Class 9 NCERT Solutions Maths Chapter 8 - Quadrilaterals

Learn to provide Mathematical proofs easily with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 8 Quadrilaterals. Can a parallelogram be a rhombus? How? Learn by practising with our solutions by Maths experts. Understand how to express your understanding of quadrilaterals, such as a parallelogram and rhombus, and provide clear explanations in your answers.

You can score full marks in questions based on quadrilaterals by revising key concepts with our CBSE Class 9 Maths textbook solutions. If you find any of the concepts to be mind-boggling, discuss your doubts with Maths experts through our ‘UnDoubt’ platform.

Ex. 8.1

Ex. 8.2

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.

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

Solution 2

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 = 90
Since ABCD is a parallelogram and one of its interior angles is 90, therefore, ABCD is rectangle.

Solution 3

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.

Solution 4

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)

ABC

DCB (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)
2

AOB = 180

AOB = 90
Hence, the diagonals of a square bisect each other at right angle.

Solution 5

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 = 90^o
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 6

(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

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

Solution 7

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.

Solution 8

(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.

Solution 9

(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.

Solution 10

(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)

Solution 11

(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)

ABC

DEF (by SSS congruence rule)

Solution 12

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 relation

CEB =

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 +

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)

ABC

BAD

AC = BD (by CPCT)

Quadrilaterals Exercise Ex. 8.2

Solution 1

(i) In

ADC, 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.

SR || AC and SR =

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 =

AC               ... (2)
      Now using equations (1) and (2), we have
      PQ || SR and PQ = SR                ... (3)

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

ABC, P and Q are mid points of sides AB and BC respectively.

PQ || AC and PQ =

AC (using mid-point theorem) ... (1)

ADC
R and S are the mid points of CD and AD respectively

RS || AC and RS =

AC (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               (

PQ || AC)
QN || OM (

QR || BD)
So, OMQN is parallelogram

MQN =

NOM

PQR =

NOM

But,

NOM = 90^o (diagonals of a rhombus are perpendicular to each other)

PQR = 90^o

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

Solution 3

Let us join AC and BD
In

ABC
P and Q are the mid-points of AB and BC respectively

PQ || AC and PQ =

AC (mid point theorem) ... (1)

Similarly in

ADC
SR || AC and SR =

AC        (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.

PS || QR and PS = QR (opposite sides of parallelogram)... (3)

Now, in

BCD, Q and R are mid points of side BC and CD respectively.

QR || BD and QR =

BD (mid point theorem) ... (4)

But diagonals of a rectangle are equal

AC = 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

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

ABD
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

EF || CD (Two lines parallel to a same line are parallel to each other)

Now, in

BCD, 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

AB || CD

So, AE || FC
Again AB = CD (opposite sides of parallelogram ABCD)

AB =

CD
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.

AF || EC (Opposite sides of a parallelogram)
Now, in

DQC, 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

DP = PQ ... (1)
Similarly, in

APB, 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

PQ = 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

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

ABD, 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 =

BD ... (1)
Similarly in

BCD
QR || BD and QR =

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

(i) In

ABC
       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,

MDC +

DCB = 180 (Co-interior angles)

MDC + 90 = 180

MDC = 90

(iii) Join MC

AMD and

CMD
AD = CD (D is the midpoint of side AC)

ADM =

CDM            (Each 90)
   DM = DM                   (common)

AMD

CMD (by SAS congruence rule)

So, AM = CM (by CPCT)
But AM =

AB (M is mid point of AB)
So, CM = MA =

Class 9 NCERT Solutions Maths Chapter 8 - Quadrilaterals

Quadrilaterals Exercise Ex. 8.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Quadrilaterals Exercise Ex. 8.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

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