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Quadrilaterals : Exercise 8.1 (Mathematics NCERT Class 9th)


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

Let the angles be (3x)º, (5x)º, (9x)º and (13x)º
Then ,       3x + 5x + 9x + 13 x = 360              [The sum of the angles of a quadrilatral is 360º]
\Rightarrow  30x = 360
\Rightarrow    x = {{360} \over {30}} = 12
Therefore, the  angles are (3 × 12)º, (5× 12)º, (9 × 12)º and (13 × 12)º i.e., 36º, 60º, 108º and 156º.


Q.2      If the diagonals of a parallelogram are equal then show that it is a rectangle.
Sol.       A parallelogram ABCD in which  AC = BD.

1To prove : ABCD is a rectangle.
Proof : In  \Delta s\,\,ABC\,and\,DCB,\,we\,have

AB = DC                       [Opp. sides of a || gm]
BC = BC                       [Common]
and     AC = DB         [Given]
Therefore, by  SSS criterion  of congruence.
 \Delta ABC \cong \,\Delta \,DCB
 \Rightarrow   \angle ABC = \angle DCB                       ... (1)
[Corresponding parts of congruent triangles are equal]

But AB || DC and BC cuts them.
Therefore,      \angle ACB + \angle DCB = 180^o
\Rightarrow      2\angle ABC = 180^o  ... (2) [Sum of consecutive interior angles is 180°]
\Rightarrow      \angle ABC = 90^o
Thus,     \angle ABC = \angle DCB = 90^o
ABCD is a parallelogram one of whose angle is 90º .
Hence , ABCD is a rectangle.


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

A quadrilateral ABCD in which  the diagonals AC and BD intersect at O such that AO = OC , BO = OD and AC \bot BD.

2To prove : ABCD is a rhombus.
Proof : Since the diagonals  AC and BD of quadrilateral ABCD bisect each other  at right  angles.

Therefore,  AC is the perpendicular bisector of the segment BD.
\Rightarrow   A and C both are equidistant from B and D.
\Rightarrow   AB = AD and CB = CD        ... (1)
Also , BD is the perpendicular bisector of line segment  AC.
 \Rightarrow B and D both  are equidistant from A and C.
 \Rightarrow   AB = BC and AD = DC        ... (2)
From (1) and (2), we get
AB = BC = CD = AD
Thus , ABCD is a quadrilateral  whose diagonals bisect each other  at right angles and all four sides are equal.
Hence , ABCD is a rhombus.
Second Proof : First we shall prove that ABCD is a || gm.

In  \Delta s\,AOD\,and\,COB , we have
AO = OC                                                        [Given]
OD = OB                                                        [Given]
 \angle AOD = \angle COB       [Vertically opp. angles]
By SAS criterion of congruence,
 \Delta AOD \cong \Delta COB
\Rightarrow    \angle OAD =\angle OCB             ... (1)
[Corresponding parts of congruent triangles are equal]
Now, line AC intersects AD and BC at A and C respectively such  that
 \angle OAD = \angle OCB     [From (1)]
i.e. alternate interior angles are equal.
Therefore,            AD || BC
Similarly,             AB || CD
Hence , ABCD is a parallelogram.
Now, we shall  prove that || gm ABCD is a rhombus.
In  \Delta s\,AOD\,and\,COD , we have
OA = OC                                                         [Given]
\angle AOD = \angle COD         [Both are right angles ]
OD = OD
Therefore, by SAS criterion of congruence
\Delta AOD \cong \Delta COD
\Rightarrow         AD = CD                 ... (2)
[Corresponding  parts of congruent triangles are equal]
Now, ABCD is a || gm             [Proved above]
 \Rightarrow     AB = CD and AD  = BC                      [Opp. sides of a || gm are equal]
 \Rightarrow     AB = CD = AD = BC                            [Using (2)]
Hence, quadrilateral ABCD is a rhombus.


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

Given  : A square ABCD.
To prove : AC =  BD,  AC  \bot BD and OA = OC, OB = OD.
Proof : Since  ABCD is a square.

Therefore,    AB || DC and  AD || BC.
Now,  AB || DC and transversal AC intersects them at A and C respectively.

