# NCERT Solutions for Class 9 Maths Chapter 7 – Triangles

## NCERT Solutions for Class 9 Maths Chapter 7 โ Free PDF Download

**NCERT Solutions for Class 9 Maths Chapter 7** **Triangles** provides the answers and questions related to the chapter. Triangle, the word itself describes its meaning. โTriโ means โthreeโ so a closed figure formed by three intersecting lines is known as a Triangle. Students must have already studied the angle sum property of a triangle in Chapter 6 of NCERT Class 9 Maths. Now, inconsequential to it, Chapter 7 of NCERT Solutions for Class 9 Maths will further brief students about the congruence of triangles and rules of congruence. Also, they will learn a few more properties of triangles and inequalities in a triangle.

Here we have provided the complete **Class 9 Maths NCERT Solutions Chapter 7** Triangles in PDF format solved by experienced teachers. Students can download the free PDF of these NCERT Solutions for Class 9 by clicking on the link below to keep it handy for future reference.Chapter 7 โ Triangles

## Download NCERT Solutions for Class 9 Maths Chapter 7 โ Triangles

**List of Exercises in NCERT Class 9 Maths Chapter 7**

Exercise 7.1 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)

Exercise 7.2 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)

Exercise 7.3 Solution 5 Questions (3 Short Answer Questions, 2 Long Answer Question)

Exercise 7.4 Solution 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 7.5 (Optional) Solution 4 Questions

## Access Answers of Maths NCERT Class 9 Chapter 7 โ Triangles

**Exercise: 7.1 (Page No: 118)**

**1. In quadrilateral ACBD, AC = AD and AB bisect A (see Fig. 7.16). Show that ฮABC ฮABD. What can you say about BC and BD?**

**Solution:**

It is given that AC and AD are equal i.e. AC = AD and the line segment AB bisects A.

We will have to now prove that the two triangles ABC and ABD are similar i.e. **ฮABC ฮABD**

**Proof:**

Consider the triangles ฮABC and ฮABD,

(i) AC = AD (It is given in the question)

(ii) AB = AB (Common)

(iii) CAB = DAB (Since AB is the bisector of angle A)

So, by **SAS congruency criterion**, ฮABC ฮABD.

For the 2^{nd} part of the question, BC and BD are of equal lengths by the rule of C.P.C.T.

**2. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see Fig. 7.17). Prove that**

(i) ฮABD ฮBAC

(ii) BD = AC

(iii) ABD = BAC.

**Solution:**

The given parameters from the questions are DAB = CBA and AD = BC.

(i) ฮABD and ฮBAC are similar by SAS congruency as

AB = BA (It is the common arm)

DAB = CBA and AD = BC (These are given in the question)

So, triangles ABD and BAC are similar i.e. ฮABD ฮBAC. (Hence proved).

(ii) It is now known that ฮABD ฮBAC so,

BD = AC (by the rule of CPCT).

(iii) Since ฮABD ฮBAC so,

Angles ABD = BAC (by the rule of CPCT).

**3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.**

**Solution:**

It is given that AD and BC are two equal perpendiculars to AB.

We will have to prove that **CD is the bisector of AB**

Now,

Triangles ฮAOD and ฮBOC are similar by AAS congruency since:

(i) A = B (They are perpendiculars)

(ii) AD = BC (As given in the question)

(iii) AOD = BOC (They are vertically opposite angles)

โด ฮAOD ฮBOC.

So, AO = OB (by the rule of CPCT).

Thus, CD bisects AB (Hence proved).

**4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ฮABC ฮCDA.**

**Solution:**

It is given that p q and l m

**To prove:**

Triangles ABC and CDA are similar i.e. ฮABC ฮCDA

**Proof**:

Consider the ฮABC and ฮCDA,

(i) BCA = DAC and BAC = DCA Since they are alternate interior angles

(ii) AC = CA as it is the common arm

So, by **ASA congruency criterion, **ฮABC ฮCDA.

**5. Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see Fig. 7.20). Show that:**

(i) ฮAPB ฮAQB

(ii) BP = BQ or B is equidistant from the arms of A.

**Solution:**

It is given that the line โ*l*โ is the bisector of angle A and the line segments BP and BQ are perpendiculars drawn from *l*.

