# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals| PDF Download

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**Chapter 8 Quadrilaterals Class 9 Maths NCERT Solutions**are available that will be helpful in understanding the key concepts in an easy way. Here you can download PDF of NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals that will serve as beneficial tool that can be used to recall various questions any time. You can also complete your homework on time through the help of these**Chapter 8 NCERT Solutions**and able to solve the difficult problems in a given in a exercise.For application of concepts, an individual should first need to focus on the

**Quadrilaterals NCERT Solutions**as it will tell you about the difficulty of questions. These NCERT Solutions will help an individual to increase concentration and you can solve questions of supplementary books easily.**Exercise 8.1**

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

**Answer**

Let x be the common ratio between the angles.

Sum of the interior angles of the quadrilateral = 360Â°

Now,

3x + 5x + 9x + 13x = 360Â°

â‡’ 30x = 360Â°

â‡’ x = 12Â°

Angles of the quadrilateral are:

3x = 3Ã—12Â° = 36Â°

5x = 5Ã—12Â° = 60Â°

9x = 9Ã—12Â° = 108Â°

13x = 13Ã—12Â° = 156Â°

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

**Answer**

AC = BD

To show,

To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.

Proof,

In Î”ABC and Î”BAD,

BC = BA (Common)

AC = AD (Opposite sides of a parallelogram are equal)

AC = BD (Given)

Therefore, Î”ABC â‰… Î”BAD by SSS congruence condition.

âˆ A = âˆ B (by CPCT)

also,

âˆ A + âˆ B = 180Â° (Sum of the angles on the same side of the transversal)

â‡’ 2âˆ A = 180Â°

â‡’ âˆ A = 90Â° = âˆ B

Thus ABCD is a rectangle.

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

Â

**Answer**

Given,

OA = OC, OB = OD and âˆ AOB = âˆ BOC = âˆ OCD = âˆ ODA = 90Â°

To show,

ABCD is parallelogram and AB = BC = CD = AD

Proof,

In Î”AOB and Î”COB,

OA = OC (Given)

âˆ AOB = âˆ COB (Opposite sides of a parallelogram are equal)

OB = OB (Common)

Therefore, Î”AOB â‰… Î”COB by SAS congruence condition.

Thus, AB = BC (by CPCT)

Similarly we can prove,

AB = BC = CD = AD

Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.

Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

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

**Answer**

To show,

Â AC = BD, AO = OC and âˆ AOB = 90Â°

Proof,

In Î”ABC and Î”BAD,

BC = BA (Common)

âˆ ABC = âˆ BAD = 90Â°

AC = AD (Given)

Therefore, Î”ABC â‰… Î”BAD by SAS congruence condition.

Thus, AC = BD by CPCT. Therefore, diagonals are equal.

Now,

In Î”AOB and Î”COD,

âˆ BAO = âˆ DCO (Alternate interior angles)

âˆ AOB = âˆ COD (Vertically opposite)

AB = CD (Given)

Therefore, Î”AOB â‰… Î”COD by AAS congruence condition.

Thus, AO = CO by CPCT. (Diagonal bisect each other.)

Now,

In Î”AOB and Î”COB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

Therefore, Î”AOB â‰… Î”COB by SSS congruence condition.

also, âˆ AOB = âˆ COB

âˆ AOB + âˆ COB = 180Â° (Linear pair)

Thus, âˆ AOB = âˆ COB = 90Â° (Diagonals bisect each other at right angles)

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

**Answer**

Given,

Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.

To prove,

Quadrilateral ABCD is a square.

Proof,

In Î”AOB and Î”COD,

AO = CO (Diagonals bisect each other)

âˆ AOB = âˆ COD (Vertically opposite)

OB = OD (Diagonals bisect each other)

Therefore, Î”AOB â‰… Î”COD by SAS congruence condition.

Thus, AB = CD by CPCT. --- (i)

also,

âˆ OAB = âˆ OCD (Alternate interior angles)

â‡’ AB || CD

Now,

In Î”AOD and Î”COD,

AO = CO (Diagonals bisect each other)

âˆ AOD = âˆ COD (Vertically opposite)

OD = OD (Common)

Therefore, Î”AOD â‰… Î”COD by SAS congruence condition.

