; 31 min
        • Lecture15.3
          Rolling a dice; Miscellaneous Example; 27 min
        • Lecture15.4
          Chapter Notes – Probability
        • Lecture15.5
          NCERT Solutions – Probability Exercise 15.1
        • Lecture15.6
          R D Sharma Solutions Probability
        • Lecture15.7
          R S Aggarwal Solutions Probability
        • Lecture15.8
          Revision Notes Probability

        NCERT Solutions – Circles Exercise

        Exercise 10.1

        Q.1     Fill in the blanks :
        (i)The center of a circle lies in ______ of the circle.(exterior/interior)
        (ii) A point , whose distance from the center of a circle is greater than its radius lies in _______of the
        circle (exterior/interior)
        (iii) The longest chord of a circle is a _________ of the circle.
        (iv) An arc is a ______ when its ends are the ends of a diameter.
        (v) Segment of a circle is the region between an arc and _____of the circle.
        (vi) A circle divides the plane, on which it lies, in _______ parts.

        Sol.

        (i) interior
        (ii) exterior
        (iii) diameter.
        (iv) semi- circle
        (v) chord
        (vi) three parts.


        Q.2     Write true or false : Give reasons for your answers :
        (i) Line segment joining the centre to any point on the circle is a radius of the circle.
        (ii) A circle has only finite number of equal chords.
        (iii) If a circle is divided into three equal arcs each is a major arc.
        (iv) A chord of a circle, which is twice as long as its radius is a diameter of the circle.
        (v) Sector is the region between the chord and its corresponding arc.
        (vi) A circle is a plane figure.

        Sol.

        (i) True : As ‘radius’ is used in two sense – in the sense of a line segment and also in the sense of a length.
        (ii) False : As an infinite number of points lie on the circle so we get infinite number of chord on joining any two points.
        (iii) False : Since these arcs are equal in length so they are called equal arc.
        (iv) True : As diameter is the longest chord and its length is twice the radius of the circle.
        (v) False : As the region between an arc and the two radii, joining the center to end points of the are is known as a sector.
        (vi) True : Since a circle is the locus of a point which moves in a plane in such a way that its distance from a given fixed point in the plane is always constant. So, a circle lies in a plane.

        Exercise 10.2

        Q.1     Recall  that two circles are congruent  if they have the same radii. Prove that equal  chords of  congruent circles subtend equal  angles at their  centers.
        Sol.circle exercise 10.2 figure 1

        Given : AB and  CD are two equal  chords of a circle  with center at O.
        To prove : ∠AOB=∠COD
        Proof : In Δs AOB and COD,  we have
        AO = CO                       [Radii of the same circle]
        BO = DO                       [Radii of the same circle]
        and     AB = CD            [Given]
        Therefore  by  SSS criterion  of congruence, we have
        ΔAOB≅ΔCOD
        ⇒    ∠AOB=∠COD                   [C.P.C.T.]


        Q.2      Prove that if chords of congruent circles subtend  equal  angles at their  centres, then the chords are equal.
        Sol. 

        Given : AB and CD are two chords such that angles subtended by these chords at the centre of the  circle are equal.

        Circle Exercise 1-.2 figure 2i.e.          ∠AOB=∠COD
        To prove :         AB = CD
        Proof  : In Δs AOB and  COD we have
        AO = CO         [Radii of the same  circle]
        BO = DO         [Radii of the same circle]
        and ∠AOB=∠COD
        Therefore   By SAS criterion  of congruence,  we have
        ΔAOB≅ΔCOD
        ⇒        AB = CD                         [C.P.C.T.]

        Exercise 10.3

        Q.1     Draw  different  pairs of circles. How  many points  does each pair  have  in common? What is the  maximum number of common points?
        Sol.

