• 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

        NCERT Solutions – Quadrilaterals Exercise 8.1 8.2

        Exercise 8.1

        Q.1     The  angles of quadrilateral  are in the ratio 3 : 5 : 9 : 13. Find  all the angles of the quadrilateral.
        Sol.

        Let the angles be (3x)º, (5x)º, (9x)º and (13x)º
        Then ,       3x + 5x + 9x + 13 x = 360              [The sum of the angles of a quadrilatral is 360º]
        ⇒  30x = 360
        ⇒  x=36030=12
        Therefore, the  angles are (3 × 12)º, (5× 12)º, (9 × 12)º and (13 × 12)º i.e., 36º, 60º, 108º and 156º.


        Q.2      If the diagonals of a parallelogram are equal then show that it is a rectangle.
        Sol.       A parallelogram ABCD in which  AC = BD.

        1To prove : ABCD is a rectangle.
        Proof : In ΔsABCandDCB,wehave
        AB = DC                       [Opp. sides of a || gm]
        BC = BC                       [Common]
        and     AC = DB         [Given]
        Therefore, by  SSS criterion  of congruence.
        ΔABC≅ΔDCB
        ⇒  ∠ABC=∠DCB                      … (1)
        [Corresponding parts of congruent triangles are equal]
        But AB || DC and BC cuts them.
        Therefore,     ∠ACB+∠DCB=180o
        ⇒    2∠ABC=180o  … (2) [Sum of consecutive interior angles is 180°]
        ⇒    ∠ABC=90o
        Thus,    ∠ABC=∠DCB=90o
        ABCD is a parallelogram one of whose angle is 90º .
        Hence , ABCD is a rectangle.


        Q.3     Show that if the diagonals of a quadrilateral bisect each other at right  angles , then it is a rhombus.
        Sol. 

        A quadrilateral ABCD in which  the diagonals AC and BD intersect at O such that AO = OC , BO = OD and AC ⊥ BD.

        2To prove : ABCD is a rhombus.
        Proof : Since the diagonals  AC and BD of quadrilateral ABCD bisect each other  at right  angles.
        Therefore,  AC is the perpendicular bisector of the segment BD.
        ⇒   A and C both are equidistant from B and D.
        ⇒   AB = AD and CB = CD        … (1)
        Also , BD is the perpendicular bisector of line segment  AC.
        ⇒ B and D both  are equidistant from A and C.
        ⇒   AB = BC and AD = DC        … (2)
        From (1) and (2), we get
        AB = BC = CD = AD
        Thus , ABCD is a quadrilateral  whose diagonals bisect each other  at right angles and all four sides are equal.
        Hence , ABCD is a rhombus.
        Second Proof : First we shall prove that ABCD is a || gm.
        In ΔsAODandCOB, we have
        AO = OC                                                        [Given]
        OD = OB                                                        [Given]
        ∠AOD=∠COB       [Vertically opp. angles]
        By SAS criterion of congruence,
        ΔAOD≅ΔCOB
        ⇒   ∠OAD=∠OCB            … (1)
        [Corresponding parts of congruent triangles are equal]
        Now, line AC intersects AD and BC at A and C respectively such  that
        ∠OAD=∠OCB    [From (1)]
        i.e. alternate interior angles are equal.
        Therefore,            AD || BC
        Similarly,             AB || CD
        Hence , ABCD is a parallelogram.
        Now, we shall  prove that || gm ABCD is a rhombus.
        In ΔsAODandCOD, we have
        OA = OC                                                         [Given]
        ∠AOD=∠COD        [Both are right angles ]
        OD = OD
        Therefore, by SAS criterion of congruence
        ΔAOD≅ΔCOD
        ⇒         AD = CD                 … (2)
        [Corresponding  parts of congruent triangles are equal]
        Now, ABCD is a || gm             [Proved above]
        ⇒     AB = CD and AD  = BC                      [Opp. sides of a || gm are equal]
        ⇒     AB = CD = AD = BC                            [Using (2)]
        Hence, quadrilateral ABCD is a rhombus.


        Q.4      Show that the diagonals of a square are equal and bisect each other  at right angles.
        Sol.   

