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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 – Constructions Exercise

        Q.1      Construct an angle of 90º at the  initial  point of a given ray and justify the construction.
        Sol.         Steps of construction :   
        1

        1.  Draw  a ray OA.
        2. With its initial point  O as  centre and any radius,  draw an arc CDE, cutting OA at C.
        3. With  centre  C and same  radius  (as in step 2), draw  an arc, cutting  the arc CDE at D.
        4. With D as centre and the same radius, draw  an arc cutting the arc CDE at E.
        5. With D and E as centres, and any convenient radius (morethan12DE), draw two arcs intersecting at P.
        6. Join OP. Then  ∠AOP=90o
        Justification : –
        By  construction , OC = CD = OD
        Therefore ΔOCD is an  equilateral triangle. So, ∠COD=60o
        Again OD  = DE = EO
        Therefore Δ ODE is also an equilateral triangle. So ∠DOE=60o
        Since OP bisects ∠DOE,so∠POD=30o.
        Now, ∠AOP=∠COD+∠DOP=60∘�+30∘�=90o


        Q.2      Construct an angle of 45º at the initial  point of a  given ray and justify the construction.
        Sol. 

        Steps of Construction :
        1.  Draw a ray OA.
        2. With O as centre and any suitable  radius  draw an arc cutting OA at B.
        3. With B as centre and same radius cut the previous drawn arc at C and then with C as centre  and same radius  cut the arc at D.
        4. With C as centre and radius  more  than half CD draw an arc.
        5. With D as centre  and same  radius  draw another  arc to cut the  previous arc at E.
        6. Join  OE. Then ∠AOE=90o
        7. Draw the bisector OF of  ∠AOE.Then∠AOF=45o

        2Justification :
        By construction ∠AOE=90o and OF is the bisector of ∠AOE
        Therefore, ∠AOF=12∠AOE=12×90∘�=45o


        Q.3       Construct the angles of the following  measurements :
                      (i) 30º          (ii) 2212∘           (iii) 15º
        Sol.          (i) Steps of Construction :
        3

        1. Draw  a ray OA.
        2. With its initial  point O as centre and any radius, draw an  arc,cutting  OA at C.
        3. With  centre C and Same radius (as in step 2). Draw an arc,cutting the arc of step 2 in D.
        4. With C and D as centres, and any convenient radius (morethan12CD),draw two arcs intersecting at B.
        5. Join  OB. Then ∠AOB=30o

        (ii)  Steps of Construction :
        4
        1. Draw an angle AOB  = 90º
        2. Draw the bisector OC of  ∠AOB,then∠AOC=45o
        3. Bisect ∠AOC, such  that ∠AOD=∠COD=22.5o
        Thus ∠AOD=22.5o

        (iii) Steps of Construction :
        5
        1. Construct an ∠AOB=60o
        2. Bisect ∠AOB so that ∠AOC=∠BOC=30o.
        3. Bisect ∠AOC, so  that ∠AOD=∠COD=15o
        Thus ∠AOD=15o


        Q.4       Construct the following  angles and verify by  measuring  them by a protractor :
                       (i)  75º           (ii) 105º           (iii) 135º
        Sol.          (i)  Steps of Construction :
        6

        1.  Draw a ray OA.
        2.  Construct ∠AOB=60o
        3.  Construct ∠AOP=90o
        4.  Bisect ∠BOP so that
        ∠BOQ=12∠BOP
        =12(∠AOP−∠AOB)
        =12(90o−60o)=12×30o=15o
        So, we obtain
        ∠AOQ=∠AOB+∠BOQ
        = 60º + 15º = 75º

        Verification :
        On  measuring  ∠AOQ, with  the protractor, we find ∠AOQ=75o

        (ii) Steps of Construction :
        1. Draw a line  segment  XY.
        2. Construct ∠XYT=120o and ∠XYS=90o , so that
        ∠SYT=∠XYT−∠XYS
        = 120º – 90º
        = 30º
        3.  Bisect angle SYT, by  drawing its bisector YZ.
        Then  ∠XYZ is the   required angle  of 105º

         

        7(iii) Steps  of Construction :

        8
        1. Draw ∠AOE=90∘
        Then ∠LOE=90∘
        2. Draw   the bisector of  ∠LOE.
        Then∠AOF=135o


        Q.5       Construct an equilateral triangle, given its side  and justify the construction.
        Sol.