        On drawing a different pairs of circles, we find that they have two points in common. Now  we shall prove that a pair of circle  cannot intersect each other at  more than two points.
        Let the two  circles cut each other at three points. But  from the three  points only one circle passes. Hence, two circles if intersect each other then they will intersect at two points. These two points are the ends of their common chords. This chord can be a common chord of a number of circles which will pass through the end points of this common chord.


        Q.2    Suppose you are given a circle. Give a  construction  to find  its centre.
        Sol.      Steps of Construction :

        51.   Take 3  points A, B and C on the  circumference of the circle.
        2.   Join  AB and BC.
        3.   Draw PQ and RS, the perpendicular bisectors of AB and BC, which intersect each other at O. Then,O is the centre  of the circle.


        Q.3     If two circles intersec at two  points, prove that their  centres lie on the perpendicular bisector of the common  chord.
        Sol. 

        Given : Two  circles, with  centres O and O’ intersect, at two points A and B so that AB is the  common chord of the two circles and OO’ is the line segment joining the centres of the two circles. Let OO’ intersect AB at M.
        To prove : OO’ is the  perpendicular  bisector of AB.
        Construction : Draw line segment OA, OB, O’A and O’B.

        figure_4
        Proof : Δs OAO’ and OBO’ we have
        OA = OB             [Radii of the same circle]
        O’A = O’B             [Radii of the same circle]
        and     OO’ = OO’
        By  SSS criterion of  congruence, we have
        ΔOAO′≅ΔOBO′
        ⇒    ∠AOO′=∠BOO′
        ⇒    ∠AOM=∠BOM        … (1)
        [Since ∠AOO′=∠AOMand∠BOM=∠BOO′]
        In Δs AOM and BOM,  we have
        OA = OB             [Radii of the same circle]
        ∠AOM=∠BOM        [From (1)]
        and      OM = OM             [Common]
        Therefore  by SAS criterion of congruence, we have
        ΔAOM≅ΔBOM
        ⇒    AM = BM and ∠AMO=∠BMO
        But  ∠AMO+∠BMO=180∘
        Therefore 2∠AMO=180∘⇒∠AMO=90∘
        Thus ,     AM = BM  and ∠AMO=∠BMO=90∘
        Hence , OO’ is the perpendicular  bisector of AB.

        Exercise 10.4

        Q.1     Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
        Sol.

        Let O and O’ be the centres of the circles of radii 5 cm and 3 cm respectively and let PQ be their common chord.
        We have OP = 5 cm, O’P = 3 cm and OO’ = 4 cm
        Since OP2=PO′2+O′O2 [52=32+42]
        ⇒ OO’P is a right ∠dΔ, right angled as O’.

        4
        Area of ΔOO′P=12×O′P×OO′
        =12×3×4
        =6sq.units...(1)
        Also ,
        Area of ΔOO′P=12×OO′×PL
        =12×4×PL=2PL … (2)
        From (1) and (2), we have
        2 × PL = 6 ⇒ PL = 3
        We know that when two circles intersect at two points, then their centre lie on the perpendicular bisector of the common chord i.e., OO’ is the perpendicular bisector of PQ.
        Therefore PQ = 2 × PL = (2 × 3) cm = 6 cm


        Q.2      If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
        Sol.

        Given : AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.
        To prove : (i) AP = PD (ii) PB = CP.
        Construction : Draw OM ⊥ AB , ON ⊥ CD.
        Join OP,

        2
        AM = MB = 12 AB [Perpendicular from centre bisects the chord]
        CN = ND = 12 CD [Perpendicular from centre bisects the chord]
        AM = ND and MB = CN [Since AB = CD (given)]
        In Δs OMP and ONP , we have
        OM = ON                                                [Equal chords of a circle are equidistant from the centre]
        ∠OMP=∠ONP [Since Each = 90º]
        OP = OP [Common]
        By RHS’ criterion of congruence,
        ΔOMP≅ΔONP
        ⇒ MP = PN … (2) [C.P.C.T.]
        Adding (1) and (2) , we have
        AM + MP = ND + PN ⇒ AP = PD
        Subtracting (2) from (1), we have
        MB – MP = CN – PN ⇒ PB = CP
        Hence (i) AP = PD and (ii) PB = CP


        Q.3     If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with th chlords.
        Sol.