        Given  : A square ABCD.
        To prove : AC =  BD,  AC ⊥ BD and OA = OC, OB = OD.
        Proof : Since  ABCD is a square.
        Therefore,    AB || DC and  AD || BC.
        Now,  AB || DC and transversal AC intersects them at A and C respectively.

        3Therefore,    ∠BAC=∠DCA            [Alternate interior angles are equal]
        ⇒    ∠BAO=∠DCO        … (1)
        Again AB || DC and BD intersects them at B and D respectively.
        Therefore, ∠ABD=∠CDB  [Since Alternate interior angles are equal]
        ⇒    ∠ABO=∠CDO        … (2)
        Now, in ΔsAOBand∠COD, we have
        ∠BAO=∠DCO        [From (1)]
        AB = CD     [Oppositge sides of a ||gm are equal]
        and      ∠ABO=∠CDO        [From (2)]
        Therefore by ASA congruence criterion
        ΔAOB≅ΔCOD
        ⇒        OA = OC and OB = OD
        [Corresponding parts of congruent Δs are equal]
        Hence , the  diagonals bisect each other.
        In Δs ADB and BCA,  we have
        AD = BC         [Sides of a square are equal]
        ∠BAD=∠ABC    [Each equal to 90º]
        and         AB = BA         [Common]
        Therefore by  SAS criterion of congruence
        ΔADB≅ΔBCA
        ⇒        AC =  BD
        [Since Corresponding parts of congruent Δs are equal]
        Hence, the diagonals are equal .
        Now in Δs AOB and AOD we have
        OB = OD                     [Since diagonals of || gm bisect each other]
        AB = AD                     [Since sides of a square are equal]
        and,   AO = AO         [Common]
        Therefore by  SSS criterion of congruence
        ΔAOB≅ΔAOD
        ⇒     ∠AOB=∠AOD………..(3)
        [Corresponding parts of congruent Δs are equal]
        But  ∠AOB+∠AOD=180o
        ⇒∠AOB+∠AOB=180o
        ⇒ 2 ∠AOB=180o
        ⇒∠AOB=90o
        Therefore  ∠AOB=∠AOD=90o
        ⇒    AO ⊥ BD
        ⇒    AC ⊥ BD
        Hence, diagonals intersect at right  angles.


        Q.5     Show that if the diagonals of a quadrilateral are equal and bisect each other at right , angles then it is  a square.
        Sol.

        Given : A quadrilateral  ABCD in which  the diagonals AC = BD , AO= OC , BO = OD and AC ⊥ BD.

        3To prove : Quadrilateral  ABCD is a square.
        Proof : First we shall prove that ABCD is a parallelogram.
        In Δs AOD and  COB, we have
        AO = OC             [Given]
        OD = OB             [Given]
        ∠AOD=∠COB        [Vertically opp. angles]
        By SAS criterion  of congruence,
        ΔAOD≅ΔCOB
        ⇒    ∠OAD=∠OCB          [Corresponding parts of congruent triangles are equal]
        Now, line AC intersects AD and BC at A and C respectively such that
        ∠OAD=∠OCB    [From (1)]
        i.e.,  alternate interior  angles are equal
        Therefore      AD || BC
        Similarly,     AB || CD
        Hence, ABCD is a parallelogram.
        Now, we shall prove that it is a square.
        In ΔsAOBandΔAOD,  we have
        AO = AO             [Common]
        ∠AOB=∠AOD        [Each = 90º, given]
        and         OB = OD             [Since diagonals of  a ||  gm bisect each  other]
        Therefore SAS criterion of congruence
        ΔAOB≅ΔAOD
        ⇒    AB = AD                           [Corresponding  parts of congruent triangles are equal]
        But   AB = CD and AD = BC                              [Opp. sides of a || gm are equal]
        Therefore   AB = BC = CD = AD            … (2)
        Now in ΔsABDandBAC, we have
        AB = BA
        AD = BC             [Opp. sides of a ||gm are equal]
        and         BD = AC                 [Given]
        Therefore by SSS criterion  of congruence
        ΔABD≅ΔBAC
        ⇒    ∠DAB=∠CBA   [Corresponding  parts of congruent Δs  are equal]


        Q.6      Diagonal AC of  parallelogram ABCD bisects ∠A (see figure). Show that
                   
        (i)  it bisects ∠C also           (ii) ABCD is a rhombus.

