NCERT Solutions for Class 9 Mathematics Chapter 8 -Quadrilaterals

Mathematics is becoming more and more significant  due to its various  applications in our daily lives.. It has been applied in many fields of sciences and technology, which has accelerated  its growth in  various applications. As a result, professionals suggest  focusing on Mathematics  helps us to improvise our reasoning, analytical skills  and  problem solving skills.  It is turning out to be a crucial subject in the primary and secondary levels. Hence, it has become part of many competitive examinations.

The main topics covered in this chapter are the basics of the parallelogram and its properties, the key concepts of rhombus and all the properties associated, the applications of the rectangle as well as square and their properties and so on..

NCERT Solutions Class 9 Mathematics Chapter 8  is based on the latest CBSE syllabus.   They are effective in clearing concepts and provide authentic, reliable and to the point answers. Students are advised to follow NCERT books first and then refer to NCERT Solutions.  The solutions include everything you might be searching for during last minute preparation. It saves time to a great extent. . Hence, many teachers advise students to follow NCERT Solutions for Class 9 Mathematics Chapter 8 provided by Extramarks.

Extramarks’ website is one of the fastest growing online platforms for all primary school and secondary school studies. It is trusted by lakhs of teachers and students through its relentless services. You can find all the study material right from learning to developing skills to revising on the Extramarks website. It’s a one stop solution to all your problems.

Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 8

An enclosed figure with four sides is called quadrilaterals. You have learnt about the different types of quadrilaterals in your lower grades. Students are familiar with basic geometrical shapes such as  square, rhombus, rectangle and parallelogram. By now, you can easily differentiate between different quadrilaterals by just looking at them. But have you ever wondered how is it possible for you to do so?

The answer lies in the shape and size of the quadrilaterals.. It’s because you know that there are certain properties governing each quadrilateral which differentiate one from the other. For example, you already know from the previous classes that ‘A square has all the four sides equal’.

In Advanced Mathematics, you need to  learn about  its properties  and find proof of the quadrilaterals.  You will find everything in detail and get more practice  regarding this topic by following  the NCERT Solutions for Class 9 Mathematics Chapter 8 available on the Extramarks’ website.

This chapter will upgrade students’ logical and analytical skills. The way the solutions are written will help you to grasp the topics completely and will lay a strong foundation for Mathematics in  higher classes.Hence, they will learn to solve any problem in a smart way. Students are advised to make full use of resources to make the most of it.

 In a way Extramarks promotes learning by encouraging the students to be great learners and try to feed their insatiable curiosity through NCERT Solutions

Introduction 

In the previous chapter, we learnt about the triangle and its properties. In the chapter, we will cover quadrilaterals,  types of quadrilaterals and their properties.

When four non-collinear points make a closed figure, then it is called a Quadrilateral. It contains four sides, four angles and four vertices. 

Angle sum property of a Quadrilateral 

In this section, we will learn about the angles of the Quadrilateral. You can recall from the previous classes that the sum of all angles of a Quadrilateral is 360 degrees.

To take your learning beyond books and understand better , we recommend NCERT Solutions for Class 9 Mathematics Chapter 8 available on the Extramarks’ website.

Type of Quadrilaterals  

In this section, we will cover the types of the quadrilaterals covered in this chapter

  • Trapezium 

When one pair of the opposite side of the quadrilateral is parallel, it is called a trapezium.

  • Parallelogram 

When both pairs of the opposite sides of the quadrilateral are parallel, it is called a parallelogram.

 Types of parallelograms like

  • Rectangular. 

When one of the angles of a parallelogram is the right angle, it is called a  rectangle.

  • Rhombus

When all sides are equal in the parallelogram, it is called a  rhombus.

  • Square

When one of the angles is a right angle, and all sides are equal in the parallelogram, it is called a square.

  • Kite

 When two pairs of adjacent sides are equal in a quadrilateral, it is called a  kite. It is not a parallelogram.

For more details, please visit our Extramarks’ website and refer to NCERT Solutions for Class 9 Mathematics Chapter 8..

