**NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.1**

NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.1 are the part of NCERT Solutions for Class 8 Maths (Rationalised Contents). Here you can find the
NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.1**.**

**Ex 5.1 Class 8 Maths Question 1.**

### What will be the unit digit of the squares of the following numbers?

**(i)** 81

**(ii)** 272

**(iii)** 799

**(iv)** 3853

**(v)** 1234

**(vi)** 26387

**(vii)** 52698

**(viii)** 99880

**(ix)** 12796

**(x)** 55555

**Solution:**

The unit digits of the squares of the given numbers are shown against the numbers in the table given below:

**Ex 5.1 Class 8 Maths Question 2.**

### The following numbers are obviously not perfect squares. Give reason.

**(i)** 1057

**(ii)** 23453

**(iii)** 7928

**(iv)** 222222

**(v)** 64000

**(vi)** 89722

**(vii)** 222000

**(viii)** 505050

**Solution:**

A number which ends with 2, 3, 7 or 8 cannot be a perfect square. Again, a number which ends with odd number of zero(s) cannot be a perfect square.**(i)** Since the number 1057 ends with 7, so it cannot be a perfect square.**(ii)** Since the number 23453 ends with 3, so it cannot be a perfect square.**(iii)** Since the number 7928 ends with 8, so it cannot be a perfect square.**(iv)** Since the number 222222 ends with 2, so it cannot be a perfect square.**(v)** Since the number 64000 ends with an odd number of zeros, so it cannot be a perfect square.**(vi)** Since the number 89722 ends with 2, so it cannot be a perfect square.**(vii)** Since the number 222000 ends with an odd number of zeros, so it cannot be a perfect square.**(viii)** Since the number 505050 ends with an odd number of zeros, so it cannot be a perfect square.

**Ex 5.1 Class 8 Maths Question 3.**

### The squares of which of the following would be odd numbers?

**(i)** 431

**(ii)** 2826

**(iii)** 7779

**(iv)** 82004

**Solution:****(i)** The number 431 is an odd number, so its square would be odd.**(ii)** The number 2826 is an even number, so its square would be even.**(iii)** The number 7779 is an odd number, so its square would be odd.**(iv)** The number 82004 is an even number, so its square would be even.

Hence, the squares of the numbers 431 and 7779 would be odd numbers.

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**Ex 5.1 Class 8 Maths Question 4.**

### Observe the following pattern and find the missing digits:

11^{2} = 121

101^{2} = 10201

1001^{2} = 1002001

100001^{2} = 1 ………….. 2 ……….. 1

10000001^{2} = ………………..

**Solution:**

The missing digits are as follows:

100001^{2} = __10000200001__

10000001^{2} = __100000020000001__

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**Ex 5.1 Class 8 Maths Question 5.**

### Observe the following pattern and supply the missing numbers:

11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} = ………..

……………. ^{2} = 10203040504030201

**Solution:**

The missing numbers are as follows:

1010101^{2} = __1020304030201____101010101__^{2} = 10203040504030201

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**Ex 5.1 Class 8 Maths Question 6.**

### Using the given pattern, find the missing numbers.

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2}+ 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} + ____^{2} = 21^{2}

5^{2 }+ ____^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + ____^{2} = ____^{2}

**Solution:**

The missing numbers are as follows:

4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + 42^{2} = 43^{2}

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**Ex 5.1 Class 8 Maths Question 7.**

### Without adding, find the sum.

**(i)** 1 + 3 + 5 + 7 + 9

**(ii)** 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

**(iii)** 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

**Solution:****(i)** 1 + 3 + 5 + 7 + 9 = Sum of first 5 odd numbers

= 5^{2} = 25**(ii)** 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

= Sum of first 10 odd numbers

= 10^{2} = 100**(iii)** 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

= Sum of first 12 odd numbers

= 12^{2} = 144

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**Ex 5.1 Class 8 Maths Question 8.**

**(i)** Express 49 as the sum of 7 odd numbers.

**(ii)** Express 121 as the sum of 11 odd numbers.

**Solution:****(i)** 49 = 7^{2} = 1 + 3 + 5 + 7 + 9 + 11 + 13**(ii)** 121 = 11^{2} = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

** **

**Ex 5.1 Class 8 Maths Question 9.**

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

**(i)** 12 and 13

**(ii)** 25 and 26

**(iii)** 99 and 100

**Solution:**

**We** know that there are 2n numbers lie between n^{2} and (n + 1)^{2}.**(i)** Therefore, between 12^{2} and 13^{2}, there are 24 (i.e., 2 × 12) numbers.**(ii)** Therefore, between 25^{2} and 26^{2}, there are 50 (i.e., 2 × 25) numbers.**(iii)** Therefore, between 99^{2} and 100^{2}, there are 198 (i.e., 99 × 2) numbers.

**Related Links:**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**