## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.2) Exercise 6.2

NCERT Solutions for Class 7 Maths Chapter 6 are offered by Vedantu which are available in PDF form to help students get access to quality study materials. Our Maths NCERT Solutions are compiled by professional faculties to help you gain an in-depth knowledge of the topics concerned. Students can access these solutions to NCERT Maths Class 7 Exercise 6.2 for faster and reliable comprehension of information on Triangles and their properties. You can download NCERT Solutions PDF for their future reference in revision for exams. Register Online for NCERT Solutions Class 7 Science tuition on Vedantu.com to score more marks in CBSE board examination.

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## Download PDF of NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.2) Exercise 6.2

## Access NCERT Solutions for Class 7 Maths Chapter 6 – Triangles and its properties

Exercise 6.2

Refer to page 6 for exercise 6.2 in the PDF.

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

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

(Image Will Be Updated Soon)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 70^\circ + 50^\circ \\ $

$= 120^\circ \\ $

The value of \[x\] is $120^\circ $.

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

(Image Will Be Updated Soon)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 65^\circ + 45^\circ \\ $

$= 110^\circ \\ $

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

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

(Image Will Be Updated Soon)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$ x = 30^\circ + 40^\circ \\ $

$= 70^\circ \\ $

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

(iv) find the value of \[x\] in the following diagram.

(Image Will Be Updated Soon)

Hence,

$x = 60^\circ + 60^\circ \\ $

$= 120^\circ \\ $

The value of \[x\] is $120^\circ $.

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

(Image Will Be Updated Soon)

Hence,

$x = 50^\circ + 50^\circ \\ $

$= 100^\circ \\ $

The value of \[x\] is ${100^ \circ }$.

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

(Image Will Be Updated Soon)

Hence,

$x = 30^\circ + 60^\circ \\ $

$= 90^\circ \\ $

The value of \[x\] is $90^\circ $.

2. Find the value of the unknown exterior angle \[x\] in the following diagrams.

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

(Image Will Be Updated Soon)

Hence,

$x + 50^\circ = 115^\circ $

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

$x = 115^\circ - 50^\circ \\ $

$= 65^\circ \\ $

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

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

(Image Will Be Updated Soon)

Ans: According to the exterior angle property, the sum of the oppositenon-adjacent interior angles equals the exterior angle of a triangle.

$x + {70^ \circ } = 100^\circ $

Subtract $70^\circ $ from both sides of the equation.

$x = 100^\circ - 70^\circ \\ $

$= 30^\circ \\ $

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

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

(Image Will Be Updated Soon)

$x + 90^\circ = 125^\circ $

Subtract $90^\circ $ from both sides of the equation.

$ x + 90^\circ - 90^\circ = 125^\circ - 90^\circ \\ $

$x = 35^\circ \\ $

The value of \[x\] is $35^\circ $.

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

(Image Will Be Updated Soon)

$x + 60^\circ = 120^\circ $

Subtract $60^\circ $ from both sides of the equation.

$x + 60^\circ - 60^\circ = 120^\circ - 60^\circ \\ $

$x= 60^\circ \\ $

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

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

(Image Will Be Updated Soon)

$x + 30^\circ = 80^\circ $

Subtract $30^\circ $ from both sides of the equation.

$ x + 30^\circ - 30^\circ = 80^\circ - 30^\circ \\ $

$x= 50^\circ \\ $

The value of \[x\] is $50^\circ $.

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

(Image Will Be Updated Soon)

$x + 35^\circ = 75^\circ $

Subtract $35^\circ $ from both sides of the equation.

$ x + 35^\circ - 35^\circ = 75^\circ - 35^\circ \\ $

$ x= 40^\circ \\ $

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

## Class 7 Maths Chapter 6 Exercise 6.2 – In a Nutshell

The concept of Triangles is crucial for students and studying it from the initial years of high school builds a strong foundation of the fundamentals. It is an essential chapter for Class 7 students, and once you understand the basics of this chapter, it will help you to gain an in-depth idea of the topics after that effortlessly.

