• Home
  • Courses
  • Online Test
  • Contact
    Have any question?
    +91-8287971571
    contact@dronstudy.com
    Login
    DronStudy
    • Home
    • Courses
    • Online Test
    • Contact

      Class 9 Maths

      • Home
      • All courses
      • 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 – Area of Parallelogram

        (1) Prove that a diagonal of a parallelogram divides it into two triangles of equal area.
        Given:  A parallelogram ABCD in which BD is one of the diagonals.
        To prove:   
        Proof: Since two congruent geometrical figures have equal area. Therefore, in order to prove that  it is sufficient to show that
        ΔABD≅ΔCDB
        In Δs ABD and CDB, we have
        AB=CD
        AD=CB
        And, BD=DB
        So, by SSS criterion of congruence, we have
        ΔABD≅ΔCDB
        Hence, ar(ΔABD)=ar(ΔCDB)

        (2) Prove that parallelograms on the same base and between the same parallels are equal in area.
        Given: Two parallelograms ABCD and ABEF, which have the same base AB and which are between the same parallel lines AB and FC.
        To prove: ar(parallelogramABCD)=ar(parallelogramABCD)Proof: In Δs ADF and BCE, we have
        AD=BC
        AF=BE
        And, ∠DAF=∠CBE         [ ⸪ AD∥BC and AF∥BE]
        So, by SAS criterion of congruence, we have
        ΔADF≅ΔBCE
        ar(ΔADF)=ar(ΔBCE)   …..(i)
        Now, ar(parallelogram ABCD)=ar(sq.ABED)+ar(ΔBCE)
        ar(parallelogram ABCD)=ar(sq.ABED)+ar(ΔADF)    [Using(i)]
        ar(parallelogram ABCD)=ar(parallelogram ABEF)
        Hence, ar(parallelogram ABCD)=ar(parallelogram ABEF)

         

        (3) Prove that the area of a parallelogram is the product of its base and the corresponding altitude.
        Given: A parallelogram ABCD in which AB is the base and AL the corresponding altitude.
        To prove: ar(parallelogram ABCD)=AB×AL
        Construction: Complete the rectangle ALMB by drawing BM⊥CD.Proof: Since ar(parallelogram ABCD) and rectangle ALMB are on the same base and between the same parallels.
        ar(parallelogram ABCD)
        =ar(rect.ALMB)
        =AB×AL    [By rect. Area axiom area of a rectangle = Base X Height]
        Hence, ar(parallelogram ABCD)=AB×AL

        (4) Prove that parallelograms on equal bases and between the same parallels are equal in area.
        Given: Two parallelograms ABCD and PQRS with equal bases AB and PQ and between the same parallels AQ and DR.
        To prove: ar(parallelogram ABCD)=ar(parallelogram PQRS)
        Construction: Draw AL⊥DR and PM⊥DRProof: Since AB⊥DR, AL⊥DR and PM⊥DR
        AL=PM
        Now, ar(parallelogram ABCD)=AB×AL
        ar(parallelogram ABCD)=PQ×PM  [AB=PQ and AL=PM]
        ar(parallelogram ABCD)=ar(parallelogram PQRS)

        (5) Prove that triangles on the same bases and between the same parallels are equal in area.
        Proof: We have,
        BD∥CA
        And, BC∥DA
        sq.BCAD is a parallelogram.Similarly, sq.BCQP is a parallelogram.
        Now, parallelograms ECQP and BCAD are on the same base BC, and between the same parallels.
        ar(parallelogram BCQP)=ar(parallelogram BCAD)    ….(i)
        We know that the diagonals of a parallelogram divides it into two triangles of equal area.
        ar(ΔPBC)=12ar(parallelogram BCQP)   …..(ii)
        And, ar(ΔABC)=12ar(parallelogram BCAQ)  ….(iii)
        Now, ar(parallelogram BCQP)=ar(parallelogram BCAD) [ From (i)]
        12ar(parallelogram BCQP)=12ar(parallelogram BCAD)
        ar(ΔABC)=ar(ΔPBC)    [From (ii) and (iii)]
        Hence, ar(ΔABC)=ar(ΔPBC)

         

