**NCERT Solutions for Class 9 Maths Chapter 8** Quadrilaterals are an educational aid for students that help them solve and learn simple and difficult tasks. It includes a complete set of questions organized with advanced level of difficulty, which provide students ample opportunity to apply combinations and skills. Get free NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals devised according to the latest update on term-wise CBSE Syllabus for 2022-23. These NCERT Solutions will help the students to understand the concept of Quadrilaterals mainly basics, properties and some important theorems. These solutions can not only help students to clear their doubts but also to prepare more efficiently for the second term examination.

### Download NCERT Solutions for Class 9 Maths Chapter 8 â€“ Quadrilaterals

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### Access Answers to NCERT Class 9 Maths Chapter 8 â€“ Quadrilaterals

## Exercise 8.1 Page: 146

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

Solution:

Let the common ratio between the angles be = x.

We know that the 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.**

Solution:

Given that,

AC = BD

To show that, ABCD is a rectangle if the diagonals of a parallelogram are equal

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

Proof,

In Î”ABC and Î”BAD,

AB = BA (Common)

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

AC = BD (Given)

Therefore, Î”ABC â‰… Î”BAD [SSS congruency]

âˆ A = âˆ B [Corresponding parts of Congruent Triangles]

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

Therefore, ABCD is a rectangle.

Hence Proved.

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

Solution:

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

Given that,

OA = OC

OB = OD

and âˆ AOB = âˆ BOC = âˆ OCD = âˆ ODA = 90Â°

To show that,

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

i.e., we have to prove that 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 [SAS congruency]

Thus, AB = BC [CPCT]

Similarly we can prove,

BC = CD

CD = AD

AD = AB

, AB = BC = CD = AD

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

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

Hence Proved.

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

Solution:

Let ABCD be a square and its diagonals AC and BD intersect each other at O.

To show that,

AC = BD

AO = OC

and âˆ AOB = 90Â°

Proof,

In Î”ABC and Î”BAD,

AB = BA (Common)

âˆ ABC = âˆ BAD = 90Â°

BC = AD (Given)

Î”ABC â‰… Î”BAD [SAS congruency]

Thus,

AC = BD [CPCT]

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)

, Î”AOB â‰… Î”COD [AAS congruency]

Thus,

AO = CO [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)

, Î”AOB â‰… Î”COB [SSS congruency]

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.**

Solution:

Given that,

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

To prove that,

The 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)

, Î”AOB â‰… Î”COD [SAS congruency]

Thus,

AB = CD [CPCT] â€” (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)

, Î”AOD â‰… Î”COD [SAS congruency]

Thus,

AD = CD [CPCT] â€” (ii)

also,

AD = BC and AD = CD

â‡’ AD = BC = CD = AB â€” (ii)

also,Â âˆ ADC = âˆ BCDÂ [CPCT]

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

â‡’ 2âˆ ADC = 180Â°

â‡’âˆ ADC = 90Â° â€” (iii)

One of the interior angles is right angle.

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

Hence Proved.

**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.**

Solution:

(i) In Î”ADC and Î”CBA,

AD = CB (Opposite sides of a parallelogram)

DC = BA (Opposite sides of a parallelogram)

AC = CA (Common Side)

, Î”ADC â‰… Î”CBA [SSS congruency]

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 above)

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

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

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.**

Solution:

Given that,

ABCD is a rhombus.

AC and BD are its diagonals.

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

, AC bisects âˆ C.

Similarly,

We can prove that diagonal AC bisects âˆ A.

Following the same method,

We can prove that the diagonal BD bisects âˆ B and 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.**

Solution:

(i) âˆ DAC = âˆ DCA (AC bisects âˆ A as well as âˆ C)

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

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

,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 B as well asÂ âˆ D.

**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**

Solution:

(i) In Î”APD and Î”CQB,

DP = BQ (Given)

âˆ ADP = âˆ CBQ (Alternate interior angles)

AD = BC (Opposite sides of a parallelogram)

Thus, Î”APD â‰… Î”CQB [SAS congruency]

(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 = CD (Opposite sides of a parallelogram)

Thus, Î”AQB â‰… Î”CPD [SAS congruency]

(iv) As Î”AQB â‰… Î”CPD

AQ = CP [CPCT]

(v) From the questions (ii) and (iv), it is clear that APCQ has equal opposite sides and also has equal and opposite angles. , APCQ is a parallelogram.

**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**

Solution:

(i) In Î”APB and Î”CQD,

âˆ ABP = âˆ CDQ (Alternate interior angles)

âˆ APB = âˆ CQD (= 90^{o} as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

, Î”APB â‰… Î”CQD [AAS congruency]

(ii) As Î”APB â‰… Î”CQD.

, AP = CQ [CPCT]

**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.**

Solution:

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

Two opposite sides of a quadrilateral are equal and parallel to each other.

Thus, quadrilateral ABED is a parallelogram

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

Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.

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

, AD = CF.

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

, 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.

