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

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

Edited By Vishal kumar | Updated on Oct 06, 2023 02:32 PM IST

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

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2- Welcome to NCERT Solutions for Class 9 Maths, Chapter 7 - Triangles, Exercise 7.2. This exercise delves deeper into the subject, helping you master the intricacies of triangle geometry. In 9th class maths exercise 7.2 answers, we encounter a range of thought-provoking problems that challenge your understanding and problem-solving skills. Our expert-crafted solutions offer clear explanations and step-by-step guidance to assist you in tackling these challenges. Furthermore, you can easily download the PDF version of these class 9 maths chapter 7 exercise 7.2 for free, ensuring accessibility wherever and whenever you need them.

Along with NCERT book Class 9 Maths, chapter 9 exercise 7.2 the following exercises are also present.

## Access Triangles Class 9 Chapter 7 Exercise: 7.2

In the triangle ABC,

Since AB = AC, thus $\angle B\ =\ \angle C$

or $\frac{1}{2}\angle B\ =\ \frac{1}{2}\angle C$

or $\angle OBC\ =\ \angle OCB$ (Angles bisectors are equal)

Thus $\small OB=OC$ as sides opposite to equal are angles are also equal.

Consider $\Delta AOB$ and $\Delta AOC$ ,

(i) $AB\ =\ AC$ (Given)

(ii) $AO\ =\ AO$ (Common in both the triangles)

(iii) $OB\ =\ OC$ (Proved in previous part)

Thus by SSS congruence rule, we can conclude that :

$\Delta AOB\ \cong \ \Delta AOC$

Now, by c.p.c.t.,

$\angle BAO\ =\ \angle CAO$

Hence AO bisects $\angle A$ .

Consider $\Delta$ ABD and $\Delta$ ADC,

(i) $AD\ =\ AD$ (Common in both the triangles)

(ii) $\angle ADB\ =\ \angle ADC$ (Right angle)

(iii) <BD=<CD

Consider $\Delta AEB$ and $\Delta AFC$ ,

(i) $\angle A$ is common in both the triangles.

(ii) $\angle AEB\ =\ \angle AFC$ (Right angles)

(iii) $AB\ =\ AC$ (Given)

Thus by AAS congruence axiom, we can conclude that :

$\Delta AEB\ \cong \Delta AFC$

Now, by c.p.c.t. we can say : $BE\ =\ CF$

Hence these altitudes are equal.

Consider $\Delta ABE$ and $\Delta ACF$ ,

(i) $\angle A$ is common in both the triangles.

(ii) $\angle AEB\ =\ \angle AFC$ (Right angles)

(iii) $BE\ =\ CF$ (Given)

Thus by AAS congruence, we can say that :

$\small \Delta ABE \cong \Delta ACF$

From the prevoius part of the question we found out that : $\Delta ABE\ \cong \Delta ACF$

Now, by c.p.c.t. we can say that : $AB\ =\ AC$

Hence $\Delta \ ABC$ is an isosceles triangle.

Consider $\Delta ABD$ and $\Delta ACD$ ,

(i) $AD\ =\ AD$ (Common in both the triangles)

(ii) $AB\ =\ AC$ (Sides of isosceles triangle)

(iii) $BD\ =\ CD$ (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that :

$\small \angle ABD\ \cong \ \angle ACD$

Consider $\Delta$ ABC,
It is given that AB = AC

So, $\angle ACB = \angle ABC$ (Since angles opposite to the equal sides are equal.)

Similarly in $\Delta$ ACD,

and $\angle ADC = \angle ACD$
So,

$\angle CAB + \angle ACB + \angle ABC = 180^{\circ}$

$\angle CAB\ +\ 2\angle ACB = 180^{\circ}$
or $\angle CAB\ = 180^{\circ}\ -\ 2\angle ACB$ ...........................(i)

And in $\Delta$ ADC,
$\angle CAD\ = 180^{\circ}\ -\ 2\angle ACD$ ..............................(ii)

Adding (i) and (ii), we get :
$\angle CAB\ +\ \angle CAD\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

or $180^{\circ}\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

and $\angle BCD\ =\ 90^{\circ}$

In the triangle ABC, sides AB and AC are equal.

We know that angles opposite to equal sides are also equal.

Thus, $\angle B\ =\ \angle C$

Also, the sum of the interior angles of a triangle is $180^{\circ}$ .

So, we have :

$\angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}$

or $90^{\circ} +\ 2\angle B\ =\ 180^{\circ}$

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

Hence $\angle B\ =\ \angle C\ =\ 45^{\circ}$

Consider a triangle ABC which has all sides equal.

