NCERT Solutions for Exercise 7.1 Class 9 Maths Chapter 7

NCERT Solutions for Exercise 7.1 Class 9 Maths Chapter 7 - Triangles

Edited By Vishal kumar | Updated on Oct 06, 2023 10:31 AM IST

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1- Welcome to NCERT Solutions for 9th class maths exercise 7.1 answers. In this chapter, we'll explore the basic concept of triangles, one of the basic shapes in geometry. This class 9 maths chapter 7 exercise 7.1 helps you understand the properties of triangles and how to work with them. Our easy-to-follow solutions provide step-by-step explanations, and you can also download the free PDF version for offline use. Let's get started and learn about triangles together. Along with NCERT book Class 9 Maths, chapter 7 exercise 7.1 the following exercises are also present.

New: Get up to 90% Scholarship on NEET/JEE Coaching from top Coaching Institutes

NCERT Solutions for Exercise 7.1 Class 9 Maths Chapter 7 - Triangles

NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.1

Download PDF

Access Triangles Class 9 Chapter 7 Exercise: 7.1

Q1 In quadrilateral $A B C D$ , $A C = A D$ and $A B$ bisects $∠ A$ (see Fig.). Show that $Δ A B C ≅ Δ A B D$ . What can you say about $B C$ and $B D$ ?

1640157021241

Answer:

In the given triangles we are given that:-

(i) $A C = A D$

(ii) Further, it is given that AB bisects angle A. Thus $∠$ BAC $= ∠$ BAD.

(iii) Side AB is common in both the triangles. $A B = A B$

Hence by SAS congruence, we can say that : $Δ A B C ≅ Δ A B D$

By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that $B C = B D$

Q2 (i) $A B C D$ is a quadrilateral in which $A D = B C$ and $∠ D A B = ∠ C B A$ (see Fig. ). Prove that

$Δ A B D ≅ Δ B A C$ .

1640157040178

Answer:

It is given that :-

(i) AD = BC

(ii) $∠ D A B = ∠ C B A$

(iii) Side AB is common in both the triangles.

So, by SAS congruence, we can write :

$Δ A B D ≅ Δ B A C$

Q2 (ii) $A B C D$ is a quadrilateral in which $A D = B C$ and $∠ D A B = ∠ C B A$ (see Fig.). Prove that $B D = A C$

1640157066047

Answer:

In the previous part, we have proved that $Δ A B D ≅ Δ B A C$ .

Thus by c.p.c.t. , we can write : $B D = A C$

Q2 (iii) $A B C D$ is a quadrilateral in which $A D = B C$ and $∠ D A B = ∠ B A C$ (see Fig.). Prove that $∠ A B D = ∠ B A C$ .

1640157095232

Answer:

In the first part we have proved that $Δ A B D ≅ Δ B A C$ .

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

$∠ A B D = ∠ B A C$

Q3 $A D$ and $B C$ are equal perpendiculars to a line segment $A B$ (see Fig.). Show that $C D$ bisects $A B$ .

1640157122932

Answer:

In the given figure consider $Δ$ AOD and $Δ$ BOC.

(i) AD = BC (given)

(ii) $∠$ A = $∠$ B (given that the line AB is perpendicular to AD and BC)

(iii) $∠$ AOD = $∠$ BOC (vertically opposite angles).

Thus by AAS Postulate, we have

$Δ A O D ≅ Δ B O C$

Hence by c.p.c.t. we can write : $A O = O B$

And thus CD bisects AB.

Q4 $l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig. ). Show that $Δ A B C ≅ Δ C D A$ .

1640157162694

Answer:

In the given figure, consider $Δ$ ABC and $Δ$ CDA :

(i) $∠ B C A = ∠ D A C$

(ii) $∠ B A C = ∠ D C A$

(iii) Side AC is common in both the triangles.

Thus by ASA congruence, we have :

$Δ A B C ≅ Δ C D A$

Q5 (i) Line $l$ is the bisector of an angle $∠ A$ and B is any point on $l$ . $B P$ and $B Q$ are perpendiculars from $B$ to the arms of $∠ A$ (see Fig.). Show that: $Δ A P B ≅ Δ A Q B$

1640157183354

Answer:

In the given figure consider $Δ A P B$ and _{$Δ A Q B$} ,

(i) $∠ P = ∠ Q$ (Right angle)

(ii) $∠ B A P = ∠ B A Q$ (Since it is given that I is bisector)

(iii) Side AB is common in both the triangle.

Thus AAS congruence, we can write :

$Δ A P B ≅ Δ A Q B$

Q5 (ii) Line $l$ is the bisector of an angle $∠ A$ and $B$ is any point on $l$ . $B P$ and $B Q$ are perpendiculars from $B$ to the arms of $∠ A$ (see Fig. ). Show that: $B P = B Q$ or $B$ is equidistant from the arms of $∠ A$ .

1640157219645

Answer:

In the previous part we have proved that _{$Δ A P B ≅ Δ A Q B$} .