        Given : AB and CD are chords of a circle with centre O, AB and CD intersect at P and AB = CD.
        To prove : ∠OPE=∠OPF

        3
        Construction : Draw OE ⊥ AB and OF ⊥ CD. Join OP.
        In Δs OEP and OFP , we have
        ∠OEP=∠OFP [Since Each = 90º]
        OP = OP [Common]
        OE = OF [ Equal chords of a circle are equidistant from the centre]
        Therefore by RHS criterion of congruence
        ΔOEP≅ΔOFP
        ⇒ ∠OPE=∠OPF [C.P.C.T.]


        Q.4      If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure)

        33
        Sol.

        Let OM be perpendicular from O on line l. We know that the perpendicular from the centre of a circle to a chord, bisects the chord.
        Since BC is a chord of the smaller circle and OM ⊥ BC

        8
        Therefore BM = CM … (1)
        Again AD is a chord of the larger circle and OM ⊥ AD.
        Therefore AM = DM … (2)
        Subtracting (1) from (2) we get
        AM – BM = DM – CM ⇒ AB = CD.


        Q.5     Three girls Reshma , Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
        Sol.

        Let the three girls Reshma, Salma and Mandip are standing on the circle of radius 5 cm at points B, A and C respectively.

        10
        We know that if AB and AC are two equal chords of a circle, then the centre of the circles lies on the bisector of ∠BAC.
        Here AB = AC = 6 cm. So the bisector of ∠BAC passes through the centre O i.e. OA is the bisector of ∠BAC.
        Since the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle. Therefore, M divides BC in the ratio 6 : 6 = 1 : 1 i.e., M is the middle point of BC.
        Now, M is the mid- point of BC ⇒ OM ⊥ BC.
        In right ∠dΔABM , we have
        AB2=AM2+BM2
        ⇒ 36=AM2+BM2
        ⇒ BM2=36−AM2 … (1)
        In the right Δ OBM, we have
        OB2=OM2+BM2
        ⇒ 25=(OA−AM)2+BM2
        ⇒ BM2=25−(OA−AM)2
        ⇒ BM2=25−(5−AM)2 … (2)
        From (1) and (2) , we get
        36−AM2=25−(5−AM)2
        ⇒ 11−AM2+(5−AM)2=0
        ⇒ 11−AM2+25−10AM+AM2=0
        ⇒ 10AM=36
        ⇒ AM=3.6
        Putting AM = 3.6 in (1) we get
        BM2=3.6−(3.6)2=36−12.96
        ⇒ BM=36−12.96−−−−−−−−−√=23.04−−−−√=4.8cm
        ⇒ BC=2BM=2×4.8=9.6cm
        Hence, the distance between Reshma and Mandip = 9.6 cm


        Q.6      A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find th e length of the string of each phone.
        Sol.

        Let ABC is an equilateral triangle of side 2x metres.
        Clearly, BM=BC2=2x2=xmetres
        In right ∠dΔABM
        9
        AM2=AB2−BM2
        =(2x)2−x2=4x2−x2=3x2
        ⇒ AM=3–√x
        Now , OM = AM – OA (3x−−√−20)m
        In right ∠dΔOBM, we have
        OB2=BM2+OM2
        ⇒ 202=x2+(3x−−√−20)2
        ⇒ 400=x2+3x2−403x−−√+400
        ⇒ 4x2−403x−−√=0
        ⇒ 4x(x−103–√)=0
        Since x≠0 Therefore x−103–√=0
        ⇒ x=103–√
        Now, BC=2BM=2x=203–√
        Hence, the length of each string =203–√m

        Exercise 10.5

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

        11Sol.