(i) ฮAPB and ฮAQB are similar by AAS congruency because:

P = Q (They are the two right angles)

AB = AB (It is the common arm)

BAP = BAQ (As line *l *is the bisector of angle A)

So, ฮAPB ฮAQB.

(ii) By the rule of CPCT, BP = BQ. So, it can be said the point B is equidistant from the arms of A.

**6. In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.**

**Solution:**

It is given in the question that AB = AD, AC = AE, and โ BAD = โ EAC

**To prove:**

The line segment BC and DE are similar i.e. BC = DE

**Proof:**

We know that BAD = EAC

Now, by adding DAC on both sides we get,

BAD + DAC = EAC +DAC

This implies, BAC = EAD

Now, ฮABC and ฮADE are similar by SAS congruency since:

(i) AC = AE (As given in the question)

(ii) BAC = EAD

(iii) AB = AD (It is also given in the question)

โด Triangles ABC and ADE are similar i.e. ฮABC ฮADE.

So, by the rule of CPCT, it can be said that BC = DE.

**7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB (see Fig. 7.22). Show that**

**(i) ฮDAP ฮEBP**

**(ii) AD = BE**

**Solutions:**

In the question, it is given that P is the mid-point of line segment AB. Also, BAD = ABE and EPA = DPB

(i) It is given that EPA = DPB

Now, add DPE on both sides,

EPA +DPE = DPB+DPE

This implies that angles DPA and EPB are equal i.e. DPA = EPB

Now, consider the triangles DAP and EBP.

DPA = EPB

AP = BP (Since P is the mid-point of the line segment AB)

BAD = ABE (As given in the question)

So, by **ASA congruency**, ฮDAP ฮEBP.

(ii) By the rule of CPCT, AD = BE.

**8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:**

(i) ฮAMC ฮBMD

(ii) DBC is a right angle.

(iii) ฮDBC ฮACB

(iv) CM = ยฝ AB

**Solution:**

It is given that M is the mid-point of the line segment AB, C = 90ยฐ, and DM = CM

(i) Consider the triangles ฮAMC and ฮBMD:

AM = BM (Since M is the mid-point)

CM = DM (Given in the question)

CMA = DMB (They are vertically opposite angles)

So, by **SAS congruency criterion**, ฮAMC ฮBMD.

(ii) ACM = BDM (by CPCT)

โด AC BD as alternate interior angles are equal.

Now, ACB +DBC = 180ยฐ (Since they are co-interiors angles)

โ 90ยฐ +B = 180ยฐ

โด DBC = 90ยฐ

(iii) In ฮDBC and ฮACB,

BC = CB (Common side)

ACB = DBC (They are right angles)

DB = AC (by CPCT)

So, ฮDBC ฮACB by **SAS congruency**.

(iv) DC = AB (Since ฮDBC ฮACB)

โ DM = CM = AM = BM (Since M the is mid-point)

So, DM + CM = BM+AM

Hence, CM + CM = AB

โ CM = (ยฝ) AB

**Exercise: 7.2 (Page No: 123)**

**1. In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that:**

(i) OB = OC (ii) AO bisects A

**Solution:**

Given:

AB = AC and

the bisectors of B and C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

B = C

ยฝ B = ยฝ C

โ OBC = OCB (Angle bisectors)

โด OB = OC (Side opposite to the equal angles are equal.)

(ii) In ฮAOB and ฮAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ฮAOB ฮAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects A.

**2. In ฮABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ฮABC is an isosceles triangle in which AB = AC.**

**Solution:**

It is given that AD is the perpendicular bisector of BC

**To prove:**

AB = AC

**Proof:**

In ฮADB and ฮADC,

AD = AD (It is the Common arm)

ADB = ADC

BD = CD (Since AD is the perpendicular bisector)

So, ฮADB ฮADC by **SAS congruency criterion**.

Thus,

AB = AC (by CPCT)

**3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.**

**Solution:**

Given:

(i) BE and CF are altitudes.

(ii) AC = AB

**To prove:**

BE = CF

**Proof:**

Triangles ฮAEB and ฮAFC are similar by AAS congruency since

A = A (It is the common arm)

AEB = AFC (They are right angles)

AB = AC (Given in the question)

โด ฮAEB ฮAFC and so, BE = CF (by CPCT).

