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

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× 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 .

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

CBSE Class 8 Maths Chapter Squares and Square Roots Other Study Materials

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.