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      Class 10 Maths

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      • Class 10
      • Class 10 Maths
      CoursesClass 10MathsClass 10 Maths
      • 01. Real Numbers
        9
        • Lecture1.1
          Real Numbers and Fundamental Theorem of Arithmetic 59 min
        • Lecture1.2
          Divisibility and Euclid’s Division Lemma 49 min
        • Lecture1.3
          Finding HCF and LCM Using Fundamental Theorem of Arithmetic 02 hour
        • Lecture1.4
          Miscellaneous Questions 31 min
        • Lecture1.5
          Chapter Notes – Real Numbers
        • Lecture1.6
          NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
        • Lecture1.7
          Revision Notes Real Numbers
        • Lecture1.8
          R S Aggarwal Real Numbers
        • Lecture1.9
          R D Sharma Real Numbers
      • 02. Polynomials
        11
        • Lecture2.1
          Introduction to Polynomials and Its Zeroes 44 min
        • Lecture2.2
          Finding Zeroes of Quadratic Polynomial 01 hour
        • Lecture2.3
          Relationship b/w the Zeroes and coefficients of a Polynomial 01 hour
        • Lecture2.4
          Cubic Polynomial 35 min
        • Lecture2.5
          Division and Division Algorithm for Polynomials 01 hour
        • Lecture2.6
          Geometrical Meaning of Zeroes of Polynomial 15 min
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.4
        • Lecture2.9
          Revision Notes Polynomials
        • Lecture2.10
          R S Aggarwal Polynomials
        • Lecture2.11
          R D Sharma Polynomials
      • 03. Linear Equation
        9
        • Lecture3.1
          Solution of linear Equation in Two Variable by Graphical and Algebraic Methods 01 hour
        • Lecture3.2
          Some Basic Questions on Solving Pair of Linear Equations 01 hour
        • Lecture3.3
          Some Basics Problems on Numbers and Ages 40 min
        • Lecture3.4
          Problems on Money Matters, Time, Distance, Speed, Time and Work 01 hour
        • Lecture3.5
          Chapter Notes – Linear Equation
        • Lecture3.6
          NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
        • Lecture3.7
          Revision Notes Linear Equation
        • Lecture3.8
          R S Aggarwal Linear Equation
        • Lecture3.9
          R D Sharma Linear Equation
      • 04. Quadratic Equation
        8
        • Lecture4.1
          Introduction and Finding The Roots of a Quadratic Equation 37 min
        • Lecture4.2
          Miscellaneous Questions 1 02 hour
        • Lecture4.3
          Miscellaneous Questions 2 02 hour
        • Lecture4.4
          Chapter Notes – Quadratic Equation
        • Lecture4.5
          NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4
        • Lecture4.6
          Revision Notes Quadratic Equation
        • Lecture4.7
          R S Aggarwal Quadratic Equation
        • Lecture4.8
          R D Sharma Quadratic Equation
      • 05. Arithmetic Progressions
        11
        • Lecture5.1
          Introduction to Arithmetic Progression 59 min
        • Lecture5.2
          Miscellaneous Questions Based on nth Term Formula of A.P. 55 min
        • Lecture5.3
          Middle term (s) of Finite A.P. and Arithmetic Mean 01 hour
        • Lecture5.4
          Selection of the Terms and Sum of nth Terms of A.P. 01 hour
        • Lecture5.5
          Miscellaneous Questions Based on Sum of nth Term of A.P. 01 hour
        • Lecture5.6
          Word Problems related to nth term and sum of nth terms of A.P. 01 hour
        • Lecture5.7
          Chapter Notes – Arithmetic Progressions
        • Lecture5.8
          NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
        • Lecture5.9
          Revision Notes Arithmetic Progressions
        • Lecture5.10
          R S Aggarwal Arithmetic Progressions
        • Lecture5.11
          R D Sharma Arithmetic Progressions
      • 06. Some Applications of Trigonometry
        7
        • Lecture6.1
          Angle of Elevation and Depression; Problems involving Elevation and Double Elevation 01 hour
        • Lecture6.2
          Miscellaneous Problems Involving Double Elevation, Depression and Elevation 01 hour
        • Lecture6.3
          Miscellaneous Questions (Level-1, 2, 3, 4) 49 min
        • Lecture6.4
          Chapter Notes – Some Applications of Trigonometry
        • Lecture6.5
          NCERT Solutions – Some Applications of Trigonometry
        • Lecture6.6
          Revision Notes Some Applications of Trigonometry
        • Lecture6.7
          R D Sharma Some Applications of Trigonometry
      • 07. Coordinate Geometry
        17
        • Lecture7.1
          Introduction to Terms related to Coordinate Geometry 49 min
        • Lecture7.2
          Locating Coordinates of Point on the Axes 37 min
        • Lecture7.3
          Finding vertices of a Geometrical Fig. and Distance b/w Two Points 58 min
        • Lecture7.4
          Distance b/w Two Points Formula Based Examples 44 min
        • Lecture7.5
          Finding the type of Triangle and Quadrilateral 01 hour
        • Lecture7.6
          Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure 42 min
        • Lecture7.7
          Section Formula and Corollary 01 hour
        • Lecture7.8
          Finding the Section Ratio and Involving Equation of Line 53 min
        • Lecture7.9
          Proof Related to Mid-points and Centroid of a Triangle 59 min
        • Lecture7.10
          Finding Area of Triangle and Quadrilateral 53 min
        • Lecture7.11
          Collinearity 34 min
        • Lecture7.12
          Examples Based on If Areas are Given 34 min
        • Lecture7.13
          Chapter Notes – Coordinate Geometry
        • Lecture7.14
          NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
        • Lecture7.15
          Revision Notes Coordinate Geometry
        • Lecture7.16
          R S Aggarwal Coordinate Geometry
        • Lecture7.17
          R D Sharma Coordinate Geometry
      • 08. Triangles
        15
        • Lecture8.1
          Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem 52 min
        • Lecture8.2
          Questions Based on Thales’s Theorem, Converse Thales’s Theorem 01 hour
        • Lecture8.3
          Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse 59 min
        • Lecture8.4
          Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity 52 min
        • Lecture8.5
          SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity 02 hour
        • Lecture8.6
          Questions Based on Criteria for Similarity Conti. 01 hour
        • Lecture8.7
          Questions Based On Area of Similar Triangles 02 hour
        • Lecture8.8
          Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1 01 hour
        • Lecture8.9
          Pythagoras Theorem Level-1 cont. 01 hour
        • Lecture8.10
          Miscellaneous Questions 01 hour
        • Lecture8.11
          Chapter Notes – Triangles
        • Lecture8.12
          NCERT Solutions – Triangles Exercise 8.1 – 8.6
        • Lecture8.13
          Revision Notes Triangles
        • Lecture8.14
          R S Aggarwal Triangles
        • Lecture8.15
          R D Sharma Triangles
      • 09. Circles
        8
        • Lecture9.1
          Introduction to Circle and Tangent and Theorems-1, 2 01 hour
        • Lecture9.2
          Property of Chord of Circle and Theorem-3 52 min
        • Lecture9.3
          Theorem 4 and Angle in Semicircle 01 hour
        • Lecture9.4
          Chapter Notes – Circles
        • Lecture9.5
          NCERT Solutions – Circles Exercise
        • Lecture9.6
          Revision Notes Circles
        • Lecture9.7
          R S Aggarwal Circles
        • Lecture9.8
          R D Sharma Circles
      • 10. Areas Related to Circles
        10
        • Lecture10.1
          Area and Perimeter of Circle, Semicircle and Quarter circle 56 min
        • Lecture10.2
          Touching Circles and Example on Bending Wire, Wheel Rotation 58 min
        • Lecture10.3
          Sector of A Circle 51 min
        • Lecture10.4
          Segment of Circle and Area of Combination of Plane Figures 01 hour
        • Lecture10.5
          Questions Based on Area of Combination of Plane Figures 01 hour
        • Lecture10.6
          Chapter Notes – Areas Related to Circles
        • Lecture10.7
          NCERT Solutions – Areas Related to Circles
        • Lecture10.8
          Revision Notes Areas Related to Circles
        • Lecture10.9
          R S Aggarwal Areas Related to Circles
        • Lecture10.10
          R D Sharma Areas Related to Circles
      • 11. Introduction to Trigonometry
        7
        • Lecture11.1
          Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions 01 hour
        • Lecture11.2
          Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems 48 min
        • Lecture11.3
          Complimentary angles in Inverse Multiplication, Miscellaneous Problems 01 hour
        • Lecture11.4
          Miscellaneous Questions 01 hour
        • Lecture11.5
          Chapter Notes – Introduction to Trigonometry
        • Lecture11.6
          NCERT Solutions – Introduction to Trigonometry
        • Lecture11.7
          Revision Notes Introduction to Trigonometry
      • 12. Surface Areas and Volumes
        9
        • Lecture12.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid 01 hour
        • Lecture12.2
          Volume Related Questions, Based On Embarkment 48 min
        • Lecture12.3
          Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume 52 min
        • Lecture12.4
          Quantity, Percentage, Very Short Answer Types Questions 44 min
        • Lecture12.5
          Chapter Notes – Surface Areas and Volumes
        • Lecture12.6
          NCERT Solutions – Surface Areas and Volumes
        • Lecture12.7
          Revision Notes Surface Areas and Volumes
        • Lecture12.8
          R S Aggarwal Surface Areas and Volumes
        • Lecture12.9
          R D Sharma Surface Areas and Volumes
      • 13. Statistics
        12
        • Lecture13.1
          Introduction and Mean of Ungrouped Data 46 min
        • Lecture13.2
          Mean of Grouped Data 58 min
        • Lecture13.3
          Median of Ungrouped and grouped Data 01 hour
        • Lecture13.4
          Mode of Ungrouped and Grouped Data 49 min
        • Lecture13.5
          Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type 53 min
        • Lecture13.6
          Cumulative Frequency Curve (Ogive)-More Than Type 52 min
        • Lecture13.7
          Miscellaneous Questions 38 min
        • Lecture13.8
          Chapter Notes – Statistics
        • Lecture13.9
          NCERT Solutions – Statistics
        • Lecture13.10
          Revision Notes Statistics
        • Lecture13.11
          R S Aggarwal Statistics
        • Lecture13.12
          R D Sharma Statistics
      • 14. Probability
        9
        • Lecture14.1
          Activities, Understanding term Probability, Experiments 54 min
        • Lecture14.2
          Calculations 43 min
        • Lecture14.3
          Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card 01 hour
        • Lecture14.4
          Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability 58 min
        • Lecture14.5
          Chapter Notes – Probability
        • Lecture14.6
          NCERT Solutions – Probability
        • Lecture14.7
          Revision Notes Probability
        • Lecture14.8
          R S Aggarwal Probability
        • Lecture14.9
          R D Sharma Probability
      • 15. Construction
        7
        • Lecture15.1
          Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles 44 min
        • Lecture15.2
          Construction of Different Angles, Construction of Angle Bisector 41 min
        • Lecture15.3
          Construction of Tangent of Circle at a Given Point 35 min
        • Lecture15.4
          Chapter Notes – Construction
        • Lecture15.5
          NCERT Solutions – Construction
        • Lecture15.6
          R S Aggarwal Construction
        • Lecture15.7
          R D Sharma Construction