Besides, you can take help of our NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2, which have solutions to every question given in the exercises. We provide step by step solutions to every critical numerical along with several shortcut techniques for specific lengthy questions.

We always recommend you to follow a proper strategy while studying any particular chapter starting from the basics is one of them. Our NCERT Solutions to Class 7 Maths Ex 6.2 follows this rule and needs every student to start with the fundamentals and then gradually move on to the advanced numerical.

In the introductory section, we list out the various terminologies you are likely to come across while studying this chapter.

Next, we define each of them for your convenience.

Then, we provide diagrams to help you understand those topics and terms easily.

Pictures always aid in remembering information for a longer time than mere texts. Therefore, we try to present information in charts, tables or lists wherever possible in our NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2.

Students should be well versed with specific important properties like:

Sum of all interior angles of a triangle.

Types of angles - Alternate, corresponding, vertically opposite, adjacent.

Sum of two angles lying on the same plane.

Types of angles based on degrees – acute, obtuse and right-angled.

Sign conventions.

Parallel lines and transversal.

Comparison between interior and exterior angles.

Properties of angles in a triangle.

Sum of vertically opposite angles.

Concept of area and perimeter of any two-dimensional figure.

Difference between a triangle and a quadrilateral.

Calculation of area and perimeter of both triangles and other regular figures, for example, quadrilaterals like a rhombus.

Our NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2 aid in your understanding of these topics in a more straightforward manner. You can keep our notes downloaded in your devices as well so that you can have a look at them even on the go. Also, they are free of cost and available online at your convenience.

Features like these make our study materials stand out in the crowd. Also, with the presence of simple language for explanations, further eases the understanding of students.

### NCERT Solutions for Class 7 Maths Chapter 6 – All Questions

Class 7 Maths Exercise 6.2 – Question 1

Here is the exercise 6.2 explained in details as it holds the basis of the chapter triangles and explains most of the properties of a triangle too. It has six sets of figures here, in which one of the angles needs to be calculated.

You are required to have a sound knowledge of triangles before you proceed to solve these. To gain a streamlined learning experience, you can always refer to our study materials. They are specially designed to make learning easier and enjoyable.

Figure (i)

The first figure has an external angle denoted by ‘x’. You have to find the third angle of the triangle at first. Then you have to calculate its adjacent angle ‘x’.

Figure (ii)

It is same as the previous one, only that the figure is presented in a titled manner. Our solutions to Class 7 Maths Chapter 6.2 will help you have to solve such numerical that require a more application-based approach.

Figure (iii)

Alternate angles are equal, and the angle you are asked to determine falls under this criteria. So, you can learn this trick to identify this feature from our NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2 and obtain your answer.

Figure (iv) & (v)

Both of these figures are similar to the first two figures. You need to calculate the third angle of the triangle at first. Then the adjacent angle is calculated based on the property of triangles.

Figure (vi)

The last figure seems similar to the previous one. However, on a closer look, you will find that the third angle represents a right angle which is equal to 90o. To learn these tricks, refer to our NCERT Solutions for Class 7 Maths Chapter 6, Exercise 6.2 and save time during exams without compromising any marks.

### NCERT Class 7 Maths Chapter 6 Exercise 6.2 - Question 2

The second set of figures in question 2 requires students to be aware of the properties of triangles and their calculations. These below-mentioned figures are quite similar to the previous set and require you to follow the same procedure. However, you will need to find one of the interior angles in these cases

Figure (i)

Here, you find the adjacent angles at first. Then, based on the property of the sum of all angles, you should calculate the required angles represented by ‘x’.

You should know the fact that a triangle has three sides and that two adjacent sides form an angle. The sides may be regarded as vertices too. So, do not get confused between these two terms. Make sure you have enough clarity on the terminologies used with the help of our NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2.