Properties of a Parallelogram 

 An essential prerequisite to learning the  properties of the parallelogram , it is necessary to brush up these  theorems: 

  • Theorem1: A diagonal is a parallelogram that divides it into two congruent triangles.
  • Theorem 2: In a parallelogram, opposite sides are equal.
  • Theorem 3: If each pair of the opposite side of the quadrilateral is equal, then it is a parallelogram. 
  • Theorem 4: In a parallelogram, opposite angles are equal.
  • Theorem 5: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
  • Theorem 6: The diagonals of a parallelogram bisect each other.
  • Theorem 7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. 

Another condition for a Quadrilateral to be a Parallelogram

 We have already  studied  the properties of a  parallelogram.  A quadrilateral must meet certain criteria to be a parallelogram. They can be best understood with the help of the following theorem:  

  • Theorem: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

The  Mid-point Theorem

 So far, you have learnt about the triangles and the quadrilaterals. Now we will turn to  the midpoint of the sides of the triangle. You can  understand these concepts better  with the help of the theorems given  below-

  • Theorem1: The line segment joining the mid- point of the two sides of a triangle is parallel to the third side.
  • Theorem 2: The line drawn through the midpoint point of one side of a triangle parallel to another side bisects the third side.

Summary

In this chapter, we looked at  the quadrilaterals, their definitions, the sum of the angles of the quadrilateral,   its properties  and  key theorems required to understand  geometry.  We can sum up the following concepts: 

  1. Quadrilateral:  When four non-collinear points make a closed figure, then it is called a quadrilateral. It contains four sides, four angles and four vertices.
  2.  The sum of all the angles of the quadrilateral is 360 degrees.
  3.  Types of quadrilaterals are:
  1.  Trapezium
  2. Parallelogram
    1. Rectangle
    2. Rhombus
    3. Square
  3. Kite
  • The properties of a  parallelogram 

 This chapter has seven important theorems.  .

  •  The conditions for a quadrilateral to be a parallelogram
  •  The mid-point theorem

NCERT Solutions for Class 9 Mathematics Chapter 8: Exercise &  Solutions

 It requires a lot of practice to be a Mathematics wizard to solve advanced levels of questions correctly. The exercises given in the NCERT textbook have different types  of questions to check your understanding. You need endless practice and patience   to be strong in this subject.

The complete Class 9 Mathematics Chapter 8 exercises and solutions have been included in the NCERT Solutions for Class 9 Mathematics Chapter 8, available on the Extramarks website. It has a brief overview of all the exercises along with the solutions designed as per the  CBSE curriculum and guidelines.

Extramarks which tries to do away with rote learning and supplements their studies  with experiential learning and other innovative educational materials. Click on the  links below have to exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 8:

  • Chapter 8: Exercise 8.1 Question and answers
  • Chapter 8: Exercise 8.2 Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 8, students can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5 
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT solutions Class 10
  • NCERT solutions Class 11
  • NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics 

NCERT Exemplar helps in laying the foundation to all the basic as well as advanced concepts in such a way that the answers are self-explanatory, meaning students may not always have to depend on  teachers to clarify their doubts while studying, especially during the last minute preparation making  it easier to understand the concepts quickly and thoroughly.   NCERT Exemplar has a repository of  NCERT related questions. As a result, teachers and mentors recommend  students to include NCERT Exemplar books as an integral part of  their study material. It is  a great guide for students to step up their preparation to get excellent results.

The book is specially designed by subject matter experts, explained in an easy to understand language, all the theorems, formulas and exercises covered in e NCERT Class 9 Mathematics textbook. You can get the NCERT Exemplar for Class 9 Mathematics  from the Extramarks’ website.

Students can find questions ranging from basic to advanced level; thus, they become capable of solving all the levels of questions. After referring to Mathematics Class 9 Chapter 8 and NCERT Exemplar, students can be rest assured  that nothing remains untouched and every example solution, exercise  has been covered in the chapter and hence they are  confident of their preparation and ace the exam with excellent results.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 8

A strong mindset always gives rise to better results. Hence, we have prepared NCERT Solutions for Class 9 Mathematics Chapter 8 in such a way that helps students to develop analytical  mindset.   Some of the key features are:  

  • The academic notes are designed in a way that helps students to manage their time efficiently
  • Students learn to quickly grasp concepts, formulas, definitions, theorems, proofs , postulates and calculations. The in-text and  end- text questions in the chapter , students will get enough practice to improvise their mathematical skills and enjoy solving mathematical problems.
  • It enhances  the confidence of the students and hence they  will be able to leverage their performance.Understanding concepts  would be easier if you have followed NCERT Solutions to clarify your concepts to solve long and short answer questions, MCQs, and intext and end text questions. Also, don’t forget to take notes and practise with sample papers. will make it easier to grasp topics  mathematical skills
  •  The students will be able to interpret proofs and draw conclusions to all  geometrical applications easily.

