NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1.

- Squares and Square Roots Class 8 Ex 6.2
- Squares and Square Roots Class 8 Ex 6.3
- Squares and Square Roots Class 8 Ex 6.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Squares and Square Roots |

Exercise |
Ex 6.1 |

Number of Questions Solved |
9 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

**Ex 6.1 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 digit of the squares of the given numbers is shown against the numbers in the following table :

**Ex 6.1 Class 8 Maths Question 2.**

The following numbers are not perfect squares. Give reason,

**(i)** 1057

**(ii)** 23453

(iii) 7928

**(iv)** 222222

**(v)** 64000

**(vi)** 89722

**(vii)** 222000

**(viii)** 505050

**Solution:**

A number that ends either with 2, 3, 7 or 8 cannot be a perfect square. Also, a number that ends with odd number of zero(s) cannot be a perfect square.

**(i)** Since the given number 1057 ends with 7, so it cannot be a perfect square.

**(ii)** Since the given number 23453 ends with 3, so it cannot be a perfect square.

**(iii)** Since the given number 7928 ends with 8, so it cannot be a perfect square.

**(iv)** Since the given number 222222 ends with 2, so it cannot be a perfect square.

**(v)** Since the number 64000 ends in an odd number of zeros, so it cannot be a perfect square.

**(vi)** Since the number 89722 ends in 2, so it cannot be a perfect square.

**(vii)** Since the number 222000 ends in an odd number of zeros, so it cannot be a perfect square.

**(viii)** Since the number 505050 ends in an odd number of zeros, so it cannot be a perfect square.

**Ex 6.1 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 given number 431 being odd, so its square must be odd.

**(ii)** The given number 2826 being even, so its square must be even.

**(iii)** The given number 7779 being odd, so its square must be odd.

**(iv)** The given number 82004 being even, so its square must be even.

Hence, the numbers 431 and 7779 will have squares as odd numbers.

**Ex 6.1 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 under :

100001^{2} = 10000200001

10000001^{2} = __100000020000001__

**Ex 6.1 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 under :

1010101^{2} = __1020304030201__

101010101^{2} = 10203040504030201

**Ex 6.1 Class 8 Maths Question 6.**

Using the given pattern, find the missing

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

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

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

42^{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 under :

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

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

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

**Ex 6.1 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

**Ex 6.1 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 6.1 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:**

**(i)** Between 12^{2} and 13^{2} there are twenty four (i.e., 2 x 12) numbers.

**(ii)** Between 25^{2} and 26^{2} there are fifty (i.e., 2 x 25) numbers.

**(iii)** Between 99^{2} and 100^{2} there are one hundred ninety-eight (i. e., 99 x 2) numbers.

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