Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 7 Maths Chapter 6: The Triangle and its Properties - Exercise 6.3

ffImage
Last updated date: 25th Apr 2024
Total views: 571.2k
Views today: 12.71k
MVSAT offline centres Dec 2023

CBSE Class 7 NCERT Solutions Maths Chapter 6 Exercise 6.3

NCERT Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.3 is an extremely crucial chapter for any student. It marks the basis of your understanding of Geometry. The chapter deals with the different types of triangles and how they are differentiated from each other. NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.3 provided to you by Vedantu aims at making the chapter interesting to study. Download the Class 7 Chapter 6 Exercise 6.3 PDF now. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 7 Science, Maths solutions and solutions of other subjects.


Class:

NCERT Solutions for Class 7

Subject:

Class 7 Maths

Chapter Name:

Chapter 6 - The Triangle and Its Properties

Exercise:

Exercise - 6.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

You will learn about Triangles NCERT Maths Solution Class 7 Chapter 6 Exercise 6.3

Exercise 6.3

1. Find the value of \[x\] in the following diagrams.

i. Find the value of \[x\] in the following diagram.

Triangle ABC

 

Ans: The sum of the internal angles of a triangle is $180^\circ $. In $\vartriangle ABC$, $\angle BAC + \angle ACB + \angle ABC = 180^\circ $.

$ \Rightarrow x + 60^\circ  + 50^\circ  = 180^\circ  $

$ \Rightarrow x + 110^\circ  = 180^\circ  \\  $

Subtract $110^\circ $ from both sides and simplify.

$\Rightarrow x + 110^\circ  - 110^\circ  = 180^\circ  - 110^\circ  $ 

 

$\Rightarrow x = 70^\circ $

 

The value of \[x\] is $70^\circ $.

 

ii. Find the value of \[x\] in the following diagram.

Triangle PQR


Ans: The sum of the internal angles of a triangle is $180^\circ $. In $\vartriangle PQR$, $\angle RPQ + \angle PQR + \angle QRP = 180^\circ $.

 

$ \Rightarrow 90^\circ  + 30^\circ  + x = 180^\circ  $

 

$ \Rightarrow x + 120^\circ  = 180^\circ $

 

Subtract $120^\circ $ from both sides and simplify.

 

\[ \Rightarrow x + 120^\circ  - 120^\circ  = 180^\circ  - 120^\circ \]

 

\[ \Rightarrow x = 60^\circ  \]

 

The value of \[x\] is \[60^\circ \].

 

iii. Find the value of \[x\] in the following diagram.

Triangle XYZ

 

Ans: The sum of the internal angles of a triangle is $180^\circ $. In $\vartriangle XYZ$, $\angle ZXY + \angle XYZ + \angle YZX = 180^\circ $.

 

\[ \Rightarrow 30^\circ  + 110^\circ  + x = 180^\circ \]

 

  \[ \Rightarrow x + 140^\circ  = 180^\circ  \]

 

Subtract \[140^\circ \] from both sides and simplify.

 

\[\Rightarrow x + 140^\circ  - 140^\circ  = 180^\circ  - 140^\circ \] 

 

\[   \Rightarrow x = 40^\circ \]

 

The value of \[x\] is \[40^\circ \].

 

iv. Find the value of \[x\] in the following diagram.

Triangle with 50 degrees as one of the angle

 

Ans: The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ  + x + x = 180^\circ \].

 

Hence,

 

\[50^\circ  + 2x = 180^\circ \]

 

Subtract \[50^\circ \] from both sides and simplify.

 

\[\Rightarrow 50^\circ  - 50^\circ  + 2x = 180^\circ  - 50^\circ \] 

 

\[\Rightarrow 2x = 130^\circ \]

 

Divide both sides by 2 and simplify.

 

\[ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{130^\circ }}{2} \]

 

 \[\Rightarrow x = 65^\circ \]

 

The value of \[x\] is \[65^\circ \].

v. Find the value of \[x\] in the following diagram.