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

Ans.

Let four angles of quadrilateral be 3x, 5x, 9x and 13x. Then, by angle sum property in quadrilateral

3x + 5x + 9x + 13x = 360°

30x = 360°
x = 360°/30
= 12°

So, the first angle of quadrilateral
= 3(12°)
= 36°

The second angle of quadrilateral
= 5(12°)
= 60°

The third angle of quadrilateral
= 9(12°)
= 108°

The fourth angle of quadrilateral
= 13(12°)
= 156°

Q.2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Ans.

Given: Let ABCD be a parallelogram, in which AC=BD.To prove: ABCD is a rectangle.Proof: In ΔADC and ΔBCD,  AD=BC          [Opposite sides of parallelogram are equal.]              DC=CD          [Common]              AC=BD          [Given]       ΔADCΔBCD     [By S.S.S.]        ADC=BCD     [By C.P.C.T.]but     ADC+BCD=180°   [Cointerior angles]       ADC=BCD=90°Since, one angle of parallelogram is 90°.So, ABCD is a rectangle.         Hence proved.

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

Ans.

Given: Let ABCD be a quadrilateral, in which AO=OC, BO=OD. AC and BD bisect each other at 90°.To prove: ABCD is a rhombus.       Proof:​  In ΔAOB and ΔCOB                AO=OC          [Given]             AOB=COB      [Each 90°]                OB=OB          [Common]            ΔAOBΔCOB      [By SAS]              AB=BC   ...(i)   [By C.P.C.T.]        In ΔBOC and ΔDOC               BO=OD         [Given]          BOC=DOC     [Each 90°]             OC=OC          [Common]             BOCΔDOC       [By SAS]              BC=DC   ...(ii)  [By C.P.C.T.]            In ΔCOD and ΔDOA                CO=OA          [Given]             COD=AOD         [Each 90°]                OD=OD         [Common]             ΔCODΔDOA      [By SAS]              DC=DA   ...(iii)[By C.P.C.T.]From equation(i),(ii) and (iii), we get              AB=BC=CD=DASince, all sides of quadrilateral ABCD are equal, soABCD is a rhombus.              Hence proved.

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

Ans.

Given: Let ABCD be a square.To prove:AC=BD, AO=OC, BO=OD and ACBD.        Proof: In ΔABC and ΔDCB           AB=DC                    [Sides of square]        ABC=DCB              [Each 90°]           BC=BC                    [Common]        ΔABCΔDCB              [By S.A.S.]                   AC=BD           [By C.P.C.T.]In ΔAOB and ΔCOD           AB=CD           [Sides of square]       AOB=COD    [Vertical opposite angles]       BAO=DCO    [Alternate interior angles]       ΔAOBΔCOD     [By S.A.S.]           AO=OC           [By C.P.C.T.]          BO=OD           [By C.P.C.T.]In ΔAOB and ΔCOB           AB=CB           [Sides of square]           OB=OB           [Common]           AO=OC           [Proved above]           ΔAOBΔCOB     [By S.A.S.]           AOB=COB     [By C.P.C.T.]and      AOB+COB=180°                 AOB=90°Thus, diagonals AC and BD are equal and bisect each other at 90°.         Hence proved.

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

Ans.

Given:Let ABCD be a quadrilateral. AC=BD and ACBD,    OA=OC and OB=OD.To prove: ABCD is a square.        Proof:In ΔAOB and ΔCOB   AO=OC [Given]     AOB=COB [Each 90°]            OB=OB [Common]      ΔAOBΔCOB [By S.A.S.]            AB=BC   ...(i)      [By C.P.C.T.]  In ΔBOC and ΔCOD   BO=OD [Given]     BOC=DOC [Each 90°]            OC=OC [Common]      ΔBOCΔCOD [By S.A.S.]            BC=CD   ...(ii)        [By C.P.C.T.]In ΔCOD and ΔDOA   CO=OA [Given]     DOC=DOA [Each 90°]            OD=OD [Common]      ΔCODΔDOA [By S.A.S.]            CD=DA   ...(iii)     [By C.P.C.T.]From equation(i),(ii) and (iii), we haveAB=BC=CD=DAThus, ABCD is a square.Hence proved.

