NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

 

NCERT solutions for class 9 maths chapter 8 Quadrilaterals: A figure designed through four straight lines is called Quadrilaterals. It has four sides and four angles and the sum of all the four angles is always 360 degrees. There are many types of quadrilaterals which you will study in this particular chapter. Solutions of NCERT class 9 maths chapter 8 Quadrilaterals is covering the problems related to all the concepts in a very comprehensive manner. The topic is easy to relate to real-life because many shapes around us are in the form of quadrilaterals such as top of a table, paper, wall, roof, etc. To deal with problems related to this chapter, the awareness about the properties of different types of quadrilaterals is necessary. CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals covers all the questions to give you guidance while solving the questions. In this particular chapter, you will get to know about angles sum property, rhombus, parallelogram, square, rectangle, trapezium, kite, etc. The chapter is starting with recollecting the angle sum property of a quadrilateral. In total there are 2 exercises that consist of 37 questions. NCERT solutions for class 9 maths chapter 8 Quadrilaterals has the solutions to all the 37 questions. If I talk apart from this chapter, then NCERT solutions for other classes and subjects can also be downloaded using the link given.

NCERT solutions for class 9 maths chapter 8 Quadrilaterals Excercise: 8.1

Q1 The angles of quadrilateral are in the ratio  \small 3:5:9:13. Find all the angles of the quadrilateral.

Answer:

Given :  The angles of a quadrilateral are in the ratio  \small 3:5:9:13
Let the angles of quadrilateral be   3x,5x,9x ,13x.

Sum of all angles is 360.

Thus,   3x+5x+9x +13x=360 \degree

          \Rightarrow 30x=360 \degree

        \Rightarrow x=12 \degree

All four  angles are : 3x=3\times 12=36\degree

                                5x=5\times 12=60 \degree

                               9x=9\times 12=108 \degree

                            13x=13\times 12=156 \degree

Q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:

Given: ABCD is a parallelogram with AC=BD.

To prove: ABCD is a rectangle.

Proof : In \triangleABC and \triangleBAD,   

           

                BC= AD            (Opposite sides of parallelogram)

                AC=BD              (Given )

               AB=AB               (common)

             \triangleABC \cong \triangleBAD   (By SSS)

        \angle ABC=\angle BAD             (CPCT)

and     \angle ABC+\angle BAD=180 \degree      (co - interior angles)

             2\angle BAD=180 \degree

               \angle BAD= 90 \degree

Hence, it is a rectangle.

Q3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:

Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.

To prove: ABCD is a rhombus.

Proof : In \triangleAOB and \triangleAOD,   

           

                \angle AOB=\angle AOD        (Each 90 \degree)

                BO=OD              (Given )

               AO=AO               (common)

             \triangleAOB \cong \triangleAOD   (By SAS)

                   AB=AD           (CPCT)

Similarly, AB=BC and BC=CD

Hence, it is a rhombus.

Q4 Show that the diagonals of a square are equal and bisect each other at right angles.

Answer:

Given : ABCD is a square i.e. AB=BC=CD=DA.

To prove : the diagonals of a square are equal and bisect each other at right angles i.e. AC=BD,AO=CO,BO=DO and \angle COD=90 \degree

Proof : In \triangleBAD and \triangleABC,   

           

                \angle BAD = \angle ABC        (Each 90 \degree)

                AD=BC              (Given )

               AB=AB               (common)

             \triangleBAD \cong \triangleABC   (By SAS)

                   BD=AC           (CPCT)

In \triangleAOB and \triangleCOD, 

           \angleOAB=\angleOCD        (Alternate angles)

                AB=CD               (Given )

             \angleOBA=\angleODC       (Alternate angles)

             \triangleAOB \cong \triangleCOD   (By AAS)

                   AO=OC ,BO=OD           (CPCT)

In \triangleAOB and \triangleAOD, 

               OB=OD        (proved above)

                AB=AD               (Given )

               OA=OA          (COMMON)

             \triangleAOB \cong \triangleAOD   (By SSS)

                   \angleAOB=\angleAOD           (CPCT)

            \angleAOB+\angleAOD =180 \degree

                   2. \angleAOB = 180 \degree

                         \angleAOB =90 \degree 

Hence,  the diagonals of a square are equal and bisect each other at right angles.