        Since are ABC makes ∠AOC=∠AOB+∠BOC = 60º + 30º = 90º at the centre of the circle and ∠ADC at a point on the remaining part of the circle.
        Therefore ∠ADC=12(∠AOC)=12×90o=45o


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

        Let PQ be chord. Join OP and OQ.
        It is given that PQ = OP = OQ (Since Chord = radius)
        Therefore Δ OPQ is equilateral.
        ⇒ ∠POQ = 60º13Since are PBQ makes reflex POQ = 360º – 60º = 300º at centre of the circle and ∠PBQ at a point in the minor arc of the circle.
        Therefore ∠PBQ=12(reflex∠POQ)
        =12×300o=150o
        Similarly, ∠PAQ=12(∠POQ)=12 (60º) = 30º
        Hence, angle subtended by the chord on the minor are 150º and on the major chord = 30º.


        Q.3      In figure ∠PQR = 100º, there P, Q and R are points on a circle with centre O. Find ∠OPR.
        12Sol.

        Since the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same are at a point on the circumference.
        Therefore Reflex ∠POR=2∠PQR
        ⇒ Reflex ∠POR=2×100∘=200∘
        ⇒ ∠POR = 360º – 200º = 160º
        In Δ OPR, OP = OR [Radii of the same circle]
        ⇒ ∠OPR=∠ORP
        [Angles opp. to equal sides are equal]
        and ∠POR= 160º ……… (1) [Proved above]
        In ΔOPR,
        ∠OPR+∠ORP+∠PQR=180∘ [Angle sum property]
        ⇒ 160∘�+2∠OPR=180∘
        ⇒ 2∠OPR=180∘�−160∘
        ⇒ 2∠OPR=20∘
        ⇒ ∠OPR=10∘


        Q.4      In figure ∠ABC= 69º, ∠ACB= 31º, Find ∠BDC.

        14Sol.

        In Δ ABC,
        ∠BAC+∠ABC+∠BCA=180o
        ⇒ ∠BAC+69o+31o=180∘
        ⇒ ∠BAC=180o−(69o+31o)
        =180o−100o=80o
        Since angles in the same segment are equal
        Therefore ∠BDC=∠BAC=80o


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

        ∠CED+∠CEB = 180º [Linear pair]
        ⇒ ∠CED+ 130º = 180º
        ⇒ ∠CED= 180º – 130º
        = 50º
        In Δ ECD, ∠EDC+∠CED+∠ECD = 180º
        ⇒ ∠EDC+ 50º + 20º = 180º
        ⇒ ∠EDC= 180º – 50º – 20º
        = 110º
        ⇒ ∠BDC=∠EDC = 110º
        Since angles in the same segment are equal
        Therefore ∠BAC=∠BDC= 110º.


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

        ∠BDC=∠BAC [Angles in the same segment]
        ⇒ ∠BDC= 30º [Since ∠BAC= 30º (given]

        16In Δ BCD, we have
        ∠BDC+∠DBC+∠BCD = 180º [Sum of ∠s of a Δ]
        ⇒ 30º + 70º + ∠BCD = 180º [Since ∠DBC = 70º, ∠BDC = 30º]
        ⇒ ∠BCD = 180º – 30º – 70º = 80º
        If AB = BC, than ∠BCA=∠BAC = 30º [Angles opp. to equal sides in a Δ are equal]
        Now , ∠ECD=∠BCD−∠BCE
        = 80º – 30º = 50º
        [Since ∠BCD= 80º (found above) and ∠BCE=∠BCA= 30º]
        Hence, ∠BCD= 80º and ∠ECD= 50º


        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.
        Sol.

        Diagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quad. ABCD.
        To prove : Quadrilateral ABCD is a rectangle.

        15Solution : – Since all the radii of the same circle are equal
        Therefore OA = OB = OC = OD
        ⇒ OA=OC=12AC
        and OB=OD=12BD
        ⇒ AC = BD
        Therefore the diagonals of the quadrilateral ABCD are equal and bisect each other.
        ⇒ Quadrilateral ABCD is a rectangle.