**4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that**

(i) ฮABE ฮACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

**Solution:**

It is given that BE = CF

(i) In ฮABE and ฮACF,

A = A (It is the common angle)

AEB = AFC (They are right angles)

BE = CF (Given in the question)

โด ฮABE ฮACF by **AAS congruency condition**.

(ii) AB = AC by CPCT and so, ABC is an isosceles triangle.

**5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ABD = ACD.**

**Solution:**

In the question, it is given that ABC and DBC are two isosceles triangles.

We will have to show that ABD = ACD

**Proof:**

Triangles ฮABD and ฮACD are similar by SSS congruency since

AD = AD (It is the common arm)

AB = AC (Since ABC is an isosceles triangle)

BD = CD (Since BCD is an isosceles triangle)

So, ฮABD ฮACD.

โด ABD = ACD by CPCT.

**6. ฮABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that BCD is a right angle.**

**Solution:**

It is given that AB = AC and AD = AB

We will have to now prove BCD is a right angle.

**Proof:**

Consider ฮABC,

AB = AC (It is given in the question)

Also, ACB = ABC (They are angles opposite to the equal sides and so, they are equal)

Now, consider ฮACD,

AD = AB

Also, ADC = ACD (They are angles opposite to the equal sides and so, they are equal)

Now,

In ฮABC,

CAB + ACB + ABC = 180ยฐ

So, CAB + 2ACB = 180ยฐ

โ CAB = 180ยฐ โ 2ACB โ (i)

Similarly, in ฮADC,

CAD = 180ยฐ โ 2ACD โ (ii)

also,

CAB + CAD = 180ยฐ (BD is a straight line.)

Adding (i) and (ii) we get,

CAB + CAD = 180ยฐ โ 2ACB+180ยฐ โ 2ACD

โ 180ยฐ = 360ยฐ โ 2ACB-2ACD

โ 2(ACB+ACD) = 180ยฐ

โ BCD = 90ยฐ

**7. ABC is a right-angled triangle in which A = 90ยฐ and AB = AC. Find B and C.**

**Solution:**

In the question, it is given that

A = 90ยฐ and AB = AC

AB = AC

โ B = C (They are angles opposite to the equal sides and so, they are equal)

Now,

A+B+C = 180ยฐ (Since the sum of the interior angles of the triangle)

โด 90ยฐ + 2B = 180ยฐ

โ 2B = 90ยฐ

โ B = 45ยฐ

So, B = C = 45ยฐ

**8. Show that the angles of an equilateral triangle are 60ยฐ each.**

**Solution:**

Let ABC be an equilateral triangle as shown below:

Here, BC = AC = AB (Since the length of all sides is same)

โ A = B =C (Sides opposite to the equal angles are equal.)

Also, we know that

A+B+C = 180ยฐ

โ 3A = 180ยฐ

โ A = 60ยฐ

โด A = B = C = 60ยฐ

So, the angles of an equilateral triangle are always 60ยฐ each.

**Exercise: 7.3 (Page No: 128)**

**1. ฮABC and ฮDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that**

(i) ฮABD ฮACD

(ii) ฮABP ฮACP

(iii) AP bisects A as well as D.

(iv) AP is the perpendicular bisector of BC.

**Solution:**

In the above question, it is given that ฮABC and ฮDBC are two isosceles triangles.

(i) ฮABD and ฮACD are similar by SSS congruency because:

AD = AD (It is the common arm)

AB = AC (Since ฮABC is isosceles)

BD = CD (Since ฮDBC is isosceles)

โด ฮABD ฮACD.

(ii) ฮABP and ฮACP are similar as:

AP = AP (It is the common side)

PAB = PAC (by CPCT since ฮABD ฮACD)

AB = AC (Since ฮABC is isosceles)

So, ฮABP ฮACP by SAS congruency condition.

(iii) PAB = PAC by CPCT as ฮABD ฮACD.

AP bisects A. โ (i)

Also, ฮBPD and ฮCPD are similar by SSS congruency as

PD = PD (It is the common side)

BD = CD (Since ฮDBC is isosceles.)

BP = CP (by CPCT as ฮABP ฮACP)

So, ฮBPD ฮCPD.

Thus, BDP = CDP by CPCT. โ (ii)

Now by comparing (i) and (ii) it can be said that AP bisects A as well as D.

(iv) BPD = CPD (by CPCT as ฮBPD ฮCPD)

and BP = CP โ (i)

also,

BPD +CPD = 180ยฐ (Since BC is a straight line.)