        NCERT Solutions – Triangles Exercise 8.1 – 8.6

        Q.1    Fill in the blanks using the correct word given in brackets :
                      (i) All circles are ____ (congruent, similar)
                      (ii) All squares are ____ (similar, congruent)
                      (iii) All ____ triangles are similar. (isosceles, equilateral)
                      (iv) Two polygons of the same number of sides are similar, if
                        (a) their corresponding angles are ___ and
                        (b)their corresponding sides are ___ (equal , proportional)

        Sol.

        (i) similar
        (ii) similar
        (iii) equilateral
        (iv) equal, proportional


        Q.2      Give two different examples of pair of
                          (i) similar figures
                       (ii) non-similar figures.

        Sol.

        (i) Two different examples of pair of similar figures are :
        (a) any two rectangles (b) any two squares
        (ii) Two different examples of pair of non-similar figures are :
        (a) a scelene and an equilateral triangle.
        (b) an equilateral triangle and a right angle triangle


        Q.3       State whether the following quadrilaterals are similar or not :

        1
        Sol.          On looking at the given figures of the quadrilaterals, we can say that they are not similar.

         

        Q.1      In figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
        2Sol.

        (i) In fig. (i),
        since DE || BC,
        ADDB=AEEC⇒153=1EC
        ⇒ EC=315=3×1015= 2 cm.
        (ii) In fig. (ii),
        since DE || BC,
        ADDB=AEEC⇒AD7.2=1.85.4
        ⇒ AD=1854×7210= 2.4 cm.


        Q.2       E and F are points on the sides PQ and PB respectively of a Δ PQR. For each of the following cases, state whether EF || QR :
                        (i) PE = 3.9 cm, EQ = 4 cm, PF = 3.6 cm and FR = 2.4 cm.
                        (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
                        (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
        Sol.

        (i) We have,
              PE = 3.9 cm, EQ = 4 cm,
              PF = 3.6 cm and FR = 2.4 cm
              Now,  PEEQ=3.94=0.97 cm
              and, PFFR=3.62.4+32=1.2 cm
              ⇒ PEEQ=PFFR

        3      ⇒ EF does not divide the sides PQ and PR of ΔPQR in the same ratio.
              Therefore, EF is not parallel to QR.
        (ii) We have, PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
               Now, PEEQ=44.5=4045=89
               and, PFFR=89
               ⇒ PEEQ=PFEQ
               Thus, EF divides sides PQ and PR of ΔPQR in the same ratio.
                Therefore, by the converse of BasicProportionality Theorem we have EF || QR.
        (iii) We have, PQ = 1.28 cm, PR = 2.56 cm
                PF = 0.18 cm and, PF = 0.36 cm
                Therefore EQ = PQ – PE = (1.28 – 0.18) cm.
                                         = 1.10 cm
                 and, ER = PR – PF = (2.56 – 0.36)
                               = 2.20 cm
                 Now, PEEQ=0.181.10=18110=955
                 and, PFFR=0.362.20=36220=955
                 ⇒ PEEQ=PFFR
                 Thus, EF divides sides PQ and PR of ΔPQR in the same ratio.
                 Therefore, by the converse of Basic Proportionality Theorem, we have EF || QR


        Q.3        In figure, if LM || CB and LN || CD, prove that AMAB=ANAD.
        4Sol.