Q5 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer:

Given : ABCD is a quadrilateral with  AC=BD,AO=CO,BO=DO,\angleCOD =90 \degree

 To prove: ABCD is a square.

Proof: Since the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a rhombus.

           Thus, AB=BC=CD=DA

           

               In \triangleBAD and \triangleABC,   

                AD=BC              (proved above )

               AB=AB               (common)

               BD=AC

             \triangleBAD \cong \triangleABC   (By SSS)

           \angle BAD = \angle ABC    (CPCT)

            \angleBAD+\angleABC =180 \degree           (Co-interior angles)

                   2. \angleABC = 180 \degree

                         \angleABC =90 \degree 

Hence,  the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Q6 (i) Diagonal AC of a parallelogram ABCD bisects \small \angle A (see Fig. \small 8.19). Show that 

it bisects \small \angle C also.

                

Answer:

Given: \angleDAC=\angleBAC ................1 

              \angleDAC=\angleBCA.................2  (Alternate angles)

             \angleBAC=\angleACD .................3  (Alternate angles)

From equation  1,2 and 3, we get

            \angleACD=\angleBCA...................4

Hence, diagonal AC bisect angle C also.

Q6 (ii) Diagonal AC of a parallelogram ABCD bisects \small \angle A (see Fig. \small 8.19). Show that

ABCD is a rhombus.

Answer:

Given: \angleDAC=\angleBAC ................1 

              \angleDAC=\angleBCA.................2  (Alternate angles)

             \angleBAC=\angleACD .................3  (Alternate angles)

From equation  1,2 and 3, we get

            \angleACD=\angleBCA...................4

From 2 and 4, we get

            \angleACD=\angleDAC

    In \triangle ADC,

                   \angleACD=\angleDAC       (proved above )

                        AD=DC         (In a triangle,sides opposite to equal angle are equal)

A parallelogram whose adjacent sides are equal , is a rhombus.

Thus, ABCD is a rhombus.

Q7 ABCD is a rhombus. Show that diagonal AC bisects \small \angle A as well as \small \angle C and diagonal BD bisects \small \angle B as well as \small \angle D.

Answer:

In \triangle ADC,

   AD = CD          (ABCD is a rhombus)

 \angle3=\angle1.................1(angles opposite to equal sides are equal )

 \angle3=\angle2.................2 (alternate angles)

From 1 and 2, we have 

                       \angle1=\angle2.................3

   and        \angle1=\angle4.................4 (alternate angles)

From 1 and 4, we get 

                        \angle3=\angle4.................5

Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.

In \triangle ADB,

   AD = AB          (ABCD is a rhombus)

 \angle5=\angle7.................6(angles opposite to equal sides are equal )

 \angle7=\angle6.................7 (alternate angles)

From 6 and 7, we have 

                       \angle5=\angle6.................8

   and        \angle5=\angle8.................9(alternate angles)

From 6 and 9, we get 

                        \angle7=\angle8.................10

Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.

Q8 (i) ABCD is a rectangle in which diagonal AC bisects \small \angle A as well as \small \angle C. Show that:

ABCD is a square 

Answer:

Given: ABCD is a rectangle with AB=CD and BC=AD \angle1=\angle2 and \angle3=\angle4.

 To prove: ABCD is a square.

Proof :     \angle1=\angle4 .............1(alternate angles)

                \angle3=\angle4 ................2(given )

       From 1 and 2, \angle1=\angle3.....................................3

         In \triangleADC,

               \angle1=\angle3          (from  3 )   

            DC=AD             (In a triangle, sides opposite to equal angle are equal )

A rectangle whose adjacent sides are equal is a square.

Hence, ABCD is a square.

Q8 (ii) ABCD is a rectangle in which diagonal AC bisects \small \angle A as well as \small \angle C. Show that:

diagonal BD bisects \small \angle B as well as \small \angle D.