Thus, AD = CD by CPCT. --- (ii)

also,

AD = BC and AD = CD

â‡’ AD = BC = CD = AB --- (ii)

also,Â âˆ ADC = âˆ BCDÂ by CPCT.

and âˆ ADC + âˆ BCD = 180Â° (co-interior angles)

â‡’ 2âˆ ADC = 180Â°

â‡’ âˆ ADC = 90Â° --- (iii)

One of the interior ang is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects âˆ A (see Fig. 8.19). Show that

(i) it bisects âˆ C also,

(ii) ABCD is a rhombus.

**Answer**

(i)

In Î”ADC and Î”CBA,

AD = CB (Opposite sides of a ||gm)

DC = BA (Opposite sides of a ||gm)

AC = CA (Common)

Therefore, Î”ADC â‰… Î”CBA by SSS congruence condition.

Thus,

âˆ ACD = âˆ CAB by CPCT

and âˆ CAB = âˆ CAD (Given)

â‡’ âˆ ACD = âˆ BCA

Thus, AC bisects âˆ C also.

(ii) âˆ ACD = âˆ CAD (Proved)

â‡’ AD = CD (Opposite sides of equal angles of a triangle are equal)

Also, AB = BC = CD = DA (Opposite sides of a ||gm)

Thus, ABCD is a rhombus.

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

**Answer**

Proof,

AD = CD (Sides of a rhombus)

âˆ DAC = âˆ DCA (Angles opposite of equal sides of a triangle are equal.)

also, AB || CD

â‡’ âˆ DAC = âˆ BCA (Alternate interior angles)

â‡’ âˆ DCA = âˆ BCA

Therefore, AC bisects âˆ C.

Similarly, we can prove that diagonal AC bisects âˆ A.

Also, by preceding above method we can prove that diagonal BD bisects âˆ B as well as âˆ D.

8. ABCD is a rectangle in which diagonal AC bisects âˆ A as well as âˆ C. Show that:

(i) ABCD is a square

(ii) diagonal BD bisects âˆ B as well as âˆ D.

**Answer**

â‡’ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

Therefore, AB = BC = CD = AD

Thus, ABCD is a square.

(ii) In Î”BCD,

BC = CD

â‡’ âˆ CDB = âˆ CBD (Angles opposite to equal sides are equal)

also, âˆ CDB = âˆ ABD (Alternate interior angles)

â‡’ âˆ CBD = âˆ ABD

Thus, BD bisects âˆ B

Now,

âˆ CBD = âˆ ADB

â‡’ âˆ CDB = âˆ ADB

Thus, BD bisects âˆ D

Page No: 147

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:

(i) Î”APD â‰… Î”CQB

(ii) AP = CQ

(iii) Î”AQB â‰… Î”CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

**Answer**

(i) In Î”APD and Î”CQB,

DP = BQ (Given)

âˆ ADP = âˆ CBQ (Alternate interior angles)

AD = BC (Opposite sides of a ||gm)

Thus, Î”APD â‰… Î”CQB by SAS congruence condition.

(ii) AP = CQ by CPCT as Î”APD â‰… Î”CQB.

(iii) In Î”AQB and Î”CPD,

BQ = DP (Given)

âˆ ABQ = âˆ CDP (Alternate interior angles)

AB = BCCD (Opposite sides of a ||gm)

Thus, Î”AQB â‰… Î”CPD by SAS congruence condition.

(iv) AQ = CP by CPCT as Î”AQB â‰… Î”CPD.

(v) From (ii)Â and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.

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

(i) Î”APB â‰… Î”CQD

(ii) AP = CQ

**Answer**

(i) In Î”APB and Î”CQD,

âˆ ABP = âˆ CDQ (Alternate interior angles)

âˆ APB = âˆ CQD (equal to right angles as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

Thus, Î”APB â‰… Î”CQD by AAS congruence condition.

(ii) AP = CQ by CPCT as Î”APB â‰… Î”CQD.

11. In Î”ABC and Î”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 Fig. 8.22).

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) Î”ABC â‰… Î”DEF.

**Answer**

(i) AB = DE and AB || DE (Given)

Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.

(iii)Â Since ABED and BEFC are parallelograms.

â‡’ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)

â‡’ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)

Thus, AD = CF.

Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel)

Thus, AD || CF

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(v) AC || DF and AC = DF because ACFD is a parallelogram.

(vi) In Î”ABC and Î”DEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, Î”ABC â‰… Î”DEF by SSS congruence condition.