        4Sol.

        (i) Given : A parallelogram ABCD in which  diagonal AC bisects ∠A
        To prove : That AC bisects ∠C.
        Proof : Since ABCD is a || gm.

        5Therefore  AB || DC.
        Now AB||DC and AC intersects them
        Therefore          ∠1=∠3....(1)    [Alternate interior angles]
        Again AD|| BC and AC intersects them
        Therefore          ∠2=∠4....(2)    [Alternate interior  angles]
        But  it is given  that AC is the bisector  of ∠A
        Therefore          ∠1=∠2            …. (3)
        From  (1) , (2) and (3) , we have
        ∠3=∠4
        Hence , AC bisects ∠C

        (ii) To prove : That ABCD is a rhombus.
        From  part (i) : (1) (2) and (3) give ∠1=∠2=∠3=∠4
        Now  in ΔABC,
        ∠1=∠4
        ⇒     BC = AB      [Sides opp. to equal  angles in a Δ are equal]
        Similarly, in Δ ADC, we have
        AD = DC
        Also ,  ABCD is a ||gm
        Therefore         AB = CD, AD = BC                 [Opp. sides of a || gm are equal]
        Combining  these, we get
        AB = BC = CD = DA
        Hence , ABCD is a rhombus.


        Q.7     ABCD is a rhombus. Show  that diagonal  AC bisects ∠A as well as ∠C and diagonal BD bisects  ∠B as well as ∠D.
        Sol.

        Given : A rhombus ABCD.
        To prove :
        (i)  Diagonal AC bisects ∠A as well ∠C
        (ii) Diagonal  BD bisects ∠B as well  as ∠D
        Proof : In  ΔADC        AD = DC        [Sides of a rhombus are equal]
        ⇒ ∠DAC=∠DCA               … (1)  [Angles opp. to equal sides of a triangle are equal]

        6Now AB || DC and AC intersects them
        ∠BCA=∠DAC            … (2)            [Alternate  angles]
        From (1) and (2), we have
        ∠DCA=∠BCA
        ⇒    AC bisects ∠C
        In ΔABC,  AB = BC                   [Sides of a rhombus  are equal]
        ⇒    ∠BCA=∠BAC  …… (3) [Angles opp. to equal sides of a triangle are equal]
        From (2) and (3) , we have
        ∠BAC=∠DAC
        ⇒    AC bisects ∠A
        Hence , diagonal  AC bisects ∠A as well as ∠C.
        Similarly, diagonal  BD bisects ∠B as well  as ∠D


        Q.8     ABCD is a rectangle in which  diagonal AC bisects ∠A as well as ∠C. Show  that :
                        (i) ABCD is a square
                 (ii) diagonal BD bisects ∠B as well as ∠D

        Sol.

        Given: ABCD is a rectangle in which  diagonal AC bisects ∠A as well as ∠C.
        To prove :
        (i) ABCD is a square.
        (ii) Diagonal BD bisects ∠B as well as ∠D.
        Proof : (i) Since AC bisects ∠A as well  as ∠C in the rectangle ABCD.

        7∠A=∠C           [All four angles of a rectangle are 90º]
        Therefore,    ∠1=∠2=∠3=∠4[Each=90o2=45o]
        Therefore,        In ΔADC,∠2=∠4
        ⇒          AD = CD                                                  [Sides opposite  to equal  angles]
        Thus, the rectangle ABCD is a square.

        (ii) In a square , diagonals bisect the angles.
        So, BD bisects ∠B  as well as ∠D


        Q.9     In parallelogram ABCD, two  points P and Q are taken  on diagonal  BD such that DP = BQ (see figure). Show that :
                  (i) ΔAPD≅ΔCQB
                  (ii) AP = CQ                   
                  (iii) ΔAQB≅ΔCPD
                  (iv)  AQ = CP
                  (v)  APCQ is a parallelogram
        9Sol.      ABCD is a parallelogram. P and Q are  points on the  diagonal BD such  that DP = BQ.

        10To prove :(i) ΔAPD≅ΔCQB
        (ii) AP = CQ
        (iii) ΔAQB≅ΔCPD
        (iv)  AQ = CP
        (v)  APCQ is a parallelogram.