โ 2BPD = 180ยฐ

โ BPD = 90ยฐ โ(ii)

Now, from equations (i) and (ii), it can be said that

AP is the perpendicular bisector of BC.

**2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that**

(i) AD bisects BC (ii) AD bisects A.

**Solution:**

It is given that AD is an altitude and AB = AC. The diagram is as follows:

(i) In ฮABD and ฮACD,

ADB = ADC = 90ยฐ

AB = AC (It is given in the question)

AD = AD (Common arm)

โด ฮABD ฮACD by RHS congruence condition.

Now, by the rule of CPCT,

BD = CD.

So, AD bisects BC

(ii) Again, by the rule of CPCT, BAD = CAD

Hence, AD bisects A.

**3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ฮPQR (see Fig. 7.40). Show that:**

(i) ฮABM ฮPQN

(ii) ฮABC ฮPQR

**Solution:**

Given parameters are:

AB = PQ,

BC = QR and

AM = PN

(i) ยฝ BC = BM and ยฝ QR = QN (Since AM and PN are medians)

Also, BC = QR

So, ยฝ BC = ยฝ QR

โ BM = QN

In ฮABM and ฮPQN,

AM = PN and AB = PQ (As given in the question)

BM = QN (Already proved)

โด ฮABM ฮPQN by SSS congruency.

(ii) In ฮABC and ฮPQR,

AB = PQ and BC = QR (As given in the question)

ABC = PQR (by CPCT)

So, ฮABC ฮPQR by SAS congruency.

**4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.**

.

**Solution:**

It is known that BE and CF are two equal altitudes.

Now, in ฮBEC and ฮCFB,

BEC = CFB = 90ยฐ (Same Altitudes)

BC = CB (Common side)

BE = CF (Common side)

So, ฮBEC ฮCFB by RHS congruence criterion.

Also, C = B (by CPCT)

Therefore, AB = AC as sides opposite to the equal angles is always equal.

**5. ABC is an isosceles triangle with AB = AC. Draw AP โฅ BC to show that B = C.**

**Solution:**

In the question, it is given that AB = AC

Now, ฮABP and ฮACP are similar by RHS congruency as

APB = APC = 90ยฐ (AP is altitude)

AB = AC (Given in the question)

AP = AP (Common side)

So, ฮABP ฮACP.

โด B = C (by CPCT)

**Exercise: 7.4 (Page No: 132)**

**1. Show that in a right-angled triangle, the hypotenuse is the longest side.**

**Solution:**

It is known that ABC is a triangle right angled at B.

We know that,

A +B+C = 180ยฐ

Now, if B+C = 90ยฐ then A has to be 90ยฐ.

Since A is the largest angle of the triangle, the side opposite to it must be the largest.

So, AB is the hypotenuse which will be the largest side of the above right-angled triangle i.e. ฮABC.

**2. In Fig. 7.48, sides AB and AC of ฮABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.**

**Solution:**

It is given that PBC < QCB

We know that ABC + PBC = 180ยฐ

So, ABC = 180ยฐ-PBC

Also,

ACB +QCB = 180ยฐ

Therefore ACB = 180ยฐ -QCB

Now, since PBC < QCB,

โด ABC > ACB

Hence, AC > AB as sides opposite to the larger angle is always larger.

**3. In Fig. 7.49, B < A and C < D. Show that AD < BC.**

**Solution:**

In the question, it is mentioned that angles B and angle C is smaller than angles A and D respectively i.e. B < A and C < D.

Now,

Since the side opposite to the smaller angle is always smaller

AO < BO โ (i)

And OD < OC โ(ii)

By adding equation (i) and equation (ii) we get

AO+OD < BO + OC

So, AD < BC

**4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).**

**Show that A > C and B > D.**

**Solution:**

In ฮABD, we see that

AB < AD < BD

So, ADB < ABD โ (i) (Since angle opposite to longer side is always larger)

Now, in ฮBCD,

BC < DC < BD

Hence, it can be concluded that

BDC < CBD โ (ii)

Now, by adding equation (i) and equation (ii) we get,

ADB + BDC < ABD + CBD

ADC < ABC

B > D

Similarly, In triangle ABC,

ACB < BAC โ (iii) (Since the angle opposite to the longer side is always larger)

Now, In ฮADC,

DCA < DAC โ (iv)