        (6) Prove that the area of a triangle is half the product of any of its sides and the corresponding altitude.
        Given: A ΔABC in which AL is the altitude to the side BC.
        To prove: ar(ΔABC)=12(BC×AL)
        Construction: Through C and A draw CD∥BA and AD∥BC respectively, intersecting each other at D.Proof: We have,
        BA∥CD
        And, AD∥BC
        BCDA is a parallelogram.
        Since AC is a diagonal of parallelogram BCDA.
        ar(ΔABC)=12ar(parallelogram BCAD)
        ar(ΔABC)=12(BC×AL) [BC is the base and AL is the corresponding altitude of parallelogram BCDA]

        (7) Prove that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram.
        Given: A ΔABC and a parallelogram BCDE on the same base BC and between the same parallel BC and AD.
        To prove: ar(ΔABC)=12ar(parallelogram BCDE)
        Construction: Draw AL⊥BC and DM⊥BC, meeting BC produced in M.Proof: Since A, E and D are collinear and BC∥AD
        AL=DM    …..(i)
        Now,
        ar(ΔABC)=12(BC×AL)
        ar(ΔABC)=12(BC×DM)   [AL=DM (from (i)]
        ar(ΔABC)=12ar(parallelogram BCDE)

        (8) Prove that the area of a trapezium is half the product of its height and the sum of parallel sides.
        Given: A trapezium ABCD in which AB∥DC; AB=a, DC=b and AL=CM=h, where AL⊥DC and CM⊥AB.
        To prove: ar(trap. ABCD)=12h×(a+b)
        Construction: Join ACProof: We have,
        ar(trap. ABCD)=ar(ΔABC)+ar(ΔACD)
        ar(trap. ABCD)=12(AB×CM)+12(DC×AL)
        ar(trap. ABCD)=12ah×12bh  [AB=a and DC=b]
        ar(trap. ABCD)=12h×(a+b)

        (9) Prove that triangles having equal areas and having one side of one of the triangles, equal to one side of the other, have their corresponding altitudes equal.
        Given: Two triangles ABC and PQR such that:
        ar(ΔABC)=ar(ΔPQR)
        AB=PQ
        CN and RT are the altitudes corresponding to AB and PQ respectively of the two triangles.
        To prove: CN=RTProof: In ΔABC, CN is the altitude corresponding to side AB.
        ar(ΔABC)=12(AB×CN)    ….(i)
        Similarly, we have,
        ar(ΔPQR)=12(PQ×RT)   …..(ii)
        Now, ar(ΔABC)=ar(ΔPQR)
        12(AB×CN)=12(PQ×RT)
        (AB×CN)=(PQ×RT)
        (PQ×CN)=(PQ×RT)    [ AB=PQ (Given)]
        CN=RT

         

        (10) Prove that if each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram.
        Given: A quadrilateral ABCD such that its diagonals AC and BD are such that
        ar(ΔABD)=ar(ΔCDB) and ar(ΔABC)=ar(ΔACD)
        To prove: Quadrilateral ABCD is a parallelogram.Proof: Since diagonal AC of the quadrilateral ABCD separates it into two triangles of equal area. Therefore,
        ar(ΔABC)=ar(ΔACD)    …..(i)
        But, ar(ΔABC)+ar(ΔACD)=ar(quad.ABCD)
        2ar(ΔABC)=ar(quad.ABCD)    [Using (i)]
        ar(ΔABC)=12ar(quad.ABCD) ….(ii)
        Since diagonal BD of the quadrilateral ABCD separates it into triangles of equal area.
        ar(ΔABD)=ar(ΔBCD)   ….(iii)
        But, ar(ΔABD)+ar(ΔBCD)=ar(quad.ABCD)
        2ar(ΔABD)=ar(quad.ABCD)    [Using(iii)]
        ar(ΔABD)=12ar(quad.ABCD)   …..(iv)
        From (ii) and (iv), we get
        ar(ΔABC)=ar(ΔABD)
        Since Δs ABC and ABD are on the same base AB. Therefore they must have equal corresponding altitudes.
        i.e. Altitude from C of ΔABC = Altitude from D of ΔABD
        DC∥AB
        Similarly, AD∥BC
        Hence, quadrilateral ABCD is a parallelogram.

