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 reallife 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. Two exercises of this chapter are explained beloow.
NCERT solutions for class 9 maths chapter 8 Quadrilaterals Excercise: 8.1
Q1 The angles of quadrilateral are in the ratio . Find all the angles of the quadrilateral.
Answer:
Given :
The angles of a quadrilateral are in the ratio
.
Let the angles of quadrilateral be
.
Sum of all angles is 360.
Thus,
All four angles are :
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 ABC and BAD,
BC= AD (Opposite sides of parallelogram)
AC=BD (Given )
AB=AB (common)
ABC BAD (By SSS)
(CPCT)
and (co  interior angles)
Hence, it is a rectangle.
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 AOB and AOD,
(Each )
BO=OD (Given )
AO=AO (common)
AOB AOD (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
Proof : In BAD and ABC,
(Each )
AD=BC (Given )
AB=AB (common)
BAD ABC (By SAS)
BD=AC (CPCT)
In AOB and COD,
OAB= OCD (Alternate angles)
AB=CD (Given )
OBA= ODC (Alternate angles)
AOB COD (By AAS)
AO=OC ,BO=OD (CPCT)
In AOB and AOD,
OB=OD (proved above)
AB=AD (Given )
OA=OA (COMMON)
AOB AOD (By SSS)
AOB= AOD (CPCT)
AOB+ AOD =
2. AOB =
AOB =
Hence, the diagonals of a square are equal and bisect each other at right angles.
Answer:
Given : ABCD is a quadrilateral with AC=BD,AO=CO,BO=DO, COD =
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 BAD and ABC,
AD=BC (proved above )
AB=AB (common)
BD=AC
BAD ABC (By SSS)
(CPCT)
BAD+ ABC = (Cointerior angles)
2. ABC =
ABC =
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 (see Fig. ). Show that
Answer:
Given: DAC= BAC ................1
DAC= BCA.................2 (Alternate angles)
BAC= ACD .................3 (Alternate angles)
From equation 1,2 and 3, we get
ACD= BCA...................4
Hence, diagonal AC bisect angle C also.
Q6 (ii) Diagonal AC of a parallelogram ABCD bisects (see Fig. ). Show that
Answer:
Given: DAC= BAC ................1
DAC= BCA.................2 (Alternate angles)
BAC= ACD .................3 (Alternate angles)
From equation 1,2 and 3, we get
ACD= BCA...................4
From 2 and 4, we get
ACD= DAC
In ADC,
ACD= DAC (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 as well as and diagonal BD bisects as well as .
Answer:
In ADC,
AD = CD (ABCD is a rhombus)
3= 1.................1(angles opposite to equal sides are equal )
3= 2.................2 (alternate angles)
From 1 and 2, we have
1= 2.................3
and 1= 4.................4 (alternate angles)
From 1 and 4, we get
3= 4.................5
Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.
In ADB,
AD = AB (ABCD is a rhombus)
5= 7.................6(angles opposite to equal sides are equal )
7= 6.................7 (alternate angles)
From 6 and 7, we have
5= 6.................8
and 5= 8.................9(alternate angles)
From 6 and 9, we get
7= 8.................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 as well as . Show that:
Answer:
Given: ABCD is a rectangle with AB=CD and BC=AD 1= 2 and 3= 4.
To prove: ABCD is a square.
Proof : 1= 4 .............1(alternate angles)
3= 4 ................2(given )
From 1 and 2, 1= 3.....................................3
In ADC,
1= 3 (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 as well as . Show that:
diagonal BD bisects as well as .
Answer:
In ADB,
AD = AB (ABCD is a square)
5= 7.................1(angles opposite to equal sides are equal )
5= 8.................2 (alternate angles)
From 1 and 2, we have
7= 8.................3
and 7 = 6.................4(alternate angles)
From 1 and 4, we get
5= 6.................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 (see Fig. ). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ADP= CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
Q9 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig. ). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ADP= CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
(CPCT)
Q9 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig. ). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ABQ= CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