Answer:

In \triangle ADB,

   AD = AB          (ABCD is a square)

 \angle5=\angle7.................1(angles opposite to equal sides are equal )

 \angle5=\angle8.................2 (alternate angles)

From 1 and 2, we have 

                       \angle7=\angle8.................3

   and        \angle7 =\angle6.................4(alternate angles)

From 1 and 4, we get 

                        \angle5=\angle6.................5

Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.

Q9 (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ (see Fig. \small 8.20). Show that: \small \Delta APD\cong \Delta CQB

                

Answer:

Given:   In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ.

To prove :\small \Delta APD\cong \Delta CQB

Proof :

In \small \Delta APD\, \, and\, \, \Delta CQB,

   DP=BQ       (Given )

 \angleADP=\angleCBQ     (alternate angles)

    AD=BC       (opposite sides of a parallelogram)

\small \Delta APD\cong \Delta CQB                (By SAS)

Q9 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that  \small DP=BQ  (see Fig. \small 8.20). Show that: \small AP=CQ

             

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ.

To prove :\small AP=CQ

Proof :

In \small \Delta APD\, \, and\, \, \Delta CQB,

   DP=BQ       (Given )

 \angleADP=\angleCBQ     (alternate angles)

    AD=BC       (opposite sides of a parallelogram)

\small \Delta APD\cong \Delta CQB                (By SAS)

    \small AP=CQ      (CPCT)

Q9 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that  \small DP=BQ (see Fig. \small 8.20). Show that: \small \Delta AQB\cong \Delta CPD

              

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ.

To prove :\small \Delta AQB\cong \Delta CPD

Proof :

In \small \Delta AQB\, \, and\, \, \Delta CPD,

   DP=BQ       (Given )

 \angleABQ=\angleCDP     (alternate angles)

    AB=CD       (opposite sides of a parallelogram)

\small \Delta AQB\cong \Delta CPD                (By SAS)

Q9 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that  \small DP=BQ (see Fig. \small 8.20). Show that: \small AQ=CP

                

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ.

To prove :\small AQ=CP 

Proof :

In \small \Delta AQB\, \, and\, \, \Delta CPD,

   DP=BQ       (Given )

 \angleABQ=\angleCDP     (alternate angles)

    AB=CD       (opposite sides of a parallelogram)

\small \Delta AQB\cong \Delta CPD                (By SAS)          

   \small AQ=CP                       (CPCT)

Q9 (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that  \small DP=BQ (see Fig. \small 8.20). Show that: APCQ is a parallelogram

                

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ.

To prove: APCQ is a parallelogram

Proof :

In \small \Delta APD\, \, and\, \, \Delta CQB,

   DP=BQ       (Given )

 \angleADP=\angleCBQ     (alternate angles)

    AD=BC       (opposite sides of a parallelogram)

\small \Delta APD\cong \Delta CQB                (By SAS)

    \small AP=CQ       (CPCT)...............................................................1

Also,

In \small \Delta AQB\, \, and\, \, \Delta CPD,

   DP=BQ       (Given )

 \angleABQ=\angleCDP     (alternate angles)

    AB=CD       (opposite sides of a parallelogram)

\small \Delta AQB\cong \Delta CPD                (By SAS)          

   \small AQ=CP                       (CPCT)........................................2

From equation 1 and 2, we get 

\small AP=CQ

\small AQ=CP

Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.

Q10 (i) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. \small 8.21). Show that \small \Delta APB\cong \Delta CQD

                    

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A  and C on diagonal BD.

To prove : \small \Delta APB\cong \Delta CQD

Proof: In\small \Delta APB\, \, and\, \, \Delta CQD,

           \angleAPB=\angleCQD             (Each 90 \degree)

           \angleABP=\angleCDQ            (Alternate angles)

               AB=CD                (Opposite sides of a parallelogram )

Thus, \small \Delta APB\cong \Delta CQD               (By SAS)

Q10 (ii) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. \small 8.21). Show that \small AP=CQ

                    

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A  and C on diagonal BD.

