NCERT Solutions for Maths Class 7 Chapter 6 Exercise 6.3 - FREE PDF Download
NCERT Class 7 Maths Chapter 6 Exercise 6.3 Solutions of, The Triangle and its Properties are designed to help students understand the fundamental concepts of triangles. This exercise focuses on important properties such as the angle sum property, types of triangles based on angles, and the criteria for congruence of triangles. These concepts are crucial for building a strong foundation in geometry.
- 3.1Exercise 6.3
The detailed NCERT Solutions for Class 7 Maths solutions provided by Vedantu’s expert teachers simplify complex problems, making it easier for students to follow and learn. Focus on understanding each property and how to apply them to different problems. Students’ understanding and problem-solving talents in this CBSE Class 7 Maths Syllabus can be improved by practicing these solutions.
Glance on NCERT Solutions Maths Chapter 6 Exercise 6.3 Class 7 | Vedantu
Class 7 Exercise 6.3, The Angle Sum Property of a Triangle simply means that when you add up all the angles inside any triangle, the total always equals 180 degrees.
This rule applies to all triangles, whether they are big or small, and helps solve problems involving angles in geometry.
Calculating unknown angles within a triangle based on given angles.
Understanding the relationship between the angles in different types of triangles.
Applying geometric principles to prove theorems and solve problems involving triangles.
Establishing a foundational understanding of geometric concepts in mathematics.
Providing a basis for further exploration into more complex geometrical figures and properties.
This article contains exercise notes, important questions, exemplar solutions, exercises, and video links for Exercise 6.3 - The Triangle and its Properties, which you can download as PDFs.
There are 2 fully solved questions in Class 7 Maths Chapter 6 Exercise 6.3 Solutions.
Exercise 6.3
1. Find the value of \[x\] in the following diagrams.
i.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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 \[60^\circ \].
Conclusion
NCERT Chapter 6 Class 7 Exercise 6.3 - The Triangle and Its Properties is necessary for understanding triangle properties. Learning the Angle Sum Property, which maintains that the interior angles of any triangle add up to 180 degrees, is important. This activity helps to improve geometry skills by solving triangle-related questions. Understanding these concepts is important for handling more advanced maths topics and performing well on tests. Vedantu's solutions provide detailed explanations, helping people understand and apply geometric ideas better. Download a FREE PDF for easy step-by-step solutions for Class 7 Exercise 6.3.
Class 7 Maths Chapter 6: Exercises Breakdown
CBSE Class 7 Maths Chapter 6 Other Study Materials
Chapter-Specific NCERT Solutions for Class 7 Maths
Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
Important Related Links for NCERT Class 7 Maths
Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.
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 Exercise 6.3 Solutions?
NCERT solutions for Class 7 Maths Chapter 6 Exercise 6.3 Solutions 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).
2. How many Questions are there in Class 7 Maths Chapter 6 Exercise 6.3 Solutions The Triangle and its Properties?
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 Class 7 Maths Exercise 6.3 Solutions 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 Class 7 Maths Exercise 6.3 Solutions 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 Maths Exercise 6.3 Solutions. 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 Exercise 6.3 Solutions?
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.
6. What is the Angle Sum Property of a triangle in Class 7 Exercise 6.3?
According to Class 7 Exercise 6.3, The Angle Sum Property states that the sum of the interior angles of any triangle is always 180 degrees.
7. How can you classify a triangle based on its angles in Class 7 Ex 6.3?
Triangles can be classified as acute (all angles less than 90 degrees), obtuse (one angle greater than 90 degrees), or right-angled (one angle exactly 90 degrees), for more refer NCERT Solutions for Class 7 Ex 6.3.
8. What are the exterior angles of a triangle according to Class 7 Ex 6.3?
In Class 7 Maths Ch 6 Ex 6.3, Exterior angles of a triangle are formed by extending one side of the triangle, and their sum is always 360 degrees.
9. How do you find the third angle of a triangle if you know the other two in Class 7 Maths Ch 6 Ex 6.3?
Subtract the sum of the two known angles from 180 degrees to find the measure of the third angle, use Class 7 Maths Ch 6 Ex 6.3 for solutions for these topics.
10. What is the difference between an equilateral and an isosceles triangle in Maths Class 7 Chapter 6 Exercise 6.3?
In Maths Class 7 Chapter 6 Exercise 6.3, an equilateral triangle has all three sides and angles equal, while an isosceles triangle has at least two sides and two angles equal.
11. How can you prove two triangles are congruent in Maths Class 7 Chapter 6 Exercise 6.3?
Two triangles are congruent if their corresponding sides and angles are equal in Maths Class 7 Chapter 6 Exercise 6.3.
12. Why is the Angle Sum Property important in Class 7 Maths Ex 6.3?
It helps in calculating unknown angles in triangles and forms the basis for many geometric proofs and theorems.
13. How does understanding triangle properties from Class 7 Maths Ex 6.3 help in real-life applications?
Knowing triangle properties from Class 7 Maths Ex 6.3 helps in fields like architecture, engineering, and navigation, where geometric principles are essential for design and calculations.
