## NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.1) Exercise 6.1

NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1 has been converted into the simplest forms of notes by Vedantu. Since Math is a tough subject and requires a practical approach to learn, the panel of the experts at NCERT Solutions dig deep to extract simple formulas. Chapter Square and Square Roots Class 8 material present in the notes can help you mark optimized points in exams. Download the CBSE NCERT Solution Class 8 Maths Chapter 6 so that you learn and practice in less time.

Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.

**Study without Internet (Offline)**

## Access NCERT Solutions for Class 8 Maths Chapter 6 – Squares and Square Roots

### Exercise 6.1

1. What will be the unit digit of the square of the following numbers?

i. 81

Ans: It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a× a’.

Now, in the given number, the unit’s place digit is 1,its square will end with the

unit digit of the multiplication (1×1 = 1) i.e., 1.

## ii. 272

Ans:It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a× a’.

Now, in the given number, the unit’s place digit is 2, its square will end with the unit digit of the multiplication (2 × 2 = 4)i.e., 4.

## iii. 799

Ans:It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a× a’.

Now, in the given number, the unit’s place digit is 9, its square will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.

## iv. 3853

Ans:It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a× a’.

Now, in the given number, the unit’s place digit is 3, its square will end with the unit digit of the multiplication (3× 3 = 9) i.e., 9.

## v. 1234

Now, in the given number, the unit’s place digit is 4 ,its square will end with the

unit digit of the multiplication (4 × 4 = 16) i.e., 6.

## vi. 2638

Now, in the given number, the unit’s place digit is 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.

## vii.52698

Now, in the given number, the unit’s place digit is 8, its square will end with the

unit digit of the multiplication (8 × 8 = 64) i.e., 4.

## viii. 99880

Now, in the given number, the unit’s place digit is 0 ,its square will have two zeroes at the end. Hence, the unit digit of the square of the given number is 0 .

## ix. 12796

Now, in the given number, the unit’s place digit is 6, its square will end with the

unit digit of the multiplication (6 × 6 = 36) i.e., 6.

## x. 55555

Now, in the given number, the unit’s place digit is 5, its square will end with the

unit digit of the multiplication (5 × 5 = 25) i.e., 5.

## 2. Give reason why the following numbers are not perfect squares.

## i. 1057

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9. Also, a perfect square has even number of zeroes at the end of it. We can see that 1057 has its unit place digit as 7 .

Hence, 1057 cannot be a perfect square.

## ii. 23453

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9

Also, a perfect square has even number of zeroes at the end of it.

We can see that 23453 has its unit place digit as 3 .

Hence, 23453 cannot be a perfect square.

## iii.7928

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9

Also, a perfect square has even number of zeroes at the end of it. We can see that 7928 has its unit place digit as 8 .

Hence, 7928 cannot be a perfect square.

## iv. 222222

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9

Also, a perfect square has even number of zeroes at the end of it. We can see that 222222 has its unit place digit as 2 .

Hence, 222222 cannot be a perfect square.

## v. 64000

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9

Also, a perfect square has even number of zeroes at the end of it. We can see that 64000 has three zeroes at the end of it.

Since a perfect square cannot end with odd number of zeroes, therefore, 64000 is not a perfect square.

## vi. 89722

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9

Also, a perfect square has even number of zeroes at the end of it. We can see that 89722 has its unit place digit as 2 .

Hence, 89722 cannot be a perfect square.

## vii. 222000

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9

Also, a perfect square has even number of zeroes at the end of it. We can see that 222000 has three zeroes at the end of it.

Since a perfect square cannot end with odd number of zeroes, therefore, 222000 is not a perfect square.

## viii. 505050

Ans:The square of numbers may end with any one of the digits 0 , 1 ,4 , 5 , 6 , or 9

Also, a perfect square has even number of zeroes at the end of it. We can see that 505050 has three zeroes at the end of it.

Since a perfect square cannot end with odd number of zeroes, therefore, 505050 is not a perfect square.

## 3. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

i. 243

Ans:First, we’ll break the number into its factors.