To prove : \small AP=CQ

Proof: In\small \Delta APB\, \, and\, \, \Delta CQD,

           \angleAPB=\angleCQD             (Each 90 \degree)

           \angleABP=\angleCDQ            (Alternate angles)

               AB=CD                (Opposite sides of a parallelogram )

Thus, \small \Delta APB\cong \Delta CQD               (By SAS)

           \small AP=CQ                 (CPCT)

Q11 (i) In \small \Delta ABC and \small \Delta DEF,  \small AB=DE,AB\parallel DE,BC=EF and  \small BC\parallel EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22). Show that quadrilateral ABED is a parallelogram

            

Answer:

Given :  In \small \Delta ABC and \small \Delta DEF,  \small AB=DE,AB\parallel DE,BC=EF and    \small BC\parallel EF.

To prove :  quadrilateral ABED is a parallelogram

Proof : In ABED,    

                  AB=DE          (Given)

                 AB||DE            (Given )

Hence, quadrilateral ABED is a parallelogram.

Q11 (ii) In \small \Delta ABC and \small \Delta DEF,  \small AB=DE,AB\parallel DE,BC=EF and   \small BC\parallel EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22). Show that quadrilateral BEFC is a parallelogram.

                 

Answer:

Given: In \small \Delta ABC and \small \Delta DEF,  \small AB=DE,AB\parallel DE,BC=EF and                              \small BC\parallel EF.

To prove: quadrilateral BEFC is a parallelogram

Proof: In BEFC,

                  BC=EF          (Given)

                 BC||EF            (Given )

Hence, quadrilateral BEFC is a parallelogram.

Q11 (iii) In \small \Delta ABC and \small \Delta DEF, \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22). Show that \small AD\parallel CF and \small AD=CF

            

Answer:

To prove: \small AD\parallel CF and \small AD=CF

Proof : 

In ABED,

   AD=BE.................1(ABED is a parallelogram)    

 AD||BE .................2(ABED is a parallelogram)      

In BEFC,

   BE=CF.................3(BEFC is a parallelogram)    

 BE||CF .................4(BEFC is a parallelogram)      

From 2 and 4 , we get

    AD||CF

From 1 and 3, we get

   AD=CF

Q11 (iv) In \small \Delta ABC and \small \Delta DEF, \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF..  Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22). Show that quadrilateral ACFD is a parallelogram.

               

Answer:

To prove :  quadrilateral ACFD is a parallelogram

Proof : 

In ABED,

   AD=BE.................1(ABED is a parallelogram)    

 AD||BE .................2(ABED is a parallelogram)      

In BEFC,

   BE=CF.................3(BEFC is a parallelogram)    

 BE||CF .................4(BEFC is a parallelogram)      

From 2 and 4 , we get

    AD||CF...........................5

From 1 and 3, we get

   AD=CF...........................6

From 5 and 6, we get

 AD||CF            and              AD=CF

Thus, quadrilateral ACFD is a parallelogram

Q11 (v) In \small \Delta ABC and \small \Delta DEF,  \small AB=DE,AB\parallel DE,BC = EF and  \small BC\parallel EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22). Show that \small AC=DF

            

Answer:

In ACFD,

    AC=DF     (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)

Q11 (vi) In \small \Delta ABC and \small \Delta DEF, \small AB=DE,AB\parallel DE,BC=EF and\small BC\parallel EF.  Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22). Show that \small \Delta ABC\cong \Delta DEF

Answer:

In \small \Delta ABC and \small \Delta DEF,

            AB=DE                   (Given )

           BC=EF                    (Given )

          AC=DF                ( proved in (v) part)

     \small \Delta ABC\cong \Delta DEF        (By SSS rule)

Q12 (i) ABCD is a trapezium in which  \small AB\parallel CD and  \small AD=BC  (see Fig. \small 8.23). Show that \small \angle A=\angle B

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

            

Answer:

Given: ABCD is a trapezium in which  \small AB\parallel CD and  \small AD=BC

To prove :\small \angle A=\angle B

Proof:      Let \angle A be \angle1, \angleABC be \angle2, \angle EBC be \angle3, \angle BEC be \angle4.

