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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 – Area of Parallelogram Exercise 9.1 – 9.4

        Exercise 9.1

        Q.1     Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

        1
        (i)
        2
        (ii)

         

        3
        (iii)

         

        (iv)
        (V)
        6
        (vi)


        Sol
        .

        The figures mentioned above lie on the same base and between the same parallels as indicated against them :
        (i) Base DC, parallels DC and AB.
        (ii) Base QR, parallels QR and PS.
        (iii) Base AD, parallels AD and BQ.

        Exercise 9.2

        Q.1     In figure ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
        7Sol.

        We have,
        Area of a || gm = Base × Height
        Therefore, Area of || gm ABCD = AB × AE
        =(16×8)cm2=128cm2            …(1)
        Also area of || gm ABCD = AD × CF
        =(AD×10)cm2                                                     … (2)
        From (1) and (2) we get
        128 = AD × 10
        ⇒ AD=12810cm=12.8cm


        Q.2      If E, F, G and H are respectively the mid- points of the sides of a parallelogram ABCD, show that ar(EFGH) =12ar(ABCD)
        Sol.        Δ HGF and || gm HDCF stand on the same base HF and lie between the same parallels HF and DC.

        8Therefore,   ar(ΔHGF)=12ar(HDCF)          … (1)
        Similarly, Δ HEF and ||gm ABFH stand on the same base HF and lie between the same parallels
        HF and AB.
        Therefore ar(ΔHEF)=12ar(ABFH)          … (2)
        Therefore Adding (1) and (2), we get
        ar(ΔHGF)+ar(ΔHEF)=12ar(HDCF)+ar(ABFH)
        ⇒ ar(EFGH)=12ar(ABCD)


        Q.3    P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).
        Sol.        Δ APB and || gm ABCD stand on the same base AB and lie between the same parallels AB and DC.

        9Therefore ar(ΔAPB)=12ar(ABCD)…. (1)
        Similarly , Δ BQC and || gm ABCD stand on the same base BC and lie between the same parallels BC and AD.
        Therefore ar(ΔBQC)=12ar(ABCD) …. (2)
        From (1) and (2) , we have
        ar(ΔAPB) = ar (ΔBQC)


        Q.4     In figure, P is a point in the interior of a parallelogram ABCD. Show that 
                     (i) ar (APB) + ar (PCD) = 12ar(ABCD)
                     (ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD).

        10
        Sol.       Draw EPF parallel to AB or DC and GPH parallel to AD or BC.

        11Now AGHD is a|| gm
        [Since GH || DA and AG|| DH]
        Similarly, HCBG, EFCD and ABFE are parallelograms.
        (i) Δ APB and || gm ABFE stand on the same base AB and lie between the same parallels AB and EF.
        Therefore  ar(APB)=12ar(ABFE) … (1)
        Similarly,  ar(PCD)=12ar(EFCD) … (2)
        Adding (1) and (2) , we get
        ar(APB)+ar(PCD)=12[ar(ABFE)+ar(EFCD)]
        ar(APB)+ar(PCD)=12ar(ABCD)....(3)

        (ii) Δ APD and || gm AGHD are on the same base AD and lie between the same parallels AD and HG.
        Therefore ar(APD)=12ar(AGHD)....(4)
        Similarly, ar(PCB)=12ar(GBCH)....(5)
        Adding (4) and (5), we get
        ar(APD)+ar(PCB)=12[ar(AGHD)+ar(GBCH)]
        ar(APD)+ar(PCB)=12ar(ABCD)...(6)
        From (3) and (6) we get
        ar (APD) + ar (PBC) = ar (APB) + ar(PCD).


        Q.5      In figure PQRS and ABRS are parallelograms and X is any point on side BR. Show that 
                       (i) ar (PQRS) = ar (ABRS)
                       (ii) ar (AXS) =12 ar (PQRS).

        12Sol.        

        (i) || gm PQRS and || gm ABRS stand on the same base RS and lie between the same parallels SR and PAQB.
        Therefore ar(PQRS) = ar (ABRS) … (1)
        (ii) Δ AXS and || gm ABRS stand on the same base AS and lie between the same parallels AS and RB.
        Therefore ar (AXS) =12 ar (ABRS)
        ⇒ ar (AXS) =12 ar (PQRS)                [Using (1)]


        Q.6      A farmer was having a field in the form of a parallelogram PQRS. He took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should he do it?
        Sol.       Clearly, the field i.e., || gm PQRS is divided into 3 parts. Each part is of the shape of triangle.