243 can be written as 243 = 3× 3× 3× 3× 3

Here, two 3 s are left which are not in a triplet. So, we need one more 3 to make

243 a cube.

If 243 is multiplied by 3 , then we get,

243× 3 = 3× 3× 3× 3× 3× 3 = 729 (which is a perfect cube).

Thus, 3 is the smallest number by which 243 must be multiplied to obtain a perfect cube.

## ii. 256

Ans:First, we’ll break the number into its factors.

256 can be written as 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, two 2 s are left which are not in a triplet. So, we need one more 2 to make

256 a cube.

If 256 is multiplied by 2 , then we get,

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 (which is a perfect cube).

Thus, 2 is the smallest number by which 256 must be multiplied to obtain a perfect cube.

## iii. 72

Ans:First, we’ll break the number into its factors.

72 can be written as 72 = 2 × 2 × 2 × 3× 3

Here, two 3 s are left which are not in a triplet. So, we need one more 3 to make 72

a cube.

If 72 is multiplied by 3 , then we get,

72 × 3 = 2 × 2 × 2 × 3× 3× 3 = 216 (which is a perfect cube).

Thus, 3 is the smallest number by which 72 must be multiplied to obtain a perfect cube.

## iv. 675

Ans:First, we’ll break the number into its factors.

675 can be written as 675 = 3× 3× 3× 5× 5

Here, two 5 s are left which are not in a triplet. So, we need one more 5 to make

675 a cube.

If 675 is multiplied by 5 , then we get,

675× 5 = 3× 3× 3× 5× 5× 5 = 3375

(which is a perfect cube).

Thus, 5 is the smallest number by which 675 must be multiplied to obtain a perfect cube.

## v. 100

Ans:First, we’ll break the number into its factors.

100 can be written as 100 = 2× 2× 5× 5

Here, two 2 s and two 5 s are left which are not in a triplet. So, we need one more

2 and one more 5 to make 100 a cube.

If 100 is multiplied by 2 and 5 , then we get

100 × 2 × 5 = 2 × 2 × 2 × 5× 5× 5 =1000 (which is a perfect cube)

Thus, 2×5 =10 is the smallest number by which 100 must be multiplied to obtain a perfect cube.

## 4. Find the missing digits after observing the following pattern.

112 =121

1012 =10201

10012=1002001

1000012 =1...2...1

100000012 =...

Ans:It can be observed from the given pattern that after doing the square of the number, there are same number of zeroes before the digit and same number of zeroes after the digit as there are in the original number.

So, the square of the number 100001 will have four zeroes before 2 and four zeroes after 2 .

Similarly, the square of the number 10000001 will have six zeroes before 2 and six zeroes after 2 .

Hence,

1000012 = 10000200001

100000012 = 100000020000001

## 5. Find the missing number after observing the following pattern.

112 =121

1012 =10201

101012 =102030201

10101012=...

...2=10203040504030201

Ans: It can be observed from the given pattern tha

The square of the numbers has odd number of digits

The first and the last digit of the square of the numbers is 1

The square of the numbers is symmetric about the middle digit

Since there are four 1 in 1010101, so the square of this number will have natural numbers up to 4 with 0 in between every consecutive number and then making the number symmetric about 4

That is, 10101012 = 1020304030201

Now, 10203040504030201 has natural numbers up to 5 and the number is symmetric about.

So, the number whose square is 10203040504030201, is 101010101

That is, 1010101012 = 10203040504030201

Hence,

10101012 = 1020304030201

1010101012 = 10203040504030201

## 6. Find the missing numbers using the given pattern.

12+22+22=32

22 +32 +62 =72

32 +42 +122 =132

42 +52+...2=212

52+...2+302=312

62+72+...2=...2

Ans:It can be observed from the given pattern that:

The third number in the addition is the product of first two numbers.

