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NCERT Solutions for Class 6 Maths Chapter 5: Understanding Elementary Shapes - Exercise 5.1

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NCERT Solutions for Class 6 Maths Chapter 5 (Ex 5.1)

Free PDF download of NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.1 (Ex 5.1) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 6

Subject:

Class 6 Maths

Chapter Name:

Chapter 5 - Understanding Elementary Shapes

Exercise:

Exercise - 5.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English,Hindi will become easy to study if you have access to NCERT Solution for Class 6 Science , Maths solutions and solutions of other subjects.

Access NCERT Solutions for Maths Class 6 Chapter 5 – Understanding Elementary Shapes

Exercise 5.1

1. What is the disadvantage in comparing line segments by mere observation?

Ans: Line segments are parts of the lines which have two distinct endpoints.

Two or more line segments can be compared by mere observation. However, there is always a chance of error in comparison. This is because of improper viewing. Human eyes can judge significant differences in length. To measure the lengths precisely is not possible just by looking 


2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?

Ans: A ruler is a tool that is calibrated to measure the lengths of the line segments. It is better to use a divider instead of a ruler for the measurement of the line segments. This is because the thickness of the ruler can cause difficulties in reading off the lengths for the various line segments. However, the divider gives up accurate measurements as the arms of the divider measure accurately.


3. Draw any line segment, say $\overline {AB} $. Take any point $C$ lying in between $A$ and $B$. Measure the lengths of $AB$, $BC$ and $AC$. Is $AB = AC + CB$?

(Note: If $A$, $B$, $C$ are any three points on a line, such that $AC + CB = AB$, then we can be sure that $C$ lies between $A$ and $B$.)

Ans: A line segment $\overline {AB} $ is drawn as follows,

(Image will be uploaded soon)

The line segment $\overline {AB} $ with any point $C$ lying in between $A$and$B$ is as follows,

(Image will be uploaded soon)

The lengths of$AB$, $BC$and$AC$is measured as follows,

(Image will be uploaded soon)

From the above figure it can be seen that$AB = 13\,{\text{cm}}$,

$AC = 5\,{\text{cm}}$, and

$BC = 8\,{\text{cm}}$.

Let us now check whether $AB = AC + CB$ is true or not.

Substitute the values of $AB$, $BC$and$AC$ in the equation $AB = AC + CB$ and verify.

$AB = AC + CB$

$AC + CB = 5 + 8$

$AC + CB = 13$

$AC + CB = AB$

Therefore, $AB = AC + CB$ is proved true. 


4. If $A$, $B$, $C$ are three points on a line such that $AB = 5\,{\text{cm}}$, $BC = 3\,{\text{cm}}$ and $AC = 8\,{\text{cm}}$, which one of them lies between the other two?

Ans: Three points $A$,$B$, and$C$are given. The length of $AB$ is given as $5\,{\text{cm}}$. The length of $BC$is given as $3\,{\text{cm}}$. And the length of $AC$is given as $8\,{\text{cm}}$.

From the given measurements it can be seen that the longest measure is of the side $AC$ which is $8\,{\text{cm}}$.

Also, it can be seen that,

$AB + BC = 5 + 3$

$AB + BC = 8$

$AB + BC = AC$

Since $AC$ is the longest side and $AB + BC = AC$, therefore, the point $B$lies between the points $A$and $C$.


5. Verify whether $D$ is the mid-point of $\overline {AG} $.

(Image will be uploaded soon)

Ans: To verify that $D$ is the midpoint of $\overline {AG} $, we will find the lengths of $\overline {AD} $and $\overline {DG} $ and then check whether they are equal or not.

From the given number line, it is clear that the distance between two consecutive points is 1 unit.

The distance between the points $A$ and $D$ is 3 units. This means that the length of $\overline {AD} $ is 3 units.

Also, the distance between the points $D$ and $G$ is 3 units. This means that the length of $\overline {DG} $ is 3 units.

It can be seen that 

$\overline {AD}  = \overline {DG} $

Therefore, it is verified that $D$ is the midpoint of $\overline {AG} $.


6. If $B$ is the mid-point of $AC$ and $C$ is the mid-point of $BD$, where $A$, $B$,$C$, $D$ lie on a straight line, say why $AB = CD$ ?