        Construction : Join AC to meet BD in O.
        Proof : We know that the diagonals of a parallelogram bisect each other. Therefore AC and BD bisect each other at O.
        Therefore,            OB = OD
        But                         BQ = DP                 [Given ]
        ⇒    OB – BQ = OD – DP
        ⇒  OQ  = OP
        Thus, in  quadrilateral APCQ diagonals  AC and PQ are such that OQ = OP and OA = OC. i.e.,  the diagonals AC and PQ bisects each other.
        (v) Hence, APCQ is a parallelogram, which  prove the (υ) part.

        (i)  In Δs APD and CQB we have
        AD = CB         [Opp. sides of a ||gm ABCD]
        AP = CQ         [Opp. sides of a||gm APCQ]
        DP = BQ         [Given]
        Therefore,  by  SSS criterion  of congruence
        ΔAPD≅ΔCQB

        (ii) AP = CQ        [Opp. sides of a ||gm APCQ]

        (iii) In Δs AQB and CPD,  we have
        AB = CD      [Opp. sides of a ||gm ABCD]
        AQ = CP      [Opp. sides of a ||gm APCQ]
        BQ = DP      [Given]
        Therefore, by SSS criterion of congruence
        ΔAQB≅ΔCPD

        (iv)    AQ = CP         [Opp. sides of a ||gm APCQ]


        Q.10    ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (See figure). Show that
                   
        (i) ΔAPB≅ΔCQD      (ii) AP = CQ
        11Sol.

        (i)  Since ABCD is a parallelogram. Therefore, DC || AB.
        Now DC|| AB and transversal  BD intersects them at B and D.
        Therefore     ∠ABD=∠BDC        [Alternate interior angles]
        Now  in Δs APB and CQD we have
        ∠ABP=∠QDC        [∠ABD=∠BDC]
        ∠APB=∠CQD            [Each = 90º]
        and         AB = CD             [Opp. sides of a || gm]
        Therefore,  by AAS criterion  of congruence
        ΔAPB≅ΔCQD

        (ii)  Since ΔAPB≅ΔCQD
        Therefore,     AP = CQ         [Since corresponding  parts of congruent triangles are equal]


        Q.11   In Δs ABC and Δ DEF, AB =  DE , AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined  to vertices D, E and F respectively (see figure). Show that
                   
        (i)  Quadrilateral  ABED is a parallelogram
                   
        (ii)  Quadrilateral BEFC is a parallelogram
                   (iii) AD|| CF and AD = CF
                   
        (iv) Quadrilateral  ACFD is a parallelogram.
                   (v)  AC = DF
                   (vi) ΔABC≅ΔDEF

        12Sol.

        Given:  Two Δs ABC and DEF such that AB = DE and AB || DE. Also BC = EF and BC || EF.
        To prove : (i) quadrilateral ABED is parallelogram.
        (ii) quadrilateral BEFC is a parallelogram.
        (iii) AD|| CF and AD = CF.
        (iv) quadrilateral ACFD is a parallelogram.
        (v) AC|| DF and AC = DF
        (vi) ΔABC≅ΔDEF

        Proof : (i) Consider the quadrilateral ABED
        We  have , AB = DE and AB || DE
        ⇒ One pair of opposite  sides are equal and parallel.
        ⇒ ABED is a parallelogram.

        (ii) Now , consider  quadrilateral  BEFC , we have
        BC = EF and BC || EF
        ⇒ One pair of opposite  sides are equal  and parallel.
        ⇒ BEFC is a parallelogram.

        (iii)  Now ,  AD = BE and AD || BE        [Since ABED is a ||gm]    … (1)
        and          CF = BE and CF|| BE                 [Since BEFC is a ||gm]     … (2)
        From  (1) and (2) , we have
        AD = CF and AD|| CF

        (iv) Since AD = CF and AD || CF
        ⇒ One pair  of opposite  sides  are equal  and parallel.
        ⇒  ACFD is a parallelogram.