        13Since Δ APQ and || gm PQRS stand on the same base PQ and lie between the same parallels PQ and SR.
        Therefore ar (APQ) =12 ar (PQRS) … (1)
        Clearly, ar (APS) + ar (AQR) = ar (PQRS) – ar (APQ)
        = ar (PQRS) −12ar(PQRS)                 [Using (1)]
        =12ar(PQRS)....(2)
        From (1) and (2), we get
        ar (APS) + ar (AQR) = ar (APQ)
        Thus , the farmer should sow wheat and pulses either as [(Δs APS and AQR) or Δ APQ] or as
        [Δ APQ or (Δs APS and AQR )]

        Exercise 9.3

        Q.1     In figure , E is any point on median AD of a Δ ABC. Show that ar (ABE) = ar (ACE).

        14
        Sol.

        Given : AD is a medium of Δ ABC and E is any point on AD.
        To prove : ar (ABE) = ar (ACE)
        Proof :   Since AD is the median of Δ ABC
        Therefore ar (ABD) = ar (ACD) … (1)
        Also , ED is the median of Δ EBC
        Therefore ar (BED) = ar (CED) … (2)
        Subtracting (2) from (1) , we get
        ar (ABD) – ar (BED) = ar (ACD) – ar (CED)
        ⇒ ar (ABE) = ar (ACE).


        Q.2      In a triangle ABC, E is the mid- point of median AD. Show that ar (BED) = 14 ar (ABC).
        Sol.

        Given : A Δ ABC, E is the mid- point of the median AD.
        To prove : ar (BED) = 14 ar (ABC)

        15
        Proof :  Since AD is a median of Δ ABC and median divides a triangle into two triangles of equal area.
        Therefore ar (ABD) = ar (ADC)
        ⇒ ar (ABD) = 12 ar (ABC) … (1)
        In Δ ABD, BE is the median
        Therefore ar(BED) = ar (BAE) … (2)
        ⇒ ar(BED)=12ar(ABD)     [ar(BAE)=12ar(ABD) ]
        ⇒ ar (BED) = 12×12 ar (ABC)                 [Using (1)]
        ⇒ ar (BED) = 14 ar (ABC) .


        Q.3      Show that the diagonals of a parallelogram divide it into four triangles of equal area.
        Sol.

        Given : A parallelogram ABCD.
        To prove : The diagonals AC and BD divide the || gm ABCD into four triangles of equal area.
        Construction : Draw BL ⊥ AC.
        Proof : Since ABCD is a || gm and so its diagonals AC and BD bisect each other at O.

        16
        Therefore AO = OC and BO = OD
        Now, ar (AOB) = 12 × AO × BL
        ar (OBC) = 12 × OC × BL
        But AO = OC
        Therefore ar (AOB) = ar (OBC)
        Similarly, we can show that
        ar (OBC) = ar (OCD) ; ar (OCD) = ar(ODA) ;
        ar (ODA) = ar (OAB) ; ar (OAB) = ar (OBC)
        ar (OCD) = ar (ODA)
        Thus , ar (OAB) = ar (OBC) = ar(OCD) = ar (OAD)

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        Q.4      In figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
        17
        Sol.

        Given : ABC and ABD are two triangles on the same base AB. A line segment CD is bisected by AB at O i.e., OC = OD.
        To prove : ar (ABC) = ar (ABD).

        Proof : In Δ ACD, we have
        OC = OD [Given]
        Therefore AO is the median.
        Therefore ar (AOC) = ar (AOD)         [Since median divides a Δ in two Δs of equal area]
        Similarly, in Δ BCD, BO is the median
        Therefore ar (BOC) = ar (BOD)
        Adding (1) and (2), we get
        ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)
        ⇒ ar (ABC) = ar (ABD)


        Q.5     D, E and F are respectively the mid- points of the side BC, CA and AB of a Δ ABC. Show that
                         (i) BDEF is a parallelogram
                         (ii) ar (DEF) = 14 ar (ABC)
                         (iii) ar (BDEF) = 12 ar (ABC).
        Sol.