        In ΔABC , we have
        LM || CB [Given]
        Therefore by Proportionality Theorem, we have
        AMAB=ALAC
        In ΔACD, we have
        LN || CD [Given]
        Therefore By Basic Proportionality Theorem, we have
        ALAC=ANAD ………….. (2)
        From (1) and (2), we obtain that
        AMAB=ANAD


        Q.4      In figure, DE || AC and DF || AE. Prove that BFFE=BEEC
        5Sol.

        In ΔBCA, we have
        DE || AC [Given]
        Therefore By Basic Proportionality Theorem, we have
        BEEC=BDDA ………… (1)
        In Δ BEA, we have
        DF || AE [Given]
        Therefore by Basic Proportionality Theorem, we have
        BFFE=BDDA …………. (2)
        From (1) and (2), we obtain that
        BFFE=BEEC


        Q.5 In figure, DE || OQ and DF || OR. Show that EF || QR. 6Sol.

        In ΔPQO, we have
        DE || OQ [Given]
        Therefore By Basic Proportionality Theorem, we have
        PEEQ=PDDO ………… (1)
        In ΔPOR, we have
        DF || OR [Given]
        Therefore By Basic Proportionality Theorem, we have
        PDDO=PFFR ………….. (2)
        From (1) and (2), we obtain that
        PEEQ=PFFR
        ⇒EF||QR [By the converse of BPT]


        Q.6      In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
        7Sol.

        Given : O is any point within ΔPQR, AB || PQ and AC || PR
        To prove : BC || QR
        Construction : Join BC
        Proof : In ΔOPQ, we have
        AB || PQ [Given]
        Therefore By Basic Proportionality Theorem, we have
        OAAP=OBBQ ……… (1)
        In Δ OPR, we have
        AC || PR [Given]
        Therefore by Basic Proportionality Theorem, we have
        OAAP=OCCR …………. (2)
        From (1) and (2), we obtain that
        OBBQ=OCCR
        Thus, in ΔOQR, B and C are points dividing the sides OQ and OR in the same ratio. Therefore, by
        the converse of Basic Proportionality Theorem, we have,
        BC || QR


        Q.7      Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
        Sol.

        Given : A ΔABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E. 8To Prove : AE = EC
        Proof : Since DE || BC
        Therefore by Basic Proportionality Theorem, we have
        ADDB=AEEC ………… (1)
        But, AD = DB [Because D is the mid-point of AB]
        ⇒ ADDB = 1
        From (1), AEEC = 1 ⇒ AE = EC
        Hence, E is the mid-point of the third side AC.


        Q.8      Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is, parallel to the third side. (Recall that you have done it in Class IX).
        Sol.

        Given : A ΔABC, in which D and E are the mid-points of sides AB and AC respectively.
        9To prove : DE || BC
        Proof : Since, D and E are the mid-points of AB and AC respectively.
        Therefore AD = DB and AE = EC
        Since, AD = DB
        Therefore ADDB = 1 and AE = EC
        Therefore AEEC = 1
        Therefore ADDB=AEEC=1
        i.e. ADDB=AEEC
        Thus, in ΔABC, D and E are points dividing the sides AB and AC in the same ratio, Therefore, by the converse
        of Basic Proportionality Theorem, we have,
        DE || BC.


        Q.9      ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOBO=CODO
        Sol.

        Given : A trapezium ABCD, in which AB || DC and its diagonals AC and BD intersect each other at O.
        To prove : AOBO=CODO
        Construction : Through O, draw OE || AB i.e. OE || DC
        10Proof : In ΔADC, we have OE || DC [Construction]
        Therefore by Basic Proportionality Theorem, we have
        AEED=AOCO …………….. (1)
        Again, in ΔABD, we have
        OE || AB [Construction]
        Therefore by Basic Proportionality Theorem, we have
        EDAE=DOBO⇒AEED=BODO …………… (2)
        From (1) and (2), we obtain that
        AOCO=BODO⇒AOBO=CODO


        Q.10    The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO. Show that ABCD is a trapezium.
        Sol.

        Given : A quadrilateral ABCD in which its diagonals AC and BD intersect each other at O such that
        AOBO=CODO i.e. AOCO=BODO
        To prove : Quadrilateral ABCD is a trapezium.
        Construction : Through O draw OE || AB meeting AD in E.
        Proof : In ΔADB, we have
        OE || AB [Construction]
        Therefore by Basic Proportionality Theorem, we have
        DEEA=ODBO
        11⇒EADE=BODO
        ⇒EADE=BODO=AOCO[ThereforeAOCO=BODO(given)]
        ⇒EADE=AOCO
        Thus, in ΔADC, E and O are points dividing the sides AD and AC in the same ratio. Therefore, by
        the converse of Basic Proportionality Theorem, we have
        EO || DC
        But, EO || AB [Construction]
        Hence, AB || DC
        Therefore Quadrilateral ABCD is a trapezium.

         

        Q.1    State which pairs of triangle in figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :121314151617Sol.

        (i) In Δs ABC and PQR, we observe that
        ∠A=∠P=60∘, ∠B=∠Q=80∘ and ∠C=∠R=40∘
        Therefore by AAA criterion of similarity,
        ΔABC∼ΔPQR
        (ii) In Δs ABC and PQR, we observe that
        ABQR=BCRP=CAPQ=12
        Therefore by SSS criterion of similarity, ΔABC∼ΔQRP
        (iii) In Δs LMP and DEF, we observe that the ratio of the sides of these triangles are not equal.
        So, the two triangles are not similar.
        (iv) In Δs MNL and QPR, we observe that
        ∠M=∠Q=70∘
        But, MNPQ≠MLQR
        Therefore These two triangles are not similar as they do not satisfy SAS criterion of similarity.
        (v) In Δs ABC and FDE, we have
        ∠A=∠F=80∘
        But, ABDF≠ACEF [Because AC is not given]
        So, by SAS criterion of similarity, these two triangless are not similar.
        (vi) In Δs DEF and PQR
        ∠D=∠P=70∘ [Because ∠P=180∘−80∘−30∘=70∘]
        ∠E=∠Q=80∘
        So, by AAA criterion of similarity,
        ΔDEF∼ΔPQR


        Q.2      In figures, ΔODC∼ΔOBA, ∠BOC=125∘ and ∠CDO=70∘. Find ∠DOC, ∠DCO and ∠OAB.
        18Sol.

        Since BD is a line and OC is a ray on it,
        Therefore ∠DOC+∠BOC=180∘
        ⇒ ∠DOC+125∘=180∘
        ⇒ ∠DOC=180∘−125∘=55∘
        In ΔCDO, we have
        ∠CDO+∠DOC+∠DCO=180∘
        ⇒ 70∘+55∘+∠DCO=180∘
        ⇒ ∠DCO=180∘−125∘=55∘
        It is given that ΔODC∼ΔOBA
        ∠OBA=∠ODC,
        ∠OAB=∠OCD
        ⇒ ∠OBA=70∘ and ∠OAB=55∘
        Hence, ∠DOC=55∘, ∠DCO=55∘
        and ∠OAB=55∘


        Q.3     Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC=OBOD.
        Sol.

        Given : ABCD is a trapezium in which AB || DC.
        To prove : OAOC=OBOD
        Proof : In Δs OAB and OCD, we have
        ∠AOB=∠COD [Vert. opp. ∠s]
        ∠OAB=∠OCD [Alternate ∠s]
        and ∠OBA=∠ODC [Alternate ∠s]
        19Therefore by AAA criterion of similarity,
        ΔOAB∼ΔODC
        Hence, OAOC=OBOD
        [Because in case of two similar triangles, the ratio of their corresponding sides are equal]


        Q.4      In figure, QRQS=QTPR and ∠1=∠2. Show that ΔPQS∼ΔTQR
        20Sol.

