You can Download KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions

Try These (Page 90)

Question 1.

Find the perfect square numbers between

(i) 30 and 40

(ii) 50 and 60

Solution:

(i) The perfect square number between 30 and 40 is 36

(ii) There is no perfect square number between 50 to 60.

Try These (Pages 90-91)

Question 1.

Can we say whether the following numbers are perfect squares? How do we know?

(i) 1057

(ii) 23453

(iii) 7928

(iv) 222222

(v) 1069

(vi) 2061

Write five numbers that you can decide by looking at their one’s digit that they are not square numbers.

Solution:

To know whether the number is perfect square’ the unit digit should be either 0, 1, 4, 5, 6, or 9.

(i) 1057 cannot be a perfect square because its unit digit is 7.

(ii) 23453 is not a perfect square as its unit digits are 3.

(iii) 7928 is not a perfect square.

(iv) 222222 is not a perfect square.

(v) 1069 may be a perfect square.

(vi) 2061 may be a perfect square.

Question 2.

Write five numbers that you cannot decide just by looking at their unit’s digit (or one’s place) whether they are square numbers or not.

Solution:

Numbers are 279, 364, 905, 1680, 581

Try These (Page 91)

Question 1.

Which of 123^{2}, 77^{2}, 82^{2}, 161^{2}, 109^{2}, would end with digit 1?

Solution:

If a number at unit’s digit is 1 or 9, then the square of it will end with digit 1.

∴ 161^{2} and 109^{2} will end with zero.

Try These (Page 91)

Question 1.

Which of the following numbers would have digit 6 at the unit’s place.

(i) 19^{2}

(ii) 24^{2}

(iii) 26^{2}

(iv) 36^{2}

(v) 34^{2}

Solution:

A number whose unit’s either 4 or 6 would end with 6 at the unit’s place.

∴ (24)^{2}, (26)^{2}, (36)^{2}, (34)^{2} would end with digit 6 at the unit’s place.

Try These (Page 92)

Question 1.

What will be the “one’s digit” in the square of the following numbers?

(i) 1234

(ii) 26387

(iii) 52698

(iv) 99880

(v) 21222

(vi) 9106

Solution:

One’s digit of the square of the following would be:

(i) 6

(ii) 9

(iii) 4

(iv) 0

(v) 4

(vi) 6

Try These (Page 92)

Question 1.

The square of which of the following numbers would be an odd number / an even number? Why?

(i) 727

(ii) 158

(iii) 269

(iv) 1980

Solution:

Since the square of the even number is even and the odd number odd.

Hence, Square of

727 is odd

158 is even

269 is odd

1980 is even

Question 2.

What will be the number of zeros in the square of the following numbers?

(i) 60

(ii) 400

Solution:

(i) The square of 60 would have two zeros.

(ii) The square of 400 would have four zeros.

Try These (Page 94)

Question 1.

How many natural numbers lie between 9^{2} and 10^{2}? Between 11^{2} and 12^{2}?

Solution:

We know that between n^{2} and (n + 1)^{2} there lies 2n, numbers.

Hence, between 9^{2} and 10^{2} there lies 2 × 9 = 18 numbers

between 11^{2} and 12^{2} there lies 2 × 11 = 22 numbers.

Question 2.

How many non-square numbers lie between the following pairs of numbers?

(i) 100^{2} and 101^{2}

(ii) 90^{2} and 91^{2}

(iii) 1000^{2} and 1001^{2}

Solution:

(i) Non square numbers (natural numbers) between 100^{2} and 101^{2} are = 2 × 100 = 200

(ii) between 90^{2} and 91^{2}, square numbers are = 2 × 90 = 180

(iii) between 1000^{2} and 1001^{2}, non square numbers are = 2 × 1000 = 2000

Try These (Page 94)

Question 1.

Find whether each of the following numbers is a perfect square or not?

(i) 121

(ii) 55

(iii) 81

(iv) 49

(v) 69

Solution:

A number is a perfect square if it can be expressed as a sum of successive odd natural numbers starting with 1.

(i) 121

121 – 1 = 120

120 – 3 = 117

117 – 5 = 112

112 – 7 = 105

We will get zero in the end after successive subtraction of odd numbers. Hence, it is a perfect square.

(ii) 55 is not a perfect square

(iii) 81 is a perfect square

(iv) 49 is a perfect square

(v) 69 is not a perfect square

Try These (Page 95)

Question 1.

Express the following as the sum of two consecutive integers.

(i) 21^{2}

(ii) 13^{2}

(iii) 11^{2}

(iv) 19^{2}

Solution:

21^{2} where we take 21 = n

n^{2} = \(\left[\frac{n^{2}-1}{2}+\frac{n^{2}+1}{2}\right]\)

(i) 21^{2} = 441 = \(\left[\frac{21^{2}-1}{2}+\frac{21^{2}+1}{2}\right]\)

= 220 + 221

(ii) 13^{2} = 169 = \(\left[\frac{13^{2}-1}{2}+\frac{13^{2}+1}{2}\right]\)

= 81 + 82

(iii) 11^{2} = 121 = \(\left[\frac{11^{2}-1}{2}+\frac{11^{2}+1}{2}\right]\)

= 110 + 111

(iv) 19^{2} = 361 = \(\left[\frac{19^{2}-1}{2}+\frac{19^{2}+1}{2}\right]\)

= 180 + 181

Question 2.