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

        Given : Non – parallel sides AD and BC of a trapezium are equal.
        To prove : ABCD is a cyclic trapezium.
        Construction : Draw DE ⊥ AB and CF ⊥ AB.

        17Proof : In order to prove that ABCD is a cyclic trapezium it is sufficient to prove that ∠B+∠D = 180º.
        In Δs DEA and CFB, we have
        AD = BC [Given]
        ∠DEA=∠CFB [Each = 90º]
        and DE = CF [Distance between two|| lines is always equal]
        Therefore by RHS criterion of congruence, we have
        DEA≅CFB
        ⇒ ∠A=∠Band∠ADE=∠BCF
        (Corresponding parts of congruent triangles are equal)
        Now, ∠ADE=∠BCF
        ⇒ 90º + ∠ADE= 90º + ∠BCF
        ⇒ ∠EDC+∠ADE=∠FCD+∠BCF
        [Since ∠EDC= 90º and ∠FCD= 90º]
        ⇒ ∠ADC=∠BCD
        ⇒ ∠D=∠C
        Thus , ∠A=∠Band∠C=∠D
        Therefore , ∠A+∠B+∠C+∠D= 360º
        [Since sum of the angles of a quad. is 360º]
        ⇒ 2∠B+2∠D= 360º
        ⇒ ∠B+∠D= = 180º
        Hence, ABCD is a cyclic trapezium.


        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 figure). Prove that ∠ACP=∠QCD

        18Sol.

        Since angles in the same segment are equal.
        Therefore ∠ACP=∠ABP … (1)
        and ∠QCD=∠QBD … (2)
        Also ∠ABP=∠QBD … (3) [Vertically opp. angles]
        Therefore From (1), (2) and (3) , we have
        ∠ACP=∠QCD


        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.
        Sol.

        Given : Two circles are drawn with sides AB and AC of Δ ABC as diameters. The circle intersect at D.

        20To prove : D lies on BC.
        Construction : Join A and D.
        Proof : Since AB and AC are the diameters of the two circles. [Given]
        Therefore ∠ADB= 90º [Angles in a semi- circle]
        and, ∠ADC= 90º [Angles in a semi-circle]
        Adding we get ∠ADB+∠ADC = 90º + 90º = 180º
        ⇒ BDC is a straight line.
        Hence, D lies on BC.


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

        Δ ABC and ADC are right ∠d with common hypotenuse AC. Draw a circle with AC as diameter passing through B and D. Join BD.
        21Clearly, ∠CAD=∠CBD [Since Angles in the same segment are equal]


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

        Given : ABCD is a parallelogram inscribed in circle.
        To prove : ABCD is a rectangle.
        Proof : Since ABCD is a cyclic parallelogram.
        22Therefore ∠A+∠C = 180º … (1)
        But ∠A=∠C … (2)
        From (1) and (2), we have
        ∠A=∠C = 90º
        Similarly, ∠B=∠D = 90º
        Therefore Each angle of ABCD is of 90º
        Hence, ABCD is a rectangle.

        Exercise 10.6

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

        1
        Given : : Two circles with centres A and B, which intersect each other at C and D.
        To prove : ∠ACB=∠ADB
        Construction : Join AC, AD, BD and BC.
        Proof : In Δs ACB and ADB, we have
        AC = AD [Radii of the same circle]
        BC = BD [Radii of the same circle]
        AB = AB [Common]
        Therefore by SSS criterion of congruence,
        ΔACB≅ΔADB
        ⇒ ∠ACB=∠ADB [C.P.C.T.]


        Q.2     Two chords AB and CD of lengths 5 cm and 11 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.
        Sol.

