NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

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Chapter 8 - Quadrilaterals Exercise Ex. 8.1

Solution 1
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.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 3x + 5x + 9x + 13x = 360

30x = 360
    x = 12
Hence, the angles are
3x = 3 Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 12 = 36
5x = 5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 12 = 60
9x = 9 Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 12 = 108
13x = 13 Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 12 = 156o

Solution 2
(i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAPB and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCQD
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAPB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCQD         (each 90o)
     AB = CD             (opposite sides of parallelogram ABCD)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABP = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCDQ         (alternate interior angles for AB || CD)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals APB Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCQD        (by AAS congruency)

(ii) By using the result obtained as above
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAPB Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCQD, we have
     AP = CQ             (by CPCT)

Solution 3
 
(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)
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 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.
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AC || DF and AC = DF

(vi)   Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDEF.
        AB = DE                                (given)
        BC = EF                                (given)
        AC = DF                                (ACFD is a parallelogram)  
        Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABCNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDEF                (by SSS congruence rule)


Solution 4
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
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCEB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBE         (angle opposite to equal sides are also equal)
     Now consider parallel lines AD and CE. AE is transversal line for them
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCEB = 180         (angles on the same side of transversal)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA+ Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBE = 180         (using the relationNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCEB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBE)    ... (1)  
     But Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsB + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBE = 180     (linear pair angles)            ... (2)
     From equations (1) and (2), we have
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsB  
(ii)  AB || CD
      Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsD = 180            (angles on the same side of transversal)
      Also Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsC + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsB = 180          (angles on the same side of transversal)  
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsD = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsC + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsB

      But Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsB             [using the result obtained proved in (i)]
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals  Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsD

(iii)  In ABC and BAD
       AB = BA                 (common side)
       BC = AD                 (given)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA                 (proved before)
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals BAD            (SAS congruence rule)

(iv)  Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals ABCNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBAD  
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AC = BD                 (by CPCT)
 

Solution 5
                                             Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
Let ABCD be a parallelogram. To show ABCD a rectangle, only we need to prove one of its interior angle is 90.
In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCB
AB = DC             (opposite sides of a parallelogram are equal)
BC = BC             (common)
AC = DB             (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals DCB         (by SSS Congruence rule)

Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCB         
We know that sum of measures of angles on the same side of transversal is 180º.
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCB = 180     (AB || CD)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC = 180     
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC = 180  
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC = 90
Since ABCD is a parallelogram and one of its interior angles is 90, therefore, ABCD is rectangle.  

Solution 6
                                             Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle
 i.e. OA = OC, OB = OD and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBOC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOD = 90
To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOD and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD
OA = OC                 (Diagonal bisects each other)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOD = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD            (given)
OD = OD                 (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAODNcert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals COD             (by SAS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAD = 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.

Solution 7
                                          Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = 90 
  Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCB
  AB = DC                          (sides of square are equal to each other)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCB                 (all interior angles are of 90 )
  BC = BC                          (common side)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABCNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCB           (by SAS congruency)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AC = DB                        (by CPCT)
 
Hence, the diagonals of a square are equal in length
    
Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD                        (vertically opposite angles)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABO = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCDO                        (alternate interior angles)
AB = CD                             (sides of square are always equal)
 Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD          (by AAS congruence rule)
 Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAO = CO and OB = OD     (by CPCT)             

Hence, the diagonals of a square bisect each other
Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOB
Now as we had proved that diagonals bisect each other
So, AO = CO                                 
AB = CB                              (sides of square are equal)
BO = BO                             (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOBNcert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals COB           (by SSS congruence)
 Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOB             (by CPCT)

But, Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOB = 180        (linear pair)
2Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = 180
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = 90
Hence, the diagonals of a square bisect each other at right angle.  


