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NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.1 - Squares and Square Roots

Last updated date: 14th Aug 2024
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NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.1 - FREE PDF Download

Table of Content
1. NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.1 - FREE PDF Download
2. Glance on  NCERT Solutions Maths Chapter 5 Exercise 5.1 Class 8 | Vedantu
3. Access NCERT Solutions for Maths Class 8 Chapter 5 - Squares and Square Roots Exercise 5.1
4. Class 8 Maths Chapter 5: Exercises Breakdown
5. CBSE Class 8 Maths Chapter Squares and Square Roots Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 6 Maths
FAQs

NCERT Solutions of Class 8 Maths Chapter 5, "Squares and Square Roots," introduces students to the fascinating world of perfect squares and their properties. Exercise 5.1 focuses on understanding and identifying perfect squares, which are numbers obtained by squaring whole numbers. This exercise helps build a solid foundation in recognizing patterns and properties related to squares.

The important aspects to focus on include learning the definition of a square number, understanding how to calculate squares of numbers, and identifying square numbers up to a certain limit. Students should pay attention to the various techniques for finding squares and practice these methods to gain confidence.

Glance on  NCERT Solutions Maths Chapter 5 Exercise 5.1 Class 8 | Vedantu

• Chapter 5 Exercise 5.1 Class 8 explains the concepts and properties of squares and square roots, including methods to find them and solve related problems.

• A number is called a perfect square if it is the square of an integer. For example, 1, 4, 9, 16, etc., are perfect squares.

• A number ending in 2, 3, 7, or 8 is never a perfect square.

• A perfect square never ends in an odd number of zeros.

• The square of an even number is always even, and the square of an odd number is always odd.

• Recognizing perfect squares (numbers obtained by squaring an integer) based on their unit digit (0, 1, 4, 5, 6, or 9).

• The last two digits of a square are always even or the sum of the digits of an odd perfect square is always odd.

• This exercise is about developing a better intuition for squares and square roots, which are essential building blocks in mathematics.

• Class 8 Math Chapter 5 Exercise 5.1 Solutions NCERT Solutions has overall 9 fully solved questions.

Access NCERT Solutions for Maths Class 8 Chapter 5 - Squares and Square Roots Exercise 5.1

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

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 4 ,its square will end with the

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

vi. 2638

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 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.

vii. 52698

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 8, its square will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.

viii. 99880

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 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

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 6, its square will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.

x. 55555

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 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

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 = 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 = 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× 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× 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 .

Conclusion

NCERT Chapter 5 of Class 8 Maths, "Squares and Square Roots," introduces students to the fundamental concepts of squares, square roots, and their properties. Exercise 5.1 focuses on identifying perfect squares, understanding the properties of square numbers, and solving related problems. Class 8 Maths Exercise 5.1 Solutions is pivotal for understanding the basics of squares and square roots. It lays the groundwork for more advanced mathematical concepts and helps students develop strong problem-solving skills. Understanding squares and square roots is essential for more complex topics in algebra and geometry.

Class 8 Maths Chapter 5: Exercises Breakdown

 Exercise Number of Questions Exercise 5.2 2 Questions and Solutions Exercise 5.3 10 Questions and Solutions Exercise 5.4 9 Questions and Solutions

Chapter-Specific NCERT Solutions for Class 6 Maths

Given below are the chapter-wise NCERT Solutions for Class 6 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.1 - Squares and Square Roots

1. What do you mean by square and square root in class 8?

When a value is multiplied by itself, squares are the resulting numbers. In contrast, the square root of a number is a quantity that, when multiplied by itself, equals the original amount. Both are, therefore, vice-versa methods. For instance, 2 is squared to give 4, and 2 is the square root of 4, giving 2.

2. What do you mean by Perfect square in NCERT class 8 maths?

A perfect square is a number that can be written as the product of an integer multiplied by itself or as the integer's second exponent. For instance, the number 25, which is the product of the integer 5 multiplied by itself, is a perfect square because 5 × 5 = 25. Due to the fact that it cannot be calculated as the product of two identical integers, 21 is not a perfect square number.

3. Where can I find accurate NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 5.1) Exercise 5.1?

In NCERT Chapter 8, which deals with the square and square roots, you learn about the perfect square and get an introduction to square numbers and their properties in a given space. Vedantu, India's No. 1 E-Learning platform, offers NCERT Solutions for all Class 8 Math chapters. Vedantu’s subject experts specially designed these solutions as per CBSE guidelines. These solutions are 100% accurate and easy to understand. You can visit Vedantu’s official website or mobile app to download these study materials in free PDF format.

4. What are the topics in Class 8 Maths Chapter 6 Squares and Square Roots (EX 5.1) Exercise 5.1?

Class 8 Maths, Chapter 6 Squares and Square Roots (EX 5.1). Exercise 5.1 covers the following topics:

1. Introduction to squares and square roots.

2. Introduction to square numbers

3. Properties of square numbers

4. Some intriguing patterns

• Numbers between square numbers

• successive natural numbers added together

5. How many Questions are there in Chapter 6 Squares and Square Roots Exercise 5.1?

Exercise 5.1 from Class 8 Math's Chapter 6: Squares and Square Roots contains a total of 9 problems. Generally speaking, every question in Chapter 6 involves a perfect square, adding triangular numbers, square numbers between numbers, and the addition of odd numbers. The premier online resource in India, Vedantu, is where you can find the NCERT answers for Class 8 Math. At Vedantu, all of the chapter exercises are collected in one location and solved by a qualified instructor in accordance with the recommendations of the NCERT books. The solutions are complete, step-by-step, and 100% correct.

6. What are some properties of square numbers discussed in ex 5.1 class 8?

Some properties of square numbers include in ex 5.1 Class 8:

• A number ending in 2, 3, 7, or 8 is never a perfect square.

• The square of an even number is always even, and the square of an odd number is always odd.

• The square of a number ending in 0 will always end in 00.

7. Why is understanding perfect squares important in Class 8 Maths Exercise 5.1?

Understanding perfect squares IN  Class 8 Maths Exercise 5.1 is important because it forms the basis for learning square roots and solving more complex algebraic problems. It also helps in understanding the properties of numbers and their applications in various mathematical concepts.

8. Why is Class 8 Maths ex 5.1 important for exams?

Class 8 Maths ex 5.1 is important for exams because it lays the groundwork for understanding more complex concepts in algebra and geometry. Mastery of squares and square roots is essential for solving various mathematical problems, making it a critical part of the curriculum.