We know that angles opposite to equal sides are equal.

Thus we can write : $\angle A\ =\ \angle B\ =\ \angle C$

Also, the sum of the interior angles of a triangle is $180 ^{\circ}$ .

Hence, $\angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}$

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

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

So, all the angles of the equilateral triangle are equal ( $60^{\circ}$ ).

## More About NCERT Solutions for Class 9 Maths Exercise 7.2

In NCERT Solutions for exercise 7.2 class 9 maths, we will begin by understanding the definitions and important terms related to triangles. It's essential to revise the basics before diving into any topic. A triangle is a closed figure formed by three intersecting lines (the prefix "tri" signifies "three"). A triangle consists of three sides, three angles, and three vertices.

Triangle Congruence: Two triangles are considered congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other. When PQR and ABC are congruent, we denote it as PQR = ABC.

The NCERT syllabus for Class 9 Maths Exercise 7.2 covers different criteria for establishing the congruence of triangles, which include: Side-Angle-Side, Angle-Side-Angle, Side-Side-Side, and Right-angle-Hypotenuse-Side.

Theorem 1: In an isosceles triangle, the angles opposite the equal sides are equal.

Theorem 2: In a triangle, the sides opposite equal angles are equal.

### Some very important properties of triangles which we should know and are helpful in solving this exercise are:

• The entire angles of a triangle sum up to 180o.
• The sum of two inside opposite angles forms the external angle formed by a triangle's side.
• Angles opposing equal sides of a triangle are equal.
• The sides of a triangle that have equal angles are equal.
• Each of an equilateral triangle's angles is 60.

Also Read| Triangles Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 7.2

• Exercise 7.2 Class 9 Maths, is related to triangles.

• From Class 9 Maths chapter 7 exercise 7.2 introduces us to theorems related to isosceles triangles.

• Understanding the principles from Class 9 Maths Chapter 7 Exercise 7.2 will assist us in understanding the notions of congruency and isosceles triangle characteristics.

## Key Features of Class 9 Maths Chapter 7 Exercise 7.2

1. Comprehensive Coverage: The 9th class maths exercise 7.2 answers covers various aspects of triangles, including congruence criteria and theorems related to triangle properties.

2. Clear Explanations: The class 9 maths ex 7.2 solutions provide clear and concise explanations, making it easier for students to understand and apply the concepts.

3. Step-by-Step Approach: Each ex 7.2 class 9 problem is solved step by step, allowing students to follow the thought process and replicate it in similar problems.

4. Variety of Problems: The class 9 ex 7.2 includes a range of problems based on different congruence criteria, ensuring that students get a well-rounded practice.

5. Application of Theorems: The exercise 7.2 class 9 maths helps students apply theorems related to triangles, enhancing their problem-solving skills.

6. Conceptual Clarity: By solving these problems, students gain a deeper understanding of the properties and congruence criteria of triangles.

7. Aligned with Curriculum: The content aligns with the Class 9 mathematics curriculum, preparing students for examinations and future topics in geometry.

8. Free Resource: These solutions are available to students at no cost, making quality math resources accessible to all.

Also, See

## Subject Wise NCERT Exemplar Solutions

1. What is the definition of an isosceles triangle?

An isosceles triangle is one with two equal sides. It has two equal angles as well.

2. How many sides and angles of an isosceles triangle are equal?

An isosceles triangle has two sides and two angles equal to each other.

3. Can we consider an equilateral triangle as an isosceles triangle?

Yes, every equilateral triangle is an isosceles triangle, but every isosceles triangle is not an equilateral triangle.

4. Find the unequal side of a right-angled triangle with equal sides of 10cm?

Yes, every equilateral triangle is an isosceles triangle, but every isosceles triangle is not an equilateral triangle.

5. The ratio in which the angle bisector of the vertex angle divides the base is?

The angle bisector of the vertex angle divides the base into 1:1 ratio which means into two equal parts i.e., it passes through the midpoint of the base.

6. Find the equal angels of the isosceles triangle if the vertex angle is 80 degrees?

The angle bisector of the vertex angle divides the base into 1:1 ratio which means into two equal parts i.e., it passes through the midpoint of the base.

7. Find the angles of the triangles if they are in a ratio of 1 : 2 : 2 ?

The angle bisector of the vertex angle divides the base into 1:1 ratio which means into two equal parts i.e., it passes through the midpoint of the base.

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