        (v)  Since ACFD is parallelogram.
        Therefore    AC = DF                 [Opp. sides of a|| gm ACFD]

        (vi)  In Δs ABC and DEF, we have
        AB = DE                [Opp. sides of a|| gm ABED]
        BC = EF                 [Opp. sides of a|| gm BEFC]
        and     CA = FD                 [Opp. sides of || gm ACFD]
        Therefore by SSS criterion of congruence.
        ΔABC≅ΔDEF


        Q.12   ABCD is a trapezium  in which AB || CD and AD = BC (see figure) Show that
                   
        (i)  ∠A=∠B
                   (ii) ∠C=∠D
                   (iii) ΔABC≅ΔBAD
                   (iv)  diagonal AC = diagonal  BD

        13Sol.

        Given : ABCD is a trapezium  in which AB || CD and AD = BC
        To prove : (i) ∠A=∠B
        (ii) ∠C=∠D
        (iii) ΔABC≅ΔBAD
        (iv) Diagonal AC = diagonal BD.

        Construction : Produce AB and draw a line CE|| AD.
        Proof : (i) Since AD || CE and transversal AE cuts them  at  A and E respectively.
        Therefore,     ∠A+∠E=180∘  … (1) (Consecutive interior angles are supplementary)
        Since  AB || CD and AD || CE. Therefore, AECD is parallelogram.
        ⇒    AD = CE
        ⇒    BC = CE         [Since  AD = BC (given)]
        Thus,  in Δ BCE , we have
        BC = CE (by Angle sum property)
        ⇒    ∠CBE=∠CEB
        ⇒    180−∠B=∠E
        ⇒    180−∠E=∠B            … (2)
        From  (1) and (2) , we get
        ∠A=∠B

        (ii)  Since     ∠A=∠B⇒∠BAD=∠ABD
        ⇒    180∘−∠BAD=180∘−∠ABD
        ⇒    ∠ADB=∠BCD
        ⇒    ∠D=∠Ci.e.∠C=∠D

        (iii)  In Δs ABC and BAD, we have
        BC = AD             [Given]
        AB = BA             [Common]
        ∠A=∠B    [Proved]
        Therefore  by SAS criterion  of congruence
        ΔABC≅ΔBAD
        (iv)  Since     ΔABC≅ΔBAD
        Therefore,         AC = BD                                    [Corresponding  parts of congruent triangles are equal]

        Exercise 8.2

        Q.1     ABCD is a quadrilateral  in which P, Q, R and S are mid- points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that :
                   
        (i) SR||AC and SR=12AC
                   (ii) PQ = SR
                   (iii) PQRS is a parallelogram.  

        14Sol.

        Given : A quadrilateral ABCD in which  P, Q, R, and  S are  respectively the mid- points of the sides AB, BC, CD and DA. Also  AC is its diagonal.
        To prove :
        (i) SR || AC and  SR=12AC
        (ii)  PQ = SR
        (iii) PQRS is a parallelogram.

        Proof : (i) In ΔACD, we have S is the mid- point of AD and R is the mid- point of CD.
        Then     SR||ACandSR=12AC        [Mid- point theorem]

        (ii) In Δ ABC, we have P is the mid- point of the side AB and Q is the mid- point the side BC.
        Then,     PQ || AC
        and,         PQ=12AC                                          [Mid- point theorem]
        Thus,  we have proved that :
        PQ||ACSR||AC}⇒PQ||SR
        Also   PQ=12ACSR=12AC}⇒PQ||SR

        (iii) Since PQ = SR and PQ|| SR
        ⇒  One pair  of opposite  sides are equal  and parallel.
        ⇒   PQRS is a parallelogram.


        Q.2    ABCD is a rhombus and P, Q, R and S are respectively the mid- points of the sides AB, BC, CD and DA respectively. Show  that the quadrilateral  PQRS is rectangle.
        Sol.

        Given : ABCD is rhombus in which  P, Q, R and  S are the mid points of AB, BC, CD and DA respectively. PQ,  OR, RS and SP are joined to obtain a quadrilateral PQRS.