        Let O be the centre of the given circle and let its radius be r cm. Draw OP⊥AB and OQ⊥CD. Since OP⊥AB, OQ⊥CD and AB || CD. Therefore, points P, O and Q are collinear. So PQ = 6 cm.
        2
        Let OP = x. Then, OQ = (6 – x) cm.
        Join OA and OC. Then, OA = OC = r.
        Since the perpendicular from the centre to a chord of the circle bisects the chord.
        Therefore AP = PB = 2.5 cm and CQ = QD = 5.5 cm.
        In right Δs QAP and OCQ, we have
        OA2=OP2+AP2 and OC2=OQ2+CQ2
        ⇒ r2=x2+(2.5)2 …………… (1)
        and r2=(6−x)2+(5.5)2 ………….. (2)
        ⇒ x2+(2.5)2=(6−x)2+(5.5)2
        ⇒ x2+6.25=36−12x+x2+30.25
        ⇒ 12x = 60 ⇒ x = 5
        Putting x = 5 in (1), we get
        r2=52+(2.5)2=25+6.25=31.25
        ⇒ r=31.25−−−−√ = 5.6 (approx.)
        Hence, the radius of the circle is 5.6 cm (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 centre, what is the distance of the other chord from the centre.
        Sol.

        3
        Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm.
        Let the radius of the circle be r cm.
        Draw OP⊥AB and OQ⊥CD. Since AB || CD and OP⊥AB, OQ⊥CD.
        Therefore, points O, Q and P are collinear. Clearly OP = 4 cm, and P, Q are mid-points of AB and CD respectively.
        Therefore AP=PB=12AB=3 cm
        and, CQ=QD=12CD=4 cm
        In rt. ∠d ΔOAP, we have
        OA2=OP2+AP2
        ⇒ r2=42+32=16+9=25
        ⇒ r = 5
        In rt. ∠d Δ OCQ, we have
        OC2=OQ2+CQ2
        ⇒ r2=OQ2+42
        ⇒ 25=OQ2+16
        ⇒ OQ2=9
        ⇒ OQ = 3
        Hence, the distance of chord CD from the centre 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.
        Sol.

        4

        Since an exterior angle of a triangle is equal to the sum of the opposite angles.
        In ΔBDC, we have
        ∠ADC=∠DBC+∠DCB ………… (1)
        Since angle at the centre is twice the angle at a point on the remaining part of circle.
        Therefore ∠ADC=12∠AOC
        and ∠DCB=12∠DOE …………….. (2)
        From (1) and (2), we have
        12∠AOC=∠ABC+12∠DOE [Since ∠DBC=∠ABC]
        ⇒ ∠ABC=12(∠AOC−∠DOE)
        Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.


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

        5
        Given : ABCD is a rhombus. AC and BD are its two diagonals which bisect each other at right angles.
        To prove : A circle drawn on AB as diameter will pass through O.
        Construction : From O draw PQ || AD and EF || AB.
        Proof : AB = DC ⇒ 12AB=12DC
        AQ = DP
        [Since Q and P are mid-points of AB and CD]
        Similarly AE = OQ
        ⇒ AQ = OQ = QB
        ⇒ A circle drawn with Q as centre and radius AQ passes through A, O and B.
        The circle thus obtained is the required circle.


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

        6
        In order to prove that AE = AD
        i.e., ΔAED is an isosceles triangle it is sufficient to prove that ∠AED=∠ADE. Since ABCE is a cyclic quadrilateral.
        Therefore ∠AED+∠ABC=180∘ ……. (1)
        Now, CDE is a straight line.
        ⇒ ∠ADE+∠ADC=180∘ ………… (2)
        [Since ∠ADC and ∠ABC are opposite angles of a parallellogram i.e. ∠ADC=∠ABC
        From (1) and (2), we get
        ∠AED+∠ABC=∠ADE+∠ABC
        ⇒ ∠AED=∠ADE
        Therefore In ΔAED, we have
        ∠AED=∠ADE
        ⇒ AD = AE.