12. ABCD is a trapezium in which AB || CD and

AD = BC (see Fig. 8.23). Show that

(i) âˆ A = âˆ B

(ii) âˆ C = âˆ D

(iii) Î”ABC â‰… Î”BAD

(iv) diagonal AC = diagonal BD

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

**Answer**

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

(i) CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefor, BC = CE

â‡’ âˆ CBE = âˆ CEB

also,

âˆ AÂ

âˆ B + âˆ CBE = 180Â° (Linear pair)

â‡’ âˆ A = âˆ B

(ii) âˆ A

â‡’ âˆ A + âˆ D = âˆ A + âˆ C (âˆ A = âˆ B)

â‡’ âˆ D = âˆ C

(iii) In Î”ABC and Î”BAD,

AB = AB (Common)

âˆ DBA = âˆ CBA

AD = BC (Given)

Thus, Î”ABC â‰… Î”BAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as Î”ABC â‰… Î”BA.

Page No: 150

**Exercise 8.2**

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

(i) SR || AC and SR = 1/2 AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

**Answer**

(i) In Î”DAC,

R is the mid point of DC and S is the mid point of DA.

Thus by mid point theorem, SR || AC and SR = 1/2 AC

(ii) In Î”BAC,

P is the mid point of AB and Q is the mid point of BC.

Thus by mid point theorem, PQ || AC and PQ = 1/2 AC

also, SR = 1/2 AC

Thus, PQ = SR

(iii) SR || AC - from (i)

and, PQ || AC - from (ii)

â‡’ SR || PQ - from (i) and (ii)

also, PQ = SR

Thus, PQRS is a parallelogram.

2. 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**

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,

AC and BD are joined.

Proof,

In Î”DRS and Î”BPQ,

DS = BQ (Halves of the opposite sides of the rhombus)

âˆ SDR = âˆ QBP (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

Thus, Î”DRS â‰… Î”BPQ by SAS congruence condition.

RS = PQ by CPCT --- (i)

In Î”QCR and Î”SAP,

RC = PA (Halves of the opposite sides of the rhombus)

âˆ RCQ = âˆ PAS (Opposite angles of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

Thus, Î”QCR â‰… Î”SAP by SAS congruence condition.

RQ = SP by CPCT --- (ii)

Now,

In Î”CDB,

R and Q are the mid points of CD and BC respectively.

â‡’ QR || BDÂ

also,

P and S are the mid points of AD and AB respectively.

â‡’ PS || BD

â‡’ QR || PS

Thus, PQRS is a parallelogram.

also, âˆ PQR = 90Â°

Now,

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

âˆ Q = 90Â°

Thus, PQRS is a rectangle.

3. 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**

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

Construction,

AC and BD are joined.

To Prove,

PQRS is a rhombus.

Proof,

In Î”ABC

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

Thus, PQ || AC and PQ = 1/2 AC (Mid point theorem) --- (i)

In Î”ADC,

SR || AC and SR = 1/2 AC (Mid point theorem) --- (ii)

So, PQ || SR and PQ = SR

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.

PS || QR and PS = QR (Opposite sides of parallelogram) --- (iii)

Now,

In Î”BCD,

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

Thus, QR || BD and QR = 1/2 BD (Mid point theorem) --- (iv)

AC = BD (Diagonals of a rectangle are equal) --- (v)

From equations (i), (ii), (iii), (iv) and (v),

PQ = QR = SR = PS

So, PQRS is a rhombus.

4. 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.

**Answer**

Given,

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.

To prove,

F is the mid-point of BC.

Proof,

BD intersected EF at G.

In Î”BAD,

E is the mid point of AD and also EG || AB.

Thus, G is the mid point of BD (Converse of mid point theorem)

Now,

In Î”BDC,

G is the mid point of BD and also GF || AB || DC.

Thus, F is the mid point of BC (Converse of mid point theorem)

Page No: 151

5. 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.

**Answer**

Given,

ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.

To show,

AF and EC trisect the diagonal BD.

Proof,

ABCD is a parallelogram

Therefor, AB || CD

also, AE || FCTherefor, AB || CD

Now,

AB = CD (Opposite sides of parallelogram ABCD)

â‡’ 1/2 AB = 1/2 CD

â‡’ AE = FC (E and F are midpoints of side AB and CD)

AECF is a parallelogram (AE and CF are parallel and equal to each other)

AF || EC (Opposite sides of a parallelogram)

Now,

In Î”DQC,

F is mid point of side DC and FP || CQ (as AF || EC).

