# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

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 complete set of questions organized with advanced level of difficulty, which provide students ample opportunity to apply combination and skills. Get free NCERT Solutions for Class 9 Maths, Chapter 8 Quadrilaterals. 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.Chapter 8 โ Quadrilaterals

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## 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 = 360ยฐ

Now,

3x+5x+9x+13x = 360ยฐ

โ 30x = 360ยฐ

โ x = 12ยฐ

, Angles of the quadrilateral are:

3x = 3ร12ยฐ = 36ยฐ

5x = 5ร12ยฐ = 60ยฐ

9x = 9ร12ยฐ = 108ยฐ

13x = 13ร12ยฐ = 156ยฐ

**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 ฮABC and ฮBAD,

AB = BA (Common)

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

AC = BD (Given)

Therefore, ฮABC โ ฮBAD [SSS congruency]

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

also,

โ A+โ B = 180ยฐ (Sum of the angles on the same side of the transversal)

โ 2โ A = 180ยฐ

โ โ A = 90ยฐ = โ 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 โ AOB = โ BOC = โ OCD = โ ODA = 90ยฐ

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 ฮAOB and ฮCOB,

OA = OC (Given)

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

OB = OB (Common)

Therefore, ฮAOB โ ฮCOB [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 โ AOB = 90ยฐ

Proof,

In ฮABC and ฮBAD,

AB = BA (Common)

โ ABC = โ BAD = 90ยฐ

BC = AD (Given)

ฮABC โ ฮBAD [SAS congruency]

Thus,

AC = BD [CPCT]

diagonals are equal.

Now,

In ฮAOB and ฮCOD,

โ BAO = โ DCO (Alternate interior angles)

โ AOB = โ COD (Vertically opposite)

AB = CD (Given)

, ฮAOB โ ฮCOD [AAS congruency]

Thus,

AO = CO [CPCT].

, Diagonal bisect each other.

Now,

In ฮAOB and ฮCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

, ฮAOB โ ฮCOB [SSS congruency]

also, โ AOB = โ COB

โ AOB+โ COB = 180ยฐ (Linear pair)

Thus, โ AOB = โ COB = 90ยฐ

, 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 ฮAOB and ฮCOD,

AO = CO (Diagonals bisect each other)

โ AOB = โ COD (Vertically opposite)

OB = OD (Diagonals bisect each other)

, ฮAOB โ ฮCOD [SAS congruency]

Thus,

AB = CD [CPCT] โ (i)

also,

โ OAB = โ OCD (Alternate interior angles)

โ AB || CD

Now,

In ฮAOD and ฮCOD,

AO = CO (Diagonals bisect each other)

โ AOD = โ COD (Vertically opposite)

OD = OD (Common)

, ฮAOD โ ฮCOD [SAS congruency]

Thus,

AD = CD [CPCT] โ (ii)

also,

AD = BC and AD = CD

โ AD = BC = CD = AB โ (ii)

also, โ ADC = โ BCD [CPCT]

and โ ADC+โ BCD = 180ยฐ (co-interior angles)

โ 2โ ADC = 180ยฐ

โโ ADC = 90ยฐ โ (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 โ A (see Fig. 8.19). Show that**

**(i) it bisects โ C also,**

**(ii) ABCD is a rhombus.**

Solution:

(i) In ฮADC and ฮCBA,

AD = CB (Opposite sides of a parallelogram)

DC = BA (Opposite sides of a parallelogram)

AC = CA (Common Side)

, ฮADC โ ฮCBA [SSS congruency]

Thus,

โ ACD = โ CAB by CPCT

and โ CAB = โ CAD (Given)

โ โ ACD = โ BCA

Thus,

AC bisects โ C also.

(ii) โ ACD = โ CAD (Proved above)

โ 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 โ A as well as โ C and diagonal BD bisects โ B as well as โ D.**

Solution:

Given that,

ABCD is a rhombus.

AC and BD are its diagonals.

Proof,

AD = CD (Sides of a rhombus)

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

also, AB || CD

โโ DAC = โ BCA (Alternate interior angles)

โโ DCA = โ BCA

, AC bisects โ C.

Similarly,

We can prove that diagonal AC bisects โ A.

Following the same method,

We can prove that the diagonal BD bisects โ B and โ D.

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

**(i) ABCD is a square**

**(ii) Diagonal BD bisects โ B as well as โ D.**

Solution:

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

โ 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 ฮBCD,

BC = CD

โ โ CDB = โ CBD (Angles opposite to equal sides are equal)

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

โ โ CBD = โ ABD

Thus, BD bisects โ B

Now,

โ CBD = โ ADB

โ โ CDB = โ ADB

Thus, BD bisects โ 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) ฮAPD โ
ฮCQB**

**(ii) AP = CQ**

**(iii) ฮAQB โ
ฮCPD**

**(iv) AQ = CP**

**(v) APCQ is a parallelogram**

Solution:

(i) In ฮAPD and ฮCQB,

DP = BQ (Given)

โ ADP = โ CBQ (Alternate interior angles)

AD = BC (Opposite sides of a parallelogram)

Thus, ฮAPD โ ฮCQB [SAS congruency]

(ii) AP = CQ by CPCT as ฮAPD โ ฮCQB.