Q.6 Diagonal AC of a parallelogram ABCD bisects ∠A . Show that
(i) it bisects ∠C and
(ii) ABCD is a rhombus.

Ans.

     Given: ABCD is a parallelogram in which diagonal AC    bisects A.To prove:(i) it bisects  C also,    (ii) ABCD is a rhombus        Proof:(i)Since, ADBC so, DAC=BCA ...(i) [Alternate interior angles]and ABDCso,  BAC=DCA ...(ii) [Alternate interior angles]But       BAC=DAC  ...(iii) [Given]So,      DCA=BCA [From equation(i) and (ii)]AC bisects C.(ii) From equation(i) and equation(iii), we have       BAC=BCA  BC=BA [Sides opposite to equal angles are equal.]Since, adjacent sides of parallelogram are equal, so ABCDis a rhombus. Hence proved.

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

Ans.

   Given:ABCD is a rhombus.To prove: AC bisects A as well as C.  BD bisects B as well as D.       Proof:In ΔABC, AB=BCSo, BAC=BCA ...(i) [Opposite angles of equal sidesare equal.]Since, ABDCSo,             BAC=DCA. ..(ii) [Alternate interior angles.]From equation(i) and equation(ii), we have        BCA=DCAdiagonal AC bisects C.Since, AD∥BCSo,             DAC=BCA (iii) [Alternate interior angles.]From equation(i) and equation(iii), we have        BAC=DACdiagonal AC bisects A.Thus, AC bisects A as well as C.Similarly, BC bisects B as well as D.Hence proved.

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

Ans.

   Given:ABCD is a rectangle. Diagonal AC bisects A as well   as C.To prove:(i) ABCD is a square.(ii)Diagonal BD bisects B as well as D.        Proof:(i)Since, 1=2 and 3=4ABCD     2=3 [Alternate interior angles]     1=3    CD=AD [Opposite sides of equal angles are equal.]

Since,adjecent sides of a rectangle are equal, so ABCD is a square.(ii)As ABCD is a square. So,    AB=AD      ADB=ABD ...(i) [Opposite angles of equal sides are equal.]Since,ABCDSo,    ABD=CDB ...(ii) [Alternate interior angles]      ADB=CDB

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Q.9

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ . Show that:(i)ΔAPDΔCQB(ii)AP=CQ (iii)ΔAQBΔCPD (iv)AQ=CP (v)APCQ is a parallelogram

Ans.

Given:ABCD is a parallelogram and DP=BQ.To prove: (i)ΔAPDΔCQB            (ii)AP=CQ             (iii)ΔAQBΔCPD             (iv)AQ=CP             (v)APCQ is a parallelogram Proof: In ΔAPD and ΔCQB              AD=BC               [Opposite sides of parallelogram.]           ADP=CBQ           [Alternater interior angles.]                 DP=BQ               [Given]             ΔAPDΔCQB           [By S.A.S.](ii)              AP=CQ               [By C.P.C.T.](iii) In ΔAQB and ΔCPD                 AB=CD               [Opposite sides of parallelogram.]           ABQ=CDP           [Alternater interior angles.]                 BQ=DP               [Given]             ΔAQBΔCPD           [By S.A.S.](iv)           AQ=PC               [By C.P.C.T.](v)Since, AP=CQ and AQ=PC So, APCQ is a parallelogram. [Opposite sides are parallel.]                                          Hence proved.

Q.10

ABCD  is a parallelogram and AP and CQ are perpendicularsfrom vertices A and C on diagonal BD see the figure below.Show thati ΔAPB  ΔCQDii   AP=CQ

Ans.

Given:ABCD is a parallelogram and AP  BD and CQ  BD.To prove : i ΔAPB  ΔCQDii AP = CQProof : i In ΔAPB and ΔCQD               ABP = CDQ           Alternate interior angles               APB = CQD           [Each 90°]                AB = CD                Opposite sides of parallelogram.              ΔAPB  ΔCQD           By A.A.S.ii AP = CQ              By C.P.C.T. Hence proved.

