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NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2

## NCERT Solutions for Class 10 Maths Chapter 6 - Triangles

We started learning about triangles right from the 1st standard. However, the concepts grew with every passing year. The CBSE Solutions of Maths exercise 6.2 Class 10 provides an overview of the concept of similarity of triangles. Along with this, chapter 6 of mathematics also uses the knowledge of Pythagoras theorem which the students learnt earlier. Experts at Vedantu provide you with the comprehensive ex 6.2 Class 10 Maths NCERT Solution to help you understand the concepts quickly and easily. Moreover, our maths experts have crafted the solutions well according to the latest CBSE syllabus and patterns. You can also download the NCERT Solution for Class 10 Science free PDFs on our website.

## NCERT Ch 6 Maths Class 10 Exercise 6.2

### Triangles Class 10 Exercise 6.2 PDF Download

There are several questions available in the NCERT textbook which the students can solve to become thorough with their concepts. It is important to continually practise mathematics to get hold of a variety of questions and understand it better. We provide you with Class 10 ex 6.2 solutions in a PDF downloadable format. This allows the students to study whenever they want without having to worry about stable internet connection or other issues.Â

### Class 10 Maths Exercise 6.2 Solution

This exercise of the chapter based on triangles consists of questions related to the similarity of triangles. It has a total of 10 questions with each one covering different fundamentals. To understand it better, let us walk you through the concepts covered in every question.Â

### The Solution of ex 6.2 Class 10

### Question 1:

The first question contains two triangles, each of them having the same condition as DE||BC. We need to find out the values of EC and AD in these respective figures. The students can find out the values easily by using the basic proportionality theorem and properties of similar triangles. For any further issues, you can check the solution of ex 6.2 Class 10.Â

### Question 2:

The second question consists of a triangle having sides PQ and QR, respectively. The measurements of the sides are provided. Following this, the students need to find out whether EF||QR. The question can again be solved using properties of similar triangles. If the ratios of sides are similar, then, they are parallel to each other. Else, they are not.Â

### Question 3:

In this question, the students are required to prove the similarity of ratios between the sides using the given conditions that LM||CB and LN||CD. They can easily do this by the basic proportionality theorem.Â

### Question 4:

This question is quite similar to the previous one. The students can solve this in three steps by the properties of similarity of triangles. The given conditions in the question are DE||AC and DF||AE.Â

### Question 5 and 6:

Both these questions in the exercise are based on the same concept of basic proportionality theorem and similarity of triangles. Some parallel conditions are given, and students need to prove that one of the lines in the triangle is parallel to the base of the triangle. Use the concepts wisely to prove the given questions.Â

### Question 7:

While solving this question, the students need to recall the concept of mid-point theorem they learnt in Class 9. The question can easily be solved by combining both the mid-point theorem and the basic proportionality theorem.Â

### Question 8:

This can be solved using theorem 6.2 which states that, â€˜If a line divides any two sides of a triangle in the same ratio, then the given line is parallel to the third side.â€™ It can be solved by the converse of basic proportionality theorem and again the mid-point theorem learnt in Class 9.Â

### Question 9:

This question is a little tricky. The students have to prove that in the given trapezium, the point at which diagonals intersect are divided in the same ratio. For this, the students must be well aware of the properties of trapezium as well as the triangle. In this question, the students need to draw a line parallel to AB and CD touching AD from O. Following this; they can use the basic proportionality theorem to prove the question.Â

### Question 10:

Again, in this question, the students need to prove that the given quadrilateral is a trapezium. They can also use the concept from the previous question and consider the conditions in the question as well. The students can easily prove it using the basic proportionality theorem and its converse for the same.Â

### Key Features for Ex 6.2 Class 10 NCERT Solutions

The students must understand the fundamentals and concepts of every exercise to be able to solve the questions easily. We have crafted the solutions to help the students understand the answers clearly. Some of the key features include:

The solutions are well following the CBSE guidelines and patterns, allowing the students to prepare well for the examinations.Â

The solutions are provided in a streamlined manner, making it easier for the students to study and revise from the notes quickly.Â

They can also download the PDF to access the complete solutions anytime they want.Â

**Study without Internet (Offline)**

**App**

1. There is a vertical pole with length 6m casting a shadow of 4m on the ground. At the same time, the other tower casts a shadow 28m long. What is the height of the other tower?

(Image to be added soon)

Given:

AB= 6mÂ

BC= 4m

Similarly, DF= h

EF= 28mÂ

In the given triangles,Â

âˆ C = âˆ E

âˆ B = âˆ F (right angles)

By the angle-angle similarity,Â

Triangles ABC and DEF are similar.Â

Following this, the ratio of the length of their sides will be proportional.Â

AB/DF = BC/EF

6/h = 4/48

h= 42m

Therefore, the height of the tower is 42m.Â

2. The angles of a given quadrilateral are in the ratio 3:5:9:13. What are the angles of the quadrilateral?

Consider the common ratio between the angles as x

Then,Â

3x + 5X + 9x + 13x= 360Â°

x= 12Â°

Therefore, the angles of the quadrilateral will be:

36Â°, 60Â°, 108Â° and 156Â°.Â

The figure is given with the conditions PS/SQ = PT/TR, and âˆ PST = âˆ PRQ. Prove that PQR is an isosceles triangle.Â

(Image to be added soon)

If a line divides the given sides of a triangle in the same ratio, the line is parallel to the third side.

Therefore, ST||QR

And also, âˆ PST = âˆ PQR (being corresponding angles)

âˆ PST = âˆ PRQ (given)

Therefore,Â

âˆ PRQ = âˆ PQR

So, the sides opposite to equal angles are also equal,

Therefore, PQ = PR.Â

So, PQR is an isosceles triangle.Â

Hence, proved.