The R.H.S can be obtained by adding to the third number. That is, in the first three patterns, it can be observed that

12 + 22 + (1× 2)2 = (2 + 1)2

22 + 32 + (2 × 3)2 = (6 + 1)2

32 + 42 + (3× 4)2 = (12 + 1)2

Hence, according to the pattern, the missing numbers are as follows:

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

## 7. Find the sum without adding.

## i. 1+3+5+7+9

Ans:Since, the sum of first n odd natural numbers is n2.

So, the sum of the first five odd natural numbers is (5)2 = 25

Thus, 1+ 3 + 5 + 7 + 9 = (5)2 = 25

## ii. 1+3+5+7+9+11+13+15+17+19

Ans:Since, the sum of first n odd natural numbers is n2.

So, the sum of the first ten odd natural numbers is (10)2 =100

Thus, 1+ 3 + 5 + 7 + 9 +11+13 +15 +17 +19 = (10)2 = 100

## iii. 1+3+5+7+9+11+13+15+17+19+21+23

Ans:Since, the sum of first n odd natural numbers is n2.

So, the sum of the first twelve odd natural numbers is (12)2 =144

Thus, 1+ 3 + 5 + 7 + 9 +11+13 +15 +17 +19 + 21+ 23 = (12)2 =144

## 8.

i. Express 49 as the sum of 7 odd numbers.

Ans: Since, the sum of first n odd natural numbers is n2.

We know that 49 = (7)2

49 = Sum of 7 odd natural numbers Hence, 49 =1+ 3 + 5 + 7 + 9 +11+13

## ii. Express 121 as the sum of 11odd numbers.

Ans: Since, the sum of first n odd natural numbers is n2. We know that 121 = (11)2

121 = Sum of 11 odd natural numbers

Hence, 121=1+ 3 + 5 + 7 + 9 +11+13 +15 +17 +19 + 21

## 9. How many numbers lie between squares of the following numbers?

i. 12 and 13

Ans: Between the squares of the numbers n and (n+1), there will be 2n numbers. So, there will be 2 ×12 = 24 numbers between (12)2 and (13)2 .

## ii. 25 and 26

Ans: Between the squares of the numbers n and (n+1), there will be 2n numbers.

So, there will be 2× 25 = 50 numbers between (25)2 and (26)2 .

## iii. 99 and 100

Ans: Between the squares of the numbers n and (n+1), there will be 2n numbers.

So, there will be 2×99 =198 numbers between (99)2 and (100)2 .

### NCERT Solution Class 8 Maths Chapter 6 Other Exercises

Chapter 6 Squares and Square Roots Exercises in PDF Format | |

2 Questions and Solutions | |

10 Questions and Solutions | |

9 Questions and Solutions |

### Important Topics Covered in Exercise 6.1 of Class 8 Maths NCERT Solutions

Exercise 6.1 of Class 8 Maths NCERT Solutions is based on the below topics:

Introduction to the squares and square roots

Introduction to square numbers

Properties of square numbers

Some interesting patterns

Adding triangular numbers

Numbers between square numbers

Adding odd numbers

A sum of consecutive natural numbers

Below are some of the key learnings from Exercise 6.1 of Class 8 Maths NCERT Solutions.

If we multiply a number by itself, the result that we get is called the square of that original number. For example, 9 is the square of 3, 16 is the square of 4, and so on.

All the square numbers end with 0, 1, 4, 5, 6, or 9 at the units' place and they have an even number of zeroes in the end.

The sum of the first ‘n’ odd natural numbers is equal to n2. If a natural number cannot be expressed as a sum of successive odd natural numbers starting with 1, then that number is not a perfect square.

The questions given in this exercise are based on finding the sum of the consecutive natural numbers and the product of two consecutive even or odd natural numbers and also on the above-mentioned concepts. Concepts such as finding squares of big numbers by the expansion method and other patterns observed in square numbers are also discussed in this chapter. The NCERT Solutions provided for this exercise help students to build a deep understanding of the concepts and form a solid base for many higher Maths topics ahead.