        remain
        Given : D, E and F are the mid- points of the sides BC, CA and AB respectively of Δ ABC.
        To prove : (i) BDEF is a || gm
        (ii) ar (DEF) = 14 ar (ABC)
        (iii) ar (BDEF) = 12 ar (ABC)
        (i) We have in Δ ABC,
        EF || BC [By mid- point theorem, since E and F are the mid- points AC and AB respectively]
        Therefore EF || BD… (1)
        Also ED || AB [By mid- point theorem, since E and D are the mid- points AC and BC respectively]
        Therefore ED || BF… (2)
        From (1) and (2), BDEF is a || gm.
        (ii) Similarly, FDCE and AFDE are || gms.
        Therefore ar (FBD) = ar (DEF)   [Since FD is a diagonal of || gm BDEF]… (3)
        ar (DEC) = ar (DEF)          [Since ED is a diagonal of || gm FDCE]… (4)
        and , ar (AFE) = ar (DEF)       [Since FE is a diagonal of || gm AFDE… (5)]
        From (3), (4) and (5)
        ar (FBD) = ar (DEC) = ar (AFE) = ar (DEF) …. (6)
        ar (ABC) = ar (AFE) + ar (FBD) + ar (DEC) + ar (DEF)
        from equation (6) –
        ar(ABC) = 4ar(DEF)
        ⇒ ar (DEF) = 14 ar (ABC).
        (iii) Also, ar (BDEF) = 2 ar (DEF)
        =2×14ar(ABC)=12ar(ABC)


        Q.6     In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that :
                   (i) ar (DOC) = ar (AOB)
                   (ii) ar (DCB) = ar (ACB)
                   (iii) DA || CB or ABCD is a parallelogram.
        18Sol.

        (i) Draw DN ⊥ AC and BM ⊥ AC.
        In Δs DON and BOM
        ∠DNO=∠BMO [Each = 90º]
        ∠DON=∠BOM [Vert. opp ∠s]
        OD = OB [Given]

        19
        By AAS criterion of congruent.
        ΔDON≅ΔBOM … (1)
        In Δs DCN and BAM
        ∠DNC=∠BMA [Each = 90º]
        DC = AB [Given ]
        DN = BM [Since ΔDON≅ΔBOM⇒DN=BM]
        Therefore by RHS criterion of congruence,
        ΔDCN≅ΔBAM … (2)
        From (1) and (2), we get
        ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)
        ⇒ ar (DOC) = ar (AOB)

        (ii) Since ar (DOC) = ar (AOB)
        Therefore ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC)
        ⇒ ar (DCB) = ar (ACB)

        (iii) Δs DCB and ACB have equal areas and have the same base. So, these Δs lie between the same parallels.
        ⇒ DA || CB i.e., ABCD is a|| gm

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        Q.7    D and E are points on sides AB and AC respectively of Δ ABC such that ar(DBC) = ar(EBC). Prove that DE|| BC.
        Sol.      Since Δs DBC and EBC are equal in area and have a same base BC.

        20
        Therefore altitude from D of Δs DBC = Altitude from E of Δs EBC.
        ⇒ Δs DBC and EBC are between the same parallels.
        ⇒ DE || BC


        Q.8     XY is a line parallel to side BC of a triangle ABC. If BE|| AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).
        Sol.

        Since XY|| BC and BE || CY
        Therefore BCYE is a || gm.
        Since Δ ABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.

        21
        Therefore ar(ABE) =12 ar(BCYE) … (1)
        Now, CF || AB and XY || BC
        ⇒ CF || AB and XF || BC
        ⇒ BCFX is a || gm
        Since Δ ACF and || gm BCFX are on the same base CF and between the same parallel AB and FC .
        Therefore ar (ACF) =12 ar (BCFX) … (2)
        But ||gm BCFX and || gm BCYE are on the same base BC and between the same parallels BC and EF.
        Therefore ar (BCFX) = ar(BCYE) … (3)
        From (1) , (2) and (3) , we get
        ar (Δ ABE) = ar(Δ ACF)


        Q.9     The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see figure). Show that ar(ABCD) = ar (PBQR).

        22Sol.

        Join AC and PQ. Since AC and PQ are diagonals of || gm ABCD and ||gm BPQR respectively.
        Therefore ar(ABC) = 12 ar (ABCD) … (1)
        and ar (PBQ) = 12 ar (BPRQ) … (2)

        23
        Now, Δs ACQ and AQP are on the same base AQ and between the same parallels AQ and CP.
        Therefore ar(ACQ) = ar(AQP)
        ⇒ ar(ACQ) – ar(ABQ) = ar(AQP) – ar(ABQ)             [Subtracting ar(ABQ) from both sides]
        ⇒ ar(ABC) = ar(BPQ)
        ⇒ 12 ar(ABCD) = 12 ar(BPRQ)             [Using (1) and (2)]
        ⇒ ar(ABCD) = ar (BPRQ).

