NCERT Solutions for class 8 Maths chapter 6 Square and square roots

Exercise 6.1  Exercise 6.2  Exercise 6.3  Exercise 6.4

Exercise 6.4

Question 1

Find the square root of each of the following numbers by division method.

(i) 2304

Sol :

(i) The square root of 2304 can be calculated as follows.

 

48

4

88

704

704

 

0

 

(ii) 4489

Sol :

(ii) The square root of 4489 can be calculated as follows.

 

67

6

127

889

889

 

0

 

(iii) 3481

Sol :

(iii) The square root of 3481 can be calculated as follows.

 

59

5

109

981

981

 

0

Therefore,

 

(iv) 529

Sol :

(iv) The square root of 529 can be calculated as follows.

 

23

2

43

129

129

 

0

 

 

(v) 3249

Sol :

(v) The square root of 3249 can be calculated as follows.

 

57

5

107

749

749

 

0

 

(vi) 1369

Sol :

(vi) The square root of 1369 can be calculated as follows.

 

37

3

67

469

469

 

0

 

 

(vii) 5776

Sol :

(vii) The square root of 5776 can be calculated as follows.

 

76

7

146

876

876

 

0

 

(viii) 7921

Sol :

(viii) The square root of 7921 can be calculated as follows.

 

89

8

169

1521

1521

 

0

 

 

(ix) 576

Sol :

(ix) The square root of 576 can be calculated as follows.

 

24

2

44

176

176

 

0

 

(x) 1024

Sol :

(x) The square root of 1024 can be calculated as follows.

 

32

3

62

124

124

 

0

 

 

(xi) 3136 

Sol :

(xi) The square root of 3136 can be calculated as follows.

 

56

5

106

636

636

 

0

 

(xii) 900

Sol :

(xii) The square root of 900 can be calculated as follows.

 

30

3

60

00

00

 

0

 

Question 2

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

(i) 64

Sol :

(i) By placing bars, we obtain

Since there is only one bar, the square root of 64 will have only one digit in it.

 

(ii) 144

Sol :

(ii) By placing bars, we obtain

Since there are two bars, the square root of 144 will have 2 digits in it.

 

 

(iii) 4489

Sol :

(iii) By placing bars, we obtain

Since there are two bars, the square root of 4489 will have 2 digits in it.

 

(iv) 27225

Sol :

(iv) By placing bars, we obtain

Since there are three bars, the square root of 27225 will have three digits in it.

 

(v) 390625

Sol :

(v) By placing the bars, we obtain

Since there are three bars, the square root of 390625 will have 3 digits in it.

 

Question 3

Find the square root of the following decimal numbers.

(i) 2.56

Sol :

(i) The square root of 2.56 can be calculated as follows.

 

1. 6

1

26

156

156

 

0

 

(ii) 7.29

Sol :

(ii) The square root of 7.29 can be calculated as follows.

 

2. 7

2

47

329

329

 

0

 

 

(iii) 51.84

Sol :

(iii) The square root of 51.84 can be calculated as follows.

 

7.2

7

142

284

284

 

0

 

(iv) 42.25

Sol :

(iv) The square root of 42.25 can be calculated as follows.

 

6.5

6

125

625

625

 

0

 

 

(v) 31.36

Sol :

(v) The square root of 31.36 can be calculated as follows.

 

5.6

5

106

636

636

 

0

 

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

Sol :

(i) The square root of 402 can be calculated by long division method as follows.

 

20

2

40

02

00

 

2

The remainder is 2. It represents that the square of 20 is less than 402 by 2. Therefore, a perfect square will be obtained by subtracting 2 from the given number 402.

Therefore, required perfect square = 402 − 2 = 400

And,

 

(ii) 1989

Sol :

(ii) The square root of 1989 can be calculated by long division method as follows.

 

44

4

84

389

336

 

53

The remainder is 53. It represents that the square of 44 is less than 1989 by 53. Therefore, a perfect square will be obtained by subtracting 53 from the given number 1989.

Therefore, required perfect square = 1989 − 53 = 1936

And,

 

(iii) 3250

Sol :

(iii) The square root of 3250 can be calculated by long division method as follows.

 

57

5

107

750

749

 

1

The remainder is 1. It represents that the square of 57 is less than 3250 by 1. Therefore, a perfect square can be obtained by subtracting 1 from the given number 3250.

Therefore, required perfect square = 3250 − 1 = 3249

And,

 

(iv) 825

Sol :

(iv) The square root of 825 can be calculated by long division method as follows.