Q.11

In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the figure below). 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)ΔABCΔDEF.

Ans.

   Given:In ΔABC and ΔDEF,AB=DE, AB||DE, BC=EF and BC||EF.To prove:(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)ΔABCΔDEF.       Proof:(i) Since, AB= DE and ABDESo, ABED is a parallelogram.[One pair of opposite sides is parallel and equal.]    (ii) Since, BC=EF and BC||EFSo, BEFC is a parallelogram.[One pair of opposite sides is parallel and equal.]  (iii) Since, ABED is a prallelogram.So, AD=BE and ADBE  ...(i)[One pair of opposite sides is parallel and equal.]Since, BEFC is a prallelogram.So, BE=CF and BECF  ...(ii)[One pair of opposite sides is parallel and equal.]From equation(i) and equation(ii), we haveAD=CF and ADCF(iv) Since,  AD=CF and ADCFSo, ADFC is a prallelogram.[One pair of opposite sides is parallel and equal.](v) Since, ADFC is a prallelogram. So, AC=DF[Opposite sides of parallelogram.](vi)  In ΔABC and ΔDEF   AB=DE[Given]BC=EF[Given]AC=DF[Proved above]   ΔABCΔDEF[ByS.S.S.]Hence proved.

Q.12

ABCD is a trapezium in which AB||CD and AD=BC(see the figure below).Show that(i)A=B(ii)C=D(iii)ΔABCΔBAD(iv)diagonal AC=diagonal BD

Ans.

       Given:In trapezium ABCD, ABCD and AD=BC.To prove: (i)A=B  (ii)C=D(iii)ΔABCΔBAD(iv)diagonal AC=diagonal BDConstruction: Draw CEDA. Join AC.   Proof:Since, ABCD AECDand CEDASo, AECD is a prallelogram.Then, AD=CEBut, AD=BCCE=BC

   CBE=CEB [Opposite sides of equal angles are equal in ΔCEB.]         CBE+CBA=180°   ...(i)[Linear pair of angles]         CEA+DAE=180°  [Cointerior angles]         CBE+DAE=180°    ...(ii)[CBE=CEA]From equation(i) and equation  (ii), we have         CBE+CBA=CBE+DAE      CBA=DAEB=AA=B(ii) Since, ABDC So, A+D=180°   and          B+C=180°A+D=B+C           D=C[A=B](iii) In ΔABC and ΔBADAB=BA[Common]A=B[Proved above]AD=BC[Given]ΔABCΔBAD[By S.A.S.](iv) AC=BD[By C.P.C.T.]Thus, diagonal AC = diagonal BD.Hence proved.

Q.13 ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the figure below). AC is a diagonal. Show that :
(i) SR||AC and SR = (1/2)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Ans.

Given:ABCD is a quadrilateral in which P, Q, R and S  are  mid-points of the sides AB, BC, CD and DA.To prove:(i) SR||AC and SR=12AC            (ii) PQ=SR            (iii) PQRS is a parallelogram.Proof:  (i) In ΔADC,S and R are the mid-points of DA and DC                respectively. So, by mid-point theorem                     SRAC and SR=12AC         ...(i)            (ii) In ΔABC,P and Q are the mid-points of AB and BC                 respectively. So, by mid-point theorem                     PQAC and PQ=12AC        ...(ii)               From equation (i) and equation(ii), we get                       PQ=SR          (iii) From equation(i) and equaton(ii),                  PQ=SR and PQSR                 PQSR is a prallelogram. [One pair of opposite sides is equal and parallel.]                                                Hence proved.

Q.14 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.

Ans.

Given:ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.To prove:PQRS is a rectangle.Construction: Draw AC and BD.Proof: In ΔABC, P and Q are the mid-points of AB and BC           respectively.So, by mid-point theorem          PQ=12AC and PQAC        ...(i)            In ΔADC, S and R are the mid-points of AD and DC         respectively. So, by mid-point theorem      SR=12AC and SRAC          ...(ii)      From equation(i) and equation(ii), we have          PQ=SR and PQSR So, PQRS is a parallelogram.          [If one pair of opposite sides of a quadrilateral is paralleland equal, then it is a parallelogram.]            In ΔABD, P and S are the mid-points of AB and AD         respectively. So, by mid-point theorem         PS=12BD and PSBD   ...(iii) From equation(i) and equation(iii), we have PNMO and MPON So, PMON is a parallelogram. P=MON     [Opposite angles of parallelogram are equal.]         =90° Since, P=90° in parallelogram PQRS, so PQRS is a rectangle.