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        Q.10    Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(AOD) = ar (BOC).
        Sol.

        Diagonals AC and BD of a trapezium ABCD with AB|| DC intersect each other at O.
        Therefore Δs ABC and ABD are on the same base AB and between the same parallels AB and DC.

        24
        Therefore ar(ABD) = ar(ABC)
        ⇒ ar(ABD) – ar(AOB) = ar(ABC) – ar (AOB)  [Subtracting ar (AOB) from both sides]
        ⇒ ar (AOD) = ar(BOC)


        Q.11    In figure , ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
                         (i) ar (ACB) = ar (ACF)
                         (ii) ar (AEDF) = ar (ABCDE)
        25Sol.

        (i) Since Δ s ACB and ACF are on the same base AC and between the same parallels AC and BF.
        Therefore ar(ACB) = ar (ACF)

        (ii) Adding ar(ACDE) on both sides we get
        ar(ACF) + ar(ACDE) = ar(ACB) + ar(ACDE)
        ⇒ ar(AEDF) = ar(ABCDE)


        Q.12    A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
        Sol.       

        Let ABCD be the quadrilateral plot. Produce BA to meet CD drawn parallel to CA at E. Join EC.

        26
        Then , Δ s EAC and ADC lie on the same parallels DE and CA
        Therefore ar(EAC) = ar (ADC)
        Now ar(ABCD) = ar(ABC) + ar(ACD)
        = ar(ABC) + ar(ADC)
        = ar (ABC) + ar(EAC) = ar(EBC)
        i.e., quad. ABCD = Δ EBC
        Which is the required explain to the suggested proposal.

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        Q.13   ABCD is a trapezium with AB|| DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
        Sol.       ABCD is a trapezium with AB || DC and XY|| AC, is drawn. Join XC.

        27
        ar(ACX) = ar(ACY) … (1) [Since Δ s ACX and ACY have same base AC and are between same parallels AC and XY]
        But ar (ACX) = ar(ADX) [Since Δs ACX and ADX have same base AX and are between same parallels AB and DC]
        From (1) and (2), we have
        ar(ADX) = ar (ACY).


        Q.14    In figure , AP || BQ|| CR . Prove that ar (AQC) = ar(PBR)

        28
        Sol. 

        From the figure, we have
        ar (AQC) = ar (AQB) + ar (BQC) … (1)
        and ar (PBR) = ar (PBQ) + ar (QBR) … (2)
        But ar (AQB) = ar(PBQ)  [Since these Δs are on the same base BQ and between same parallel line AP and BQ]
        Also, ar (BQC) = ar (QBR) [Since these Δs are on the same base BQ and between same parallel lines BQ and CR]
        Using (3) and (4) in (1) and (2) , we get :
        ar (AQC) = ar (PBR).


        Q.15    Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
        Sol.        Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that

        29
        ar (AOD) = ar (BOC). … (1)   [Given]
        Adding ar (ODC) on both sides , we get
        ar(AOD) + ar (ODC) = ar (BOC) + ar (ODC)
        ⇒ ar(ADC) = ar (BDC)
        ⇒ 12×DC×AL=12×DC×BM
        ⇒ AL = BM
        ⇒ AB || DC
        Hence ABCD is a trapezium.


        Q.16    In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
        30
        Sol.

        From the figure,
        ar (BDP) = ar (ARC)             [Given]
        and ar (DPC) = ar (DRC)       [Given]
        On subtracting, we get
        ar (BDP) – ar (DPC) = ar (ARC) – ar (DRC)
        ⇒ ar (BDC) = ar (ADC)
        ⇒ DC || AB    [ar(BDC) and ar(ADC) have equal areas and have the same base. So, these Δs lie between the same parallels.]
        Hence , ABCD is a trapezium.
        ar (DRC) = ar (DPC) [Given]
        On subtracting ar (DLC) from both sides, we get
        ar (DRC) – ar (DLC) = ar (DPC) – ar (DLC)
        ⇒ ar (DLR) = ar (CLP)
        ON adding ar (RLP) to both sides, we get
        ar (DLR) + ar(RLP) = ar(CLP) + ar(RLP)
        ⇒ ar (DRP) = ar (CRP)
        ⇒ RP || DC
        Hence, DCPR is a trapezium.

