NCERT Solutions for Exercise 7.3 Class 9 Maths Chapter 7

NCERT Solutions for Exercise 7.3 Class 9 Maths Chapter 7 - Triangles

Edited By Vishal kumar | Updated on Oct 07, 2023 09:21 AM IST

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3- In this chapter of NCERT Solutions, we explore triangles and their properties, and Exercise 7.3 focuses on practical problem-solving using these concepts. Our free PDF 9th class maths exercise 7.3 answers break down complex concepts, making it easier for students to understand and practice. These resources are designed to help your learning and are accessible at your convenience. Whether you need to revise, prepare for exams, or strengthen your grasp of triangle properties, these NCERT class 9 maths chapter 7 exercise 7.3 solutions are here to assist you on your academic journey. Along with NCERT book Class 9 Maths, chapter 7 exercise 7.3 the following exercises are also present.

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3- Download Free PDF
NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.3
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Key Features of Exercise 7.3 Class 9 Maths
NCERT Solutions of Class 10 Subject Wise
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NCERT Solutions for Exercise 7.3 Class 9 Maths Chapter 7 - Triangles

NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.3

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Access Triangles Class 9 Chapter 7 Exercise: 7.3

Q1 (i) $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that $\small \Delta ABD\cong \Delta ACD$

1640157485851

Answer:

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

(i) $AD\ =\ AD$ (Common)

(ii) $AB\ =\ AC$ (Isosceles triangle)

(iii) $BD\ =\ CD$ (Isosceles triangle)

Thus by SSS congruency we can conclude that :

$\small \Delta ABD\cong \Delta ACD$

Q1 (ii) Triangles ABC and Triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that $\small \Delta ABP \cong \Delta ACP$

1640157528618

Answer:

Consider $\Delta ABP$ and $\Delta ACP$ ,

(i) $AP$ is common side in both the triangles.

(ii) $\angle PAB\ =\ \angle PAC$ (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii) $AB\ =\ AC$ (Isosceles triangles)

Thus by SAS axiom, we can conclude that :

$\small \Delta ABP \cong \Delta ACP$

Q1 (iii) $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP bisects $\small \angle A$ as well as $\small \angle D$ .

1640157546671

Answer:

In the first part, we have proved that $\small \Delta ABD\cong \Delta ACD$ .

So, by c.p.c.t. $\angle PAB\ =\ \angle PAC$ .

Hence AP bisects $\angle A$ .

Now consider $\Delta BPD$ and $\Delta CPD$ ,

(i) $PD\ =\ PD$ (Common)

(ii) $BD\ =\ CD$ (Isosceles triangle)

(iii) $BP\ =\ CP$ (by c.p.c.t. from the part (b))

Thus by SSS congruency we have :

$\Delta BPD\ \cong \ \Delta CPD$

Hence by c.p.c.t. we have : $\angle BDP\ =\ \angle CDP$

or AP bisects $\angle D$ .

Q1 (iv) $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP is the perpendicular bisector of BC.

1640157712234

Answer:

In the previous part we have proved that $\Delta BPD\ \cong \ \Delta CPD$ .

Thus by c.p.c.t. we can say that : $\angle BPD\ =\ \angle CPD$

Also, $BP\ =\ CP$

SInce BC is a straight line, thus : $\angle BPD\ +\ \angle CPD\ =\ 180^{\circ}$

or $2\angle BPD\ =\ 180^{\circ}$

or $\angle BPD\ =\ 90^{\circ}$

Hence it is clear that AP is a perpendicular bisector of line BC.

Q2 (i) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$ . Show that AD bisects BC

Answer:

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

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

(ii) $AD\ =\ AD$ (Common in both triangles)

(iii) $\angle ADB\ =\ \angle ADC\ =\ 90^{\circ}$

Thus by RHS axiom we can conclude that :

$\Delta ABD\ \cong \ \Delta ACD$

Hence by c.p.c.t. we can say that : $BD\ =\ CD$ or AD bisects BC.

Q2 (ii) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$ . Show that AD bisects $\small \angle A$ .

Answer:

In the previous part of the question we have proved that $\Delta ABD\ \cong \ \Delta ACD$

Thus by c.p.c.t., we can write :

$\angle BAD\ =\ \angle CAD$

Hence $AD$ bisects $\angle A$ .