Ans: It is given that $B$ is the mid-point of $AC$.

From the above information, we can conclude that

$AB = CB\,\,\,\,\,\,\,\,\,\,\left( 1 \right)$

Also, it is given that $C$ is the mid-point of $BD$.

From the above information, we can conclude that

$BC = CD\,\,\,\,\,\,\,\,\,\,\left( 2 \right)$

From equation $\left( 1 \right)$ and $\left( 2 \right)$, we can conclude that

$AB = CD$


7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.

Ans: Draw the first triangle.

(Image will be uploaded soon)

We will now add the lengths of the sides and check whether their sum is greater than the third side or not in each case.

Consider the sides $AB$ and $BC$.  

$AB + BC = 5.2 + 5.2$

$AB + BC = 10.4$

$AB + BC > AC$

Consider the sides $BC$ and $CA$. 

$BC + CA = 5.2 + 5.2$

$BC + CA = 10.4$

$BC + CA > AB$

Consider the sides $AB$ and $AC$. 

$AB + AC = 5.2 + 5.2$

$AB + AC = 10.4$

$AB + AC > BC$

Therefore, we can see that the sum of the lengths of any two sides is always greater than the third side in the considered triangle. 

Draw the second triangle.

(Image will be uploaded soon)

We will now add the lengths of the sides and check whether their sum is greater than the third side or not in each case.

Consider the sides $AB$ and $BC$.  

$AB + BC = 17 + 8$

$AB + BC = 25$

$AB + BC > AC$

Consider the sides $BC$ and $CA$. 

$BC + CA = 8 + 15$

$BC + CA = 23$

$BC + CA > AB$

Consider the sides $AB$ and $AC$. 

$AB + AC = 17 + 15$

$AB + AC = 32$

$AB + AC > BC$

Therefore, we can see that the sum of the lengths of any two sides is always greater than the third side in the considered triangle. 

Draw the third triangle.

(Image will be uploaded soon)

We will now add the lengths of the sides and check whether their sum is greater than the third side or not in each case.

Consider the sides $AB$ and $BC$.  

$AB + BC = 9 + 10$

$AB + BC = 19$

$AB + BC > AC$

Consider the sides $BC$ and $CA$. 

$BC + CA = 10 + 15$

$BC + CA = 25$

$BC + CA > AB$

Consider the sides $AB$ and $AC$. 

$AB + AC = 9 + 15$

$AB + AC = 24$

$AB + AC > BC$

Therefore, we can see that the sum of the lengths of any two sides is always greater than the third side in the considered triangle. 

Draw the fourth triangle.

(Image will be uploaded soon)

We will now add the lengths of the sides and check whether their sum is greater than the third side or not in each case.

Consider the sides $AB$ and $BC$.  

$AB + BC = 8 + 5$

$AB + BC = 13$

$AB + BC > AC$

Consider the sides $BC$ and $CA$. 

$BC + CA = 5 + 5$

$BC + CA = 10$

$BC + CA > AB$

Consider the sides $AB$ and $AC$. 

$AB + AC = 8 + 5$

$AB + AC = 13$

$AB + AC > BC$

Therefore, we can see that the sum of the lengths of any two sides is always greater than the third side in the considered triangle. 

Draw the fifth triangle.

(Image will be uploaded soon)

We will now add the lengths of the sides and check whether their sum is greater than the third side or not in each case.

Consider the sides $AB$ and $BC$.  

$AB + BC = 10 + 10$

$AB + BC = 20$

$AB + BC > AC$

Consider the sides $BC$ and $CA$. 

$BC + CA = 10 + 14$

$BC + CA = 24$

$BC + CA > AB$

Consider the sides $AB$ and $AC$. 

$AB + AC = 10 + 14$

$AB + AC = 24$

$AB + AC > BC$

Therefore, we can see that the sum of the lengths of any two sides is always greater than the third side in the considered triangle. 

From the above examples we can see that the lengths of any two sides of the triangle is always greater than the third side for all the triangles.


NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.1

Opting for the NCERT solutions for Ex 5.1 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 5.1 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 5 Exercise 5.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 6 Maths Chapter 5 Exercise 5.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 6 Maths Chapter 5 Exercise 5.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.