        (11) Prove that the area of a rhombus is half the product of the lengths of its diagonals.
        Given: A rhombus ABCD whose diagonals AC and BD intersect at O.
        To prove: ar(rhombus ABCD) =12(AC×BD)Proof: Since the diagonals of a rhombus intersect at right angles. Therefore,
        OB⊥AC and OD⊥AC
        ar(rhombus) =ar(ΔABC)+ar(ΔADC)
        ar(rhombus) =12(AC×BO)+12(AC×DO)
        ar(rhombus)=12(AC×(BO+DO))
        ar(rhombus) =12(AC×BD)

        (12) Prove that diagonals of a parallelogram divide it into four triangles of equal area.
        Given: A parallelogram ABCD. The diagonals AC and BD intersect at O.
        To prove: ar(ΔOAB)=ar(ΔOBC)=ar(ΔOCD)=ar(ΔAOD)Proof: Since the diagonals of a parallelogram bisect each other at the point of intersection.
        OA=OC and OB=OD
        Also, the median of a triangle divides it into two equal parts.
        Now, in ΔABC, BO is the median.
        ar(ΔOAB)=ar(ΔOBC)  ….(i)
        In ΔBCD, CO is the median
        ar(ΔOBC)=ar(ΔOCD)    …..(ii)
        In ΔACD, DO is the median
        ar(ΔOCD)=ar(ΔAOD)   ….(iii)
        From (i), (ii) and (iii), we get
        ar(ΔOAB)=ar(ΔOBC)=ar(ΔOCD)=ar(ΔAOD)

         

        (13) Prove that if the diagonals AC and BD of a quadrilateral ABCD, intersect at O and separate the quadrilateral into four triangles of equal area, then the quadrilateral ABCD is parallelogram.
        Given: A quadrilateral ABCD such that its diagonals AC and BD intersect at O and separate it into four parts such that
        ar(ΔOAB)=ar(ΔOBC)=ar(ΔOCD)=ar(ΔAOD)
        To prove: Quadrilateral ABCD is a parallelogram.Proof: We have,
        ar(ΔAOD)=ar(ΔBOC)
        ar(ΔAOD)+ar(ΔAOB)=ar(ΔBOC)+ar(ΔAOB)
        ar(ΔABD)=ar(ΔABC)
        Thus, Δs ABD and ABC have the same base AB and have equal areas. So, their corresponding altitudes must be equal.
        Altitude from ΔABD Altitude from C of ΔABC
        DC∥AB
        Similarly, we have, AD∥BC.
        Hence, quadrilateral ABCD is a parallelogram.

        (14) Prove that a median of a triangle divides it into two triangles of equal area.
        Given: A ΔABC in which AD is the median.
        To prove: ar(ΔABD)=ar(ΔADC)
        Construction: Draw AL⊥BC.Proof: Since AD is the median of ΔABC. Therefore, D is the mid point of BC.
        BD=DC
        BD×AL=DC×AL  [ Multiplying both sides by AL]
        12(BD×AL)=12(DC×AL)
        ar(ΔABD)=ar(ΔADC)
        ALITER Since Δs ABD and ADC have equal bases and the same altitude AL.
        ar(ΔABD)=ar(ΔADC)

        Prev Summary of All Concepts and Miscellaneous Questions
        Next NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4

        Leave A Reply Cancel reply

        Your email address will not be published. Required fields are marked *

        All Courses

        • Backend
        • Chemistry
        • Chemistry
        • Chemistry
        • Class 08
          • Maths
          • Science
        • Class 09
          • Maths
          • Science
          • Social Studies
        • Class 10
          • Maths
          • Science
          • Social Studies
        • Class 11
          • Chemistry
          • English
          • Maths
          • Physics
        • Class 12
          • Chemistry
          • English
          • Maths
          • Physics
        • CSS
        • English
        • English
        • Frontend
        • General
        • IT & Software
        • JEE Foundation (Class 9 & 10)
          • Chemistry
          • Physics
        • Maths
        • Maths
        • Maths
        • Maths
        • Maths
        • Photography
        • Physics
        • Physics
        • Physics
        • Programming Language
        • Science
        • Science
        • Science
        • Social Studies
        • Social Studies
        • Technology

        Latest Courses

        Class 8 Science

        Class 8 Science

        ₹8,000.00
        Class 8 Maths

        Class 8 Maths

        ₹8,000.00
        Class 9 Science

        Class 9 Science

        ₹10,000.00

        Contact Us

        +91-8287971571

        contact@dronstudy.com

        Company

        • About Us
        • Contact
        • Privacy Policy

        Links

        • Courses
        • Test Series

        Copyright © 2021 DronStudy Pvt. Ltd.

        Login with your site account

        Lost your password?

        Modal title

        Message modal