        We have, QRQS=QTPR
        ⇒ QTQR=PRQS ………….. (1)
        Also, ∠1=∠2 [Given]
        ⇒ PR = PQ ………….. (2) [Because sides opp. to equal ∠s are equal]
        From (1) and (2), we get
        QTQR=PQQS⇒PQQT=QSQR ……….. (3)
        Therefore in Δs PQS and TQR, we have
        PQQT=QSQR and ∠PQS=∠TQR=∠Q
        Therefore by SAS criterion of similarity,
        ΔPQS∼ΔTQR.


        Q.5      S and T are points on sides PR and QR of ΔPQR such that ∠P=∠RTS. Show that ΔRPQ∼ΔRTS.
        Sol.
        21

        In Δs RPQ and RTS, we have
        ∠RPQ=∠RTS [Given]
        ∠PRQ=∠TRS [Common]
        Therefore by AA criterion of similarity,
        ΔRPQ∼ΔRTS


        Q.6      In figure, if ΔABE≅ΔACD, show that ΔADE∼ΔABC.
        22Sol.

        It is given that
        ΔABE≅ΔACD
        Therefore AB = AC [Because corresponding parts of congruent triangles are equal]
        and, AE = AD
        ⇒ ABAD=ACAE
        ⇒ ABAC=ADAE ………. (1)
        Therefore In Δs ADE and ABC, we have
        ABAC=ADAE [Because of (1)]
        and, ∠BAC=∠DAE [Common]
        Thus, by SAS criterion of similarity,
        ΔADE∼ΔABC


        Q.7       In figure, altitudes AD and CE of ΔABC intersect each other at the point P, Show that :
        23             
        (i) ΔAEP∼ΔCDP
                        (ii) ΔABD∼ΔCBE
                        (iii) ΔAEP∼ΔADB
                        (iv) ΔPDC∼ΔBEC
        Sol.

        (i)   In Δs AEP and CDP, we have
               ∠AEP=∠CDP=90∘ [Because CE⊥AB and AD⊥BC]
               ∠APE=∠CPD [Vert. opp. ∠s]
               Therefore by AA – criterion of similarity, we have
               ΔAEP∼ΔCDP
        (ii)  In Δs ABD and CBE, we have
               ∠ABD=∠CBE [Common angle]
               ∠ADB=∠CEB=90∘
               Therefore by AA-criterion of similarity, we have
               ΔABD∼ΔCBE
        (iii)  In Δs AEP and ADB, we have
                 ∠AEP=∠ADB=90∘ [Becauese AD⊥BC and CE⊥AB]
                 ∠PAE=∠DAB [Common angle]
                 Therefore by AA-criterion of similarity, we have
                 ΔAEP∼ΔADB
        (iv)   In Δs PDC and BEC, we have
                 ∠PDC=∠BEC=90∘ [Because AD⊥BC and CE⊥AB]
                 ∠PCD=∠ECB [Common angle]
                 Therefore AA-criterion of similarity, we have
                 ΔPDC∼ΔBEC


        Q.8      E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE∼ΔCFB.
        Sol.
        24

        In Δs ABE and CFB, we have
        ∠AEB=∠CBF [Alt. ∠s]
        ∠A=∠C [Opp. ∠s of a ||gm]
        Therefore by AA-criterion of similarity, we have
        ΔADE∼ΔCFB


        Q.9      In figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :
                       (i)  ΔABC∼ΔAMP
                       (ii) CAPA=BCMP
        25Sol.

        (i)   In Δs ABC and AMP, we have
               ∠ABC=∠AMP=90∘ [Given]
               and, ∠BAC=∠MAP [Common ∠s]
               Therefore by AA, criterion of similarity, we have
               ΔABC∼ΔAMP
        (ii)  We have, ΔABC∼ΔAMP [As proved above]
                ⇒ CAPA=BCMP


        Q.10     CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC∼ΔFEG, show that :
                        (i) CDGH=ACFG
                        (ii) ΔDCB∼ΔHGE
                        (iii) ΔDCA∼ΔHGF
        Sol.
        q10

        We have, ΔABC∼ΔFEG
        ⇒ ∠A=∠F ………….. (1)
        and, ∠C=∠G
        ⇒ 12∠C=12∠G
        ⇒ ∠1=∠3 and ∠2=∠4 ………….. (2)
        [Because CD and GH are bisectors of ∠C and ∠G respectively.
        Therefore In Δs DCA and HGF, we have
        ∠A=∠F [From (1)]
        ∠2=∠4 [From (2)]
        Therefore by AA-criterion of similarity, we have
        ΔDCA∼ΔHGF
        which proves the (iii) part.
        We have, ΔDCA∼ΔHGF [As proved above]d
        ⇒ ACFG=CDGH
        ⇒ CDGH=ACFG
        which proves the (i) part.
        In Δs DCB and HGE, we have
        ∠1=∠3 [From (2)]
        ∠B=∠E [Because ΔABC∼ΔFEG]
        Therefore by AA-criterion of similarity, we have
        ΔDCB∼ΔHGE
        which proves, the (ii) part.


        Q.11    In figure, E is a point on side CB produced of an isosceles triangle ABC with  AB = AC. If AD⊥BC and EF⊥AC, prove that ΔABD∼ΔECF.
        26Sol.

        Here, ΔABC is isosceles with AB = AC
        ∠B=∠C
        In Δs ABD and ECF, we have
        ∠ABD=∠ECF [Because ∠B=∠C]
        ∠ADB=∠EFC=90∘ [Because AD⊥BC and EF⊥AC]
        Therefore by AA-criterion of similarity,
        ΔABD∼ΔECF.


        Q.12     Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see figure). Show that ΔABC∼ΔPQR.
        27Sol.

        Given : AD is the median of Δ ABC and PM is the median of Δ PQR such that
        ABPQ=BCQR=ADPM
        To prove : ΔABC∼ΔPQR
        Proof : BD=12BC [Given]
        and, QM=12QR [Given]
        Also, ABPQ=BCQR=ADPM [Given]
        ⇒ ABPQ=2BD2QM=ADPM
        ⇒ ABPQ=BCQR=ADPM
        By SSS-criterion of similarity, we have
        ΔABD∼ΔPQM
        ⇒ ∠B=∠Q [Similar Δs have corresponding ∠s]
        and, ABPQ=BCQR [Given]
        Therefore by SAS-criterion of similarity, we have
        ΔABC∼ΔPQR


        Q.13    D is a point on the side BC of a triangle ABC such that ∠ADC=∠BAC. Show that CA2=CB.CD.
        Sol.

        In Δs ABC and DAC, we have
        ∠ADC=∠BAC and ∠C=∠C
        Therefore by AA-criterion of similarity, we have
        ΔABC∼ΔDAC
        28⇒ ABDA=BCAC=ACDC
        ⇒ CBCA=CACD
        ⇒ CA2=CB×CD


        Q.14    Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC∼ΔPQR.
        Sol.        Same as question 12.


        Q.15    A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
        Sol.