In AECD,

AE||DC             (Given)

AD||CE             (By cnstruction)

Hence, AECD is a parallelogram.

 AD=CE...............1(opposite sides of a parallelogram)

AD=BC.................2(Given)

From 1 and 2, we get

 CE=BC

In \triangleBCE,

      \angle 3=\angle 4.................3  (opposite angles of equal sides)

       \angle 2+\angle 3=180 \degree...................4(linear pairs)

     \angle 1+\angle 4=180 \degree .....................5(Co-interior angles)

From 4 and 5, we get 

 \angle 2+\angle 3=\angle 1+\angle 4

\therefore \angle 2=\angle 1 \rightarrow \angle B=\angle A          (Since,\angle 3=\angle 4)

Q12 (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that \small \angle C=\angle D

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
 

                    

Answer:

Given: ABCD is a trapezium in which  \small AB\parallel CD and  \small AD=BC

To prove :\small \angle C=\angle D

Proof:   Let \angle A be \angle1, \angleABC be \angle2, \angle EBC be \angle3, \angle BEC be \angle4.

            \angle 1+\angle D= 180 \degree       (Co-interior angles)

          \angle 2+\angle C= 180 \degree            (Co-interior angles)

\therefore \angle 1+\angle D=\angle 2+\angle C

       Thus, \small \angle C=\angle D                (Since ,\small \angle 1=\angle 2 )

Q12 (iii) ABCD is a trapezium in which  \small AB\parallel CD and  \small AD=BC (see Fig. \small 8.23). Show that \small \Delta ABC\cong \Delta BAD

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

        

Answer:

Given: ABCD is a trapezium in which  \small AB\parallel CD and  \small AD=BC

To prove :\small \Delta ABC\cong \Delta BAD

Proof: In  \small \Delta ABC\, \, and\, \, \, \Delta BAD,

             BC=AD            (Given )

             AB=AB            (Common )

       \angle ABC=\angle BAD              (proved in (i)  )

Thus, \small \Delta ABC\cong \Delta BAD              (By SAS rule)

Q12 (iv) ABCD is a trapezium in which  \small AB\parallel CD and \small AD=BC (see Fig. \small 8.23). Show that diagonal AC \small = diagonal BD

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

         

Answer:

Given: ABCD is a trapezium in which  \small AB\parallel CD and  \small AD=BC

To prove: diagonal AC \small = diagonal BD

Proof: In  \small \Delta ABC\, \, and\, \, \, \Delta BAD,

             BC=AD            (Given )

             AB=AB            (Common )

       \angle ABC=\angle BAD              (proved in (i)  )

Thus, \small \Delta ABC\cong \Delta BAD              (By SAS rule)

diagonal AC \small = diagonal BD            (CPCT)

NCERT solutions for class 9 maths chapter 8 Quadrilaterals Excercise: 8.2

Q1 (i) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig \small 8.29). AC is a diagonal. Show that : \small SR\parallel AC  and   \small SR=\frac{1}{2}AC

                 

Answer:

Given :   ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig \small 8.29). AC is a diagonal.

To prove :  \small SR\parallel AC  and   \small SR=\frac{1}{2}AC

Proof: In \triangleACD,

      S is the midpoint of DA.                (Given)

      R  is the midpoint of DC.               (Given)

  By midpoint theorem,

                        \small SR\parallel AC  and   \small SR=\frac{1}{2}AC

Q1 (ii) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig \small 8.29). AC is a diagonal. Show that : \small PQ=SR

            

Answer:

Given :   ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig \small 8.29). AC is a diagonal.