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

$B P = B Q$

Thus B is equidistant from arms of angle A.

Q6 In Fig, _{$A C = A E, A B = A D$} and $∠ B A D = ∠ E A C$ . Show that $B C = D E$ .

1640157243651

Answer:

From the given figure following result can be drawn:-

$∠ B A D = ∠ E A C$

Adding $∠ D A C$ to the both sides, we get :

$∠ B A D + ∠ D A C = ∠ E A C + ∠ D A C$

$∠ B A C = ∠ E A D$

Now consider $Δ A B C$ and $Δ A D E$ , :-

(i) $A C = A E$ (Given)

(ii) $∠ B A C = ∠ E A D$ (proved above)

(iii) $A B = A D$ (Given)

Thus by SAS congruence we can say that :

$Δ A B C ≅ Δ A D E$

Hence by c.p.c.t., we can say that : $B C = D E$

Q7 (i) $A B$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $A B$ such that $∠ B A D = ∠ A B E$ and $∠ E P A = ∠ D P B$ (see Fig). Show that $Δ D A P ≅ Δ E B P$

1640157265740

Answer:

From the figure, it is clear that :
$∠ E P A = ∠ D P B$

Adding $∠ D P E$ both sides, we get :

$∠ E P A + ∠ D P E = ∠ D P B + ∠ D P E$

or $∠ D P A = ∠ E P B$

Now, consider $Δ D A P$ and $Δ E B P$ :

(i) $∠ D P A = ∠ E P B$ (Proved above)

(ii) $A P = B P$ (Since P is the midpoint of line AB)

(iii) $∠ B A D = ∠ A B E$ (Given)

Hence by ASA congruence, we can say that :

$Δ D A P ≅ Δ E B P$

Q7 (ii) AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that $∠ B A D = ∠ A B E$ and $∠ E P A = ∠ D P A$ (see Fig). Show that $A D = B E$

1640157291575

Answer:

In the previous part we have proved that $Δ D A P ≅ Δ E B P$ .

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

$A D = B E$

Q8 (i) 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 $D M = C M$ . Point D is joined to point B (see Fig.). Show that: $Δ A M C ≅ Δ B M D$

1640157315976

Answer:

Consider $Δ A M C$ and $Δ B M D$ ,

(i) $A M = B M$ (Since M is the mid-point)

(ii) $∠ C M A = ∠ D M B$ (Vertically opposite angles are equal)

(iii) $C M = D M$ (Given)

Thus by SAS congruency, we can conclude that :

$Δ A M C ≅ Δ B M D$

Q8 (ii) 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 $D M = C M$ . Point D is joined to point B (see Fig.). Show that: $∠ D B C$ is a right angle.

1640157364987

Answer:

In the previous part, we have proved that $Δ A M C ≅ Δ B M D$ .

By c.p.c.t. we can say that : $∠ A C M = ∠ B D M$

This implies side AC is parallel to BD.

Thus we can write : $∠ A C B + ∠ D B C = 180^{\circ}$ (Co-interior angles)

and, $90^{\circ} + ∠ D B C = 180^{\circ}$

or $∠ D B C = 90^{\circ}$

Hence $∠ D B C$ is a right angle.

Q8 (iii) 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 $D M = C M$ . Point D is joined to point B (see Fig.). Show that: $Δ D B C ≅ Δ A C B$

1640157384655

Answer:

Consider $Δ D B C$ and $Δ A C B$ ,

(i) $B C = B C$ (Common in both the triangles)

(ii) $∠ A C B = ∠ D B C$ (Right angle)

(iii) $D B = A C$ (By c.p.c.t. from the part (a) of the question.)

Thus SAS congruence we can conclude that :

$Δ D B C ≅ Δ A C B$

Q8 (iv) 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 $D M = C M$ . Point D is joined to point B (see Fig.). Show that: $C M = \frac{1}{2} A B$

1640162977049

Answer:

In the previous part, we have proved that

$Δ D B C ≅ Δ A C B$ .

Thus by c.p.c.t., we can write : $D C = A B$

$D M + C M = A M + B M$

or $C M + C M = A B$ (Since M is midpoint.)

or $C M = \frac{1}{2} A B$ .

Hence proved.

NCERT Solutions for Class 9 Maths exercise 7.1 – Triangle: A triangle is a closed figure formed by three intersecting lines ('Tri' means 'three'). Three sides, three angles, and three vertices make up a triangle. In two triangles the triangles are said to be congruent if all three corresponding sides and angles are exactly equal. The corresponding parts of congruent triangles are equal and are written as CPCT (The corresponding part of the congruent triangle).

NCERT syllabus for Class 9 Maths exercise 7.1 includes criteria for congruence of triangles. SSS Criteria for Congruency-If the three sides of one triangle equal the three sides of another, the two triangles are congruent. If all sides are the same, then their corresponding angles must be the same as well.