Q9 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig. ). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ABQ= CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
(CPCT)
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove: APCQ is a parallelogram
Proof :
In
DP=BQ (Given )
ADP= CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
(CPCT)...............................................................1
Also,
In
DP=BQ (Given )
ABQ= CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
(CPCT)........................................2
From equation 1 and 2, we get
Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove :
Proof: In ,
APB= CQD (Each )
ABP= CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, (By SAS)
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove :
Proof: In ,
APB= CQD (Each )
ABP= CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, (By SAS)
(CPCT)
Answer:
Given : In and , and .
To prove : quadrilateral ABED is a parallelogram
Proof : In ABED,
AB=DE (Given)
ABDE (Given )
Hence, quadrilateral ABED is a parallelogram.
Answer:
Given: In and , and .
To prove: quadrilateral BEFC is a parallelogram
Proof: In BEFC,
BC=EF (Given)
BCEF (Given )
Hence, quadrilateral BEFC is a parallelogram.
Answer:
To prove: and
Proof :
In ABED,
AD=BE.................1(ABED is a parallelogram)
ADBE .................2(ABED is a parallelogram)
In BEFC,
BE=CF.................3(BEFC is a parallelogram)
BECF .................4(BEFC is a parallelogram)
From 2 and 4 , we get
ADCF
From 1 and 3, we get
AD=CF
Answer:
To prove : quadrilateral ACFD is a parallelogram
Proof :
In ABED,
AD=BE.................1(ABED is a parallelogram)
ADBE .................2(ABED is a parallelogram)
In BEFC,
BE=CF.................3(BEFC is a parallelogram)
BECF .................4(BEFC is a parallelogram)
From 2 and 4 , we get
ADCF...........................5
From 1 and 3, we get
AD=CF...........................6
From 5 and 6, we get
ADCF and AD=CF
Thus, quadrilateral ACFD is a parallelogram
Answer:
In ACFD,
AC=DF (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)
Answer:
In and ,
AB=DE (Given )
BC=EF (Given )
AC=DF ( proved in (v) part)
(By SSS rule)
Q12 (i) ABCD is a trapezium in which and (see Fig. ). Show that
[ 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 and
To prove :
Proof: Let A be 1, ABC be 2, EBC be 3, BEC be 4.
In AECD,
AEDC (Given)
ADCE (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 BCE,
.................3 (opposite angles of equal sides)
...................4(linear pairs)
.....................5(Cointerior angles)
From 4 and 5, we get
(Since, )
Q12 (ii) ABCD is a trapezium in which AB  CD and AD = BC (see Fig. 8.23). Show that
[
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 and
To prove :
Proof: Let A be 1, ABC be 2, EBC be 3, BEC be 4.
(Cointerior angles)
(Cointerior angles)
Thus, (Since , )
Q12 (iii) ABCD is a trapezium in which and (see Fig. ). Show that
[ 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 and
To prove :
Proof: In ,
BC=AD (Given )
AB=AB (Common )
(proved in (i) )
Thus, (By SAS rule)
Q12 (iv) ABCD is a trapezium in which and (see Fig. ). Show that diagonal AC 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 and
To prove: diagonal AC diagonal BD
Proof: In ,
BC=AD (Given )
AB=AB (Common )
(proved in (i) )
Thus, (By SAS rule)
diagonal AC diagonal BD (CPCT)
NCERT solutions for class 9 maths chapter 8 Quadrilaterals Excercise: 8.2
Answer:
Given : ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see Fig ). AC is a diagonal.
To prove : and
Proof: In ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
and
Answer:
Given : ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see Fig ). AC is a diagonal.
To prove :
Proof : In ACD,
S is mid point of DA. (Given)
R is mid point of DC. (Given)
By mid point theorem,
and ...................................1
In ABC,
P is mid point of AB. (Given)
Q is mid point of BC. (Given)
By mid point theorem,
and .................................2
From 1 and 2,we get
and
Thus,
Answer:
Given : ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see Fig ). AC is a diagonal.
To prove : PQRS is a parallelogram.
Proof : In PQRS,
Since,
and .
So,PQRS is a parallelogram.
Answer:
Given: ABCD is a rhombus in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA . AC, BD are diagonals.
To prove: the quadrilateral PQRS is a rectangle.
Proof: In ACD,
S is midpoint of DA. (Given)
R is midpoint of DC. (Given)
By midpoint theorem,
and ...................................1
In ABC,
P is midpoint of AB. (Given)
Q is mid point of BC. (Given)
By mid point theorem,
and .................................2
From 1 and 2,we get
and
Thus, and
So,the quadrilateral PQRS is a parallelogram.
Similarly, in BCD,
Q is mid point of BC. (Given)
R is mid point of DC. (Given)
By mid point theorem,
So, QN  LM ...........5
LQ  MN ..........6 (Since, PQ  AC)
From 5 and 6, we get
LMPQ is a parallelogram.
Hence, LMN= LQN (opposite angles of the parallelogram)
But, LMN= 90 (Diagonals of a rhombus are perpendicular)
so, LQN=90
Thus, a parallelogram whose one angle is right angle,ia a rectangle.Hence,PQRS is a rectangle.
Answer:
Given: ABCD is a rectangle and P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively.
To prove: the quadrilateral PQRS is a rhombus.
Proof :
In ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
and ...................................1
In ABC,
P is the midpoint of AB. (Given)
Q is the midpoint of BC. (Given)
By midpoint theorem,
and .................................2
From 1 and 2, we get
and
Thus, and
So, the quadrilateral PQRS is a parallelogram.
Similarly, in BCD,
Q is the midpoint of BC. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
and ...................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.
Answer:
Given: ABCD is a trapezium in which , BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. ).
To prove: F is the midpoint of BC.
In ABD,
E is the midpoint of AD. (Given)
EG  AB (Given)
By converse of midpoint theorem,
G is the midpoint of BD.
In BCD,
G is mid point of BD. (Proved above)
FG  DC (Given)
By converse of midpoint theorem,
F is the midpoint of BC.
Answer:
Given: In a parallelogram ABCD, E and F are the midpoints 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)
(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 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 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.
Answer:
Given: ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA. AC, BD are diagonals.
To prove: the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
Proof: In ACD,
S is the midpoint of DA. (Given)
R is midpoint of DC. (Given)
By midpoint theorem,
and ...................................1
In ABC,
P is the midpoint of AB. (Given)
Q is the midpoint of BC. (Given)
By midpoint theorem,
and .................................2
From 1 and 2, we get
and
Thus, and
So, the quadrilateral PQRS is a parallelogram and diagonals of a parallelogram bisect each other.
Thus, SQ and PR bisect each other.
Answer:
Given: ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D.
To prove :D is mid point of AC.
Proof: In ABC,
M is mid point of AB. (Given)
DM  BC (Given)
By converse of mid point theorem,
D is the mid point of AC.
Answer:
Given: ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallels to BC intersects AC at D.
To prove :
Proof : ADM = ACB (Corresponding angles)
ADM= . ( ACB = )
Hence, .
Answer:
Given: ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D.
To prove :
Proof : In ABC,
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 AMD and CMD,
AD = DC (proved above)
ADM = CDM (Each right angle)
DM = DM (Common)
AMD CMD (By SAS)
AM = CM (CPCT)
But ,
Hence, .
NCERT solutions for class 9 maths chapter wise
Chapter No. 
Chapter Name 
Chapter 1 

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 

Chapter 7 

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 

Chapter 11 
CBSE NCERT solutions for class 9 maths chapter 11 Constructions 
Chapter 12 

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

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