To prove : \small PQ=SR

Proof : In \triangleACD,

      S is mid point of DA.                (Given)

      R  is mid point of DC.               (Given)

  By mid point theorem,

                        \small SR\parallel AC  and   \small SR=\frac{1}{2}AC...................................1

   In \triangleABC,

      P is mid point of AB.                (Given)

      Q  is mid point of BC.               (Given)

  By mid point theorem,

                        \small PQ\parallel AC  and   \small PQ=\frac{1}{2}AC.................................2

From 1 and 2,we get

     \small PQ\parallel SR          and   \small PQ=SR=\frac{1}{2}AC

Thus, \small PQ=SR

Q1 (iii)ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig \small 8.29). AC is a diagonal. Show that : PQRS is a parallelogram.
 

Answer:

Given :   ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig \small 8.29). AC is a diagonal.

To prove : PQRS is a parallelogram.

Proof : In PQRS,

    Since, 

 \small PQ\parallel SR      and       \small PQ=SR.

So,PQRS is a parallelogram.

Q2 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Answer:

Given: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC, BD are  diagonals.

To prove: the quadrilateral PQRS is a rectangle.

Proof: In \triangleACD,     

S is midpoint of DA.                (Given)

      R  is midpoint of DC.               (Given)

  By midpoint theorem,

                        \small SR\parallel AC  and   \small SR=\frac{1}{2}AC...................................1

   In \triangleABC,

      P is midpoint of AB.                (Given)

      Q  is mid point of BC.               (Given)

  By mid point theorem,

                        \small PQ\parallel AC  and   \small PQ=\frac{1}{2}AC.................................2

From 1 and 2,we get

     \small PQ\parallel SR          and   \small PQ=SR=\frac{1}{2}AC

Thus, \small PQ=SR     and \small PQ\parallel SR

So,the quadrilateral PQRS is a parallelogram.

Similarly, in \triangleBCD,

      Q is mid point of BC.                (Given)

      R  is mid point of DC.               (Given)

  By mid point theorem,

                        \small QR\parallel BD 

                    So,  QN || LM ...........5

                            LQ || MN ..........6  (Since, PQ || AC)

From 5 and 6, we get

      LMPQ is a parallelogram.

   Hence, \small \angleLMN=\small \angleLQN   (opposite angles of the parallelogram)

          But, \small \angleLMN= 90     (Diagonals of a rhombus are perpendicular)

             so,   \small \angleLQN=90

Thus, a parallelogram whose one angle is right angle,ia a rectangle.Hence,PQRS is a rectangle.

Q3 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Answer:

Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

To prove: the quadrilateral PQRS is a rhombus.

Proof : 

In \triangleACD,     

S is the midpoint of DA.                (Given)

      R  is the midpoint of DC.               (Given)

  By midpoint theorem,

                        \small SR\parallel AC  and   \small SR=\frac{1}{2}AC...................................1

   In \triangleABC,

      P is the midpoint of AB.                (Given)

      Q  is the midpoint of BC.               (Given)

  By midpoint theorem,

                        \small PQ\parallel AC  and   \small PQ=\frac{1}{2}AC.................................2

From 1 and 2, we get

     \small PQ\parallel SR          and   \small PQ=SR=\frac{1}{2}AC

Thus, \small PQ=SR     and \small PQ\parallel SR

So, the quadrilateral PQRS is a parallelogram.

Similarly, in \triangleBCD,

      Q is the midpoint of BC.                (Given)

      R  is the midpoint of DC.               (Given)

  By midpoint theorem,

                        \small QR\parallel BD                  and      \small QR=\frac{1}{2}BD...................5

                         AC = BD.......................6(diagonals )

                            

From 2,  5 and 6, we get

                     PQ=QR 

Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.

Q4 ABCD is a trapezium in which  \small AB\parallel DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. \small 8.30). Show that F is the mid-point of BC.

            

Answer:

Given: ABCD is a trapezium in which  \small AB\parallel DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. \small 8.30).

To prove: F is the mid-point of BC.

   In \triangleABD,

      E is the midpoint of AD.                (Given)

               EG || AB                          (Given)

  By converse of  midpoint theorem,

                       G is the midpoint of BD.