Triangle with x degrees

 

Ans: The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[x + x + x = 180^\circ \].

 

Hence,

 

\[3x = 180^\circ \]

 

Divide both sides by 3 and simplify.

 

\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{180^\circ }}{3} \]

 

\[ \Rightarrow x = 60^\circ   \]

 

The value of \[x\] is \[60^\circ \].


vi. Find the value of \[x\] in the following diagram.

Triangle where one angle is 90 degrees

 

Ans: The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[90^\circ  + x + 2x = 180^\circ \].

 

\[ \Rightarrow 90^\circ  + 3x = 180^\circ \]

 

Subtract \[90^\circ \] from both sides and simplify.

 

\[\Rightarrow 90^\circ  - 90^\circ  + 3x = 180^\circ  - 90^\circ  \]

 

\[\Rightarrow 3x = 90^\circ \]

 

Divide both sides by 3 and simplify.

 

\[\Rightarrow \dfrac{{3x}}{3} = \dfrac{{90^\circ }}{3} \]

 

 \[\Rightarrow x = 30^\circ  \]

 

The value of \[x\] is \[30^\circ \].


2. Find the value of \[x\] and $y$ in the following diagrams.

i. Find the value of \[x\] and $y$ in the following diagram.

Triangle where it's one its angle is 50 degrees

 

Ans: The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.

$x + 50^\circ  = 120^\circ $

 

Subtract $50^\circ $ from both sides of the equation.

 

$ \Rightarrow x = 120^\circ  - 50^\circ  $ 

 

$ \Rightarrow x = 70^\circ  $

 

The value of \[x\] is $70^\circ $.

 

The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ  + 70^\circ  + y = 180^\circ \].

 

\[ \Rightarrow 120^\circ  + y = 180^\circ \]

 

Subtract $120^\circ$ from both sides and simplify.

 

\[ \Rightarrow 120^\circ  + y - 120^\circ  = 180^\circ  - 120^\circ  \]

 

\[\Rightarrow y = 60^\circ   \]

 

The value of \[y\] is \[60^\circ \].


ii. Find the value of \[x\] and $y$ in the following diagram.

Triangle with unknown x and y angels

 

Ans: Since the vertical opposite angles are equal, $y = 80^\circ $.

 

The value of \[y\] is $80^\circ $.

 

The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ  + 80^\circ  + x = 180^\circ \].

 

\[ \Rightarrow 130^\circ  + x = 180^\circ \]

 

Subtract \[130^\circ \] from both sides and simplify.

 

\[ \Rightarrow 130^\circ  - 130^\circ  + x = 180^\circ  - 130^\circ \]

 

 \[  \Rightarrow x = 50^\circ   \]

 

The value of \[y\] is \[50^\circ \].

 

iii. Find the value of \[x\] and $y$ in the following diagram.

Triangle with 50,60 and y as its degrees

 

Ans: The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.

$50^\circ  + 60^\circ  = x$.

Hence,

$x = 110^\circ $

The value of \[x\] is $110^\circ $.

The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ  + 60^\circ  + y = 180^\circ \].

Hence,

 

\[110^\circ  + y = 180^\circ \]

 

Subtract $110^\circ $ from both sides and simplify.

 

\[\Rightarrow 110^\circ  + y - 110^\circ  = 180^\circ  - 110^\circ  \]

 

\[\Rightarrow y = 70^\circ  \]

 

The value of \[y\] is \[70^\circ \].

 

iv. Find the value of \[x\] and $y$ in the following diagram.

Triangle with 30,x and y as its internal angels

 

Ans: Since the vertical opposite angles are equal, $x = 60^\circ $.

 

The value of \[x\] is $60^\circ $.

 

The sum of the internal angles of a triangle is $180^\circ $. In the given triangle,

 

 \[\Rightarrow 60^\circ  + 30^\circ  + y = 180^\circ  \]

 

 \[ \Rightarrow 90^\circ  + y = 180^\circ  \].