Solution 8
                                            Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
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
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBOC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOD = 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD
AO = CO                              (Diagonals bisect each other)
OB = OD                              (Diagonals bisect each other)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD                   (Vertically opposite angles)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOB Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals COD           (SAS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AB = CD                           (by CPCT)        ... (1)

And Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsOAB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsOCD             (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
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AB || CD                              ... (2)

From equations (1) and (2), we have
ABCD is a parallelogram
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOD and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD
AO = CO                               (Diagonals bisect each other)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOD = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD                    (Given that each is 90)
OD = OD                               (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAOD Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCOD                         (SAS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AD = DC                            ... (3)

But, AD = BC and AB = CD         (opposite sides of parallelogram ABCD)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AB = BC = CD = DA

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

Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD
AD = BC                                 (Already proved)
AC = BD                                 (given)
DC = CD                                (Common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD              (SSS Congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD                 (by CPCT)
But Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD = 180o     (co-interior angles)
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC = 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.


Solution 9
 
(i) ABCD is a parallelogram.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCA        (Alternate interior angles)    ... (1)
    And Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCA        (Alternate interior angles)    ... (2)
    But it is given that AC bisects Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBAC                        ... (3)
    From equations (1), (2) and (3), we have
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCA = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCA                ... (4) 
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCA = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCA
     Hence, AC bisects Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsC.
(ii)
     From equation (4), we have
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCA    
      Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDA = DC                         (side opposite to equal angles are equal)

     But DA = BC and AB = CD     (opposite sides of parallelogram)
      Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAB = BC = CD = DA
 
     Hence, ABCD is rhombus

Solution 10
                                           Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
Let us join AC
In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC
BC = AB        (side of a rhombus are equal to each other)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals1 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals2         (angles opposite to equal sides of a triangle are equal)

But Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals1 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals3         (alternate interior angles for parallel lines AB and CD)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals2 =Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 3
So, AC bisects Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsC.
Also, Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals2 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals4    (alternate interior angles for || lines BC and DA)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals1 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals4
So, AC bisects Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA
Similarly, we can prove that BD bisects B and D as well.

Solution 11
                             Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
(i) Given that AC is bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsA and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsC.
    Or Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCA         
    CD = DA                             (sides opposite to equal angles are also equal)
    But DA = BC and AB = CD     (opposite sides of rectangle are equal)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAB = 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD
     BC = CD                               (side of a square are equal to each other)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCDB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBD                    (angles opposite to equal sides are equal)
     But Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCDB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABD              (alternate interior angles for AB || CD)
      Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBD = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABD

     Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals BD bisects  Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsB.
      Also Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBD = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADB        (alternate interior angles for BC || AD)
      Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCDB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADB
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals  BD bisects D.

Solution 12
 
(i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAPD and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCQB
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADP = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCBQ                   (alternate interior angles for BC || AD)
     AD = CB                       (opposite sides of parallelogram ABCD)
     DP = BQ                       (given)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAPD Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCQB         (using SAS congruence rule)

(ii)  As we had observed that Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAPD Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals CQB
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AP = CQ                     (CPCT)
 
(iii)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAQB and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCPD
       Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCDP          (alternate interior angles for AB || CD)
       AB = CD                     (opposite sides of parallelogram ABCD)
       BQ = DP                     (given)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAQB  CPD               (using SAS congruence rule)

(iv)  As we had observed that Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAQBNcert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals CPD
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals 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.


Chapter 8 - Quadrilaterals Exercise Ex. 8.2

Solution 1
(i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC, S and R are the mid points of sides AD and CD respectively.
     In a triangle the line segment joining the mid points of any two sides of the triangle is
     parallel to the third side and is half of it.
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsSR || AC and SR = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AC                ... (1)

(ii)  In ABC, P and Q are mid points of sides AB and BC respectively. So, by using
      mid-point theorem, we have
      PQ || AC and PQ =  Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAC                ... (2)   
      Now using equations (1) and (2), we have
      PQ || SR and PQ = SR                 ... (3)
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals PQ = SR
(iii)  From equations (3), we have
       PQ || SR and PQ = SR          
       Clearly one pair of opposite sides of quadrilateral PQRS is parallel and equal
       Hence, PQRS is a parallelogram.