(iii) In ฮAQB and ฮCPD,

BQ = DP (Given)

โ ABQ = โ CDP (Alternate interior angles)

AB = CD (Opposite sides of a parallelogram)

Thus, ฮAQB โ ฮCPD [SAS congruency]

(iv) As ฮAQB โ ฮCPD

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) ฮAPB โ
ฮCQD**

**(ii) AP = CQ**

Solution:

(i) In ฮAPB and ฮCQD,

โ ABP = โ CDQ (Alternate interior angles)

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

AB = CD (ABCD is a parallelogram)

, ฮAPB โ ฮCQD [AAS congruency]

(ii) As ฮAPB โ ฮCQD.

, AP = CQ [CPCT]

**11. In ฮABC and ฮDEF, 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) ฮABC โ
ฮDEF.**

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.

โ 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 ฮABC and ฮDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

, ฮABC โ ฮDEF [SSS congruency]

**12. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that**

**(i) โ A = โ B**

**(ii) โ C = โ D**

**(iii) ฮABC โ
ฮBAD**

**(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

โโ CBE = โ CEB

also,

โ A+โ CBE = 180ยฐ (Angles on the same side of transversal and โ CBE = โ CEB)

โ B +โ CBE = 180ยฐ ( As Linear pair)

โโ A = โ B

(ii) โ A+โ D = โ B+โ C = 180ยฐ (Angles on the same side of transversal)

โโ A+โ D = โ A+โ C (โ A = โ B)

โโ D = โ C

(iii) In ฮABC and ฮBAD,

AB = AB (Common)

โ DBA = โ CBA

AD = BC (Given)

, ฮABC โ ฮBAD [SAS congruency]

(iv) Diagonal AC = diagonal BD by CPCT as ฮABC โ ฮBA.

## 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 ฮDAC,

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

Thus by mid point theorem, SR || AC and SR = ยฝ AC

(ii) In ฮBAC,

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

Thus by mid point theorem, PQ || AC and PQ = ยฝ AC

also, SR = ยฝ AC

, PQ = SR

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

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

โ SR || PQ โ 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 ฮDRS and ฮBPQ,

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

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

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

, ฮDRS โ ฮBPQ [SAS congruency]

RS = PQ [CPCT]โโโโโโโ- (i)

In ฮQCR and ฮSAP,

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

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

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

, ฮQCR โ ฮSAP [SAS congruency]

RQ = SP [CPCT]โโโโโโโ- (ii)

Now,

In ฮCDB,

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

โ QR || BD

also,

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

โ PS || BD

โ QR || PS

, PQRS is a parallelogram.

also, โ PQR = 90ยฐ

Now,

In PQRS,

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

โ Q = 90ยฐ

, 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 ฮABC

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

, PQ || AC and PQ = ยฝ AC (Midpoint theorem) โ (i)

In ฮADC,

SR || AC and SR = ยฝ AC (Midpoint theorem) โ (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) โ (iii)

Now,

In ฮBCD,

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

, QR || BD and QR = ยฝ BD (Midpoint theorem) โ (iv)

AC = BD (Diagonals of a rectangle are equal) โ (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 ฮBAD,

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 ฮBDC,

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)

โยฝ AB = ยฝ CD

โ 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 ฮDQC,

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)

โ DP = PQ โ (i)

Similarly,

In ฮAPB,

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

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

โ PQ = QB โ (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 ฮACD,

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 โฅ AC(iii) CM = MA = ยฝ AB**

Solution:

(i) In ฮACB,

M is the midpoint of AB and MD || BC

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

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

also, โ ACB = 90ยฐ

, โ ADM = 90ยฐ and MD โฅ AC

(iii) In ฮAMD and ฮCMD,

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

โ ADM = โ CDM (Each 90ยฐ)

DM = DM (common)

, ฮAMD โ ฮCMD [SAS congruency]

AM = CM [CPCT]

also, AM = ยฝ AB (M is midpoint of AB)

Hence, CM = MA = ยฝ 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-Quadrilaterals is a part of Unit IV, Geometry, and holds 28 marks of weightage in the final exam. Two or three questions are asked every year in the final 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?

The main topics covered in NCERT Solutions for Class 9 Maths Chapter 8 are given below:

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

### How many questions are there in NCERT Solutions for Class 9 Maths Chapter 8?

NCERT Solutions for Class 9 Maths Chapter 8 contains two exercises. The first exercise has 12 questions and second exercise has 7 questions. Practising these exercises help you in scoring high in board exams and also help to ease the subject. These solutions are explained by subject matter experts to help you in clearing all the doubts.

### What is the meaning of quadrilaterals according to NCERT Solutions for Class 9 Maths Chapter 8?

According to NCERT Solutions for Class 9 Maths Chapter 8 quadrilateral is a plane figure that has four sides or edges, and also has four corners or vertices. Quadrilaterals will typically be of standard shapes with four sides like rectangle, square, trapezoid, and kite or irregular and uncharacterized shapes.