 

Subtract \[{90^\circ }\] from both sides and simplify.

 

\[\Rightarrow 90^\circ  - 90^\circ  + y = 180^\circ  - 90^\circ \]

 

\[\Rightarrow y = 90^\circ   \]

 

The value of \[y\] is \[90^\circ \].

 

v. Find the value of \[x\] and $y$ in the following diagram.

Triangle where vertical opposite angles are equal

 

Ans: Since the vertical opposite angles are equal, $y = 90^\circ $.

 

The value of \[y\] is \[90^\circ \].

 

The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[x + x + 90^\circ  = 180^\circ \].

 

Hence,

 

\[90^\circ  + 2x = 180^\circ \]

 

Subtract \[90^\circ \] from both sides and simplify.

 

\[ \Rightarrow 90^\circ  + 2x - 90^\circ  = 180^\circ  - 90^\circ  \]

 

\[ \Rightarrow 2x = 90^\circ   \]

 

Divide both sides by 2 and simplify.

 

\[ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{90^\circ }}{2} \]

 

\[   \Rightarrow x = 45^\circ   \]

 

The value of \[x\] is \[45^\circ \].

 

vi. Find the value of \[x\] and $y$ in the following diagram.

Triangle where its all 3 angels are unknown

 

Ans: Since the vertical opposite angles are equal, $y = x$.

The sum of the internal angles of a triangle is $180^\circ $.

In the given triangle, \[x + x + x = 180^\circ \].

Hence,

\[3x = 180^\circ \]

 

Divide both sides by 3 and simplify.

 

\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{180^\circ }}{3} \]

 

 \[ \Rightarrow x = 60^\circ   \]

 

The value of \[x\] is \[60^\circ \] and the value of \[y\] is \[60^\circ \].

 

You will Learn about Triangles NCERT Maths Solution Class 7 Chapter 6 Exercise 6.3

Having a detailed knowledge about triangles is pivotal for your understanding of geometry. Crafted by CBSE, this chapter is extremely crucial for all your examination including the competitive ones. Once you know the differences between equilateral, isosceles, and scalene triangles and get a clear idea about their properties, geometry will become very easy for you. And Vedantu will provide you with all the necessary notes on this chapter.

Makes Studying Interesting

The NCERT Class 7 Maths Chapter 6 Exercise 6.3 notes prepared by Vedantu will help you get a thorough understanding of triangles and their properties. The questions and the solutions are presented in a smart way for smart learning. This will create a lot of difference during exams. You won’t need to go through the solutions again and again to remember them. Go thoroughly through them once or twice and try solving them yourself once. You are just done with the chapter.

Grows your Interest in the Subject

The way in which the notes by Vedantu on NCERT Solution for Class 7 Maths Chapter 6 Exercise 6.3 are presented to the students, is sure to churn their interest in the subject. Even if you don’t like the subject, Vedantu will make sure that you fall in love with it. And when it comes to one of the most interesting chapters that deal with Triangles, there is no way that you won’t grow your interest. This will directly help you in scoring high marks in the examination.

 

Rules Out the Fear

Once you have gone through the NCERT solution of Class 7 Maths Chapter 6 Exercise 6.3 your fear of Maths and Maths examination will be effectively ruled out. The way in which Vedantu presents the notes makes the learning process easy. Questions and their solutions are very clear and easy to understand. There is nothing new apart from the questions and solutions provided in the notes that can surprisingly pop up. Thus when you first look at the question paper, you will surely be thanking Vedantu’s notes. Download the NCERT Solutions Class 7 Chapter 6 Exercise 6.3 PDF now.

 

The Interactive Way of Teaching

Vedantu provides all its notes in an unparalleled way. They are highly effective and interactive for students. This boosts their interest in studying maths at an unmatched level. You will get to understand what exactly is being talked about here.


