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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

        Chapter Notes – Triangles

        (1) Triangle : It is a closed figure formed by three intersecting lines. It has three sides, three angles and three vertices.
        Consider a triangle ABC shown below:The triangle ABC will be denoted as ∆ ABC. Here, ∆ ABC have three sides AB, BC, CA; three angles ∠ A, ∠ B, ∠ C and three vertices A, B, C.

        (2) Congruence of Triangles: The word ‘congruent’ means equal in all aspects or the figures whose shapes and sizes are same.
        For triangles, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle then they are said to be congruent triangles.

        For Example: Consider two ∆ ABC and ∆ PQR as shown below:Here, ∆ ABC is congruent to ∆ PQR which is denoted as ∆ ABC ≅ ∆ PQR.
        ∆ ABC ≅ ∆ PQR means sides AB = PQ, BC = QR, CA = RP; the ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R and vertices A corresponds to P, B corresponds to Q and C corresponds to R.
        Note: CPCT is short form for Corresponding Parts of Congruent Triangles.

         

        (3) Criteria for Congruence of Triangles:
        (i) SAS Congruence Rule:
        Statement: Two triangles are congruent if two sides and the included angle of one triangle are equal to the sides and the included angle of the other triangle.

        For example: Prove Δ AOD ≅ Δ BOC.From figure, we can see that
        OA = OB and OC = OD
        Also, we can see that, ∠ AOD and ∠ BOC form a pair of vertically opposite angles,
        ∠ AOD = ∠ BOC
        Now, since two sides and an included angle of triangle are equal, by SAS congruence rule, we can write that Δ AOD ≅ Δ BOC.

        (ii) ASA Congruence Rule:
        Statement: Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.
        Proof: Suppose we have two triangles ABC and DEF, such that ∠ B = ∠ E, ∠ C = ∠ F, and BC = EF.
        We need to prove that Δ ABC ≅ Δ DEF.
        Case 1: Suppose AB = DE.From the assumption, AB = DE and given that ∠ B = ∠ E, BC = EF, we can say that Δ ABC ≅ Δ DEF as per the SAS rule.
        Case 2: Suppose AB > DE or AB < DE.Let us take a point P on AB such that PB = DE as shown in the figure.
        Now, from the assumption, PB = DE and given that ∠ B = ∠ E, BC = EF, we can say that Δ PBC ≅ Δ DEF as per the SAS rule.
        Now, since triangles are congruent, their corresponding parts will be equal. Hence, ∠ PCB = ∠DFE
        We are given that ∠ ACB = ∠ DFE, which implies that ∠ ACB = ∠PCB
        This thing is possible only if P are A are same points or BA = ED.
        Thus, Δ ABC ≅ Δ DEF as per the SAS rule.
        On similar arguments, for AB < DE, it can be proved that Δ ABC ≅ Δ DEF.

        For Example: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.From the figure, we can see that,
        ∠ AOD = ∠ BOC (Vertically opposite angles)
        ∠ CBO = ∠ DAO (Both are of 90o)
        BC = AD (Given)
        Now, as per AAS Congruence Rule, we can say that Δ AOD ≅ Δ BOC.
        Hence, BO = AO which means CD bisects AB.

         

        (4) Some Properties of a Triangle:
        Theorem 1: Angles opposite to equal sides of an isosceles triangle are equal.
        Proof: Suppose we are given isosceles triangle ABC having AB = AC.
        We need to prove that ∠ B = ∠CFirstly, we will draw bisector of ∠ A which intersects BC at point D.
        For the Δ BAD and Δ CAD, given that AB = AC, from the figure ∠ BAD = ∠ CAD and AD = AD.
        Thus, by SAS rule Δ BAD ≅ Δ CAD.
        Therefore, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles.
        Hence, ∠ B = ∠C.

        For Example: In ∆ ABC, AB = AC, D and E are points on BC such that BE = CD. Show that AD = AE.From the figure, we can see that in Δ ABD and Δ ACE,
        AB = AC and
        ∠ B = ∠ C (Angles opposite to equal sides)
        Given that BE = CD.
        Subtracting DE from both the sides, we have,
        BE – DE = CD – DE i.e. BD = CE.
        Now, using SAS rule, we can say that Δ ABD ≅ Δ ACE
        Therefore, by CPCT, AD = AE.

        Theorem 2: The sides opposite to equal angles of a triangle are equal.
        For Example: In Δ ABC, the bisector AD of ∠ A is perpendicular to side BC. Show that AB = AC and Δ ABC is isosceles.From the figure, we can see that in Δ ABD and Δ ACD,
        It is given that, ∠ BAD = ∠ CAD
        AD = AD (Common side)
        ∠ ADB = ∠ ADC = 90°
        So, Δ ABD ≅ Δ ACD by ASA congruence rule.
        Therefore, by CPCT, AB = AC (CPCT) or in other words Δ ABC is an isosceles triangle.

         

        (5) Some More Criteria for Congruence of Triangles:
        (i) SSS Congruence Rule:
        Statement: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

        For Example: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR. Show that Δ ABM ≅ Δ PQN.From the figure, we can see that, AM is the median to BC.
        So, BM = ½ BC.
        Similarly, PN is median to QR. So, QN = ½ QR.
        Now, BC = QR.
        So, ½ BC = ½ QR i.e. BM = QN
        Given that, AB = PQ, AM = QN and AM = PN.
        Therefore, Δ ABM ≅ Δ PQN by SSS Congruence Rule.

        (6) Inequalities in a Triangle:
        Theorem 1: If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).
        Theorem 2: In any triangle, the side opposite to the larger (greater) angle is longer.
        Theorem 3: The sum of any two sides of a triangle is greater than the third side.

        For Example: For the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.Given, PR > PQ.
        Therefore, ∠ PQR > ∠ PRQ (As per angle opposite to larger side is larger) – (1)
        Also, PS bisects QPR, so, ∠ QPS = ∠ RPS – (2)
        Now, ∠ PSR = ∠ PQR + ∠ QPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. – (3)
        Similarly, ∠ PSQ = ∠ PRQ + ∠ RPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. – (4)
        Adding (1) and (2), we get,
        ∠ PQR + ∠ QPS > ∠ PRQ + ∠ RPS
        Now, from 3 & 4, we get,
        ∠ PSR > ∠ PSQ.

        For Example: D is a point on side BC of Δ ABC such that AD = AC. Show that AB > AD.Given that AD = AC,
        Hence, ∠ ADC = ∠ ACD as they are angles opposite to equal sides.
        Now, ∠ ADC is an exterior angle for ΔABD. Therefore, ∠ ADC > ∠ ABD or, ∠ ACD > ∠ ABD or, ∠ ACB > ∠ ABC.
        So, AB > AC since side opposite to larger angle in Δ ABC.
        In other words, AB > AD (AD = AC).

        Prev Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions
        Next NCERT Solutions – Triangles Exercise 7.1 – 7.5

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