(v) Since ACFD is a parallelogram

AC || DF and AC = DF

(vi) In Î”ABC and Î”DEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

, Î”ABC â‰… Î”DEF [SSS congruency]

**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.]**

Solution:

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

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

AD = BC (Given)

, BC = CE

â‡’âˆ CBE = âˆ CEB

also,

âˆ A+âˆ CBE = 180Â° (Angles on the same side of transversal and âˆ CBE = âˆ CEB)

âˆ B +âˆ CBE = 180Â° ( As Linear pair)

â‡’âˆ A = âˆ B

(ii) âˆ A+âˆ D = âˆ B+âˆ C = 180Â° (Angles on the same side of transversal)

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)

, Î”ABC â‰… Î”BAD [SAS congruency]

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

## Exercise 8.2 Page: 150

**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.**

Solution:

(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 = Â½ 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 = Â½ AC

also, SR = Â½ AC

, PQ = SR

(iii) SR || AC â€”â€”â€”â€”â€”â€”â€”- from question (i)

and, PQ || AC â€”â€”â€”â€”â€”â€”â€”- from question (ii)

â‡’ SR || PQ â€“ from (i) and (ii)

also, PQ = SR

, 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.**

Solution:

Given in the question,

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,

Join AC and BD.

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)

, Î”DRS â‰… Î”BPQ [SAS congruency]

RS = PQ [CPCT]â€”â€”â€”â€”â€”â€”â€”- (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)

, Î”QCR â‰… Î”SAP [SAS congruency]

RQ = SP [CPCT]â€”â€”â€”â€”â€”â€”â€”- (ii)

Now,

In Î”CDB,

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

â‡’ QR || BD

also,

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

â‡’ PS || BD

â‡’ QR || PS

, PQRS is a parallelogram.

also, âˆ PQR = 90Â°

Now,

In PQRS,

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

âˆ Q = 90Â°

, 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.**

Solution:

Given in the question,

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

Construction,

Join AC and BD.

To Prove,

PQRS is a rhombus.

Proof:

In Î”ABC

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

, PQ || AC and PQ = Â½ AC (Midpoint theorem) â€” (i)

In Î”ADC,

SR || AC and SR = Â½ AC (Midpoint theorem) â€” (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) â€” (iii)

Now,

In Î”BCD,

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

, QR || BD and QR = Â½ BD (Midpoint theorem) â€” (iv)

AC = BD (Diagonals of a rectangle are equal) â€” (v)

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

PQ = QR = SR = PS

So, PQRS is a rhombus.

Hence Proved

**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.**

Solution:

Given that,

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)

**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.**

Solution:

Given that,

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

, AB || CD

also, AE || FC

Now,

AB = CD (Opposite sides of parallelogram ABCD)

â‡’Â½ AB = Â½ 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 â€” (i)

Similarly,

In Î”APB,

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

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

â‡’ PQ = QB â€” (ii)

From equations (i) and (i),

DP = PQ = BQ

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

Hence Proved.

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

Solution:

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Now,

In Î”ACD,

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

, SR || AC.

Similarly we can show that,

PQ || AC,

PS || BD and

QR || BD

, 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 = Â½ AB**

Solution:

(i) In Î”ACB,

M is the midpoint of AB and MD || BC

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

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

also, âˆ ACB = 90Â°

, âˆ ADM = 90Â° and MD âŠ¥ AC

(iii) In Î”AMD and Î”CMD,

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

âˆ ADM = âˆ CDM (Each 90Â°)

DM = DM (common)

, Î”AMD â‰… Î”CMD [SAS congruency]

AM = CM [CPCT]

also, AM =Â Â½ AB (M is midpoint of AB)

Hence, CM = MA =Â Â½ AB

### NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 explains Angle Sum Property of a Quadrilateral, Types of Quadrilaterals and Mid-Point theorem.

Topics covered under this chapter help the students to understand the basics of a geometrical figure named as a quadrilateral, its properties and various important theorems. This chapter of NCERT Solutions for Class 9 Maths is extremely crucial as the formulas and theorem results are extensively used in several other maths concepts in higher grades.

Chapter 8 QuadrilateralsÂ is included in the second term CBSE Syllabus 2022-23 and is a part of Unit-Geometry which holds 28 marks of weightage in the term exams of CBSE Class 9 Maths. Two or three questions are asked every year in the second term examination from this chapter.

**NCERT Solutions For Class 9 Maths Chapter 8 Exercises:**

Get detailed solutions for all the questions listed under the below exercises:

Exercise 8.1 Solutions (12 Questions)

Exercise 8.2 Solutions (7 Questions) NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals NCERT Solutions for Class 9 Maths Chapter 8 is about Theorems and properties on Quadrilaterals. They are accompanied with explanatory figures and solved examples, which are explained in a comprehensive way. The main topics covered in this chapter include:

Exercise |
Topic |

8.1 | Introduction |

8.2 | Angle Sum Property of a Quadrilateral |

8.3 | Types of Quadrilaterals |

8.4 | Properties of a Parallelogram |

8.5 | Another Condition for a Quadrilateral to be a Parallelogram |

8.6 | The Mid-point Theorem |

8.7 | Summary |

Key Features of NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

- NCERT solutions have been prepared in a logical and simple language.
- Pictorial presentation of all the questions.
- Emphasizes that learning should be activity-based and knowledge-driven.
- The solutions are explained in a well-organised way.
- Step by step approach used to solve all NCERT questions.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 8

### What are the main topics covered in NCERT Solutions for Class 9 Maths Chapter 8?

8.1 Introduction of quadrilaterals

8.2 Angle Sum Property of a Quadrilateral

8.3 Types of Quadrilaterals

8.4 Properties of a Parallelogram

8.5 Another Condition for a Quadrilateral to be a Parallelogram

8.6 The Mid-point Theorem

8.7 Summary

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