**NCERT Solutions for Class 9 Maths Chapter 8** Quadrilaterals are an educational aid for students to solve and learn simple and difficult problems. It includes a complete set of questions organised with an advanced level of difficulty, which provides students ample opportunity to apply their knowledge and skills. Get free NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals devised according to the latest update on CBSE Syllabus for 2023-24.

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The NCERT Solutions will help the students to understand the concept of Quadrilaterals – mainly the basics, properties, and some important theorems. The Class 9 Maths NCERT solutions will not only help students to clear their doubts but also prepare more efficiently for the CBSE examination.

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

### Access Answers to NCERT Class 9 Maths Chapter 8 – Quadrilaterals

## Exercise 8.1 Page: 146

**1. The angles of a 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, we have to prove that ABCD is a 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

Opposite sides of a quadrilateral are equal. Hence, it is a parallelogram.

ABCD is rhombus as it is a parallelogram whose diagonals intersect at a 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 a 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] — (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] — (ii)

also,

AD = BC and AD = CD

â‡’ AD = BC = CD = AB — (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° — (iii)

One of the interior angles is a 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 ———————- from question (i)

and, PQ || AC ———————- from question (ii)

â‡’ SR || PQ – 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]———————- (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]———————- (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) — (i)

In Î”ADC,

SR || AC and SR = Â½ AC (Midpoint 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.

, QR || BD and QR = Â½ BD (Midpoint 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.

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 — (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 — (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 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

Also AccessÂ |

NCERT Exemplar for Class 9 Maths Chapter 8 |

CBSE Notes for Class 9 Maths Chapter 8 |

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

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

Topics covered in this chapter help the students understand the basics of a quadrilateral geometrical figure, 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 Quadrilaterals is included in the CBSE Syllabus 2023-24 and is a part of Unit – Geometry, which holds 28 marks of weightage in the CBSE Class 9 Maths exams. Two or three questions are asked every year in the board 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 comprehensively. 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.

**Disclaimer:**

**Dropped Topics –** 8.1 Introduction, 8.2 Angle sum property of a quadrilateral, 8.3 Types of quadrilaterals and 8.5 Another condition for a Quadrilateral to be a parallelogram.

## 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 to 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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