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# NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 - Triangles

Last updated date: 05th Aug 2024
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## NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 - FREE PDF Download

NCERT Solutions for Maths Chapter 7 Exercise 7.2 Class 9 PDF contains all chapter exercises in one place prepared by an expert teacher as per NCERT (CBSE) book guidelines. Class 9 Maths Ex 7.2 specifically focuses on understanding and applying various properties and theorems related to triangles. This exercise is designed to enhance your comprehension of concepts such as the congruence of triangles, criteria for congruence, and the properties of different types of triangles.

Table of Content
1. NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 - FREE PDF Download
2. Glance of NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 | Vedantu
3. Access PDF for Maths NCERT Chapter 7 Triangles Ex 7.2 Class 9
4. Class 9 Maths Chapter 7: Exercises Breakdown
5. CBSE Class 9 Maths Chapter 7 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs

## Glance of NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 | Vedantu

• This exercise covers the properties of triangles, including congruence criteria (SAS, ASA, SSS, and RHS), relations between sides and angles (including angles and opposite sides), and special types of triangles (isosceles, equilateral).

• In an isosceles triangle, the angles opposite the equal sides are equal.

• Conversely, if two angles in a triangle are equal, the sides opposite those angles are also equal.

• The sum of the angles in a triangle is always 180 degrees.

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NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 - Triangles
TRIANGLES L-1 (𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞 𝐨𝐟 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬 & 𝐂𝐫𝐢𝐭𝐞𝐫𝐢𝐚 𝐟𝐨𝐫 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞 𝐨𝐟 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬) CBSE 9 Maths -Term 1
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## Access PDF for Maths NCERT Chapter 7 Triangles Ex 7.2 Class 9

Exercise 7.2

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

(i) $\text{OB}=\text{OC}$

(ii) $AO$ bisects $\angle \text{A}$

Ans:

(i) It is given that in triangle $\text{ABC}$ is an isosceles triangle with $\text{AB}=\text{AC}$.

$\therefore \,\,\angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are equal)

Multiplying $\frac{1}{2}$ on both sides, we get

$\frac{1}{2}\angle \text{ACB}=\frac{1}{2}\angle \text{ABC}$

$\angle \text{OCB}=\angle \text{OBC}$

$\therefore \text{OB}=\text{OC}$ (Sides opposite to equal angles of a triangle are also equal)

(ii) In $\Delta \text{OAB}$ and $\Delta \text{OAC}$,

Side AO is present in both triangle. Therefore,

$\text{AO}=\text{AO}$ (Common)

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\text{OB}=\text{OC}$ (Proved above)

Therefore,

$\Delta \text{OAB}\cong \Delta \text{OAC}$ (By SSS axiom of congruency)

Therefore, $\angle \text{BAO}=\angle \text{CAO}$.

As they are corresponding parts of the corresponding triangle.

$\therefore$ AO bisects $\angle \text{A}$.     ($\angle \text{BAO}=\angle \text{CAO}$)

2. In $\Delta \text{ABC},\text{AD}$ is the perpendicular bisector of $\text{BC}$ (see the given figure). Show that $\vartriangle \text{ABC}$ is an isosceles triangle in which $\text{AB}=\text{AC}$.

Ans:

In $\Delta \text{ADC}$ and $\Delta \text{ADB}$,

Side AD is present in both triangle.

$\therefore \,\,\text{AD}=\text{AD}$ (Common)

Since AD is the perpendicular bisector of BC.

$\angle ADC=\angle ADB={{90}^{{}^\circ }}$ and

$\text{CD}=\text{BD}$

$\therefore \Delta \text{ADC}\cong \Delta \text{ADB}$ (By SAS axiom of congruency)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

Hence, $\Delta ABC$ is an isosceles triangle as two of it’s sides are equal, i.e., $AB=AC$.

3. $\text{ABC}$ is an isosceles triangle in which altitudes $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$ respectively (see the given figure). Show that these altitudes are equal.