From NCERT solutions for Class 9 Maths chapter 7 exercise 7.1 we get to learn SAS Criteria for Congruence-Two triangles are congruent if their two sides and included angle are the same as the corresponding sides and included angle of the other triangle.

ASA Criteria for Congruence-Two triangles are congruent if their two angles and included sides are equal to the other triangle's corresponding two angles and included side.

AAS Congruence Criteria-The two triangles are said to be congruent if two angles and one side of one triangle are equal to two angles and one side of the other triangle.

Key Features of 9th Class Maths Exercise 7.1 Answers

Comprehensive Coverage: The class 9 maths chapter 7 exercise 7.1 solution cover the fundamental concepts related to triangles, including their properties and types.
Expert-Crafted Solutions: The exercise 7.1 class 9 maths solutions are crafted by subject experts, ensuring accuracy and clarity in explanations.
Step-by-Step Explanations: Each class 9 maths ex 7.1 answer includes detailed step-by-step explanations to help students understand the concepts and solve problems with ease.
Clear and Easy Language: The ex 7.1 class 9 answers are presented in clear and easy-to-understand language, making them accessible to students of all levels.
Offline Accessibility: Students can download the PDF version of the class 9 ex 7.1 answers for offline use, making it convenient for study anytime, anywhere.
Alignment with Curriculum: The content aligns with the Class 9 mathematics curriculum, ensuring that students are well-prepared for examinations and future mathematical topics.
Free Resource: These exercise 7.1 class 9 maths answers are available to students at no cost, providing quality math resources to all.

Also, See

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 7.1 ?

This exercise is about TRIANGLES, congruence of triangles and different methods of congruency.

2. What is the definition of a triangle?

A triangle is a closed geometric object with three sides and three angles that has three sides and three angles.

3. How many right angles are possible in a triangle?

A maximum of one right angle is possible in a triangle, if it exceeds one it will nullify the criteria for triangle formation.

4. Mention all the criteria for two triangles to be congruent?

Side-Side-Side Criterion, Side-Angle-Side Criterion, Angle-Side-Angle Criterion, right angle-hypotenuse-side congruence.

5. Define angle side angle criteria of congruence?

Angle-Side-Angle is also called ASA criterion states that:

If two angles and the side which is included between them are equal for two triangles then the triangles are congruent.

6. What do you understand by the vertical opposite angle ?

When two lines intersect each other at a point, then the opposite angles formed due to this intersection are called vertically opposite angles. They are always equal to each other.

7. What is the Pythagoras theorem, exactly?

When two lines intersect each other at a point, then the opposite angles formed due to this intersection are called vertically opposite angles. They are always equal to each other.

8. Find the angles of the triangles if they are in a ratio of 1 : 2 : 3 ?

The angles are 30°, 60° and 90°

Also Read

NCERT Solutions For Class 9 Science Chapter 5 The Fundamental Unit Of Life

NCERT Syllabus for Class 9 Maths 2025-26 - Download Latest Syllabus PDF

NCERT Solutions for Class 9 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Exemplar Class 9 Science Solutions Chapter 7 Diversity in Living Organisms

NCERT Exemplar Class 9 Science Solutions Chapter 5 The Fundamental Unit of Life

NCERT Exemplar Class 9 Science Solutions Chapter 13 Why do we fall ill?

NCERT Exemplar Class 9 Science Solutions Chapter 6 Tissues

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

Articles

APRJC CET Syllabus 2025 – Check Subject-Wise Topics & Exam Pattern

Mar 27, 2025

Ophthalmology Courses After 12th - Eligibility Criteria, Duration, Institutes

Mar 27, 2025

How to Get a Scholarship in Resonance? – Eligibility, Exams & Benefits

Mar 26, 2025

KVS Lottery Result 2025-26 Class 1, Download KVS Class 1 Admission List PDF

Mar 26, 2025

BiPC Courses After 12th - Fees, Eligibility, Duration & Top Institutes

Mar 26, 2025

IB 2025 Exam Schedule: All exam zones (A, B, C)

Mar 24, 2025

IB Results 2025 – Check IB MYP & IB DP Program Scores and Updates

Mar 24, 2025

APRJC CET Application Form 2025 (Started) – Register Online for Admission

Mar 20, 2025

Upcoming School Exams

Central Board of Secondary Education Class 12th Examination

Admit Card Date:03 February,2025 - 04 April,2025

Exam Pattern

Admit Card

Result

Telangana State Secondary School Certificate Examination

Admit Card Date:07 March,2025 - 04 April,2025

Exam Pattern

Mock Test

Admit Card

Karnataka Secondary School Leaving Certificate Examination

Admit Card Date:10 March,2025 - 05 April,2025

Admit Card

Exam Pattern

Result

National Institute of Open Schooling 10th examination

Exam Date:17 March,2025 - 01 April,2025

Application Process

Exam Pattern

Mock Test

National Institute of Open Schooling 12th Examination

Exam Date:17 March,2025 - 01 April,2025

Application Process

Exam Pattern

Admit Card

View All School Exams

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is