        Exercise 9.4

        Q.1    Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
        Sol.

        Given : A || gm ABCD and a rectangle ABEF with same base AB and equal areas.
        To prove : Perimeter of || gm ABCD > Perimeter of rectangle ABEF.

        31
        Proof : Since opposite sides of a|| gm and rectangle are equal.
        Therefore AB = DC               [Since ABCD is a || gm]
        and,           AB = EF               [Since ABEF is a rectangle]
        Therefore  DC = EF                                                                     … (1)
        ⇒          AB + DC = AB + EF  (Add AB in both sides)                   … (2)
        Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.
        Therefore BE < BC and AF < AD
        ⇒ BC > BE and AD > AF
        ⇒ BC + AD > BE + AF                   … (3)
        Adding (2) and (3), we get
        AB + DC + BC + AD > AB + EF + BE + AF
        ⇒ AB + BC + CD + DA > AB + BE + EF + FA
        ⇒  perimeter of || gm ABCD > perimeter of rectangle ABEF.
        Hence,the perimeter of the parallelogram is greater than that of the rectangle.


        Q.2     In figure , D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
        Can you know answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

        32
        Sol.

        Let AL be perpendicular to BC. So, AL is the height of Δs ABD, ADE and, AEC.
        Therefore ar (ABD) =12 × BD × AL
        ar (ADE) =12 ×DE × AL
        and, ar (AEC) =12 × EC × AL
        Since BD = DE = EC
        Therefore ar (ABD) = ar (ADE) = ar(AEC)
        33 
        Yes , altitudes of all triangles are same. Budhia has use the result of this question in dividing her land in three equal parts.


        Q.3     In figure , ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF)

        34
        Sol.

        Since opposite sides of a|| gm are equal
        Therefore , AD = DC                         [Since ABCD is a || gm]
        DE = CF                                                [Since DCFE is a || gm]
        and AE = BF                                       [Since ABFE is a || gm]
        Consider Δs ADE and BCF, in which AE = BF , AD = BC and DE = CF
        Therefore by SSS criterion of congruence
        ΔADE≅ΔBCF
        ⇒ ar(ADE) = ar (BCF)


        Q.4      In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that   ar (BPC) = ar (DPQ).

        35
        Sol.

        Join AC.
        Since Δs APC and BPC are on the same base PC and between the same parallels PC and AB.

        37
        Therefore , ar(APC) = ar (BPC) … (1)
        Since AD = CQ
        and ,  AD || CQ [Given]
        Therefore in the quadrilateral ADQC, one pair of opposite sides is equal and parallel.
        Therefore ADQC is a parallelogram.
        ⇒ AP = PQ and CP = DP [Since diagonals of a|| gm bisect each other]
        In Δs APC and DPQ we have
        AP = PQ [Proved above]
        ∠APC=∠DPQ [Vertically opp. ∠s]
        and , PC = PD [Proved above]
        Therefore by SAS criterion of congruence,
        ΔAPC≅ΔDPQ
        ⇒ ar (APC) = ar (DPQ) … (2)
        [Since congruent Δs have equal area]
        Therefore ar (BPC) = ar (DPQ)        [From (1)]
        Hence , ar (BPC) = ar (DPQ)


        Q.5      In figure, ABC and BDE are two equilateral triangles such that D is the mid- point of BC. If AE intersects BC at F, Show that
                   
        (i) ar (BDE) = 14 ar (ABC)
                   (ii) ar (BDE) = 12 ar (BAE)
                   (iii) ar (ABC) = 2 ar (BEC)
                   (iv) ar (BFE) = ar (AFD)
                   (v) ar (BFE) = 2 ar (FED)
                   (vi) ar (FED) = 18 ar (AFC).
        42

        Sol.