3Therefore,     \angle BAC = \angle DCA            [Alternate interior angles are equal]
\Rightarrow      \angle BAO = \angle DCO         ... (1)
Again AB || DC and BD intersects them at B and D respectively.
Therefore,  \angle ABD = \angle CDB   [Since Alternate interior angles are equal]
\Rightarrow     \angle ABO = \angle CDO         ... (2)

Now, in  \Delta s\,AOB\,\,and\,\angle COD , we have
\angle BAO = \angle DCO         [From (1)]
AB = CD     [Oppositge sides of a ||gm are equal]
and      \angle ABO = \angle CDO         [From (2)]
Therefore by ASA congruence criterion
\Delta\,AOB\,\, \cong \Delta COD
\Rightarrow         OA = OC and OB = OD
[Corresponding parts of congruent  \Delta s are equal]

Hence , the  diagonals bisect each other.
In  \Delta s ADB and BCA,  we have       
AD = BC         [Sides of a square are equal]
\angle BAD = \angle ABC     [Each equal to 90º]
and         AB = BA         [Common]
Therefore by  SAS criterion of congruence
\Delta \,ADB\,\, \cong \Delta \,BCA
\Rightarrow         AC =  BD
[Since Corresponding parts of congruent  \Delta s are equal]
Hence, the diagonals are equal .
Now in  \Delta s AOB and AOD we have
OB = OD                     [Since diagonals of || gm bisect each other]
AB = AD                     [Since sides of a square are equal]
and,   AO = AO         [Common]

Therefore by  SSS criterion of congruence
\Delta \,AOB\,\, \cong \Delta \,AOD
\Rightarrow      \angle AOB = \angle AOD ...........(3)
[Corresponding parts of congruent  \Delta s are equal]

But  \angle AOB + \angle AOD = 180^o  
\Rightarrow \angle AOB + \angle AOB = 180^o
\Rightarrow 2 \angle AOB = 180^o
\Rightarrow \angle AOB = 90^o
Therefore   \angle AOB = \angle AOD = 90^o
\Rightarrow     AO  \bot BD
\Rightarrow     AC  \bot BD
Hence, diagonals intersect at right  angles.


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

Given : A quadrilateral  ABCD in which  the diagonals AC = BD , AO= OC , BO = OD and AC  \bot BD.

3To prove : Quadrilateral  ABCD is a square.
Proof : First we shall prove that ABCD is a parallelogram.
In \Delta s AOD and  COB, we have
AO = OC             [Given]
OD = OB             [Given]
\angle AOD = \angle COB         [Vertically opp. angles]
By SAS criterion  of congruence,
\Delta \,AOD\,\, \cong \Delta COB
\Rightarrow      \angle OAD = \angle OCB          [Corresponding parts of congruent triangles are equal]
Now, line AC intersects AD and BC at A and C respectively such that
\angle OAD = \angle OCB     [From (1)]

i.e.,  alternate interior  angles are equal
Therefore      AD || BC
Similarly,     AB || CD
Hence, ABCD is a parallelogram.
Now, we shall prove that it is a square.
In  \Delta s\,AOB\,\,and\,\,\Delta AOD ,  we have
AO = AO             [Common]
 \angle AOB = \angle AOD         [Each = 90º, given]
and         OB = OD             [Since diagonals of  a ||  gm bisect each  other]
Therefore SAS criterion of congruence

\Delta \,AOB\,\, \cong \Delta AOD
\Rightarrow     AB = AD                           [Corresponding  parts of congruent triangles are equal]
But   AB = CD and AD = BC                              [Opp. sides of a || gm are equal]
Therefore   AB = BC = CD = AD            ... (2)
Now in \Delta s\,ABD\,\,and\,\,BAC , we have
AB = BA
AD = BC             [Opp. sides of a ||gm are equal]
and         BD = AC                 [Given]
Therefore by SSS criterion  of congruence
\Delta \,ABD\,\, \cong \,\Delta \,BAC
\Rightarrow     \angle DAB = \angle CBA   [Corresponding  parts of congruent  \Delta s   are equal]


Q.6      Diagonal AC of  parallelogram ABCD bisects  \angle A (see figure). Show that
          
(i)  it bisects \angle C also           (ii) ABCD is a rhombus.

4Sol.

(i) Given : A parallelogram ABCD in which  diagonal AC bisects \angle A
To prove : That AC bisects \angle C.

Proof : Since ABCD is a || gm.

5Therefore  AB || DC.
Now AB||DC and AC intersects them
Therefore          \angle 1 = \angle 3\,\,....\left( 1 \right)    [Alternate interior angles]
Again AD|| BC and AC intersects them
Therefore          \angle 2 = \angle 4\,....\left( 2 \right)    [Alternate interior  angles]
But  it is given  that AC is the bisector  of \angle A
Therefore          \angle 1 = \angle 2\,            .... (3)
From  (1) , (2) and (3) , we have
\angle 3 = \angle 4
Hence , AC bisects \angle C

(ii) To prove : That ABCD is a rhombus.
From  part (i) : (1) (2) and (3) give \angle 1 = \angle 2\, = \angle 3 = \angle 4
Now  in \Delta \,ABC,
\angle 1 = \angle \,4
 \Rightarrow     BC = AB      [Sides opp. to equal  angles in a \Delta are equal]
Similarly, in \Delta ADC, we have
AD = DC
Also ,  ABCD is a ||gm
Therefore         AB = CD, AD = BC                 [Opp. sides of a || gm are equal]
Combining  these, we get
AB = BC = CD = DA
Hence , ABCD is a rhombus.


