# 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. NCERT solutions for 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. 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. Here you will get NCERT solutions for class 9 also.

## ** 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 = (Co-interior 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)

AB||DE (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)

BC||EF (Given )

Hence, quadrilateral BEFC is a parallelogram.

** Answer: **

To prove: and

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

** 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

** 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,

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 BCE,

.................3 (opposite angles of equal sides)

...................4(linear pairs)

.....................5(Co-interior 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.

(Co-interior angles)

(Co-interior 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 mid-points 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 mid-points 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 mid-points 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 mid-points 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 mid-points 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 mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. ).

To prove: F is the mid-point 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 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)

(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 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 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 mid-point 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 mid-point 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 mid-point 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 **

** NCERT solutions for class 9 subject wise **

** **

** 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.

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

**Question: **What are the important topics in chapter Quadrilaterals ?

**Answer: **

Angle sum property of a quadrilateral, properties of the parallelogram, another condition for a quadrilateral to be a parallelogram, and mid-point theorem are the important topics covered in this chapter.

**Question: **How does the NCERT solutions are helpful ?

**Answer: **

NCERT solutions are not only helpful for the students if they stuck while solving NCERT problems but also it will provide them new ways to solve the problems. These solutions are provided in a very detailed manner which will give them conceptual clarity.

**Question: **Does CBSE class 9 maths is tough ?

**Answer: **

CBSE class 9 maths is a building block or base for higher class maths. There are some new topics where students may find difficulties but overall class 9 maths is very simple.

**Question: **Where can I find the complete solutions of NCERT for class 9 ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 by clicking on the link.

**Question: **Where can I find the complete solutions of NCERT for class 9 maths ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link.

**Question: **Which is the official website of NCERT ?

**Answer: **

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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