        Join EC and AD. Let a be the side of Δ ABC. then,
        ar(ABC)=3√4a2=Δ(say)

        36
        (i) ar(BDE)=3√4(a2)2
        [SinceBD=12BC=a2]
        =3√16a2=Δ4
        ⇒ ar(BDE)=14ar(ABC)

        (ii) We have , ar (BDE) =12 ar (BEC) … (1)
        [Since DE is a median of Δ BEC and each median divides a triangle
        in two other Δs of equal area]
        Now, ∠EBC=60o
        and ∠BCA=60o
        ⇒ ∠EBC=∠BCA
        But these are alterante angles with respect to the line- segments BE and CA and their transversal BC.
        Hence BE || AC.
        Now, Δs BEC and BAE stand on the same base BE and lie between the same parallels BE and AC.
        Therefore ar(BEC) = ar(BAE)
        Therefore From (1) ar (Δ BDE) = 12 ar (BAE).

        (iii) Since ED is a median of Δ BEC and we know that each median divides a triangle in two other Δs of equal area.
        Therefore ar(BDE) = 12 ar (BEC)
        From part (i),
        ar (BDE) = 14 ar (ABC)
        Combining these results , we get
        14 ar (ABC) = 12 ar (BEC)
        ⇒ ar (ABC) = 2 ar (BEC)

        (iv) Now,∠ABD=∠BDE=60∘ [Given]
        But ∠ABDand∠BDE are alternate angles with respect to the line- segment BA and DE and their transversal BD.
        Hence BA || ED.
        Now, Δs BDE and AED stand on the same base ED and lie between the same parallels BA and DE
        Therefore ar (BDE) = ar (AED)
        ⇒ ar(BDE) – ar(FED) = ar (AED) – ar (FED)
        ⇒ ar (BEF) = ar (AFD)

        (v) In Δ ABC, AD2=AB2−BD2
        =(a)2−(a2)2
        =a2−a24=3a24
        38
        ⇒ AD=3√2a
        In Δ BED , EL2=DE2−DL2
        =(a2)2−(a4)2=a24−a216=3a216        [EL is median ofΔ BED]
        ⇒ EL=3√a4
        Therefore ar(AFD) = 12 × FD × AD
        = 12 × FD × 3√2a … (1)
        and ar(EFD) = 12 × FD × EL
        = 12 × FD = 3√4a … (2)
        From (1) and (2), we have ar (AFD) = 2 ar (EFD)
        Combining this result with part (iv).
        We have ar (BFE) = ar (AFD) = 2ar (EFD)

        (vi) From part (i)
        ar (BDE) = 14 ar (ABC)
        ⇒ ar (BEF) + ar (FED) = 14 × 2ar (ADC)
        ⇒ 2ar (FED) + ar (FED) = 12 (ar(AFC) – ar (AFD) [Using part (v)]
        ⇒ 3ar (FED) = 12 ar (AFC) – 12 × 2 ar (FED)
        ⇒ 4ar (FED) = 12 ar (AFC)
        ⇒ ar (FED) = 18 ar (AFC)


        Q.6     Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
        Sol.       Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

        39
        Draw AM ⊥ BD and CN ⊥ BD.
        Now, ar (APB) × ar(CPD)
        =(12×BP×AM)×(12×DP×CN)
        =14 × BP × DP × AM × CN
        and , ar (APD) × ar (BPC)
        =(12×DP×AM)×(12×BP×CN)
        =14 × BP × DP × AM × CN
        From (1) and (2), we have
        ar (APB) × ar (CPD) = ar (APD) × ar (BPC)


        Q.7     P and Q are respectively the mid- points of sides AB and BC of of a triangle ABC and R is the mid- point of AP, show that
                           (i) ar (PRQ) = 12 ar (ARC)
                   (ii) ar (RQC) = 38 ar (ABC)

                          (iii) ar (PBQ) = ar (ARC)
        Sol.

        P and Q are respectively the mid- points of sides AB and BC of Δ ABC and R is the mid- point of AP.

        40
        Join AQ and PC.
        (i) We have,
        ar (PQR) = 12 ar (APQ)              [Since QR is a median of Δ APQ and it divides the Δ into two other Δs of equal area]
        =12×12 ar (ABQ)                                 [Since QP is a median of Δ ABQ]
        =14 ar (ABQ) = 14×12 ar (ABC)
        [Since AQ is a median of Δ ABC]
        =18 ar (ABC) … (1)
        Again , ar(ARC) = 12 ar (APC)                  [Since CR is a median of Δ APC]
        =12×12 ar (ABC)                [Since CP is a median of Δ ABC]
        =14 ar (ABC) … (2)
        From (1) and (2) , we get
        ar (PQR) = 18 ar (ABC) = 12×14 ar (ABC)
        =12 ar (ARC).