Solution 2
                                          Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC, P and Q are mid points of sides AB and BC respectively.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals PQ || AC and PQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AC         (using mid-point theorem)    ... (1)
 
    In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC
    R and S are the mid points of CD and AD respectively
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals RS || AC and RS =  Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAC         (using mid-point theorem)    ... (2)

    From equations (1) and (2), we have
     PQ || RS and PQ = RS

    As in quadrilateral PQRS one pair of opposite sides are equal and parallel to each other, so, it is a parallelogram.
    Let diagonals of rhombus ABCD intersect each other at point O.
    Now in quadrilateral OMQN
    MQ || ON                ( Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals PQ || AC)
    QN || OM                 (Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals QR || BD)
    So, OMQN is parallelogram
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsMQN = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNOM
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsPQR = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNOM

    But, Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNOM = 90o     (diagonals of a rhombus are perpendicular to each other)

    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsPQR = 90o

    Clearly PQRS is a parallelogram having one of its interior angle as 90.
    Hence, PQRS is rectangle.

Solution 3
                                            Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
    Let us join AC and BD
    In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC
    P and Q are the mid-points of AB and BC respectively
    Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsPQ || AC and PQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AC         (mid point theorem)        ... (1)

    Similarly in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADC
    SR || AC and SR =   Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAC         (mid point theorem)    ...  ... (2)
    Clearly, PQ || SR and PQ = SR
    As in quadrilateral PQRS one pair of opposite sides is equal and parallel to
    each other, so, it is a parallelogram.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals PS || QR and PS = QR        (opposite sides of parallelogram)... (3)


    Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD, Q and R are mid points of side BC and CD respectively.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals QR || BD and QR = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBD         (mid point theorem)        ... (4)

    But diagonals of a rectangle are equal
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAC = BD                   ... ...  (5)

    Now, by using equation (1), (2), (3), (4), (5) we can say that
    PQ = QR = SR = PS  
    So, PQRS is a rhombus.

Solution 4
                                           Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
 
By converse of mid-point theorem a line drawn, through the mid point of any side of a triangle and parallel to another side bisects the third side.
Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABD
EF || AB and E is mid-point of AD
So, this line will intersect BD at point G and G will be the mid-point of DB.
Now as EF || AB and AB || CD
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsEF || CD    (Two lines parallel to a same line are parallel to each other)
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD, GF || CD and G is the midpoint of line BD. So, by using converse of mid-point theorem, F is the mid-point of BC.

Solution 5
ABCD is a parallelogram
Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AB || CD

So, AE || FC
Again AB = CD         (opposite sides of parallelogram ABCD)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAB = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCD
          AE = FC             (E and F are midpoints of side AB and CD)
As in quadrilateral AECF one pair of opposite sides (AE and CF) are parallel and equal to each other. So, AECF is a parallelogram.
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAF || EC            (Opposite sides of a parallelogram)
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDQC, F is mid point of side DC and FP || CQ (as AF || EC). So, by using converse of mid-point theorem, we can say that
P is the mid-point of DQ
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDP = PQ            ... (1)
Similarly, in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAPB, E is mid point of side AB and EQ || AP (as AF || EC). So, by using converse of mid-point theorem, we can say that
Q is the mid-point of PB
Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsPQ = QB            ... (2)
From equations (1) and (2), we may say that
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

Solution 6
                                           Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals BD             ... (1)  
Similarly in Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsBCD
QR || BD and QR = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals BD               ... (2)
From equations (1) and (2), we have
SP || QR and SP = QR  
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.
Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Solution 7
                                             Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
(i)     In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsABC
        Given that M is mid point of AB and MD || BC.
        So, D is the mid-point of AC.        (Converse of mid-point     theorem)

(ii)    As DM || CB and AC is a transversal line for them.
        So, Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsMDC + Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsDCB = 180         (Co-interior angles)  
        Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsMDC + 90 = 180
        Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsMDC = 90
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals MD Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AC
 
(iii)   Join MC
 
                              Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals
 
     In Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAMD and Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCMD
     AD = CD                         (D is the midpoint of side AC)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsADM = Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCDM             (Each 90)
     DM = DM                         (common)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals Ncert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsAMDNcert Solutions Cbse Class 9 Mathematics Chapter - QuadrilateralsCMD        (by SAS congruence rule)
     So, AM = CM                   (by CPCT)
     But AM = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AB              (M is mid point of AB)
     So, CM = MA = Ncert Solutions Cbse Class 9 Mathematics Chapter - Quadrilaterals AB