Q3 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\small \Delta PQR$ (see Fig). Show that:

(i) $\small \Delta ABM \cong \Delta PQN$

(ii) $\small \Delta ABC \cong \Delta PQR$

1640157733376

Answer:

(i) From the figure we can say that :

$BC\ =\ QR$

or $\frac{1}{2}BC\ =\ \frac{1}{2}QR$

or $BM\ =\ QN$

Now, consider $\Delta ABM$ and $\Delta PQN$ ,

(a) $AM\ =\ PN$ (Given)

(b) $AB\ =\ PQ$ (Given)

Thus by SSS congruence rule, we can conclude that :

$\small \Delta ABM \cong \Delta PQN$

(ii) Consider $\Delta ABC$ and $\Delta PQR$ :

(a) $AB\ =\ PQ$ (Given)

(b) $\angle ABC\ =\ \angle PQR$ (by c.p.c.t. from the above proof)

Thus by SAS congruence rule,

$\small \Delta ABC \cong \Delta PQR$

Q4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:

Using the given conditions, consider $\Delta BEC$ and $\Delta CFB$ ,

(i) $\angle BEC\ =\ \angle CFB$ (Right angle)

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

(iii) $BE\ =\ CF$ (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that : $\Delta BEC\ \cong \Delta CFB$

Hence by c.p.c.t., $\angle B\ =\ \angle C$

And thus $AB\ =\ AC$ (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

Q5 ABC is an isosceles triangle with $\small AB=AC$ . Draw $\small AP\perp BC$ to show that $\small \angle B=\angle C$ .

Answer:

Consider $\Delta ABP$ and $\Delta ACP$ ,

(i) $\angle APB\ =\ \angle APC\ =\ 90^{\circ}$ (Since it is given that AP is altitude.)

(ii) $AB\ =\ AC$ (Isosceles triangle)

(iii) $AP\ =\ AP$ (Common in both triangles)

Thus by RHS axiom we can conclude that :

$\Delta ABP\ \cong \Delta ACP$

Now, by c.p.c.t.we can say that :

$\angle B\ =\ \angle C$

NCERT Solutions for Class 9 Maths Exercise 7.3 -The SSS congruence principle expresses that if three sides of one triangle are equivalent to those of another triangle, the two triangles are congruent to one other.

The RHS Congruence Rule explained in exercise 7.3 Class 9 Maths expresses that two right triangles are congruent if the hypotenuse and one side of one triangle are equivalent to the hypotenuse and one side of the other triangle.

A point that is equidistant from two given points is found on the perpendicular bisector of the line segment that connects them, and vice versa. On the bisectors of the angles formed by the two lines, a point equidistant from two intersecting lines lies.

Benefits of NCERT Solutions for Class 9 Maths Exercise 7.3

Exercise 7.3 Class 9 Maths, is based on RHS and SSS criteria of congruent triangles.
Understanding the concepts from Class 9 Maths chapter 7 exercise 7.3 will allow us to understand the new concepts which are going to be introduced in exercise 7.4

Key Features of Exercise 7.3 Class 9 Maths

Step-by-Step Solutions: The 9th class maths exercise 7.3 answers offers step-by-step solutions to guide students through solving problems, promoting a clear understanding of the concepts.
Varied Difficulty Levels: Problems in this class 9 maths ex 7.3 come with varying levels of difficulty, allowing students to gradually build their skills and confidence.
Preparation for Exams: Ex 7.3 class 9 helps students prepare for exams by offering a range of problems that may be encountered in assessments.
Consolidation of Knowledge: It serves as a tool to consolidate knowledge acquired in previous chapters related to triangles and their properties.

Also, See

NCERT Solutions of Class 10 Subject Wise

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Frequently Asked Questions (FAQs)

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 7.3?

This exercise includes new methods of congruency of triangles like right-angle hypotenuse side, side-side- side theorem and some properties.

2. Name the new criteria for congruence of two triangles in this exercise?

We are introduced to two new criteria of congruence:

The RHS (Right angle Hypotenuse Side) and the SSS (Side Side Side) congruence.

3. What do you understand by congruent triangles?

Two triangles are said to be congruent if all three corresponding sides of the two triangles are equal and all the three corresponding angles of the triangles are equal in measurement.

4. What do you understand by RHS criteria for congruence?

It expresses that if the hypotenuse and side of one right-calculated triangle are equivalent to the hypotenuse and the relating side of another right-angled triangle, then the two triangles are congruent.

5. What do you comprehend by SSS criteria for congruence?

It expresses that two triangles are congruent if three sides of a triangle are equivalent to the relating sides of the other triangle.

6. We have two triangles with angles in the ratio of 2 : 3 : 5 can they be congruent?

No, these triangles are similar, not congruent.

7. Name the different types of triangles?

The different types of triangles are:

Scalene Triangle.

Isosceles Triangle.

Equilateral Triangle.

8. Can an equilateral triangle be a right-angled triangle?

No, an equilateral triangle can never be a right-angled triangle as each angle of an equilateral triangle is 60 degrees.

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