TOPICS
Exercise - 6.2

Question-1 :-  In Figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). triangle

Solution :-
(i) Given that DE || BC.
    Let EC = x cm
    By using basic proportionality theorem, we get
    AD/DB = AE/EC
    1.5/3 = 1/x
    1.5x = 3
    x = 3/1.5
    x = 30/15
    x = 2
    Hence, EC = x cm = 2 cm
    
(ii) Given that DE || BC.
    Let AD = x cm
    By using basic proportionality theorem, we get
    AD/DB = AE/EC
    x/7.2 = 1.8/5.4
    5.4x = 1.8 x 7.2
    5.4x = 12.96
    x = 12.96/5.4
    x = 2.4
    Hence, AD = x cm = 2.4 cm
    

Question-2 :-  E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR :
(i) PE = 3.9 cm, EQ = 3 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

Solution :-
(i) Given that PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
    triangle
    PE/EQ = 3.9/3 = 1.3
    PF/FR = 3.6/2.4 = 1.5
    Hence, PE/EQ ≠ PF/FR
    Therefore, EF is not parallel to QR.
    
(ii) Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
    triangle
    PE/EQ = 4/4.5 = 8/9
    PF/FR = 8/9
    Hence, PE/EQ = PF/FR
    Therefore, EF || QR.
    
(iii) Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
    triangle
    PE/PQ = 0.18/1.28 = 18/128 = 9/64
    PF/PR = 0.36/2.56 = 9/64
    Hence, PE/PQ = PF/PR
    Therefore, EF || QR.
    

Question-3 :-  In Figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD quadrilateral

Solution :-
  Given that : LM || CB and LN || CD
  Prove that : AM/AB = AN/AD
  Proof : LM || CB                      (Given)
          By using basic proportionality theorem, we get
          AM/AB = AL/AC .........(i)
          Similarly, LN || CD
          AN/AD = AL/AC .........(ii)
          From (i) & (ii), We get
          AM/AB = AN/AD
    

Question-4 :-  In Figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC quadrilateral

Solution :-
  Given that : DE || AC and DF || AE
  Prove that : BF/FE = BE/EC 
  Proof : In ΔABC, DE || AC                   (Given)
          By using basic proportionality theorem, we get
          BD/DA = BE/EC .........(i)
          Similarly, In ΔBAE, DF || AE
          BD/DA = BF/FE .........(ii)
          From (i) & (ii), We get
          BE/EC = BF/FE
    

Question-5 :-  In Figure, DE || OQ and DF || OR. Show that EF || QR. quadrilateral

Solution :-
  Given that : DE || OQ and DF || OR
  Prove that : EF || QR
  Proof : In Δ POQ, DE || OQ                   (Given)
          By using basic proportionality theorem, we get
          PE/EQ = PD/DO .........(i)
          Similarly, In Δ POR, DF || OR
          PF/FR = PD/DO .........(ii)
          From (i) & (ii), We get
          PE/EQ = PF/FR
          By using converse of basic proportionality theorem
          Therefore, EF || QR
    

Question-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. quadrilateral

Solution :-
  Given that : AB || PQ and AC || PR
  Prove that : BC || QR
  Proof : In Δ POQ, AB || PQ                   (Given)
          By using basic proportionality theorem, we get
          OA/AP = OB/BQ .........(i)
          Similarly, In Δ POR, AC || PR
          OA/AP = OC/CR .........(ii)
          From (i) & (ii), We get
          OB/BQ = OC/CR
          By using converse of basic proportionality theorem
          Therefore, BC || QR
    

Question-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).

Solution :-
  Given that : Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ || BC
  triangle
  Prove that : Q is the mid-point of AC.
  Proof : In Δ ABC, PQ || BC             
          By using basic proportionality theorem, we get
          AQ/QC = AP/PB 
          AQ/QC = 1/1       (P is the mid-point of AB. So, AB = PB)
          AQ = QC
          or Q is the mid point of AC.
    

Question-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).

Solution :-
  Given that : Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. 
               So, we have AP = PB and AQ = QC.
  triangle
  Prove that : PQ || BC
  Proof : In Δ ABC           
          AP/PB = 1/1       (Given AP = PB)
          AQ/QC = 1/1       (Given AQ = QC)
          Therefore, AQ/QC = AP/PB
          By using basic proportionality theorem, we get
          PQ || BC
    

Question-9 :-  ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Solution :-
  Given that : ABCD is a trapezium in which AB || DC.
  Prove that : AO/BO = CO/DO
  Construction : Draw a line EF through point O, such that EF || CD
  trapezium
  Proof : In Δ ADC, EO || CD          
          By using basic proportionality theorem, we get
          AE/ED = OA/OC  ........(i)
          In Δ ABD, OE || AB          
          By using basic proportionality theorem, we get
          ED/AE = OD/OB 
          AE/ED = OB/OD  ........(ii)
          From equations (i) & (ii), we get
          OA/OC = OB/OD
          Therefore, AO/BO = OC/OD
    

Question-10 :-  The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Solution :-
  Given that : The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO.
  Prove that : ABCD is a trapezium.
  Construction : Draw a line OE || AB.
  trapezium
  Proof : In Δ ABD, OE || AB          
          By using basic proportionality theorem, we get
          AE/ED = OB/OD  ........(i)
          AO/BO = CO/DO  ........(ii) (Given)
          From equations (i) & (ii), we get
          AE/ED = AO/OC
          EO || DC                    (By the converse of basic proportionality theorem)
          AB || OE || DC
          AB || CD
          Therefore, ABCD is a trapezium.
    
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