TOPICS
Exercise - 8.1

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

Solution :-
  Let the common ratio between the angles be x. 
  Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.   
  As the sum of all interior angles of a quadrilateral is 360°,   
  3x + 5x + 9x + 13x = 360°   
  30x = 360°
  x = 12°   
  Hence, the angles are   
  3x = 3 × 12 = 36° 
  5x = 5 × 12 = 60°   
  9x = 9 × 12 = 108° 
  13x = 13 × 12 = 156°
    

Question-2 :-  If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution :-
       parallelogram 
  Given that : Let ABCD be a parallelogram.  
  Prove that :  ABCD is a rectangle or one of its interior angles is 90°. 
  Proof : In ∆ABC and ∆DCB,   
          AB = DC (Opposite sides of a parallelogram are equal)   
          BC = BC ( Common)   
          AC = DB (Given)    
          ∆ABC ≅ ∆DCB (By SSS Congruence rule) 
          By CPCT Rule,       
          ∠ABC = ∠DCB
           
          It is known that the sum of the measures of angles on the same side of transversal is 180°.    
          ∠ABC + ∠DCB = 180° (AB || CD)   
          ∠ABC + ∠ABC = 180°   
          2 ∠ABC = 180°   
          ∠ABC = 90°   
          Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.   
    

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

Solution :-
  parallelogram 
  Given that : Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle 
               i.e., OA = OC, OB = OD, and  
               ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°.
  Prove that : ABCD is a rhombus or ABCD is a parallelogram and all the sides of ABCD are equal.
  Proof :  In ∆AOD and ∆COD,   
           OA = OC (Diagonals bisect each other)   
           ∠AOD = ∠COD (Given)   
           OD = OD (Common)   
           ∆AOD ≅ ∆COD (By SAS congruence rule) 
           By CPCT Rule,  
           AD = CD .....(1)  
         
           Similarly, it can be proved that   
           AD = AB and CD = BC .....(2)   
           From equations (1) and (2),   
           AB = BC = CD = AD   
           Since opposite sides of quadrilateral ABCD are equal, 
           It can be said that ABCD is a parallelogram. 
           Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.   
    

Question-4 :-  Show that the diagonals of a square are equal and bisect each other at right angles.

Solution :-
       parallelogram 
  Given that : Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
  Prove that : The diagonals of a square are equal and bisect each other at right angles 
               or AC = BD, OA = OC, OB = OD, and ∠AOB = 90°.   
  Proof : In ∆ABC and ∆DCB,   
          AB = DC (Sides of a square are equal to each other)   
          ∠ABC = ∠DCB (All interior angles are of 90 )   
          BC = CB (Common side)   
          ∆ABC ≅ ∆DCB (By SAS congruency)   
          AC = DB (By CPCT)   
          Hence, the diagonals of a square are equal in length. 
          
          In ∆AOB and ∆COD,   
          ∠AOB = ∠COD (Vertically opposite angles)   
          ∠ABO = ∠CDO (Alternate interior angles)   
          AB = CD (Sides of a square are always equal)    
          ∆AOB ≅ ∆COD (By AAS congruence rule)   
          AO = CO and OB = OD (By CPCT)   
          Hence, the diagonals of a square bisect each other.   

          In ∆AOB and ∆COB,   
          As we had proved that diagonals bisect each other, therefore,   
          AO = CO   
          AB = CB (Sides of a square are equal)   
          BO = BO (Common)    
          ∆AOB ≅ ∆COB (By SSS congruency)   
          ∠AOB = ∠COB (By CPCT)   
          However, AOB + COB = 180° (Linear pair)   
          2 ∠AOB = 180°   
          ∠AOB = 90°   
          Hence, the diagonals of a square bisect each other at right angles.
    

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

Solution :-
       parallelogram 
  Given that : Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. 
               The diagonals of ABCD are equal and bisect each other at right angles. 
               Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = = 90°. 
  Prove that : ABCD is a square or ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90°.
  Proof : In ∆AOB and ∆COD,   
          AO = CO (Diagonals bisect each other)   
          OB = OD (Diagonals bisect each other)   
          ∠AOB = ∠COD (Vertically opposite angles)   
          ∆AOB ≅ ∆COD (SAS congruence rule)   
          By CPCT Rule,
          AB = CD ... (1) 
          And, ∠OAB = ∠OCD  
        
          However, these are alternate interior angles for line AB and CD and 
          alternate interior angles are equal to each other only when the two lines are parallel.  
          AB || CD ... (2)   
          From equations (1) and (2), we obtain ABCD is a parallelogram.   