All Questions and Answers are Provided  

The questions and the solutions that are provided by Vedantu are enough for your exam preparation. If you go through all the solutions thoroughly it will help you get your dream score. 

 

Less Effort, More Output

Vedantu’s notes provide all the concepts along with the related problems and solutions. Students now don’t need to refer to several books to understand the concept that is taught in this chapter. This is the best solution for your pre-exam preparation. When you don’t have time and a lot of concepts and chapters to study, Vedantu can help.

Why Vedantu? 

There is no alternative to Vedantu’s notes when it comes to the quality of NCERT subject notes. All the notes of all the subjects are written to fulfil the basic requirement of understanding of the chapter.

 

Experts with years of experience in the arena prepare the notes for students. The notes help students understand the concepts better and grow an interest in the chapter along with the subject.

Vedantu knows exactly what students require and thus they provide them with the same.

You don’t need to go and spend hours at a tuition centre to learn anything anymore. With Vedantu the most difficult concepts of Maths can also be learned alone. The notes are self-explanatory for students’. It helps them learn and remember the concepts fast. They can now know what triangles are and what are their main properties by just downloading the PDF on Maths NCERT Solutions Class 7 Chapter 6 Exercise 6.3 for free. Yes, you read that right. Choose the right notes and become your own-well wisher. Download Vedantu’s NCERT Maths Class 7 Chapter 6 Exercise 6.3

FAQs on NCERT Solutions for Class 7 Maths Chapter 6: The Triangle and its Properties - Exercise 6.3

1. Where can I get NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.3?

NCERT solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.3 are available at Vedantu, India's leading online learning portal. In strict accordance with the most recent CBSE guidelines, The Triangle and its Properties Exercise 6.3 has been created by highly qualified and experienced teachers. These solutions include precise and comprehensive answers to every sum presented in the NCERT Maths textbook for Class 7. You can easily and for free download PDF versions of these study guides from Vedantu's official website (Vedantu.com). You can also get the Vedantu mobile app.

2. How many Questions are there in Chapter 6 The Triangle and its Properties Exercise 6.3  Class 7 Maths?

Chapter 6 The Triangle and its Properties Exercise 6.3 Class 7 Maths consists of two questions. Exercise 6.3 questions typically depend on the sum of a triangle's internal angles and its exterior angle property. You can consult Vedantu, India's most popular online portal if you're looking for NCERT solutions for Class 7 Math. At Vedantu, all of the chapter exercises are collected in one location and solved step-by-step by a qualified teacher in accordance with the NCERT book's instructions.

3. Is NCERT Maths Solution Class 7 Chapter 6 Exercise 6.3 important for the exams?

Your comprehension of geometry depends on how well you understand triangles. This chapter, which was created by the CBSE, is incredibly important for all of your exams, even competitive ones. Geometry will be very simple for you once you understand the distinctions between equilateral, isosceles, and scalene triangles and have a good understanding of their characteristics. Vedantu will also give you all the notes you require for this chapter.

4. Is choosing Vedantu for the NCERT Maths Solution Class 7 Chapter 6 Exercise 6.3 a good choice?

You may learn everything there is to know about triangles and their characteristics with the help of the notes Vedantu has created for NCERT Maths Solution Class 7 Chapter 6 Exercise 6.3. For smart learning, the questions and answers are presented in an intelligent manner. This will significantly alter how exams go. You won't have to review the answers repeatedly in order to remember them. Go over them thoroughly once or twice, and then attempt to solve one of the problems. You've just finished reading the chapter.

5. How can we grow our interest in studying class 7 maths?

Students' interest in the topic is bound to be piqued by the manner in which Vedantu's notes on the NCERT Solution for Class 7 Maths Chapter 6 Exercise 6.3 are presented to them. Vedantu will make sure you fall in love with the topic even if you don't like it. There is no way that your interest won't increase when it comes to one of the most fascinating chapters, which deals with triangles. You will directly benefit from this if you want to perform well on the test.