        Let AB be the vertical pole and AC be the shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF. Let DE = x metres.
        We have, AB = 6 m, AC = 4 cm and DF = 28 m
        In Δs ABC and DEF, we have
        29∠A=∠D=90∘,
        and ∠C=∠F
        [Because each is the angular elevation of the sun]
        Therefore by AA-criterion of similarity,
        ΔABC∼ΔDEF
        30⇒ABDE=ACDF
        ⇒ 6x=428
        ⇒ 6x=17
        ⇒ x=6×7=42
        Hence, the height of the tower is 42 metres.


        Q.16    If AD and PM are medians of triangles ABC and PQR respectively, where ΔABC∼ΔPQR, prove that ABPQ=ADPM
        Sol.

        Given : AD and PM are the medians of Δs ABC and PQR respectively, where
        ΔABC∼ΔPQR
        To prove : ABPQ=ADPM
        Proof : In Δs ABD and PQM, we have
        ∠B=∠Q [Because ΔABC∼ΔPQR]
        ABPQ=12BC12QR 31[Since AD and PM are the median of BC and QR respectively]
        ⇒ ABPQ=BDQM
        Therefore by SAS – criterion of similarity
        32ΔABD∼ΔPQM
        ⇒ ABPQ=BDQM=ADPM
        ⇒ ABPQ=ADPM

         

        Q.1     Let ΔABC∼ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, Find BC.
        Sol.

        We have, Area(ΔABC)Area(ΔDEF)=BC2EF2
        ⇒ 64121=BC2(15.4)2
        ⇒ 811=BC15.4
        ⇒ BC=(811×15.4) cm = 11.2 cm


        Q.2     Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
        Sol.

        In Δs AOB and COD, we have
        33∠AOB=∠COD [Vert. opp. ∠s]
        and, ∠OAB=∠OCD [Alernate ∠s]
        Therefore by AA-criterion of similarity, we have
        ΔAOB∼ΔCOD
        ⇒ Area(ΔAOB)Area(ΔCOD)=AB2DC2
        ⇒ Area(ΔAOB)Area(ΔCOD)=(2DC)2DC2=41
        Hence, area (ΔAOB) : area (ΔCOD) = 4 : 1


        Q.3     In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar(ΔABC)ar(ΔDBC)=AODO
        34Sol.

        Given : Two Δs ABC and DBC which stand on the same base but on the opposite sides of BC.
        To prove : Area(ΔABC)Area(ΔDBC)=AODO
        Construction : Draw AE⊥BC and DF⊥BC
        Proof : In Δs AOE and DOF, we have
        35∠AEO=∠DFO=90∘
        ∠AOE=∠DOF [Vertically opp. ∠s]
        Therefore by AA-criterion of similarity, we have
        ΔAOE∼ΔDOF
        ⇒ AEDF=AOOD ……….. (1)
        Now, Area(ΔABC)Area(ΔDBC)=12×BC×AE12×BC×DF
        ⇒ AEDF=AOOD [Using (1)]
        Therefore Area(ΔABC)Area(ΔDBC)=AOOD


        Q.4     If the areas of two similar triangles are equal, prove that they are congruent.
        Sol.

        Given : Two Δs ABC and DEF such that
        ΔABC∼ΔDEF
        and, Area (ΔABC) = Area (ΔDEF)
        To prove : ΔABC≅ΔDEF
        36Proof : ΔABC∼ΔDEF
        ⇒ ∠A=∠D, ∠B=∠E, ∠C=∠F
        and, ABDE=BCEF=ACDF
        To establish ΔABC≅ΔDEF, it is sufficient to prove that
        AB = DE, BC = EF and AC = DF
        Now, Area (ΔABC) = Area (ΔDEF)
        ⇒ Area(ΔABC)Area(ΔDEF)=1
        ⇒ AB2DE2=BC2EF2=AC2DF2=1
        ⇒ ABDE=BCEF=ACDF=1
        ⇒ AB = DE, BC = EF, AC = DF
        Hence, ΔABC≅ΔDEF.


        Q.5      D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC.
        Sol.

        Since, D and E are the mid-points of the sides AB and BC of ΔABC respectively.
        37Therefore DE || AC
        ⇒ DE || FC …………. (1)
        Since, D and F are the mid-points of the sides AB and AC of ΔABC respectively.
        Therefore DF || BC ⇒ DF || EC ………… (2)
        From (1) and (2), we can say that DECF is a parallelogram.
        Similarly, ADEF is  a parallelogram
        Now, in Δs DEF and ABC, we have
        ∠DEF=∠A [Opp. ∠s of ||gm ADEF]
        and, ∠EDF=∠C [Opp. ∠s of ||gm DECF]
        Therefore by AA-criterion of similarity, we have
        ΔDEF∼ΔABC
        ⇒ Area(ΔDEF)Area(ΔABC)=DE2AC2
        =(12AC)2AC2=14 [Because DE =======  ]
        Hence, Area (ΔDEF) : Area (ΔABC) = 1 : 4.


        Q.6      Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
        Sol.

        Given : ΔABC∼ΔPQR, AD and PM are the medians of Δs ABC and PQR respectively.
        To prove : Area(ΔABC)Area(ΔPQR)=AD2PM2
        38Proof : Since ΔABC∼ΔPQR
        Area(ΔABC)Area(ΔPQR)=(AB)2(PQ)2 ……………. (1)
        But, sideABsidePQ=medianADmedianPM ………….. (2)
        From (1) and (2), we have
        Area(ΔABC)Area(ΔPQR)=AD2PM2


        Q.7    Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of the diagonals.
        Sol.

        Given : A square ABCD. Equilateral Δs BCE and have been drawn on side and the diagonals AC respectively.
        To prove : Area (ΔBCE) 12(AreaΔACF)
        39Proof : ΔBCE∼ΔACF
        [Being equilateral so similar by AAA criterion of similarity]
        ⇒ Area(ΔBCE)Area(ΔACF)=BC2AC2
        ⇒ Area(ΔBCE)Area(ΔACF)=BC2(2√BC)2
        [Because Diagonal = 2–√ side ⇒ AC = 2–√ BC]
        ⇒ Area(ΔBCE)Area(ΔACF)=12


        Tick the correct answer and justify :
        Q.8    ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
                    (a) 2 : 1                      (b) 1 : 2
                    (c) 4 : 1                      (d) 1 : 4
        Sol.

        Since ΔABC and ΔBDE are equilateral, they are equiangular and hence
        ΔABC∼ΔBDE
        40⇒ Area(ΔABC)Area(ΔBDE)=BC2BD2
        ⇒ Area(ΔABC)Area(ΔBDE)=(2BD)2BD2
        [Because D is the mid-point of BC therefore BC = 2BD]
        ⇒ Area(ΔABC)Area(ΔBDE)=41
        Therefore (C) is the correct answer.


        Q.9      Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.
                      (A) 2 : 3                       (B) 4 : 9
                      (C) 81 : 16                  (D) 16 : 81
        Sol.

        Since the ratio of the areas of two similar triangles equal to the ratio of the squares of any two corresponding sides. Therefore,
        ratio of areas =  (4)2:(9)2=16:81
        Therefore (D) is the correct answer

         

        Q.1    Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
                    (i) 7 cm, 24 cm, 25 cm
                    (ii) 3 cm, 8 cm, 6 cm
                    (iii) 50 cm, 80 cm, 100 cm
                    (iv) 13 cm, 12 cm, 5 cm
        Sol.