        15
        To prove : PQRS is a rectangle.
        Construction : Join AC.

        Proof : In Δ ABC, P and Q are the mid- points of AB and BC.
        Therefore      PQ || AC and PQ=12AC            [Mid-point Theorem]
        Similarly, in Δ ADC , R and S are  the mid- points of CD and AD.
        Therefore      SR || AC and SR=12AC             [Mid-point Theorem]
        From (1) and (2) , we get
        PQ || RS and PQ = SR
        Now , in quadrilateral PQRS its one pair  of opposite  sides PQ and SR is equal  and parallel.
        Therefore  PQRS is  a parallelogram
        Therefore      AB = BC                 [Sides of a rhombus]
        ⇒    12AB=12BC
        ⇒PB=BQ
        ⇒    ∠3=∠4        [∠s opp. to equal sides of  a triangle]
        Now  in Δ APS and Δ CQR we have
        AP = CQ                 [Halves of equal sides AB, BC]
        AS = CR                [Halves of equal sides AD, CD]
        PS = QR                 [Opp. sides of parallelogram PQRS]
        Therefore     ΔAPS≅ΔCQR    [SSS Cong. Theorem]
        ⇒    ∠1=∠2        [Corresponding parts of congruent triangles are equal]
        Now , ∠1+∠SPQ+∠3=180o    [Linear pair  axiom]
        Therefore ∠1+∠SPQ+∠3=∠2+∠PQR+∠4
        But          ∠1=∠2and∠3=∠4    [Proved above]
        Therefore         ∠SPQ=∠PQR
        Since    SP|| RQ , and PQ  intersects them ,
        Therefore     ∠SPQ+∠PQR=180o      [Since consecutive interior angles are supplementary]
        From  (3) and (4), we get
        ∠PQR+∠PQR=180o
        2 ∠PQR=180o
        ∠PQR=90o
        ∠SPQ=∠PQR=90o
        Thus , PQRS is a parallelogram whose one angle  ∠SPQ=90o.
        Hence PQRS is a rectangle.


        Q.3    ABCD is a rectangle  and P, Q, R and S are  mid- points of the sides AB, BC, CD and DA respectively. Show  that the quadrilateral PQRS is a rhombus.
        Sol. 

        Given : ABCD is a rectangle in which P, Q, R, and S are the mid- points of AB, BC, CD and DA respectively.PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.

        16
        To prove : PQRS is rhombus.
        Construction : Join AC.

        Proof : In Δ ABC, P and Q are the mid- points of sides AB and BC.
        Therefore,      PQ || AC and PQ=12AC   [Mid-point theorem]           … (1)
        Similarly, in Δ ADC, R and S are the mid- points of sides CD and AD.
        Therefore      SR|| AC and  SR=12AC     [Mid-point theorem]        … (2)
        From  (1) and (2), we get
        PQ || SR and PQ = SR                                                                            … (3)
        Now  in quad. PQRS, its one pair  of opposite  side PQ and SR is parallel and equal.         [From (3)]
        Therefore PQRS is a parallelogram                                               … (4)
        Now,     AD = BC                 [Opp. sides of rect. ABCD]
        ⇒    12AD=12BC⇒AS=BQ
        In Δ APS and BPQ , we have
        AP = BP                                                   [Since P is the mid- point of AB]
        ∠PAS=∠PBQ             [Each = 90º]
        AS = BQ                                                           [Proved above]
        Therefore     ΔAPS≅ΔBPQ      [SAS congruence axiom]
        ⇒    PS = PQ                 [Corresponding parts of congruent triangles are equal]        … (5)
        From  (4) and (5) we get,
        PQ = QR = RS = PS
        PQRS is a rhombus.


        Q.4     ABCD is a trapezium in which  AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (See figure). Show  that F is the mid- point of BC.

        17
        Sol.

        Given : In trapezium ABCD, AB || DC
        E is the mid- point of AD, EF || AB.
        To prove : F is the mid- point of BC.
        Construction : Join DB. Let it intersects EF in G.

        Proof : In Δ DAB, E is the mid- point of AD         [Given]
        EG || AB                    [Since EF || AB]

        18
        Therefore by converse of mid- point theorem G is  the mid- point of DB.
        In Δ BCD, G is the  mid- point of BD                        [Proved]
        GF || DC        [Since AB|| DC, EF || AB ⇒ DC|| EF]
        Therefore  by converse of mid- point theorem –
        F is the mid- point of BC.


        Q.5    In a parallelogram ABCD, E and F are the mid- points of sides AB and CD respectively (see figure). Show  that the line segements AF and EC trisect the diagonal BD.