Ans:

In $\Delta \text{AEB}$ and $\Delta \text{AFC}$,

Since $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$. Therefore,

$\angle \text{AEB}$ and $\angle \text{AFC}$ (Each ${{90}^{{}^\circ }}$)

As $\angle A$ lies in both the triangles. Therefore,

$\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side AB is equal to AC, i.e.,$\text{AB}=\text{AC}$ (Given)

$\therefore \Delta \text{AEB}\cong \Delta \text{AFC}$ (By AAS axiom of congruency)

Therefore, $\text{BE}=\text{CF}$.

As they are corresponding parts of the corresponding triangle.

4.$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$ are equal (see the given figure). Show that

(i) $\Delta \text{ABE}\cong \Delta \text{ACF}$

(ii) $\text{AB}=\text{AC}$, ie., $\text{ABC}$ is an isosceles triangle.

Ans:

(i) In $\Delta \text{ABE}$ and $\Delta \text{ACF}$,

$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$.

Therefore, $\angle \text{ABE}$ and $\angle ACF={{90}^{{}^\circ }}$

Since $\angle A$ is present in both triangles,

$\therefore \,\,\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side BE is equal to CF, i.e., $\text{BE}=\text{CF}$ (Given)

$\therefore \Delta \text{ABE}\cong \Delta \text{ACF}$ (By AAS axiom of congruency)

(ii) Since $\Delta \text{ABE}\cong \Delta \text{ACF}$   (proved above)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

5. $\text{ABC}$ and $\text{DBC}$ are two isosceles triangles on the same base $\text{BC}$ (see the given figure). Show that $\angle \text{ABD}=\angle \text{ACD}$.

Ans:

Construction: join $\text{AD}$.

In $\Delta \text{ABD}$ and $\Delta \text{ACD}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

Length of side BD is equal to CD, i.e., $\text{BD}=\text{CD}$ (Given)

As side AD is present in both triangles. Therefore,

$\text{AD}=\text{AD}$ (Common side)

$\therefore \Delta \text{ABD}\cong \Delta \text{ACD}$ (By SSS axiom of congruency)

Therefore, $\angle \text{ABD}=\angle \text{ACD}$.

As they are corresponding parts of the corresponding triangle.

6. $\Delta \text{ABC}$ is an isosceles triangle in which $\text{AB}=\text{AC}$. Side $\text{BA}$ is produced to $\text{D}$ such that $\text{AD}=\text{AB}$ (see the given figure). Show that $\angle \text{BCD}$ is a right angle.

Ans:

In $\Delta \text{ABC}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\therefore \angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{ACD}$,

Length of side AC is equal to AD, i.e., $\text{AC}=\text{AD}$  (Given)

$\therefore \angle \text{ADC}=\angle \text{ACD}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{BCD}$,

$\angle \text{ABC}+\angle \text{BCD}+\angle \text{ADC}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{ACB}+\angle \text{ACB}+\angle \text{ACD}+\angle \text{ACD}={{180}^{{}^\circ }}$  ($\angle BCD=\angle BCA+\angle ACD$)

$2(\angle \text{ACB}+\angle \text{ACD})={{180}^{{}^\circ }}$

$2(\angle \text{BCD})={{180}^{{}^\circ }}$

$\therefore \angle \text{BCD}={{90}^{{}^\circ }}$

7. $\text{ABC}$ is a right-angled triangle in which $\angle \text{A}={{90}^{{}^\circ }}$ and $\text{AB}=\text{AC}$. Find $\angle \text{B}$ and $\angle \text{C}$.

Ans:

It is given that the side $\text{AB}=\text{AC}$

Angles opposite to equal sides are also equal

$\therefore \angle \text{C}=\angle \text{B}$  …(i)

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$  …from equation (i)

${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{B}={{180}^{{}^\circ }}$

$2\angle \text{B}={{90}^{{}^\circ }}$

$\angle \text{B}={{45}^{{}^\circ }}$

\begin{align} & \because \angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,\,\,\,...\text{from equation}\,\left( i \right) \\ & \therefore \angle \text{C}={{45}^{{}^\circ }} \\ \end{align}

8. Show that the angles of an equilateral triangle are ${{60}^{{}^\circ }}$ each.

Ans:

Consider that $\text{ABC}$ is an equilateral triangle in which all sides are equal.

