NCERT Class 9 Maths Solutions Chapter - 8 Quadrilaterals, Ex - 8.1

Ex - 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 is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since 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^o

^{Question 2.}  If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution

Let ABCD be a parallelogram. To show ABCD a rectangle, only we need to prove one of its interior angle is 90�.
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)

ABC = DCB       
We know that sum of measures of angles on the same side of transversal is 180º.
ABC + DCB = 180�     (AB || CD)
ABC + ABC = 180�   
ABC = 180�
ABC = 90�
Since ABCD is a parallelogram and one of its interior angles is 90�, therefore, ABCD is rectangle.

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

Solution

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�
To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.
Now, in AOD and COD
OA = OC                 (Diagonal bisects each other)
AOD = COD            (given)
OD = OD                 (common)

 AOD COD             (by SAS congruence rule)

AD = CD                 (1)

Similarly we can prove that
AD = AB and CD = BC        (2)
From equations (1) and (2), we can say that
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, so, we can say that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, so, we can say that ABCD is a rhombus.

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

Solution

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
 To show diagonals of a square are equal and bisect each other at right angles, we need to prove AC = BD, OA = OC, OB = OD and AOB = 90� 
  Now, in ABC and DCB
  AB = DC                          (sides of square are equal to each other)
ABC = DCB                 (all interior angles are of 90 )
  BC = BC                          (common side)
 ABCDCB           (by SAS congruency)
 AC = DB                        (by CPCT)

Hence, the diagonals of a square are equal in length
  
Now in AOB and COD
AOB = COD                        (vertically opposite angles)
ABO = CDO                        (alternate interior angles)
AB = CD                             (sides of 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
Now in AOB and COB
Now as we had proved that diagonals bisect each other
So, AO = CO                               
AB = CB                              (sides of square are equal)
BO = BO                             (common)
 AOB COB           (by SSS congruence)
 AOB = COB             (by CPCT)

But, AOB + COB = 180�        (linear pair)
2AOB = 180�
AOB = 90�
Hence, the diagonals of a square bisect each other at right angle.

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

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.
Given that the diagonals of ABCD are equal and bisect each other at right angles.

So, AC = BD, OA = OC, OB = OD     and

    AOB = BOC = COD = AOD = 90�.
To prove ABCD a square, we need to prove ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90�.

Now, 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)
 AB = CD                           (by CPCT)        ... (1)

And OAB = OCD             (by CPCT)
But these are alternate interior angles for line AB and CD and alternate interior angle are equal to each other only when the two lines are parallel
 AB || CD                              ... (2)

From equations (1) and (2), we have
ABCD is a parallelogram
Now, 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)
 AD = DC                            ... (3)

But, AD = BC and AB = CD         (opposite sides of parallelogram ABCD)
 AB = BC = CD = DA

So, all the sides quadrilateral ABCD are equal to each other

Now, in ADC and BCD
AD = BC                                 (Already proved)
AC = BD                                 (given)
DC = CD                                (Common)
 ADC BCD              (SSS Congruence rule)
 ADC = BCD                 (by CPCT)

But ADC + BCD = 180^o     (co-interior angles)
ADC + ADC = 180^o
ADC = 180^o

<sup>ADC = 90o 
One of interior angle of ABCD quadrilateral is a right angle

Now, we have ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90�. Therefore, ABCD is a square.</sup>

^{Question 6.}  Diagonal AC of a parallelogram ABCD bisects A (see the given figure).  Show that (i) It bisects C also;                                                                                    

(ii) ABCD is a rhombus.