 

28

2

48

425

384

 

41

The remainder is 41. It represents that the square of 28 is less than 825 by 41. Therefore, a perfect square can be calculated by subtracting 41 from the given number 825.

Therefore, required perfect square = 825 − 41 = 784

And,

 

(v) 4000

Sol :

(v) The square root of 4000 can be calculated by long division method as follows.

 

63

6

123

400

369

 

31

The remainder is 31. It represents that the square of 63 is less than 4000 by 31. Therefore, a perfect square can be obtained by subtracting 31 from the given number 4000.

Therefore, required perfect square = 4000 − 31 = 3969

And,

 

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

Sol :

(i) The square root of 525 can be calculated by long division method as follows.

 

22

2

42

125

84

 

41

The remainder is 41.

It represents that the square of 22 is less than 525.

Next number is 23 and 232 = 529

Hence, number to be added to 525 = 232 − 525 = 529 − 525 = 4

The required perfect square is 529 and

 

(ii) 1750

Sol :

(ii) The square root of 1750 can be calculated by long division method as follows.</p

 

41

4

81

150

81

 

69

The remainder is 69.

It represents that the square of 41 is less than 1750.

The next number is 42 and 422 = 1764

Hence, number to be added to 1750 = 422 − 1750 = 1764 − 1750 = 14

The required perfect square is 1764 and

 

 

(iii) 252

Sol :

(iii) The square root of 252 can be calculated by long division method as follows.

 

15

1

25

152

125

 

27

The remainder is 27. It represents that the square of 15 is less than 252.

The next number is 16 and 162 = 256

Hence, number to be added to 252 = 162 − 252 = 256 − 252 = 4

The required perfect square is 256 and

 

(iv) 1825

Sol :

(iv) The square root of 1825 can be calculated by long division method as follows.

 

42

4

82

225

164

 

61

The remainder is 61. It represents that the square of 42 is less than 1825.

The next number is 43 and 432 = 1849

Hence, number to be added to 1825 = 432 − 1825 = 1849 − 1825 = 24

The required perfect square is 1849 and

 

 

(v) 6412

Sol :

(v) The square root of 6412 can be calculated by long division method as follows.

 

80

8

160

012

0

 

12

The remainder is 12.

It represents that the square of 80 is less than 6412.

The next number is 81 and 812 = 6561

Hence, number to be added to 6412 = 812 − 6412 = 6561 − 6412 = 149

The required perfect square is 6561 and

 

Question 6

Find the length of the side of a square whose area is 441 m2.

Sol :

Let the length of the side of the square be x m.

Area of square = (x)2 = 441 m2

The square root of 441 can be calculated as follows.

 

21

2

41

041

41

 

0

Hence, the length of the side of the square is 21 m.

 

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

Sol :

(a) ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

AC2 = AB2 + BC2

AC2 = (6 cm)2 + (8 cm)2

AC2 = (36 + 64) cm2 =100 cm2

AC =

AC = 10 cm

(b) ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

AC2 = AB2 + BC2

(13 cm)2 = (AB)2 + (5 cm)2

AB2 = (13 cm)2 − (5 cm)2 = (169 − 25) cm2 = 144 cm2

AB =

AB = 12 cm

 

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.

Sol :

It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same.

We have to find the number of more plants that should be there, so that when the gardener plants them, the number of rows and columns are same.

That is, the number which should be added to 1000 to make it a perfect square has to be calculated.

The square root of 1000 can be calculated by long division method as follows.

 

31

3

61

100

61

 

39

The remainder is 39. It represents that the square of 31 is less than 1000.

The next number is 32 and 322 = 1024

Hence, number to be added to 1000 to make it a perfect square

= 322 − 1000 = 1024 − 1000 = 24

Thus, the required number of plants is 24.

Question 9

These 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?

Sol :

It is given that there are 500 children in the school. They have to stand for a P.T. drill such that the number of rows is equal to the number of columns.

The number of children who will be left out in this arrangement has to be calculated. That is, the number which should be subtracted from 500 to make it a perfect square has to be calculated.

The square root of 500 can be calculated by long division method as follows.

 

22

2

42

100

84

 

16

The remainder is 16.

It shows that the square of 22 is less than 500 by 16. Therefore, if we subtract 16 from 500, we will obtain a perfect square.

Required perfect square = 500 − 16 = 484

Thus, the number of children who will be left out is 16.

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