        (i)      Let a = 7 cm, b = 24 cm and c = 25 cm
                  Here the larger side is c = 25 cm
                  We have, a2+b2=72+242=49+576=625=c2
                  So, the triangle with the given sides is a right triangle. Its hypotenuse = 25 cm.
        (ii)     Let a = 3 cm, b = 8 cm and c = 6 cm
                   Here the larger side is b = 8 cm
                  We have, a2+c2=32+62=9+36=45≠b2
                   So, the triangle with the given sides is not a right triangle.
        (iii)    Let a = 50 cm, b = 80 cm and c = 100 cm
                   Here the larger side is c = 100 cm
                   We have, a2+b2=502+802=2500+6400 = 8900 ≠ c2
                   So, the triangle with the given sides is not a right triangle.
        (iv)     Let a = 13 cm, b = 12 cm and c = 5 cm
                   Here the larger side is a = 13 cm
                   We have, b2+c2=122+52=144+25=169=a2
                   So, the triangle with the given sides is a right triangle. Its hypotenuse = 13 cm.


        Q.2     PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that PM2=QM.MR.
        Sol.

        Given : PQR is a triangle right angled at P and PM⊥QR.
        To prove : PM2=QM.MR
        Proof : Since PM⊥QR
        41Therefore ΔPQM∼ΔPRM
        ⇒ PMQM=MRPM
        ⇒ PM2=QM.MR


        Q.3    In figure, ABD is a triangle right angled at A and AC⊥BD. SHow that
                     (i) AB2=BC.BD
                     (ii) AC2=BC.DC
                     (iii) AD2=BD.CD
        42Sol.

        Given : ABD is a triangle right angled at A and AC⊥BD.
        To prove :
        (i) AB2=BC.BD
        (ii) AC2=BC.DC
        (iii) AD2=BD.CD
        Proof :
        (i) Since AC⊥BD
        Therefore ΔABC∼ΔADC and each triangle is similar to ΔABD
        Because ΔABC∼ΔABD
        Therefore ABBD=BCAB ⇒ AB2=BC.BD
        (ii) Since, ΔABC∼ΔADC
        Therefore ACBC=DCAC ⇒ AC2=BC.DC
        (iii) Since, ΔACD∼ΔABD
        ADCD=BDAD ⇒ AD2=BD.CD


        Q.4     ABC is an isosceles triangle right angled at C. Prove that AB2=2AC2.
        Sol.

        Since, ABC is an isosceles right triangle, right angled at C.
        43Therefore AB2=AC2+BC2
        ⇒ AB2=AC2+AC2 [Because BC = AC, given]
        ⇒ AB2=2AC2


        Q.5     ABC is an isosceles triangle with AC = BC. If AB2=2AC2, prove that ABC is a right triangle.
        Sol.

        Since, ABC is an isosceles triangle with AC = BC and AB2=2AC2
        ⇒ AB2=AC2+AC2
        ⇒ AB2=AC2+BC2 [Because AC = BC, given]
        Therefore ΔABC is right angled at AC.


        Q.6      ABC is an equilateral triangle of side 2a. Find each of its altitudes.
        Sol.

        Let ABC be an equilateral triangle of side 2a units.
        Draw AD⊥BC. Then, D is the mid-point of BC.
        44⇒ BD=12BC=12×2a=a
        Since, ABD is a right triangle, right angled at D.
        Therefore AB2=AD2+BD2
        ⇒ (2a)2=AD2+(a)2
        ⇒ AD2=4a2−a2=3a2
        Therefore each of its altitude = 3–√a.


        Q.7      Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
        Sol.

        Let the diagonals AC and BD of rhombus ABCD interesect each other at O. Since the diagonals of a rhombus bisect each other at right angles.
        45Therefore ∠AOB=∠BOC=∠COD = ∠DOA=90∘
        and, AO = CO, BO = OD
        Since, AOB is a right triangle, right angled at O.
        Therefore AB2=OA2+OB2
        ⇒ AB2=(12AC)2+(12BD)2 [Because OA = OC and OB = OD]
        ⇒ 4AB2=AC2+BD2 ……………. (1)
        Siumilarly, we have
        4BC2=AC2+BD2 …………… (2)
        4CD2=AC2+BD2 …………… (3)
        and 4AD2=AC2+BD2 …………….(4)
        Adding all these results, we get
        4(AB2+BC2+CD2+AD)2=4(AC2+BD2)
        ⇒ AB2+BC2+CD2+DA2=AC2+BD2


        Q.8     In figure, O is a point in the interior of a triangle ABC, OD⊥BC, OE⊥AC and OF⊥AB. Show that
                    (i) OA2+OB2+OC2−OD2−OE2−OF2=AF2+BD2+CE2
                
        (ii) AF2+BD2+CE2=AE2+CD2+BF2
        46Sol.

        Join AO, BO and CO
        (i)    In right Δs OFA, ODB and OEC, we have
                OA2=AF2+OF2
                OB2=BD2+OD2
                and, OC2=CE2+OE2
        47            Adding all these, we get
                    OA2+OB2+OC2=AF2+BD2+CE2+OF2+OD2+OE2
                    ⇒ OA2+OB2+OC2−OD2−OE2−OF2
                   = AF2+BD2+CE2
        (ii)      In right Δs ODB and ODC, we have
                   OB2=OD2+BD2
                   and, OC2=OD2+CD2
                   ⇒ OB2−OC2=BD2−CD2 …………… (1)
                   Suimilarly, we have
                   OC2−OA2=CE2−AE2 ……………. (2)
                   and, OA2−OB2=AF2−BF2 …………….. (3)
                   Adding equations (1), (2) and (3), we get
                   (OB2−OC2)+(OC2−OA2)+(OA2−OB2)
                   = (BD2−CD2)+(CE2−AE2)+(AF2−BF2)
                    ⇒ (BD2+CE2+AF2)−(AE2+CD2+BF2) = 0
                    ⇒ AF2+BD2+CE2
                     = AE2+BF2+CD2.


        Q.9    A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
        Sol.

        Let AB be the ladder, B be the window and CB be the wall. Then, ABC is a right triangle, right angled at C.
        Therefore AB2=AC2+BC2
        48⇒ 102=AC2+82
        ⇒ AC2=100−64
        ⇒ AC2=36
        ⇒ AC=6
        Hence, the foot of the ladder is at a distance 6 m from the base of the wall.


        Q.10    A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
        Sol.

        Let AB (= 24 m) be a guy wire attached to a vertical pole. BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at C.
        Because AB2=AC2+BC2
        49⇒ 242=AC2+182
        ⇒ AC2=576−324
        ⇒ AC2=252
        ⇒ AC=252−−−√=67–√
        Hence, the stake may be placed at distance of 67–√ m from the base of the pole.


        Q.11    An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1½ hours ?
        Sol.

        Let the first aeroplane starts from O and goes upto A towards north where OA=(1000×32) km = 1500 km.
        50Let the second aeroplane starts from O at the same time and goes upto B towards west where OB=(1200×32) km = 1800 km.
        According to the problem the required distane = BA.
        In right angled ΔABC, by Pythagoras theorem, we have,
        AB2=OA2+OB2
                            = (1500)2+(1800)2
                            = 2250000+3240000
                            = 5490000=9×61×100×100
        ⇒ AB=3×10061−−√=30061−−√ km.


        Q.12   Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
        Sol.

        Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m.
        Draw CE⊥AB and join AC
        51Therefore CE = DB = 12 m,
        AE = AB – BE = AB – CD = (11–6) m = 5 m.
        In right angled ΔACE, by Pythagoras theorem, we have
        AC2=CE2+AE2
                           = (12)2+(5)2
                            = 144+25=169
        ⇒ AC=169−−−√=13
        Hence, the distance between the tops of the two poles is 13 m.


        Q.13    D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2+BD2=AB2+DE2.
        Sol.

        In right angled Δs ACE and DCB, we have
        AE2=AC2+CE2
        and, BD2=DC2+BC2
        ⇒ AE2+BD2=(AC2+CE2)+(DC2+BC2)
        ⇒ AE2+BD2=(AC2+BC2)+(DC2+CE2)
        ⇒ AE2+BD2=AB2+DE2
        [By Pythagoras theorem, AC2+BC2=AB2 and DC2+CE2=DE2]
        52


        Q.14   The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3 CD (see figure). Prove that 2AB2=2AC2+BC2.
        53Sol.

        We have, DB = 3CD
        Now, BC = DB + CD
        ⇒ BC = 3CD + CD [Because BD = 3CD]
        ⇒ BC = 4CD
        Therefore CD=14BC and,
        DB=3CD=34BC …………. (1)
        Since, ΔABD is a right triangle, right angled at D. Therefore, by Pythagoras theorem, we have
        AB2=AD2+DB2 …………… (1)
        Similarly, from ΔACD, we have
        AC2=AD2+CD2 ……….. (2)
        (1) – (2) gives,
        AB2−AC2=DB2−CD2
        ⇒ AB2−AC2=(34BC)2−(14BC)2 [Using (1)]
        ⇒ AB2−AC2=(916−116)BC2
        ⇒ AB2−AC2=12BC2
        ⇒ 2AB2−2AC2=BC2
        ⇒ 2AB2=2AC2+BC2
        Which is the required result.


        Q.15   In an equilateral triangle ABC, D is a point on side BC such that BD=13BC. Prove that 9AD2=7AB2
        Sol.

        Let ABC be an equilateral triangle and let D be a point on BC such that
        BD=13BC
        Draw AE⊥BC. Join AD.
        In Δs AEB and AEC, we have
        54AB = AC [Because ΔABC is equilateral]
        ∠AEB=∠AEC [Because each = 90°]
        and, AE = AE
        Therefore by SAS-criterion of similarity, we have
        ΔAEB∼ΔAEC
        ⇒ BE = EC
        Thus, we have
        BD=13BC,DC=23BC
        and, BE=EC=12BC …………….. (1)
        Since, ∠C=60∘
        Therefore ΔADC is an acute triangle.
        Therefore AD2=AC2+DC2−2DC×EC
        = AC2+(23BC)2−2×23BC×12BC [Using (1)]
        = AC2+49BC2−23BC2
        = AB2+49AB2−23AB2 [Because AB = BC = AC]
        = (9+4−6)AB29=79AB2
        ⇒ 9AD2=7AB2, which is the required result.


        Q.16    In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
        Sol.

        Let ABC be an equilateral triangle and let AD⊥BC.
        In ΔADB and ΔADC, we have
        AB = AC [Given]
        ∠B=∠C [Each = 60°]
        and, ∠ADB=∠ADC [Each = 90°]
        By RHS criterion of congruence, we have
        ΔADB≅ΔADC
        ⇒ BD=DC
        ⇒ BD=DC=12BC
        Since ΔADB is a right triangle, right angled at D, by Pythagoras theorem, we have
        55AB2=AD2+BD2
        ⇒ AB2=AD2+(12BC)2
        ⇒ AB2=AD2+14BC2
        ⇒ AB2=AD2+AB24 [Because BC = AB]
        ⇒ 34AB2=AD2
        ⇒ 3AB2=4AD2
        Hence, the result.


        Q.17    Tick the correct answer and justify : In ΔABC, AB=63–√ cm, AC = 12 cm and BC = 6 cm. The angles A and B are respectively :
                      (a) 90° and 30°
                      (b) 90° and 60°
                      (c) 30° and 90°
                      (d) 60° and 90°
        Sol.

        In ΔABC, we have
        56AB = 63–√ cm,
        AC = 12 cm
        and BC = 6 cm …………… (1)
        Now, AB2+BC2=(63–√)2+(6)2
        = 36 × 3 + 36 = 108 + 36
        = 144=(AC)2
        Thus, ΔABC is a right triangle, right angled at B.
        Therefore ∠B=90∘
        Let D be the mid-point of AC. We know that the mid-point of the hypotenuse of a right triangle is equidistant from the vertices.
        AD = BD = CD
        ⇒ CD = BD = 6 cm [Because CD=12AC]
        Also BC = 6 cm
        Therefore in ΔBDC, we have
        BD = CD = BC
        ⇒ ΔBDC is equilateral ⇒ ∠ACB=60∘
        Therefore ∠A=180∘−(∠B+∠C)
        = 180° – (90° + 60°) = 30°
        Thus, ∠A=30∘ and ∠B=90∘
        Therefore (C) is the correct choice.

         

        (*These exercises are not for examination point of view)
        Q.1     In Figure, PS is the bisector of ∠QPR of ΔPQR. Prove that QSSR=PQPR
        57Sol.

        Given PQR is a triangle and PS is the internal bisector of ∠QPR meeting QR at S.
        Therefore ∠QPS=∠SPR
        To prove : QSSR=PQPR
        Construction : Draw RT || SP to cut QP produced at T.
        Proof : Since PS || TR and PR cuts them, hence, we have
        58
        ∠SPR=∠PRT ………… (1)  [Alternate ∠s]
        and, ∠QPS=∠PTR …………… (2)  [Corresponding ∠s]
        But, ∠QPS=∠SPR [Given]
        Therefore ∠PRT=∠PTR [From (1) and (2)]
        ⇒ PT = PR ………. (3) [Because sides opp. to equal ∠s are equal]
        Now, in ΔQRT, we have
        RT || SP [By construction]
        Therefore QSSR=PQPT [By Basic Proportionality Theorem]
        ⇒ QSSR=PQPR [From (3)]


        Q.2     In figure, D is a point on hypotenuse AC of ΔABC, BD⊥AC, DM⊥BC and DN⊥AB. Prove that
                      (i) DM2=DN.MC
                      (ii) DN2=DM.AN
        59
        Sol.

        We have, AB⊥BC and DM⊥BC
        ⇒ AB || DM
        Similarly, we have
        BC⊥AB and DN⊥AB
        ⇒ CB || DN
        60
        Hence, quadrilateral BMDN is a rectangle.
        Therefore BM = ND
        (i)   In ΔBMD, we have
               ∠1+∠BMD+∠2=180∘
               ⇒ ∠1+90∘+∠2=180∘⇒∠1+∠2=90∘
              Similarly, in ΔDMC, we have
               ∠3+∠4=90∘
               Since BD⊥AC. Therefore,
               ∠2+∠3=90∘
               Now, ∠1+∠2=90∘ and ∠2+∠3=90∘
               ⇒ ∠1+∠2=∠2+∠3
               ⇒ ∠1=∠3
               Also, ∠3+∠4=90∘ and ∠2+∠3=90∘
               ⇒ ∠3+∠4=∠2+∠3⇒∠2=∠4
               Thus, in Δs BMD and DMC, we have
               ∠1=∠3 and ∠2=∠4
               Therefore By AA-criterion of similarity, we have
                ΔBMD∼ΔDMC
                ⇒ BMDM=MDMC
                ⇒ DNDM=DMMC [Because BM = ND]
                ⇒ DM2=DN×MC
        (ii)   Proceeding as in (i), we can prove that
                 ΔBND∼ΔDNA
                 ⇒ BNDN=NDNA
                 ⇒ DMDN=DNAN [Because BN = DM]
                 ⇒ DN2=DM×AN


        Q.3      In figure, ABC is a triangle in which ∠ABC>90∘ and AD⊥CB produced. Prove that AC2=AB2+BC2+2BC.BD
        61
        Sol.

