Contents

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots 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.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Squares and Square Roots |

Exercise |
Ex 6.1, Ex 6.2, Ex 6.3, Ex 6.4 |

Number of Questions Solved |
30 |

Category |
NCERT Solutions |

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

### Chapter 6 Squares and Square Roots Exercise 6.1

**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 :

**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.

**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.

**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__

**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

**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}.

**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

**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

**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.

### Chapter 6 Squares and Square Roots Exercise 6.2

**Question 1.**

Find the square of the following numbers,

**(i)** 32

**(ii)** 35

**(iii)** 86

**(iv)** 93

**(v)** 71

**(vi)** 46

**Solution:**

**(i)** 32^{2} = (30 + 2)^{2}

= 30 (30 + 2) + 2 (30 + 2)

= 30^{2} +30 x 2 + 2 x 30 + 2^{2}

= 900 + 60 + 60 + 4 = 1024

**(ii)** = (30 + 5)^{2} = 30 (30 + 5) + 5 (30 + 5)

= 30 ^{2} + 30 x 5 + 5 x 30 + 5^{2}

= 900 +150 +150 + 25 = 1225

**(iii)** = (80 + 6)^{2} = 80 (80 + 6) + 6 (80 + 6)

= 80^{2} +80 x 6 + 6 x 80 + 62

= 6400 + 480 + 480 + 36 = 7396

**(iv)** = (90 + 3)^{2} = 90 (90 + 3) + 3 (90 + 3)

= 90^{2} + 90 x 3 + 3 x 90 + 92

= 8100 + 270 + 270 + 81 = 8649

**(v)** = (70 +1)^{2} = 70 (70 +1) +1 (70 +1)

= 70^{2} +70 x 1 + 1 x 70 + 1^{2}

= 4900 + 70 + 70 +1 = 5041

**
(vi)** = (40 + 6)

^{2}= 40 (40 + 6) + 6 (40 + 6)

= 40

^{2}+ 40 x 6 + 6 x 40 +6 x 6

= 1600 +240+240 +36 =2116

**Question 2.**

Write a Pythagorean triplet whose one member is

**(i)** 6

**(ii)** 14

**(iii)** 16

**(iv)** 18

**Solution:**

**(i)** Put m = 3 in 2m, m^{2} -1, m^{2} +1

∴ 2m=6, m^{2} – l =3^{2} – 1= 9 – 1 = 8

and m^{2} + 1=9 + 1=10

Thus, 6, 8 and 10 are Pythagorean triplets.

**(ii)** Put m = 7 in 2m, m^{2} -1, m^{2} +1

∴ 2m = 14, m^{2} – 1 = 7^{2} – 1 = 49 – 1 = 48

and m^{2} + 1 = 72 +1 = 49+1 = 50

Thus, 14, 48 and 50 are Pythagorean triplets.

**(iii)** Put m = 8 in 2m, m^{2} – 1, m^{2} +1

∴ 2m =16, m^{2} – 1 = 8^{2} – 1 = 64 – 1 =63

and m^{2} + 1 =8^{2} + 1 =64 + 1 =65

Thus, 16, 63 and 65 are Pythagorean triplets.

**(iv)** Put m = 9 in 2m, m^{2} – 1, m^{2} +1

∴ 2m =18, m^{2} – 1 = 9^{2} – 1=81 – 1 = 80,

m^{2}+ 1 =9^{2} + 1 =81 + 1 =82

Thus, 18, 80 and 82 are Pythagorean triplets.

### Chapter 6 Squares and Square Roots Exercise 6.3

**Question 1.**

What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

**(i)** 9801

**(ii)** 99856

**(iii)** 998001

**(iv)** 657666025

**Solution:**

The possible ‘one’s’ digits of the square root of the numbers :

**(i)** 9801 is 1 or 9

**(ii)** 99856 is 4 or 6

**(iii)** 998001 is 1 or 9

**(iv)** 657666025 is 5

**Question 2.**

Without doing any calculation, find the numbers which are surely not perfect squares :

**(i)** 153

**(ii)** 257

**(iii)** 408

**(iv)** 441

**Solution:**

We know that a number ending with 2, 3, 7 or 8 is never a perfect square.

Therefore,

**(i)** 153 is not a perfect square.

**(ii)** 257 is not a perfect square.

**(iii)** 408 is not a perfect square.

**(iv)** 441 may be a perfect square.

So, 153, 257 and 408 are surely not perfect squares.

**Question 3.**

Find the square roots of 100 and 169 hy the method of repeated subtraction.

**Solution:**

From 100, we subtract successive odd numbers starting from 1 as under:

From 169, we subtract successive odd numbers starting from 1 as under:

**Question 4.**

Find the square roots of the following numbers by the Prime Factorisation Method :

**(i)** 729

**(ii)** 400

**(iii)** 1764

**(iv)** 4096

**(v)** 7744

**(vi)** 9604

**(vii)** 6929

**(viii)** 9216

**(ix)** 529

**(x)** 8100

**Solution:**

**Question 5.**

For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

**(i)** 252

**(ii)** 180

**(iii)** 1008

**(iv)** 2028

**(v)** 1458

**(vi)** 768

**Solution:**

**Question 6.**

For each of the following numbers. find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

**(i)** 252

**(ii)** 2925

**(iii)** 396

**(iv)** 2645

**(v)** 2800

**(vi)** 1620

**Solution:**

**Question 7.**

The students of Class VIII of a school donated ? 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

**Solution:**

Let x be the number of students in the class. Therefore, total money donated by students

Hence, the number of the students in the class is 49.