Q.7     ABCD is a rhombus. Show  that diagonal  AC bisects \angle A as well as \angle C and diagonal BD bisects  \angle B as well as \angle D.
Sol.

Given : A rhombus ABCD.
To prove :
(i)  Diagonal AC bisects \angle A as well \angle C
(ii) Diagonal  BD bisects \angle B as well  as \angle D
Proof : In  \Delta \,ADC        AD = DC        [Sides of a rhombus are equal]
 \Rightarrow \angle DAC = \angle DCA               ... (1)  [Angles opp. to equal sides of a triangle are equal]

6Now AB || DC and AC intersects them
\angle BCA = \angle DAC            ... (2)            [Alternate  angles]
From (1) and (2), we have
\angle DCA = \angle BCA
 \Rightarrow     AC bisects \angle C
In \Delta \,ABC,\,\,  AB = BC                   [Sides of a rhombus  are equal]
 \Rightarrow     \angle BCA = \angle BAC  ...... (3) [Angles opp. to equal sides of a triangle are equal]
From (2) and (3) , we have
\angle BAC = \angle DAC
 \Rightarrow     AC bisects \angle A
Hence , diagonal  AC bisects \angle A as well as \angle C.
Similarly, diagonal  BD bisects \angle B as well  as \angle D


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

Sol.

Given: ABCD is a rectangle in which  diagonal AC bisects \angle A as well as \angle C.
To prove :
(i) ABCD is a square.

(ii) Diagonal BD bisects \angle B as well as \angle D.
Proof : (i) Since AC bisects \angle A as well  as \angle C in the rectangle ABCD.

7 \angle A = \angle C           [All four angles of a rectangle are 90º]
Therefore,    \angle 1 = \angle 2 = \angle 3 = \angle 4\left[ {Each = {{90^o} \over 2} = 45^o} \right]

Therefore,        In \Delta ADC,\,\,\,\angle 2 = \angle 4
 \Rightarrow          AD = CD                                                  [Sides opposite  to equal  angles]
Thus, the rectangle ABCD is a square.

(ii) In a square , diagonals bisect the angles.
So, BD bisects \angle B  as well as \angle D


Q.9     In parallelogram ABCD, two  points P and Q are taken  on diagonal  BD such that DP = BQ (see figure). Show that :
          (i) \Delta \,APD \cong \,\Delta \,CQB
          (ii) AP = CQ                   
          (iii) \Delta \,AQB \cong \,\Delta \,CPD
          (iv)  AQ = CP
          (v)  APCQ is a parallelogram
9
Sol.      ABCD is a parallelogram. P and Q are  points on the  diagonal BD such  that DP = BQ.         

10To prove :(i) \Delta \,APD \cong \,\Delta \,CQB
(ii) AP = CQ
(iii) \Delta \,AQB \cong \,\Delta \,CPD
(iv)  AQ = CP
(v)  APCQ is a parallelogram.

Construction : Join AC to meet BD in O.
Proof : We know that the diagonals of a parallelogram bisect each other. Therefore AC and BD bisect each other at O.
Therefore,            OB = OD
But                         BQ = DP                 [Given ]
 \Rightarrow     OB – BQ = OD – DP
 \Rightarrow   OQ  = OP

Thus, in  quadrilateral APCQ diagonals  AC and PQ are such that OQ = OP and OA = OC. i.e.,  the diagonals AC and PQ bisects each other.
(v) Hence, APCQ is a parallelogram, which  prove the \left(\upsilon \right) part.

(i)  In \Delta s APD and CQB we have
AD = CB         [Opp. sides of a ||gm ABCD]
AP = CQ         [Opp. sides of a||gm APCQ]
DP = BQ         [Given]
Therefore,  by  SSS criterion  of congruence

\Delta \,APD \cong \,\Delta \,CQB

(ii) AP = CQ        [Opp. sides of a ||gm APCQ]

(iii) In \Delta s AQB and CPD,  we have
AB = CD      [Opp. sides of a ||gm ABCD]
AQ = CP      [Opp. sides of a ||gm APCQ]
BQ = DP      [Given]
Therefore, by SSS criterion of congruence
\Delta \,AQB \cong \,\Delta \,CPD

(iv)    AQ = CP         [Opp. sides of a ||gm APCQ]


Q.10    ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (See figure). Show that
          
(i) \Delta \,APB \cong \,\Delta \,CQD      (ii) AP = CQ

11Sol.