P is the mid-point of DQ (Converse of mid-point theorem)

â‡’ DP = PQ --- (i)

Similarly,

In APB,

E is mid point of side AB and EQ || AP (as AF || EC).

Q is the mid-point of PB (Converse of mid-point theorem)

â‡’ PQ = QB --- (ii)

From equations (i) and (i),

DP = PQ = BQ

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

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

**Answer**

Now,

In Î”ACD,

R and S are the mid points of CD and DA respectively.

Thus, SR || AC.

Similarly we can show that,

PQ || AC

PS || BD

QR || BD

Thus, PQRS is parallelogram.

PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

7. 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

(i) D is the mid-point of AC

(ii) MD âŠ¥ AC

(iii) CM = MA = 1/2 AB

**Answer**

(i) In Î”ACB,

M is the mid point of AB and MD || BC

Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) âˆ ACB = âˆ ADM (Corresponding angles)

also, âˆ ACB = 90Â°

Thus, âˆ ADM = 90Â° and MD âŠ¥ AC

(iii)Â In Î”AMD and Î”CMD,

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

âˆ ADM = âˆ CDM (Each 90Â°)

DM = DM (common)

Thus, Î”AMD â‰… Î”CMD by SAS congruence condition.

AM = CM by CPCT

also, AM =Â 1/2 AB (M is mid point of AB)

Hence, CM = MA =Â 1/2 AB

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## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Chapter 8 Quadrilaterals NCERT Solutions is very important for the preparation of exams. A figure formed by joining four points in an order is called a quadrilateral. A quadrilateral has four sides, four angles and four vertices. In this chapter, we will be discussing different types of quadrilaterals, their properties and about parallelograms.

â€¢ Angle Sum Property of a Quadrilateral: The sum of the angles of a quadrilateral is 360Âº. This can be verified by drawing a diagonal and dividing the quadrilateral into two triangles.

â€¢ Types of Quadrilaterals:

A square is a rectangle and also a rhombus.

A parallelogram is a trapezium.

A kite is not a parallelogram.

A trapezium is not a parallelogram (as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram).

A rectangle or a rhombus is not a square.

â€¢ Properties of a Parallelogram:

1. A diagonal of a parallelogram divides it into two congruent triangles.

2. In a parallelogram, opposite sides are equal.

3. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

4. In a parallelogram, opposite angles are equal.

5. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

6. The diagonals of a parallelogram bisect each other.

7. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

â€¢ Another Condition for a Quadrilateral to be a Parallelogram: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

â€¢ The Mid-point Theorem:

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.Â

The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

There are only two exercises in Chapter 8 Quadrilaterals NCERT Solutions which are provided below which can be helpful in completing your homework on time.

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**Class 9 Maths NCERT Solutions**so you can easily clear your doubts. These NCERT Solutions are updated as per the latest marking scheme released by CBSE.### NCERT Solutions for Class 9 Maths Chapters:

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#### What are the benefits of NCERT Solutions for Chapter 8 Quadrilaterals Class 9 NCERT Solutions?

What are the benefits of NCERT Solutions for Chapter 8 Quadrilaterals Class 9 NCERT Solutions?

Through the help of NCERT Solutions of Chapter 8 Quadrilaterals you will be able to solve difficult questions easily and revising the chapter properly. It will improve the learning behaviour of the students and learning diverse topics.

#### What is a Rhombus?

What is a Rhombus?

A parallelogram having all sides equal is called a rhombus.

#### What is mid-point theorem?

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

#### In a quadrilateral, âˆ A : âˆ B : âˆ C : âˆ D = 1 : 2 : 3 : 4, then find the measure of each angle of the quadrilateral.

Since âˆ A : âˆ B : âˆ C : âˆ D = 1 : 2 : 3 : 4

âˆ´ If âˆ A = x, then âˆ B = 2x, âˆ C = 3x and âˆ D = 4x. âˆ´ âˆ A + âˆ B + âˆ C + âˆ D = 360Â°

â‡’ x + 2x + 3x + 4x = 360Â° â‡’ 10x = 36Â°

â‡’ x= (360Â°/10)= 36Â°

âˆ´ âˆ A = x = 36Â° âˆ B = 2x = 2 x 36Â° = 72Â° âˆ C = 3x = 3 x 36Â° = 108Â° âˆ D = 4x = 4 x 36Â° = 144Â°.