        19
        Sol.

        Given : E and F are the mid- points of sides AB and CD of the parallelogram ABCD whose diagonal is BD.
        To prove : BQ = QP = PD

        Proof : ABCD is parallelogram                     [Given]
        Therefore         AB || DC and AB = DC        [Opp. sides of parallelogram]
        E is the mid- point of AB                                 [Given]
        Therefore         AE=12AB                … (1)
        F is the mid- point of CD
        Therefore          CF=12CD
        ⇒        CF=12AB        [Since CD = AB]    … (2)
        From (1) and (2) ,
        AE = CF
        Also         AE || CF                                                                          [Since AB || DC]
        Thus, a pair  of opposite  sides of a quadrilateral  AECF are parallel and equal.
        Quadrilateral AECF is a parallelogram .
        ⇒        EC || AF
        ⇒        EQ|| AP and QC || PF
        In Δ BPA , E is the mid- point  of BA         [Given]
        EQ|| AP                            [Proved]
        Therefore,         BQ = PQ        [Converse of mid- point theorem]    … (3)
        Similarly by taking  Δ CQD,  we can prove that
        ⇒  DP = QP
        From (3) and (4), we get
        ⇒     BQ = QP = PD
        Hence , AF and CE trisect the diagonal BD.


        Q.6     Show that the line segements joining  the mid-points of the opposite sides of a quadrilateral bisect each other.
        Sol. 

        Given : In a quadrilateral ABCD, P, Q, R and S are respectively the mid- points of AB, BC, CD and DA. PR and QS  intersect each other at O.

        20
        To prove : OP  = OR,  OQ = OS.
        Construction : Join PQ,  QR, RS, SP, AC and BD.

        Proof : In Δ ABC, P and Q are mid- points of AB and BC respectively.
        Therefore     PQ|| AC and PQ=12AC          [Mid-point theorem]………(1)
        Similarly, we can prove that
        RS || AC and RS=12AC     [Mid-point theorem]……………….(2)
        From (1) and (2)-
        Therefore      PQ || SR and PQ = SR
        Thus, a pair  of opposite  sides of a quadrilateral PQRS are parallel and equal.
        Therefore , quadrilateral  PQRS is a parallelogram.
        Since the diagonals of a parallelogram bisect each other.
        OP = OR = and OQ = OS
        Therefore , diagonals PR and QS of a || gm PQRS i.e.,  the line segements joining  the mid- points of opposite  sides of quadrilateral  ABCD bisect each other.


        Q.7     ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show  that
                    
        (i) D is the mid-point of AC
                    
        (ii) MD ⊥ AC
                    
        (iii) CM=MA=12AB
        Sol.

        Given: Δ ABC is right angled at C, M is the mid- point  of hypotenuse AB. Also MD||BC.

        21
        To prove that :
        (i)  D is the mid- points of AC
        (ii)  MD ⊥ AC
        (iii)  CM=MA=12AB

        Proof : (i) In Δ ABC, M is the mid- point  of AB and MD|| BC. Therefore , D is the mid- point of AC.
        i.e.,         AD = DC                                                                   … (1)

        (ii) Since MD || BC, Therefore,
        ∠ADM=∠ACB         [Corresponding angles]
        ⇒    ∠ADM=90o            [Since ∠ACB=90o (given)]
        But, ∠ADM+∠CDM=180o               [Since ∠ADMand∠CDM are angles of a linear pair]
        Therefore,      90o+∠CDM=180o⇒∠CDM=90o
        Thus         ∠ADM=∠CDM=90o        … (2)
        ⇒    MD⊥AC

        (iii) In Δs AMD and CMD, we have
        AD = CD                 [From (1)]
        ∠ADM=∠CDM        [From (2)]
        and,          MD = MD             [Common]
        Therefore,  by  SAS criterion  of congruence
        ΔAMD≅ΔCMD
        ⇒        MA  = MC          [Since corresponding  parts of congruent triangles are equal]
        Also,     MA=12AB, since M is the mid-point of AB
        Hence,     CM=MA=12AB

        Prev Chapter Notes – Quadrilaterals
        Next R D Sharma Solutions Quadrilaterals

        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