 

In \triangleBCD,

      G is mid point of BD.                (Proved above)

               FG || DC                          (Given)

  By converse of  midpoint theorem,

                       F is the midpoint of BC.

Q5 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. \small 8.31). Show that the line segments AF and EC trisect the diagonal BD.

            

Answer:

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

To prove: the line segments AF and EC trisect the diagonal BD.

Proof : In quadrilatweral ABCD,     

             AB=CD                             (Given)

   \frac{1}{2}AB=\frac{1}{2}CD

\Rightarrow AE=CF         (E and F are midpoints of AB and CD)

In quadrilateral AECF,     

             AE=CF                             (Given)

            AE || CF               (Opposite sides of a parallelogram)

Hence,  AECF is a parallelogram.

In \triangle DCQ,

      F is the midpoint of DC.      (given )

              FP || CQ           (AECF  is a parallelogram)

By converse of midpoint theorem,

          P is the mid point of DQ.      

         DP= PQ....................1

Similarly,

        In \triangle ABP,

      E is the midpoint of AB.      (given )

              EQ || AP          (AECF  is a parallelogram)

By converse of midpoint theorem,

          Q is the midpoint of PB.      

         OQ= QB....................2

From 1 and 2, we have 

         DP = PQ = QB.

Hence, the line segments AF and EC trisect the diagonal BD.

Q6 Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer:

Given: ABCD is a  quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC, BD are diagonals.

To prove: the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Proof: In \triangleACD,     

S is the midpoint of DA.                (Given)

      R  is midpoint of DC.               (Given)

  By midpoint theorem,

                        \small SR\parallel AC  and   \small SR=\frac{1}{2}AC...................................1

   In \triangleABC,

      P is the midpoint of AB.                (Given)

      Q  is the midpoint of BC.               (Given)

  By midpoint theorem,

                        \small PQ\parallel AC  and   \small PQ=\frac{1}{2}AC.................................2

From 1 and 2, we get

     \small PQ\parallel SR          and   \small PQ=SR=\frac{1}{2}AC

Thus, \small PQ=SR     and \small PQ\parallel SR

So, the quadrilateral PQRS is a parallelogram and diagonals of a parallelogram bisect each other.

Thus, SQ and PR bisect each other.

Q7 (i) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that D is the mid-point of AC.

Answer:

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse    AB and parallel to BC intersects AC at D.

To prove :D is mid point of AC.

Proof: In \triangleABC,     

       M is mid point of AB.                (Given)

                   DM || BC            (Given)

  By converse of mid point theorem,

                    D is the mid point of AC.

Q7 (ii) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that \small MD\perp AC

Answer:

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D.

To prove : \small MD\perp AC

Proof :  \angleADM = \angleACB      (Corresponding angles)

            \angleADM= 90 \degree.           (\angleACB = 90 \degree)

Hence,\small MD\perp AC.

Q7 (iii) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that \small CM=MA=\frac{1}{2}AB

Answer:

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove :  \small CM=MA=\frac{1}{2}AB

Proof : In \triangleABC,     

       M is the midpoint of AB.                (Given)

                   DM || BC            (Given)

  By converse of midpoint theorem,

                    D is the midpoint of AC  i.e.  AD = DC.

In \triangle AMD  and \triangle CMD,

        AD = DC    (proved above)

       \angleADM = \angle CDM    (Each  right angle)

            DM = DM      (Common)

           \triangle AMD \cong\triangle CMD      (By SAS)

             AM = CM         (CPCT)

       But ,\small AM=\frac{1}{2}AB

  Hence,\small CM=MA=\frac{1}{2}AB.

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 8 Quadrilaterals

  • Read and memorize the properties related to all kinds of quadrilaterals.

  • Have a glance through some solved to understand the pattern of the solution to that particular question.

  • Now check your learning on the practice exercises problems.

  • If you stuck in any question then you can assist yourself using NCERT solutions for class 9 maths chapter 8 Quadrilaterals.

  • Following the above-written steps, you can get 100% out of the chapter.

 

Recently Asked Questions

 

Related Articles

Exams
Articles
Questions