        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.
        Sol.

        7
        (i) Let AB and CD be two chords of a circle with center O.
        Let they bisect each other at O.
        Join AC, BD, AD and BC.
        In Δs AOC and BOD, we have
        OA = OB [Since O is the mid-point of AB]
        ∠AOC=∠BOD [Vert. opp. ∠s]
        and, OC = OD [Since O is the mid-point of CD]
        By SAS criterion of congruence,
        ΔAOC≅ΔBOD
        ⇒ AC = BD
        ⇒ arc(AC) = arc(BD) …………. (1)
        Similarly, from Δs AOD and BOC, we have
        arc(AD) = arc(BC)
        From (1) and (2), we have
        arc(AC) + arc(AD) = arc(BD) + arc(BC)
        ⇒ arc(CAD) = arc(CBD)
        ⇒ CD divides the circle into two parts
        ⇒ CD is a diameter.
        Similarly, AB is a diameter.
        (ii) Since ΔAOC≅ΔBOD [Proved above]
        ⇒ ∠OAC i.e., ∠BAC=∠OBD i.e., ∠ABD
        ⇒ AC || BD
        Again, ΔAOD≅ΔCOB
        ⇒ AD || CB
        ⇒ ABCD is a cyclic parallelogram.
        ⇒ ∠DAC=∠DBA ………….. (3)
        [Opp. ∠s of a parallelogram are equal]
        Also, ACBD is a cyclic quadrilateral
        Therefore ∠DAC+∠DBA=180∘ ……………. (4)
        From (3) and (4), we get
        ∠DAC=∠DBA=90∘
        Hence, ABCD is a rectangle.


        Q.8      Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90∘−12A, 90∘−12B, 90∘−12C.
        Sol.

        8
        We have ∠D = ∠EDF
        = ∠EDA+∠ADF
        = ∠EBA+∠FCA
        [Since ∠EDA and ∠EBA are the angles in the same segment of the circle]
        Therefore ∠EDA=∠EBA. Similarly, ∠ADF and ∠FCA are the angles in the same segment and hence, ∠ADF=∠FCA
        =12∠B+12∠C
        [Since BE is the internal bisector of ∠B and CF is the internal bisector ∠C
        ∠D=∠B+∠C2
        Similarly, ∠E=∠C+∠A2
        and ∠F=∠A+∠B2
        ⇒ ⇒∠D=180∘−∠A2
        ∠E=180∘−∠B2
        and ∠F=180∘−∠C2
        [Since ∠A+∠B=180∘]
        ⇒ ∠D=90∘−∠A2,
        ∠E=90∘−∠B2
        and ∠F=90∘−∠C2
        ⇒ The angles of the ΔDEF are 90∘−12A, 90∘−12B, 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.
        Sol.

        9Let O and O’ be the centre of two congruent circles.
        Since AB is a common chord of these circles.
        Therefore arc(ACB) = arc(ADB)
        ∠BPA=∠BQA
        ⇒ 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.
        Given : ABC is a triangle inscribed in a circle with centre at O, E is a point on the circle such that AE is the internal bisector of ∠BAC and D is the mid-point of BC.
        10

        To Prove : DE is the right bisector of BC i.e. ∠BDE=∠CDE=90∘.
        Construction : Join BE and EC.
        Sol.

        In ΔBDE and ΔCDE, we have
        BE = CE
        [Since ∠BAE=∠CAE, therefore arc(BE) = arc(CE) ⇒ chord BE = chord CE]
        BD = CD [Given]
        DE = DE [Common]
        Therefore by SSS criterion of congruence,
        ΔBDE≅ΔCDE
        ⇒ ∠BDE=∠CDE [C.P.C.T.]
        Also, ∠BDE+∠CDE=180∘ [Linear pari]
        Therefore ∠BDE=∠CDE=90∘
        Hence, DE is the right bisector of BC.

        Prev Chapter Notes – Circles
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