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

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.

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

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


Access Triangles Class 9 Chapter 7 Exercise: 7.1

Q1 In quadrilateral \small ABCD , \small AC=AD and \small AB bisects \small \angle A (see Fig.). Show that \small \Delta ABC\cong \Delta ABD . What can you say about \small BC and \small BD ?

1640157021241

Answer:

In the given triangles we are given that:-

(i) \small AC=AD

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

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

Hence by SAS congruence, we can say that : \small \Delta ABC\cong \Delta ABD

By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that BC\ =\ BD

Q2 (i) ABCD is a quadrilateral in which \small AD=BC and \small \angle DAB= \angle CBA (see Fig. ). Prove that

\small \Delta ABD\cong \Delta BAC .

1640157040178

Answer:

It is given that :-

(i) AD = BC

(ii) \small \angle DAB= \angle CBA

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

So, by SAS congruence, we can write :

\small \Delta ABD\cong \Delta BAC

Q2 (ii) \small ABCD is a quadrilateral in which \small AD=BC and \small \angle DAB=\angle CBA (see Fig.). Prove that \small BD=AC

1640157066047

Answer:

In the previous part, we have proved that \small \Delta ABD\cong \Delta BAC .

Thus by c.p.c.t. , we can write : \small BD=AC

Q2 (iii) \small ABCD is a quadrilateral in which \small AD=BC and \small \angle DAB= \angle BAC (see Fig.). Prove that \small \angle ABD= \angle BAC .

1640157095232

Answer:

In the first part we have proved that \small \Delta ABD\cong \Delta BAC .

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

\small \angle ABD= \angle BAC

Q3 \small AD and \small BC are equal perpendiculars to a line segment \small AB (see Fig.). Show that \small CD bisects \small AB .

1640157122932

Answer:

In the given figure consider \Delta AOD and \Delta BOC.

(i) AD = BC (given)

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

(iii) \angle AOD = \angle BOC (vertically opposite angles).

Thus by AAS Postulate, we have

\Delta AOD\ \cong \ \Delta BOC

Hence by c.p.c.t. we can write : AO\ =\ OB

And thus CD bisects AB.

Q4 \small l and \small m are two parallel lines intersected by another pair of parallel lines \small p and \small q (see Fig. ). Show that \small \Delta ABC\cong \Delta CDA .

1640157162694

Answer:

In the given figure, consider \Delta ABC and \Delta CDA :

(i) \angle\ BCA\ =\ \angle DAC

(ii) \angle\ BAC\ =\ \angle DCA

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

Thus by ASA congruence, we have :

\Delta ABC\ \cong \ \Delta CDA

Q5 (i) Line \small l is the bisector of an angle \small \angle A and B is any point on \small l . \small BP and \small BQ are perpendiculars from \small B to the arms of \small \angle A (see Fig.). Show that: \small \Delta APB\cong \Delta AQB

1640157183354

Answer:

In the given figure consider \small \Delta APB and \small \Delta AQB ,

(i) \angle P\ =\ \angle Q (Right angle)

(ii) \angle BAP\ =\ \angle BAQ (Since it is given that I is bisector)

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

Thus AAS congruence, we can write :

\small \Delta APB\cong \Delta AQB

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

1640157219645

Answer:

In the previous part we have proved that \small \Delta APB\cong \Delta AQB .

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

BP\ =\ BQ

Thus B is equidistant from arms of angle A.

Q6 In Fig, \small AC=AE,AB=AD and \small \angle BAD= \angle EAC . Show that \small BC=DE .

1640157243651

Answer:

From the given figure following result can be drawn:-

\angle BAD\ =\ \angle EAC

Adding \angle DAC to the both sides, we get :

\angle BAD\ +\ \angle DAC\ =\ \angle EAC\ +\ \angle DAC

\angle BAC\ =\ \angle EAD

Now consider \Delta ABC and \Delta ADE , :-

(i) AC\ =\ AE (Given)

(ii) \angle BAC\ =\ \angle EAD (proved above)

(iii) AB\ =\ AD (Given)

Thus by SAS congruence we can say that :

\Delta ABC\ \cong \ \Delta ADE

Hence by c.p.c.t., we can say that : BC\ =\ DE

Q7 (i) \small AB is a line segment and \small P is its mid-point. \small D and \small E are points on the same side of \small AB such that \small \angle BAD=\angle ABE and \small \angle EPA=\angle DPB (see Fig). Show that \small \Delta DAP\cong \Delta EBP

1640157265740

Answer:

From the figure, it is clear that :
\angle EPA\ =\ \angle DPB

Adding \angle DPE both sides, we get :

\angle EPA\ +\ \angle DPE =\ \angle DPB\ +\ \angle DPE

or \angle DPA =\ \angle EPB

Now, consider \Delta DAP and \Delta EBP :

(i) \angle DPA =\ \angle EPB (Proved above)

(ii) AP\ =\ BP (Since P is the midpoint of line AB)

(iii) \small \angle BAD=\angle ABE (Given)

Hence by ASA congruence, we can say that :

\small \Delta DAP\cong \Delta EBP

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 \small \angle BAD=\angle ABE and \small \angle EPA=\angle DPA (see Fig). Show that \small AD=BE

1640157291575

Answer:

In the previous part we have proved that \small \Delta DAP\cong \Delta EBP .

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

\small AD=BE

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 \small DM=CM . Point D is joined to point B (see Fig.). Show that: \small \Delta AMC\cong \Delta BMD

1640157315976

Answer:

Consider \Delta AMC and \Delta BMD ,

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

(ii) \angle CMA\ =\ \angle DMB (Vertically opposite angles are equal)

(iii) CM\ =\ DM (Given)

Thus by SAS congruency, we can conclude that :

\small \Delta AMC\cong \Delta BMD

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 \small DM=CM . Point D is joined to point B (see Fig.). Show that: \small \angle DBC is a right angle.

1640157364987

Answer:

In the previous part, we have proved that \small \Delta AMC\cong \Delta BMD .

By c.p.c.t. we can say that : \angle ACM\ =\ \angle BDM

This implies side AC is parallel to BD.

Thus we can write : \angle ACB\ +\ \angle DBC\ =\ 180^{\circ} (Co-interior angles)

and, 90^{\circ}\ +\ \angle DBC\ =\ 180^{\circ}

or \angle DBC\ =\ 90^{\circ}

Hence \small \angle DBC 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 \small DM=CM . Point D is joined to point B (see Fig.). Show that: \small \Delta DBC\cong \Delta ACB

1640157384655

Answer:

Consider \Delta DBC and \Delta ACB ,

(i) BC\ =\ BC (Common in both the triangles)

(ii) \angle ACB\ =\ \angle DBC (Right angle)

(iii) DB\ =\ AC (By c.p.c.t. from the part (a) of the question.)

Thus SAS congruence we can conclude that :

\small \Delta DBC\cong \Delta ACB

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 \small DM=CM . Point D is joined to point B (see Fig.). Show that: \small CM=\frac{1}{2}AB

1640162977049

Answer:

In the previous part, we have proved that

\Delta DBC\ \cong \ \Delta ACB .

Thus by c.p.c.t., we can write : DC\ =\ AB

DM\ +\ CM\ =\ AM\ +\ BM

or CM\ +\ CM\ =\ AB (Since M is midpoint.)

or \small CM=\frac{1}{2}AB .

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.

More About NCERT Solutions for Class 9 Maths Exercise 7.1

In NCERT solutions for Class 9 Maths exercise 7.1, we will now see types of triangles.

  • Scalene Triangle – A scalene triangle is a triangle with no two sides that are equal.
  • Equilateral Triangle - A triangle with all sides equal is known as an equilateral triangle.
  • Triangle with one right angle - A triangle with one right angle is known as a right-angled triangle.
  • A triangle's angles opposite equal sides are equal.
  • A triangle's sides opposite equal angles are equal.
  • In an equilateral triangle, each angle is 60 degrees.

Also Read| Triangles Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 7.1

  • Exercise 7.1 Class 9 Maths, is based on triangles

  • From Class 9 Maths chapter 7 exercise 7.1 we learn new techniques to find congruence among triangles

  • Understanding the concepts from Class 9 Maths chapter 7 exercise 7.1 will allow us to understand the concepts related to triangles

Key Features of 9th Class Maths Exercise 7.1 Answers

  1. Comprehensive Coverage: The class 9 maths chapter 7 exercise 7.1 solution cover the fundamental concepts related to triangles, including their properties and types.

  2. Expert-Crafted Solutions: The exercise 7.1 class 9 maths solutions are crafted by subject experts, ensuring accuracy and clarity in explanations.

  3. 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.

  4. 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.

  5. 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.
  6. Alignment with Curriculum: The content aligns with the Class 9 mathematics curriculum, ensuring that students are well-prepared for examinations and future mathematical topics.
  7. 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 Question (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°

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