Figure (ii)

On a closer look, you can see the adjacent angle to be right-angled. Refer to our Exercise 6.2 Class 7 NCERT solutions for learning shortcuts like how you can use the value of a right angle as 90o instead of deriving it every time during numerical.

With shortcut tricks at your bay, you can ace any exam at hand. It will help you develop the confidence required to handle an exam with so many questions at one go. Further, you can save time with lesser calculations and minimising the errors too. Also, you get a chance to boost your scores.

Figure (iii)

Here, you have to identify the sign convention for right angle in the figure. Based on that, you follow the process of calculating the adjacent angle first and then the other angle inside the triangle.

Our solutions have compactly listed the properties, and due to their online availability, you can download them and keep at your desk for easy reference whenever required.

Figure (iv)

It has a similar approach to the first question. Here, you find the angle adjacent to the given angle, both of which lie on the same base. It will help you to evaluate the third angle of the triangle.

Also, make sure you know about parallel lines and a transversal cutting them at different points to form angles. Students can also revise their idea about the properties to those angles from NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2 offered by Vedantu. It will aid you in scoring more in exams.

Figure (v)

This question again tests the knowledge of triangles and their properties. Sum of all the angles of a triangle is the most common and fundamental property you must be aware of. Students often tend to confuse with other figures that may not hold the same property to be true.

Besides, other figures have a difference in their number of sides as well, which is why the total measurement of all angles is also more. You can gain insights into these concepts from our Ex 6.2 Class 7 NCERT solutions and improve your hold on the concerned topic.

Figure (vi)

It is the last figure in this second set. This sum also requires a similar approach where students have to determine the adjacent angle at first, and then the other interior angle.

It is a common tendency to get confused during exams; therefore, it is crucial that you have a stronghold on the concepts from beforehand. It will minimise the risk of errors during examinations.

Avail our NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2 for shortcut techniques and standard study materials. Considering that such solutions have been penned by eminent teachers, that they are accurate to the ‘T’ is a guarantee.

### CBSE NCERT Books for Maths Class 7 – Other Exercises

You can also have a glance on other exercises included in this chapter. Here is a brief overview of those questions below.

Exercise 6.1

The first exercise has three questions:

First one is based on a diagram of a triangle provided, and you have to identify what each expression denotes.

The second question has three statements and asks you to draw figures according to that information.

The third question asks you to verify a theorem for an isosceles triangle with the help of a relevant diagram.

From our NCERT Solutions for Class 7 Maths Chapter 6

Exercise 6.2

you can have a clear idea on various types of a triangle – equilateral, isosceles and scalene along with their properties.

Exercise 6.3

The third exercise requires a similar approach as in the previous exercise. You are given twelve figures with an unknown angle in each. You have to calculate that based on various properties of a triangle.

Get to know the right tricks behind these from our solutions to Exercise 6.2 Class 7 Maths. The tricks will help you go a long way in your exams and also boost your scores.

Exercise 6.4

This exercise has six questions in all, and they are from a slightly higher perspective. Students have to deal with quadrilaterals here but have to find triangles from within them. After that, you need to apply the properties of triangles into appropriate use to determine the required answer.

Our NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2 will help you in deciphering this hidden information in the questions.

Exercise 6.5

It is the last exercise in this chapter and contains eight questions in total. Here, students will be tested for their knowledge in Pythagoras theorem as well as the fundamentals of triangles too.

Also, you will be asked about the perimeter of a rhombus. Hence, you can have a prior understanding of this kite-like figure from our NCERT solutions for Exercise 6.2 Class 7th to be able to solve these questions correctly.

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Our solutions have various shortcut techniques inculcated in them along with several short tricks and tips to guide students in apprehending the given information appropriately. These NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2 are based on the latest syllabus to ensure that students do not miss out on any latest inclusions. To ensure that you do not miss out on any essential aspects of a subject, you can avail our NCERT solutions and revise your learning from them before your exam. These solutions follow the latest syllabus for your convenience.

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