By adding equation (iii) and equation (iv) we get,

ACB + DCA < BAC+DAC

โ BCD < BAD

โด A > C

**5. In Fig 7.51, PR > PQ and PS bisect QPR. Prove that PSR > PSQ.**

**Solution:**

It is given that PR > PQ and PS bisects QPR

Now we will have to prove that angle PSR is smaller than PSQ i.e. PSR > PSQ

**Proof:**

QPS = RPS โ (ii) (As PS bisects โ QPR)

PQR > PRQ โ (i) (Since PR > PQ as angle opposite to the larger side is always larger)

PSR = PQR + QPS โ (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)

PSQ = PRQ + RPS โ (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)

By adding (i) and (ii)

PQR +QPS > PRQ +RPS

Thus, from (i), (ii), (iii) and (iv), we get

PSR > PSQ

**6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.**

**Solution:**

First, let โ*l*โ be a line segment and โBโ be a point lying on it. A line AB perpendicular to *l* is now drawn. Also, let C be any other point on *l*. The diagram will be as follows:

**To prove:**

AB < AC

**Proof:**

In ฮABC, B = 90ยฐ

Now, we know that

A+B+C = 180ยฐ

โด A +C = 90ยฐ

Hence, C must be an acute angle which implies C < B

So, AB < AC (As the side opposite to the larger angle is always larger)

## Summary of NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Triangle is a part of Geometry. However, the complete Geometry of Class 9 constitutes a weightage of 22 marks out of 80 marks. Have a look at the types of questions that are expected from Geometry in the annual exam of class 9 Maths paper.

NCERT Class 9 Maths Geometry โ Marks Distribution | |||

Topics | Multiple Choice Questions | Short Questions | Long Questions |

Introduction to Euclidโs Geometry, Lines and Angles, Triangles, Quadrilaterals, Areas, Circles, Constructions | 4 Qs of 1 mark each | 2 Qs of 3 marks each | 2 Qs of 6 marks each |

Total Marks | 4 Marks | 6 Marks | 12 Marks |

Students should practice all the questions from the exercise to score high marks in the Class 9 Maths paper. The step by step solution to all the exercises of NCERT Class 9 Maths Chapter 7 is provided below:

In previous classes, students must have used the properties of triangles for solving the questions, but now in the **NCERT Textbook** of Class 9 Maths Chapter 7, students will learn how to prove these properties. For solving the questions related to this topic, it is very necessary that students should know the congruence rule. So, firstly go through the theory and then look at the solved examples that are already there in the book. After that, start solving the exercise problems.

Students can also have a look at the **NCERT Solutions of Class 9 **for Science subject to know the answers of all the chapters with a detailed explanation.

### Important Concepts Learned in NCERT Class 9 Maths Chapter 7 โ Triangles

The aim of including this Triangle Chapter in Class 9 Maths NCERT textbook is to make students know the following concepts:

- Congruence of triangles.
- Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).
- Properties of triangles.
- Inequalities of triangle.

We hope this information on โ**NCERT Solution Class 9 Maths Chapter 7**โ is useful for students. Click on NCERT Solutions to get the solved answers of the NCERT book for all the classes. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos download BYJUโS App and subscribe to YouTube Channel.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 7

### List out the important topics covered in NCERT Solutions for Class 9 Maths Chapter 7.

The important concepts covered in NCERT Solutions for Class 9 Maths Chapter 7 are

1. Congruence of triangles.

2. Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).

3. Properties of triangles.

4. Inequalities of triangles.

### What is the meaning of congruence of triangles in NCERT Solutions for Class 9 Maths Chapter 7?

According to NCERT Solutions for Class 9 Maths Chapter 7congruence of triangles states that two triangles are said to be congruent if they are copies of each other and when superposed, they cover each other exactly. In other words, two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.

### How is NCERT Solutions for Class 9 Maths Chapter 7 helpful for board exam preparation?

Practising NCERT Solutions for Class 9 Maths Chapter 7 help the students to top the final exams and ace a subject. These solutions are devised, based on the most updated syllabus, covering all the crucial topics of the respective subjects. Hence, solving these questions will make the students more confident to face the board exams. Topics given in these solutions form the basis for top scores. It also helps the students to get familiar with answering questions of all difficulty levels. These solutions are highly recommended to the students for referencing and to practice for the board exams.