Benefits of NCERT Solutions for Class 8 Maths Chapter 6

We are offering promising results that can lead to auspicious first-class marks in the exams. You learn your terms in accessible language, and the formulas we put are in the most uncomplicated style. You are prone to gain excellent marks by following the NCERT solutions for learning Class 8, Chapter 6 Maths. We are offering examples-induced problems of Maths by which you can understand the detail of an answer. By learning from significantly shallow writing, you move a step higher than the other fellow members in your class. CBSE assure the students that the content is authentic for Class 8 Maths Chapter 6. Some of the notable benefits of using CBSE materials are as follows.

The importance of NCERT Solutions in the student's life who wish to go through CBSE exams is immense. A student learns better if he or she has the most clarified material in front. NCERT Solutions have the responsibility of providing detailed and clarified answers for Square and Square Roots Class 8 questions. It will help them manage the difficulty of solving during the CBSE exam. You have to write the statements according to CBSE in the paper, which is accessible through a thorough read of NCERT Solutions material.

NCERT Solutions do not compromise on quality when it comes to the future of a student in Maths. All the material for NCERT solutions for Class 8 Maths Chapter 6 Exercise 6.1 lies within the high level of quality. The expert teachers spend their precious time to create helping material for students. The content is helpful to make you understand the question better while maintaining the quality on top. Since the essence of the detail is essential, the specialists of learning authorize the material by giving it a thorough go through.

It becomes perfect for a student when he gets a detail that is easy to memorize and stays in memory for a long time. The material that Vedantu provides has meaningful information while it has a natural and decent style of language that forces it to become more memorable. When you memorize your answer, you can score big, and the information stays to exempt you from remembering it again. All the formulas from Class 8 Maths Chapter 6 are easily memorable in the Vedantu offered material.

Once you are ready to try a new source of learning in simple language, then consider Vedantu. They have all the material for NCERT Solutions for Class 8 Maths Chapter 6 exercise 6.1 in PDF form online. You can always download it for free. To download the free study material from Vedantu, you must become a member of the website, which is also free of cost. Once you download your file, you can find the history in your account details. This method helps you avoid downloading the same file again.

The material is not limited to only Chapter 6 for Class 8, but you can ask for all other chapters, as well. They all are free to download, as Vedantu makes a useful platform for the students. To make sure you obtain high marks in exams, you should go through all the chapters included in your course. Since you can download all of them from Vedantu for free, the risk of time and cost vanishes because you are reading while spending zero money.

Don't get wasted if you are still confused about understanding the pattern of the material. There are instructions and directions provided for free for everyone to stay in the flow of learning. Don't stop learning of you face hurdles. There is always a way out.

### Write All Your Answers Efficiently Employing the Solutions of Ncert Class 8 Chapter 6

Once you pick up the continuity of the learning with Vedantu, your next step is to download the PDF file of the material about Square and Square Roots Class 8 for free. By following the simple instructional tutorial, you can learn and understand the answers quicker and easier. We have a unique and user-friendly online platform from where you can get the desired files in PDF for free. The interface guides every user for the search and acquires the exact data.

If you don't understand the answers, you can download the instructions file from the website, as well. Our teachers have worked hard on writing the answers in a way that students grip the concept in a single read. You can learn new and unique methodologies while preparing for the CBSE exam. We have included all the key-points with every exercise so that the students can acquire back all the missed equations.

### Why Should You Choose Us?

Vedantu is an answer to the difficulties you are facing with learning the solutions. Many reasons make Vedantu special, including hundreds of teachers contributing to making the life of a student more accessible by converting the answers into the formulated form and the updated level of advancement with technology. We make learning more exciting and energetic by implementing modern tools and methods.

You can access Vedantu from your smartphone through an app by which you are just a tap away from the most straightforward learning platform. Through the app subscription, you get all the updates about exclusive materials from NCERT. Vedantu contributes to providing modern learning concepts, clarified study materials, free online classes, discussion groups, and many more.

Vedantu is a way towards smartness. By being a part of the discussion forum, you can evaluate your learning IQ and make study buddies online. You can ask your fellow students about a specific problem you are facing in Class 8 Maths Chapter 6 and find answers from other students all over the world.