        Let us draw an equilateral  triangle  of side 4.6 cm (say).
        Steps  of Construction :
        1. Draw  BC = 4.6 cm
        2. With  B and C as centres and radii  equal to BC = 4.6 cm, draw two  arcs on  the same  side of BC,    intersecting each-other at A.
        3. Join AB and AC.

        9Then, ABC is the required equilateral  triangle.
        Justification : Since by construction :
        AB = BC = CA = 4.6 cm
        Therefore  Δ ABC is an equilateral  triangle.

         

        Q.1    Construct a triangle ABC in which  BC = 7 cm, ∠B=75o and  AB + AC = 13 cm
        Sol.       Steps of Construction :

        10
        1.Draw a ray BX and cut off a line segment BC = 7 cm
        2. Construct ∠XBY=75o
        3. From BY, cut off BD = 13 cm
        4. Join CD.
        5. Draw the perpendicular  bisect of CD, intersecting BA at A.
        6. Join AC.
        The  triangle ABC thus obtained is the required triangle.


        Q.2      Construct a triangle ABC in which  BC = 8 cm, ∠B=45o and AB – AC = 3.5 cm
        Sol.

        Steps of Construction :
        1. Draw  a ray BX and cut off a line  segment BC = 8 cm from it.
        2. Construct ∠YBC=45o
        3. Cut off a line  segment BD = 3.5 cm from BY.
        4. Join CD.
        5. Draw perpendicular bisector of CD intersecting  BY at a point A.
        6. Join AC

        12
        Then  ABC is the  required triangle.


        Q.3       Construct a triangle PQR in which  QR = 6 cm ∠Q=60o and PR – PQ = 2cm
        Sol.

        Steps of Construction :
        1. Draw a ray QX and cut off a line segment QR = 6 cm  from it.
        2. Construct  a ray QY making  an angle  of 60º with QR and produce YQ to form a line YQY’
        3. Cut off  a line segment QS = 2cm  from QY’.
        4. Join RS.
        5. Draw perpendicular bisector  of RS intersecting QY at a point  P.
        6. Join  PR.

        13
        Then  PQR is the required triangle.


        Q.4       Construct a triangle XYZ in which ∠Y=30o,  ∠Z=90o and XY + YZ + ZX = 11 cm.
        Sol.

        Steps  of Construction :
        1. Draw  a line segment PQ = 11 cm

        14
        2. At P,  draw a ray PL  such  that ∠LPQ=12×30o=15o
        3. At Q, draw ray QM such that ∠MQP=12×90o=45o intersecting  PL at X.
        4. Draw perpendicular bisectors of XP and XQ intersecting  PQ in Y and Z respectively.
        Then  Δ XYZ is the required  triangle.
        Note : –  For  clarity in figure, method of drawing angles of 15º and 45º have not been shown. Students should draw these angles with the help of ruler and  compass only by the method as shown earlier.


        Q.5      Construct a right triangle  whose base is 12 cm and sum  of its hypotenuse and other  side is 18 cm.
        Sol.          Steps of Construction :

        15
        1. Draw a ray BX and cut off a line segment  BC = 12 cm
        2. Construct ∠XBY=90o
        3. From by  cut off a line segment BD = 18 cm .
        4. Join CD.
        5. Draw the perpendicular  bisector of CD intersecting BD at A.
        6. Join AC
        Then  ABC is the  required triangle.

        Prev Chapter Notes – Constructions
        Next R D Sharma Solutions Constructions

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