        Given : ABC is a triangle in which ∠ABC>90∘ and AD⊥CB produced.
        To Prove : AC2=AB2+BC2+2BC.BD
        Proof : Since ΔADB is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have
        AB2=AD2+DB2 ……….. (1)
        62
        Again, ΔADC is a right triangle, right-angled at D. Therefore, by Pythagoras theorem, we have
        AC2=AD2+DC2
        ⇒ AC2=AD2+(DB+BC)2
        ⇒ AC2=AD2+DB2+BC2+2DB.BC
        ⇒ AC2=(AD2+DB2)+BC2+2BC.BD
        ⇒ AC2=AB2+BC2+2BC.BD [Using (1)]
        which proves the required result.


        Q.4      In figures, ABC is a triangle in which ∠ABC<90∘ and AD⊥BC. Prove that AC2=AB2+BC2−2BC.BD
        63
        Sol.

        Given : ABC is a triangle in which ∠ABC<90∘ and AD⊥BC.
        To prove : AC2=AB2+BC2−2BC.BD
        Proof : Since ΔADB is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have,
        AB2=AD2+BD2 ………….. (1)
        Again, ΔADC is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have,
        AC2=AD2+DC2
        ⇒ AC2=AD2+(BC−BD)2
        ⇒ AC2=AD2+(BC2+BD2−2BC.BD)
        ⇒ AC2=(AD2+BD2)+BC2−2BC.BD
        ⇒ AC2=AB2+BC2−2BC.BD [Using (1)]
        which proves the required result.


        Q.5      In figure, AD is a median of a triangle ABC and AM⊥BC. Prove that
                        (i) AC2=AD2+BC.DM+(BC2)2
                        (ii) AB2=AD2−BC.DM+(BC2)2
                        (iii) AC2=AB2=2AD2+12BC2
        64
        Sol.

        Since ∠AMD=90∘, therefore, ∠ADM<90∘ and ∠ADC>90∘.
        Thus, ∠ADC is acute and ∠ADC is obtuse.
        (i)    In ΔADC, ∠ADC is an obtuse angle.
        65

                  Therefore AC2=AD2+DC2+2DC.DM
                  ⇒ AC2=AD2+(BC2)2+2.BC2.DM
                 ⇒ AC2=AD2+(BC2)2+BC.DM
                  ⇒ AC2=AD2+BC.DM+(BC2)2 ………….. (1)
        (ii)     In ΔABD, ∠ADM is an acute angle.
                  Therefore AB2=AD2+BD2−2BD.DM
                   ⇒ AB2=AD2+(BC2)2−2.BC2.DM
                  ⇒ AB2=AD2−BC.DM+(BC2)2 ………… (2)
        (iii)    From (1) and (2), we get
                   AB2+AC2=2AD2+12BC2


        Q.6      Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
        Sol.

        We know that if AD is a median of ΔABC, then AB2+AC2=2AD2+12BC2
        [See above question part (iii)]
        Sine the diagonals of a parallelogram bisect each other, therefore, BO and DO are medians of Δs
        ABC and ADC respectively.
        66
        Therefore AB2+BC2=2BO2+12AC2 ……….. (1)
        and, AD2+CD2=2DO2+12AC2 ………. (2)
        Adding (1) and (2), we get
        AB2+BC2+CD2+AD2=2(BO2+DO2)+AC2
        ⇒ AB2+BC2+CD2+AD2=2(14BD2+14BD2)+AC2 [Because DO=12BD]
        ⇒ AB2+BC2+CD2+AD2=AC2+BD2


        Q.7      In figure, two chords AB and CD intersect each other at the point P. Prove that
                       (i) ΔAPC∼ΔDPB
                       (ii) AP.PB = CP.DP
        67
        Sol.

        (i)   In Δs APC and DPB, we have
               ∠APC=∠DPB [Vert. opp. ∠s]
               ∠CAP=∠BDP [Angles in the same segment of a circle are equal]
               Therefore by AA-criterion of similarity, we have
               ΔAPC∼ΔDPB
        (ii)  Since ΔAPC∼ΔDPB
                Therefore APDP=CPPB
                ⇒ AP×PB=CP×DP


        Q.8      In figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle, Prove that
                       (i) ΔPAC∼ΔPDB
                       (ii) PA.PB = PC.PD
        68
        Sol.

        (i)    In Δs PAC and PDB, we have
                ∠APC=∠BPD [Common]
                ∠PAC=∠PDB [Because ∠BAC=180∘−∠PAC and ∠PDB=∠CDB=180∘−∠BAC=180∘−(180∘−∠PAC)=∠PAC]
               Therefore by AA-criterion of similarity, we have
                ΔPAC∼ΔDPB
        (ii)   Since ΔPAC∼ΔDPB
                 PAPD=PCPB
                 ⇒ PA.PB = PC.PD


        Q.9     In figure, D is a point on side BC of ΔABC such that BDCD=ABAC. Prove that AD is the bisector of ∠BAC.
        69
        Sol.

        Given : ABC is a triangle and D is a point on BC such that
        BDCD=ABAC
        To prove : AD is the internal bisector of ∠BAC.
        Construction : Produce BA to E such that AE = AC. Join CE.
        70
        Proof : In ΔAEC, since AE = AC, hence
        ∠AEC=∠ACE ……….. (1)
        [Because Angles opp. to equal sides of a Δ are equal]
        Now, BDCD=ABAC [Given]
        ⇒ BDCD=ABAE [Because AE = AC, construction]
        Therefore by converse of Basic Proportionality Theorem, we have
        DA || CE
        Now, since CA is a transeversal, we have
        ∠BAD=∠AEC ……… (2) [Corresponding ∠s]
        and, ∠DAC=∠ACE …… (3) [Alternate angles]
        Also, ∠AEC=∠ACE [From (1)]
        Hence, ∠BAD=∠DAC [From (2) and (3)]
        Thus, AD bisects, ∠BAC internally.


        Q.10 Nazima is fly fishing i a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taur, how much string does she have out (see figure) ? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds ?
        nazima
        Sol.

        In fact, we want to find AC.
        By Pythagoras theorem, we have
        AC2=(2.4)2+(1.8)2
        71
        ⇒ AC2=5.76+3.24=9.00
        ⇒ AC = 3 m
        Therefore length of string she have out = 3 m.
        Length of the string pulled at the rate of 5 cm/sec in 12 seconds.
        = (5 × 12) cm
        = 60 cm = 0.60 m
        Therefore remaining string left out = (3 – 0.6) m = 2.4 m
        In 2nd case let us find PB
        PB2=PC2−BC2
                           = (2.4)2−(1.8)2
                           = 5.76−3.24=2.52
        ⇒ PB=2.52−−−−√=1.59 (nearly)
        72
        Hence, the horizontal distance of the fly from Nazima after 12 seconds
        = (1.59 + 1.2) m
        = 2.79 m (nearly)

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