Solution

(i) ABCD is a parallelogram.
     DAC = BCA        (Alternate interior angles)    ... (1)

    And BAC = DCA        (Alternate interior angles)    ... (2)
    But it is given that AC bisects A.
     DAC = BAC                        ... (3)

    From equations (1), (2) and (3), we have
    DAC = BCA = BAC = DCA                ... (4) 
    DCA = BCA
     Hence, AC bisects C.
(ii)

     From equation (4), we have
     DAC = DCA    
      DA = DC                         (side opposite to equal angles are equal)

     But DA = BC and AB = CD     (opposite sides of parallelogram)
      AB = BC = CD = DA

     Hence, ABCD is 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

Let us join AC
In ABC
BC = AB        (side of a rhombus are equal to each other)
 1 = 2         (angles opposite to equal sides of a triangle are equal)

But 1 = 3         (alternate interior angles for parallel lines AB and CD)
2 = 3
So, AC bisects C.
Also, 2 = 4    (alternate interior angles for || lines BC and DA)
1 = 4
So, AC bisects A
Similarly, we can prove 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

(i) Given that AC is bisector of A and C.
    Or DAC = DCA       
    CD = DA                             (sides opposite to equal angles are also equal)
    But DA = BC and AB = CD     (opposite sides of 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                               (side of a square are equal to each other)
     CDB = CBD                    (angles opposite to equal sides are equal)
     But CDB = ABD              (alternate interior angles for AB || CD)
      CBD = ABD

      BD bisects  B.
      Also CBD = ADB        (alternate interior angles for BC || AD)
      CDB = ADB
        BD bisects D.

Question 9.  In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that:

(i) APD CQB                
(ii) AP = CQ
(iii) AQB  CPD
(iv) AQ = CP
(v) APCQ is a parallelogram

Solution

(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), we have
              AQ = CP and  AP = CQ
       Since opposite sides in quadrilateral APCQ are equal to each other. So, APCQ is a

       parallelogram.

Question 10.  ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that

(i)   APB  CQD               
(ii)  AP = CQ

Solution

(i)  In APB and CQD
     APB = CQD         (each 90^o) 
     AB = CD             (opposite sides of parallelogram ABCD) 
     ABP = CDQ         (alternate interior angles for AB || CD) 
      APB CQD        (by AAS congruency) 

(ii) By using the result obtained as above 
     APB CQD, we have 
     AP = CQ             (by CPCT) 

Question 11.  In ABC and DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that

(i)    Quadrilateral ABED is a parallelogram        
(ii)    Quadrilateral BEFC is a parallelogram
(iii)    AD || CF and AD = CF
(iv)    Quadrilateral ACFD is a parallelogram
(v)    AC = DF 
(vi)    ABCDEF 

Solution

(i)   Here AB = DE and AB || DE.    
      Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be

      a parallelogram.
      Therefore, quadrilateral ABED is a parallelogram.

(ii)  Again BC = EF and BC || EF.
      Therefore, quadrilateral BEFC is a parallelogram.

(iii)  Here ABED and BEFC are parallelograms.
       AD = BE, and AD || BE        
       (Opposite sides of parallelogram are equal and parallel)
       And BE = CF, and BE || CF        
       (Opposite sides of parallelogram are equal and parallel)
         AD = CF, and AD || CF

(iv)   Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and

        parallel to each other,
        so, it is a parallelogram.
(v)    As ACFD is a parallelogram, so, 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)
        ABCDEF                (by SSS congruence rule)

Question 12.  ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that

(i)   A = B                        
(ii)   C = D
(iii)   ABC BAD
(iv)    Diagonal AC = diagonal BD

Solution

Extend AB. Draw a line through C, which is parallel to AD, intersecting AE at point E.
Now, AECD is a parallelogram.

(i)   AD = CE                  (opposite sides of parallelogram AECD) 
      But AD = BC             (given)
      So, BC = CE
     CEB = CBE         (angle opposite to equal sides are also equal)
     Now consider parallel lines AD and CE. AE is transversal line for them
    A + CEB = 180�         (angles on the same side of transversal)
    A+ CBE = 180�         (using the relationCEB = CBE)    ... (1)
     But B + CBE = 180�     (linear pair angles)            ... (2)
     From equations (1) and (2), we have
     A = B
(ii)  AB || CD
      A + D = 180�            (angles on the same side of transversal)
      Also C + B = 180          (angles on the same side of transversal)
       A + D = C + B

      But A = B             [using the result obtained proved 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)   ABCBAD
       AC = BD                 (by CPCT)

Ex-7.5

Ex-8.2