(i)  Since ABCD is a parallelogram. Therefore, DC || AB.
Now DC|| AB and transversal  BD intersects them at B and D.
Therefore     \angle ABD = \angle BDC        [Alternate interior angles]
Now  in \Delta s APB and CQD we have
\angle ABP\, = \angle QDC        \left[ {\angle ABD\, = \angle BDC} \right]
\angle APB\, = \angle CQD            [Each = 90º]
and         AB = CD             [Opp. sides of a || gm]
Therefore,  by AAS criterion  of congruence
\Delta APB \cong \Delta CQD

(ii)  Since \Delta APB \cong \Delta CQD
Therefore,     AP = CQ         [Since corresponding  parts of congruent triangles are equal]


 Q.11   In \Delta s ABC and \Delta 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 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) \Delta ABC \cong \Delta DEF

12Sol.

Given:  Two \Delta s ABC and DEF such that AB = DE and AB || DE. Also BC = EF and BC || EF.
To prove : (i) quadrilateral ABED is parallelogram.
(ii) quadrilateral BEFC is a parallelogram.
(iii) AD|| CF and AD = CF.
(iv) quadrilateral ACFD is a parallelogram.
(v) AC|| DF and AC = DF
(vi) \Delta ABC \cong \Delta DEF

Proof : (i) Consider the quadrilateral ABED
We  have , AB = DE and AB || DE
 \Rightarrow One pair of opposite  sides are equal and parallel.
 \Rightarrow ABED is a parallelogram.

(ii) Now , consider  quadrilateral  BEFC , we have
BC = EF and BC || EF
 \Rightarrow One pair of opposite  sides are equal  and parallel.
 \Rightarrow BEFC is a parallelogram.

(iii)  Now ,  AD = BE and AD || BE        [Since ABED is a ||gm]    ... (1)
and          CF = BE and CF|| BE                 [Since BEFC is a ||gm]     ... (2)
From  (1) and (2) , we have
AD = CF and AD|| CF

(iv) Since AD = CF and AD || CF
 \Rightarrow One pair  of opposite  sides  are equal  and parallel.
 \Rightarrow   ACFD is a parallelogram.

(v)  Since ACFD is parallelogram.
Therefore    AC = DF                 [Opp. sides of a|| gm ACFD]

(vi)  In \Delta s ABC and DEF, we have
AB = DE                [Opp. sides of a|| gm ABED]
BC = EF                 [Opp. sides of a|| gm BEFC]
and     CA = FD                 [Opp. sides of || gm ACFD]
Therefore by SSS criterion of congruence.
\Delta ABC \cong \Delta DEF


Q.12   ABCD is a trapezium  in which AB || CD and AD = BC (see figure) Show that
          
(i)  \angle A = \angle B
           (ii) \angle C = \angle D
           (iii) \Delta ABC \cong \Delta BAD
           (iv)  diagonal AC = diagonal  BD

13Sol.

Given : ABCD is a trapezium  in which AB || CD and AD = BC
To prove : (i) \angle A = \angle B
(ii) \angle C = \angle D
(iii) \Delta ABC \cong \Delta BAD
(iv) Diagonal AC = diagonal BD.

Construction : Produce AB and draw a line CE|| AD.
Proof : (i) Since AD || CE and transversal AE cuts them  at  A and E respectively.
Therefore,     \angle A + \angle E = 180^\circ   ... (1) (Consecutive interior angles are supplementary)  
Since  AB || CD and AD || CE. Therefore, AECD is parallelogram.
 \Rightarrow     AD = CE
 \Rightarrow     BC = CE         [Since  AD = BC (given)]
Thus,  in \Delta BCE , we have
BC = CE (by Angle sum property)
 \Rightarrow     \angle CBE = \angle CEB
 \Rightarrow     180 - \angle B = \angle E
 \Rightarrow     180 - \angle E = \angle B            ... (2)
From  (1) and (2) , we get
\angle A = \angle \,B

(ii)  Since     \angle A = \angle \,B\,\,\, \Rightarrow \,\,\angle BAD = \angle ABD
 \Rightarrow     180^\circ - \angle BAD = 180^\circ - \angle ABD
 \Rightarrow     \angle ADB = \angle BCD
 \Rightarrow     \angle D = \angle C\,\,\,i.e.\,\angle C = \angle D

(iii)  In \Delta s ABC and BAD, we have
BC = AD             [Given]
AB = BA             [Common]
\angle A = \angle B    [Proved]
Therefore  by SAS criterion  of congruence
\Delta ABC \cong \Delta BAD
(iv)  Since     \Delta ABC \cong \Delta BAD
Therefore,         AC = BD                                    [Corresponding  parts of congruent triangles are equal]



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