          In ∆AOD and ∆COD,   
          AO = CO (Diagonals bisect each other)   
          ∠AOD = ∠COD (Given that each is 90°)   
          OD = OD (Common)   
          ∆AOD ≅ ∆COD (SAS congruence rule)  
          By CPCT Rule, 
          AD = DC ... (3)   
          However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)   
          AB = BC = CD = DA   
          Therefore, all the sides of quadrilateral ABCD are equal to each other.   

          In ∆ADC and ∆BCD,   
          AD = BC (Already proved)   
          AC = BD (Given)   
          DC = CD (Common)   
          ∆ADC ≅ ∆BCD (SSS Congruence rule) 
          By CPCT Rule,  
          ∠ADC = ∠BCD
          
          However, ∠ADC + ∠BCD = 180° (Co-interior angles)   
          ∠ADC + ∠ADC = 180° 
          2 ∠ADC = 180°   
          ∠ADC = 90°  
          One of the interior angles of quadrilateral ABCD is a right angle.   
          Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and 
          one of its interior angles is 90°. 
          Therefore, ABCD is a square.    
    

Question-6 :- : Diagonal AC of a parallelogram ABCD bisects ∠ A. Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus. Parallelogram

Solution :-
  parallelogram 
  Given that :  ABCD is a parallelogram and AC bisects A.   
  Prove that : (i) it bisects ∠ C also, 
               (ii) ABCD is a rhombus. 
  Proof : (i) ABCD is a parallelogram.   
          ∠DAC = ∠BCA (Alternate interior angles) ... (1)  
          And, ∠BAC = ∠DCA (Alternate interior angles) ... (2) 
          However, it is given that AC bisects A.   
          ∠DAC = ∠BAC ... (3)   
          From equations (1), (2), and (3), we obtain   
          ∠DAC = ∠BCA = ∠BAC = ∠DCA ... (4)   
          ∠DCA = ∠BCA   
          Hence, AC bisects C. 
          
          (ii)From equation (4), we obtain   
          ∠DAC = ∠DCA   
          DA = DC (Side opposite to equal angles are equal)   
          However, DA = BC and AB = CD (Opposite sides of a parallelogram)   
          AB = BC = CD = DA  
          Hence, ABCD is a rhombus.   
    

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

Solution :-
       parallelogram 
  Given that : ABCD is a rhombus.
  Prove that : Diagonal AC bisects ∠ A as well as ∠ C and 
               Diagonal BD bisects ∠ B as well as ∠ D.  
  Proof : In ∆ABC,   
          BC = AB (Sides of a rhombus are equal to each other)   
          ∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal)   
          However, ∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD)   
          ∠2 = ∠3   
          Therefore, AC bisects C.  
         
          Also, ∠2 = ∠4 (Alternate interior angles for || lines BC and DA)   
          ∠1 = ∠4   
          Therefore, AC bisects A.   
          Similarly, it can be proved that BD bisects B and D as well. 
    

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

Solution :-
       parallelogram 
  Given that : ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C.
  Prove that : (i) ABCD is a square 
               (ii) diagonal BD bisects ∠ B as well as ∠ D.
  Proof : (i) It is given that ABCD is a rectangle.   
          ∠A = ∠C     
          CD = DA (Sides opposite to equal angles are also equal)   
          However, DA = BC and AB = CD (Opposite sides of a rectangle are equal)   
          AB = BC = CD = DA   
          ABCD is a rectangle and all of its sides are equal.   
          Hence, ABCD is a square.   

          (ii) Let us join BD.   
          In ∆BCD,   
          BC = CD (Sides of a square are equal to each other)   
          ∠CDB = ∠CBD (Angles opposite to equal sides are equal)   
          However, ∠CDB = ∠ABD (Alternate interior angles for AB || CD)   
          ∠CBD = ∠ABD   
          BD bisects B.   
          Also, ∠CBD = ∠ADB (Alternate interior angles for BC || AD)   
          ∠CDB = ∠ABD   
          BD bisects D.   
    

Question-9 :- : In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) Δ APD ≅ Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram Parallelogram

Solution :-
  Given that :  ABCD is a parallelogram and AC bisects A.   
  Prove that : (i) Δ APD ≅ Δ CQB 
               (ii) AP = CQ 
               (iii) Δ AQB ≅ Δ CPD 
               (iv) AQ = CP 
               (v) APCQ is a parallelogram 
  Proof : (i) In ∆APD and ∆CQB,   
          ∠ADP = ∠CBQ (Alternate interior angles for BC || AD)   
          AD = CB (Opposite sides of parallelogram ABCD)   
          DP = BQ (Given)   
          ∆APD ≅ ∆CQB (Using SAS congruence rule) 
        
          (ii) As we had observed that ∆APD ≅ ∆CQB,   
          AP = CQ (CPCT) 
          
          (iii) In ∆AQB and ∆CPD,   
          ∠ABQ = ∠CDP (Alternate interior angles for AB || CD)   
          AB = CD (Opposite sides of parallelogram ABCD)   
          BQ = DP (Given)   
          ∆AQB ≅ ∆CPD (Using SAS congruence rule) 
        
          (iv) As we had observed that ∆AQB ≅ ∆CPD,  
          AQ = CP (CPCT)   

          (v) From the result obtained in (ii) and (iv),   
          AQ = CP and  AP = CQ Since  
          opposite sides in quadrilateral APCQ are equal  
          to each other,  APCQ is a parallelogram.   
    