        (ii) We have,
        ar (RQC) = ar(RQA) + ar (AQC) – ar (ARC) … (3)
        Now, ar (Δ RQA) = 12 ar (PQA)              [Since RQ is a median of Δ PQA]
        =12×12 ar (AQB)               [Since PQ is a median of Δ AQB]
        =14 ar (AQB)
        =14×12 ar(ABC)                               [Since AQ is a median of Δ ABC]
        =18 ar (ABC) … (4)
        ar (AQC) = 12 ar (ABC) … (5)                [Since AQ is a median of Δ ABC]
        ar (Δ ARC) = 12 ar (APC)                  [Since CR is a median of Δ APC]
        =12×12 ar(ABC) [Since CP is a median of Δ ABC]
        =14 ar (ABC) … (6)
        From (3), (4) , (5) and (6) we have
        ar (RQC) = 18 ar (ABC) + 12 ar (ABC) – 14 ar (ABC)
        =(18+12−14) ar (ABC)
        =38 ar (ABC)

        (iii) We have,
        ar (PBQ) = 12 ar (ABQ)                    [Since PQ is a median of Δ ABQ]
        =12×12ar(ABC)                [Since AQ is a median of Δ ABC]
        =14 ar (ABC)
        = ar(ARC)                               [Using (6)]


        Q.8     In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively.  Line segment AX ⊥ DE meets BC at Y. Show that :
                  (i) ΔMBC≅ΔABD
                  (ii) ar (BYXD) = 2 ar (MBC)
                  (iii) ar (BYXD) = ar (ABMN)
                  (iv) ΔFCB≅ΔACE
                  (v) ar (CYXE) = 2 ar (FCB)
                  (vi) ar (CYXE) = ar (ACFG)
                 (vii) ar (BCED) = ar (ABMN) + ar(ACFG)
        Note : Result (vii) is the famous theorem of pythagoras. You shall learn a simpler proof of this theorem in class X.

        41

        Sol.

        (i) In Δs MBC and ABD, we have
        BC = BD [Sides of the square BCED]
        MB = AB [Sides of the square ABMN]
        ∠MBC=∠ABD [Since Each = 90º + ∠ABC]
        Therefore by SAS criterion of congruence, we have
        ΔMBC≅ΔABD

        (ii) Δ ABD and square BYXD have the same base BD and are between the same parallels BD and AX.
        Therefore ar(ABD) =12 ar (BYXD)
        But ΔMBC≅ΔABD [Proved in part (i)]
        ⇒ ar(MBC) = ar (ABD)
        Therefore ar(MBC) = ar (ABD) = 12 ar (BYXD)
        ⇒ ar(BYXD) = 2 ar (MBC).

        (iii) Square ABMN and Δ MBC have the same base MB and are between same parallels MB and NAC.
        Therefore ar(MBC) = 12 ar (ABMN)
        ⇒ ar(ABMN) = 2ar (MBC)
        = ar (BYXD) [Using part (ii)]

        (iv) In Δs ACE and BCF, we have
        CE = BC [Sides of the square BCED]
        AC = CF [Sides of the square ACFG]
        and ∠ACE=∠BCF [Since Each = 90º + ∠BCA]
        Therefore by SAS criterion of congruence,
        ΔACE≅ΔBCF

        (v) Δ ACE and square CYXE have the same base CE and are between same parallels CE and AYX.
        Therefore ar(ACE) = 12 ar (CYXE)
        ⇒ ar(FCB) = 12 ar (CYXE) [Since ΔACE≅ΔBCF, part (iv)]
        ⇒ ar(CYXE) =2 ar (FCB) .

        (vi) Square ACFG and Δ BCF have the same base CF and are between same parallels CF and BAG.
        Therefore ar(BCF) = 12 ar(ACFG)
        ⇒ 12 ar(CYXE) = 12 ar (ACFG) [Using part (v)]
        ⇒ ar(CYXE) = ar (ACFG)

        (vii) From part (iii) and (vi) we have
        ar (BYXD) = ar (ABMN)
        and ar (CYXE) = ar (ACFG)
        On adding we get
        ar (BYXD) + ar(CYXE) = ar (ABMN) + ar(ACFG)
        ar (BCED) = ar (ABMN) + ar(ACFG)

        Prev Chapter Notes – Area of Parallelogram
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