**Question 8.**

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

**Solution:**

let x be the number of rows and x be the number of plants in each row.

∴ The total number of plants = x x x = x^{2}

But is it given as 2025.

Hence, the number of rows is 45 and the number of plants in each row is also 45.

**Question 9.**

Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

**Solution:**

The smallest number divisible by each one of the numbers 4, 9 and 10 is their LCM., which is (2 x 2 x 9 x 5), i.e., 180.

Now, 180 = __2 x 2__ x __3 x 3__ x 5

To make it a perfect square, it must be multiplied by 5.

∴ Required number = 180 x 5 = 900

**Question 10.**

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

**Solution:**

The smallest number divisible by each of the numbers 8, 15 and 20 is their LCM., which is

(2 x 2 x 5 x 2 x 3), i.e., 120.

Now, 120 = __2 x 2__ x 2 x 3 x 5

To make it a perfect square, it must be multiplied by

2 x 3 x 5, i.e., 30.

∴ Required number = 120 x 30 = 3600

### Chapter 6 Squares and Square Roots Exercise 6.4

**Question 1.**

Find the square root of each of the following numbers by Division method :

**(i)** 2304

**(ii)** 4489

**(iii)** 3481

**(iv)** 529

**(v)** 3249

**(vi)** 1369

**(vii)** 5776

**(viii)** 7921

**(ix)** 576

**(x)** 1024

**(xi)** 3136

**(xii)** 900

**Solution:**

**Question 2.**

Find the number of digits in the square root of each of the following numbers (without any calculation) :

**(i)** 64

**(ii)** 144

**(iii)** 4489

**(iv)** 27225

**(v)** 390625

**Solution:**

**Question 3.**

Find the square root of the following decimal numbers :

**(i)** 2.56

**(ii)** 7.29

**(iii)** 51.84

**(iv)** 42.25

**(v)** 31.36

**Solution:**

**(i)** Here, the number of decimal places is already even. So, mark off periods and proceed as under :

∴

**(ii)** Here, the number of decimal places are already even. So, mark off periods and proceed as under :

∴

**(iii)** Here, the number of decimal places are already even. So, mark off periods and proceed as under :

∴

**(iv)** Here, the number of decimal places are already even. So, mark off periods and proceed as under :

∴

**(v)** Here, the number of decimal places are already even. So, mark off periods and proceed as under :

∴

**Question 4.**

Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

**(i)** 402

**(ii)** 1989

**(iii)** 3250

**(iv)** 825

**(v)** 4000

**Solution:**

**(i)** Let us try to find the square root of 402.

This shows the (20)^{2} is less than 402 by 2. So, in order to get a perfect square, 2 must be subtracted from the given number.

∴ Required perfect square number = 402 – 2 = 400

Also,

**(ii)** Let us try to find the square root of 1989.

This shows that (44)^{2} is less than 1989 by 53. So, in order to get a perfect square, 53 must be subtracted from the given number.

∴ Required perfect square number = 1989 – 53 = 1936

Also,

**(iii)** Let us try to find the square root of 3250.

This shows that (57)^{2} is less than 3250 by 1. So, in order to get a perfect square, 1 must be subtracted from the given number.

∴ Required perfect number = 3250 -1 = 3249

Also,

**(iv)** Let us try to find the square root of 825.

This shows that (28)^{2} is less than 825 by 41. So, in order to get a perfect square, 41 must be subtracted from the given number.

∴ Required perfect square number = 825 – 41 = 784

Also,

**(v)** Let us try to find the square root of 4000.

This shows that (63)^{2} is less than 4000 by 31. So, in order to get a perfect square, 31 must be subtracted from the given number.

∴ Required perfect square number = 4000 – 31 = 3969

Also,

**Question 5.**

Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

**(i)** 525

**(ii)** 1750

**(iii)** 252

**(iv)** 1825

**(v)** 6412

**Solution:**

**(i)** We try to find out the square root of 525.

**(ii)** We try to find out the square root of 1750.

**(iii)** We try to find out the square root of 252.

**(iv)** We try to find out the square root of 1825.

**(v)** We try to find out the square root of 6412.

**Question 6.**

Find the length of the side of a square whose area is 441 m^{2}.

**Solution:**

**Question 7.**

In a right triangle ABC, ∠B = 90°.

**(a)** If AB = 6 cm, BC = 8 cm, find AC.

**(b)** If AC = 13 cm, BC = 5 cm, find AB.

**Solution:**

**Question 8.**

A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.

**Solution:**

Let us find the square root of 1000.

This shows that (31)^{2} is less than 1000 by 39 and (32)^{2} =1024. Thus, the gardener needs 1024 -1000 = 24 plants more to plant in such a way that the number of rows and the number of columns remain the same.

**Question 9.**

There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?

**Solution:**

Let us find the square root of 500.

This shows that (22)^{2} = 484 is less than 500 by 16.

∴ 16 students have to go out for others to do the P.T. practice as per condition.

We hope the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots, drop a comment below and we will get back to you at the earliest.

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