Question-10 :-  ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ Parallelogram

Solution :-
  Given that : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
  Prove that : (i) Δ APB ≅ Δ CQD 
               (ii) AP = CQ 
  Proof : (i) In ∆APB and ∆CQD,   
          ∠APB = ∠CQD (Each 90°)   
          AB = CD (Opposite sides of parallelogram ABCD)  
          ∠ABP = ∠CDQ (Alternate interior angles for AB || CD)   
          ∆APB ≅ ∆CQD (By AAS congruency)   
            
          (ii) By using the above result   
          ∆APB ≅ ∆CQD, we obtain   
          AP = CQ (By CPCT) 
    

Question-11 :-  In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively . Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) Δ ABC ≅ Δ DEF. Parallelogram

Solution :-
  Given that : In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF.
  Prove that : (i) quadrilateral ABED is a parallelogram 
              (ii) quadrilateral BEFC is a parallelogram 
              (iii) AD || CF and AD = CF 
              (iv) quadrilateral ACFD is a parallelogram 
              (v) AC = DF 
              (vi) Δ ABC ≅ Δ DEF. 
  Proof : (i) It is given that AB = DE and AB || DE.   
          If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.   
          Therefore, quadrilateral ABED is a parallelogram.   
            
          (ii) Again, BC = EF and BC || EF   
          Therefore, quadrilateral BCEF is a parallelogram. 
          
          (iii) As we had observed that ABED and BEFC are parallelograms, therefore   
          AD = BE and AD || BE (Opposite sides of a parallelogram are equal and parallel)   
          And, BE = CF and BE || CF (Opposite sides of a parallelogram are equal and parallel)   
          AD = CF and AD || CF 
          
          (iv) As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD 
          are equal and parallel to each other, therefore, it is a parallelogram.   
            
          (v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal 
          and parallel to each other.    
          AC || DF and AC = DF   

          (vi) ∆ABC and ∆DEF, AB = DE (Given)   
          BC = EF (Given)   
          AC = DF (ACFD is a parallelogram)  
          ∆ABC ≅ ∆DEF (By SSS congruence rule)   
    

Question-12 :- : ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i) ∠ A = ∠ B
(ii) ∠ C = ∠ D
(iii) Δ ABC ≅ Δ BAD
(iv) diagonal AC = diagonal BD Parallelogram

Solution :-
  Given that :  ABCD is a trapezium in which AB || CD and AD = BC .   
  Prove that : (i) ∠ A = ∠ B 
               (ii) ∠ C = ∠ D 
               (iii) Δ ABC ≅ Δ BAD 
               (iv) diagonal AC = diagonal BD 
  Construction : Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E.
  Proof : (i) AD = CE (Opposite sides of parallelogram AECD)   
          However, AD = BC (Given)   
          Therefore, BC = CE   
          ∠CEB = ∠CBE (Angle opposite to equal sides are also equal)   
          Consider parallel lines AD and CE. AE is the transversal line for them.   
          ∠A + ∠CEB = 180º (Angles on the same side of transversal)   
          ∠A + ∠CBE = 180º (Using the relation ∠CEB = ∠CBE) ... (1)   
          However, B + ∠CBE = 180º (Linear pair angles) ... (2)   
          From equations (1) and (2), we obtain 
          ∠A = ∠B   
            
          (ii) AB || CD   
          ∠A + ∠D = 180º (Angles on the same side of the transversal)   
          Also, ∠C + ∠B = 180° (Angles on the same side of the transversal)   
          ∠A + ∠D = ∠C + ∠B   
          However, ∠A = ∠B [Using the result obtained in (i)]   
          ∠C = ∠D   
             
          (iii) In ∆ABC and ∆BAD,   
          AB = BA (Common side)   
          BC = AD (Given)   
          ∠B = ∠A (Proved before)   
          ∆ABC ≅ ∆BAD (SAS congruence rule)   
      
          (iv) We had observed that, ∆ABC ≅ ∆BAD   
          AC = BD (By CPCT)   
    
CLASSES

Connect with us:

Copyright © 2015-16 by a1classes.

www.000webhost.com