Q.15 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.

Ans.

Given:ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.To prove:PQRS is a rhombus.Construction: Draw AC and BD.They are the diagonals of the quadrilateral.Proof: In ΔABC, P and Q are the mid-points of AB and BC respectively. So, by mid-point theoremPQ=12AC and PQAC(i)In ΔADC, S and R are the mid-points of AD and DC respectively. So, by mid-point theoremSR=12AC and SRAC(ii) From equation(i) and equation(ii), we have PQ=SR and PQSR So, PQRS is a parallelogram.[If one pair of opposite sides of a quadrilateral is paralleland equal, then it is a parallelogram.]In ΔABD, P and S are the mid-points of AB and ADrespectively. So, by mid-point theorem PS=12BD and PSBD(iii) Since, AC=BD[Diagonals of rectangle are equal.]PQ=PS[12AC=12BD] Since, adjecent sides of a parallelogram ABCD are equal.So, PQRS is rhombus. Hence proved.

Q.16 ABCD is a trapezium in which AB || 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. 8.30). Show that F is the mid-point of BC.

Ans.

Given: ABCD is a trapezium in which AB||DC, BD is a diagonaland E is the midpoint of AD. A line through E is prallel to AB.To prove: F is the mid-point of BC i.e., BF=CF.Proof: In ΔABD, E is mid-point of AD and EOAB So, by converse of mid-point theorem,OD=OBIn ΔBCD, O is mid-point of BD and EFCD So, by converse of mid-point theorem,CF=FBTherefore, F is the mid-point of BC. Hence proved.

Q.17 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31).
Show that the line segments AF and EC trisect the diagonal BD.

Ans.

Given: In parallelogram ABCD, E and F are the mid-points of           sides AB and CD respectively.To prove: DP=PQ=QBProof: In ΔDQC, F is mid-point of DC and PFQC,        So, by converse of mid-point theorem, we have                    DP=PQ          ...(i)       In ΔABP, E is mid-point of AB and EQAP,       So, by converse of mid-point theorem, we have                    PQ=QB          ...(ii)       From equation(i) and equation(ii), we have                    DP=PQ=QBThis implies that line segments AF and EC trisect the diagonal BD.                                        Hence proved.

Q.18 Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Ans.

Given: Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.To prove:PR and QS bisect each other.Construction: Draw AC.Proof: In ΔABC, P and Q are the mid-points of AB and BC respectively. So, by mid-point theoremPQ=12AC and PQAC(i)In ΔADC, S and R are the mid-points of AD and DC respectively. So, by mid-point theorem SR=12AC and SRAC(ii) From equation(i) and equation(ii), we have PQ=SR and PQSR So, PQRS is a parallelogram. [If one pair of opposite sides of a quadrilateral is paralleland equal, then it is a parallelogram.] Since, diagonals of a parallelogram bisect each other, so PR and QS bisect each other.  Hence proved.

Q.19

ABC is a triangle right angled at C. A line through themidpoint M of hypotenuse AB and parallel to BC intersectsAC at D. Show that(i)D is the mid-point ofAC(ii) MD⊥ACiiiCM=MA=12AB

Ans.

Given:In ΔABC, C=90° and M is mid-point of AB, MDBC.To prove: (i) D is the mid-point of AC             (ii)MD⊥AC            (iii)CM=MA=12ABProof:(i)In ΔABC, MDBC and M is mid-point of AB, then         by converse of mid-point theorem,            AD=DC i.e., D is the mid-point of AC.      (ii) Since, MDBC          So, MDA=BCA         [Corresponding angles]                          =90°          Therefore, MDAC.      (iii)  In ΔMDC and ΔMDA             DC=DA             [Proved above]        MDC=MDA        [Each 90° as MDAC]           MD=MD               [Common]        ΔMDCΔMDA         [By S.A.S.]         CM=MA             [By C.P.C.T.]          Since, AM=12AB       So,CM=MA=12AB                  Hence proved.

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