Therefore, $\text{AB}=\text{BC}=\text{AC}$

$\text{AB}=\text{AC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{C}=\angle \text{B}\,\,\,\,\,\,...\left( i \right)$

Also, $\text{AC}=\text{BC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{B}=\angle \text{A}\,\,\,\,\,\,\,\,\,...\left( ii \right)$

Therefore, from equation (i) and (ii), we get

$\angle \text{A}=\angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,...\left( iii \right)$

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$   (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{A}+\angle \text{A}+\angle \text{A}={{180}^{{}^\circ }}$     …from equation (iii)

$3\angle \text{A}={{180}^{{}^\circ }}$

$\angle \text{A}={{60}^{{}^\circ }}$

$\therefore \angle \text{A}=\angle \text{B}=\angle \text{C}={{60}^{{}^\circ }}$

From above we can conclude that all the interior angles of an equilateral triangle has an angle of ${{60}^{{}^\circ }}$.

## Conclusion

In Class 9 Maths Ch 7 Ex 7.2 it is important to understand and focus on the properties and criteria for triangle congruence, such as SAS (Side-Angle-Side), ASA (Angle-Side-Angle), and SSS (Side-Side-Side). These concepts are crucial for solving problems related to triangle congruence. Pay attention to how these criteria are applied to different questions to prove that two triangles are congruent. Understanding these principles will help you tackle various geometric problems and establish a solid foundation for future studies in geometry.

## Class 9 Maths Chapter 7: Exercises Breakdown

 Chapter 7 - Triangles Exercises in PDF Format Exercise 7.1 8 Questions (6 Short Answer Questions, 2 Long Answer Questions) Exercise 7.3 5 Questions (3 Short Answer Questions, 2 Long Answer Questions)

## CBSE Class 9 Maths Chapter 7 Other Study Materials

Other than Maths Class 9 Chapter 7 Exercise 7.2  you can also check on the additional study materials provided for Class 9 Maths Chapter 7.

## Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 - Triangles

1. How many exercises are made available for class 9 Maths Chapter 7?

Triangles is one of the most important chapters in the Class 9 syllabus. These form the basis for further chapters regarding trigonometry and related chapters. Therefore students must ensure that they know all the basic concepts of triangles in detail.

Following are the exercises provided in this chapter for the benefit of the students.

• Exercise 7.1 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

Students must ensure that they solve all the above in order to perform better in the examination.

2. What are the important concepts covered in the Triangle chapters?

Students need to learn a lot of concepts in order to be able to solve and understand concepts in higher classes. Following are the concepts that you will learn in this chapter:

• 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 the triangle.

These topics help students understand the concept of the triangle better. Understanding triangles and its properties are very important as it carries high marks percentage in the exam paper as well. Therefore students must ensure that they solve all the exercises in order to perform better in the exam.

3. What are the marks distribution for the Triangle chapter in class 9 syllabus?

The marks distribution for the triangles chapter in class 9 examination is as follows:

• Topics: Introduction to Euclid’s Geometry, Lines and Angles, Triangles, Quadrilaterals, Areas, Circles, Constructions

• Multiple Choice Questions - 4 Qs of 1 mark each

• Short Answer Questions - 2 Qs of 3 marks each

• Long Answer Questions - 2 Qs of 6 marks each

You must not just mug up the theorems and equations related to triangles. You must also be able to apply the concepts and use them to solve all kinds of triangle problems. Also, students must learn to prove the theorems as they will also be asked to prove the same in the exam paper.

4. What are the theorems covered in the class 9 maths chapter 7 exercise 7.2?

The exercise of class 9 maths chapter 7 exercise 7.2 covers the properties of Triangles along with the following theorems:

• Theorem 7.2: Angles opposite to equal sides of an isosceles triangle are equal.

• Theorem 7.3: The sides opposite to equal angles of a triangle are equal.

These theorems can be proved by the congruence rule that students have learnt in the previous exercise.  Therefore students must ensure that they learn all the theorems in detail. Also, they need to learn problems regarding these theorems. In case they are unable to understand anything, they can always refer to the NCERT solutions provided here, by Vedantu. Vedantu is the top online education platform in India. These solutions are prepared by subject matter experts in a simplified and easy manner in order to make the process of studying easy for the students. Apart from NCERT solutions, they also provide solved previous years question papers, sample papers and related syllabus to aid students in their study process.