Do you think the reverse is also true, i.e., is the sum of any two consecutive positive integers is a perfect square of a number? Give example to support your answer.

Solution:

No, reverse is not true

Take 4 + 5 = 9 = (3)^{2}

6 + 7 = 13 is not a perfect square

Try These (Page 95)

Observe the squares of numbers; 1, 11, 111, ……….etc.

1^{2} = 1

11^{2} = 1 2 1

111^{2} = 1 2 3 2 1

1111^{2} = 1 2 3 4 3 2 1

11111^{2} = 1 2 3 4 5 4 3 2 1

111111^{2} = 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1

Question 1.

Write the square, making use of the above pattern.

(i) 111111^{2}

(ii) 1111111^{2}

Solution:

(i) (111111)^{2} = 12345654321

(ii) (1111111)^{2} = 1234567654321

Try These (Page 95)

Question 1.

Can you find the square of the following numbers using the above pattern

(i) 6666667^{2}

(ii) 66666667^{2}

Solution:

(i) (6666667)^{2} = 44444448888889

(ii) (66666667)^{2} = 4444444488888889

Observe the following pattern

7^{2} = 49

67^{2} = 4489

667^{2} = 444889

Try These (Page 97)

Question 1.

Find the squares of the following numbers containing 5 in unit place.

(i) 15

(ii) 95

(iii) 105

(iv) 205

Solution:

(i) 15^{2} = (1 × 2) hundreds + 25 = 225

(ii) 95^{2} = (9 × 10) hundreds + 25 = 9025

(iii) 105^{2} = (10 × 11) hundreds + 25 = 11025

(iv) 205^{2} = (20 × 21) hundreds + 25 = 42025

Try These (Page 99)

Question 1.

(i) 11^{2} = 121. What is the square root of 121?

(ii) 14^{2} = 196. What is the square root of 196?

Solution:

(i) 11^{2} = 121

∴ Square root of 121 is 11

(ii) 14^{2} = 196

∴ Square root of 196 is 14.

Try These (Page 100)

Question 1.

By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root.

(I) 121

(II) 55

(III) 36

(IV) 49

(V) 90

Solution:

(I) (i) 121 – 1 = 120

(ii) 120 – 3 = 117

(iii) 117 – 5 = 112

(iv) 112 – 7 = 105

(v) 105 – 9 = 96

(vi) 96 – 11 = 85

(vii) 85 – 13 = 72

(viii) 72 – 15 = 57

(ix) 57 – 17 = 40

(x) 40 – 19 = 21

(xi) 21 – 21 = 0

From 121, we have substracted successive odd numbers starting from 1 and obtain 0 at 11th step

∴ √121 = 11

(II) (i) 55 – 1 = 54

(ii) 54 – 3 = 51

(iii) 51 – 5 = 46

(iv) 46 – 7 = 39

(v) 39 – 9 = 30

(vi) 30 – 11 = 19

(vii) 19 – 13 = 6

After successive subtraction of odd numbers, we did not get 0. Hence, the number is not a perfect square.

(III) (i) 36 – 1 = 35

(ii) 35 – 3 = 32

(iii) 32 – 5 = 27

(iv) 27 – 7 = 20

(v) 20 – 9 = 11

(vi) 11 – 11 = 0

From 36, we subtracted successive odd numbers starting from 1 and obtain 0 at the 6th step.

∴ √36 = 6

(IV) (i) 49 – 1 = 48

(ii) 48 – 3 = 45

(iii) 45 – 5 = 40

(iv) 40 – 7 = 33

(v) 33 – 9 = 24

(vi) 24 – 11 = 13

(vii) 13 – 13 = 0

From 49, we subtracted successive odd numbers starting from 1 and obtain 0 at the 7th step.

∴ √49 = 7

(V) (i) 90 – 1 = 89

(ii) 89 – 3 = 86

(iii) 86 – 5 = 81

(iv) 81 – 7 = 74

(v) 74 – 9 = 65

(vi) 65 – 11 = 54

(vii) 54 – 13 = 41

(viii) 41 – 15 = 26

(ix) 26 – 17 = 9

After successive subtraction, odd numbers we get 9, not 0. Hence 90 is not a perfect square.

Try These (Page 105)

Question 1.

Without calculating square roots, find the number of digits in the square root of the following numbers.

(i) 25600

(ii) 100000000

(iii) 36864

Solution:

(i) 25600

Total digit in the number is 5

∴ Number of digits in the square roots of 25600 = \(\frac{n+1}{2}\) = \(\frac{5+1}{6}\) = 3

(ii) 100000000

Total digit in the number = 9

∴ Number of digits in the square roots of 100000000 = \(\frac{n+1}{2}\) = \(\frac{9+1}{2}\) = 5

(iii) 36864

Total number of digit = 5

∴ Number of digits in the square roots of 36864 = \(\frac{n+1}{2}\) = \(\frac{5+1}{2}\) = 3

Try These (Page 107)

Question 1.

Estimate the value of the following to the nearest whole number.

(i) √80

(ii) √1000

(iii) √350

(iv) √500

Solution:

(i) We know that 80 < 81 < 100

√81 = 9

√100 = 10

9 < √80 < 10

Now, √80 is much closer to 9^{2} = 81

So √80 is approximately = 9

(ii) 1000

100 < 1000 < 10000

√100 = 10

√10000 = 100

10 < √10000 < 100

But we are not very close to square number.

then 31^{2} = 961

32^{2} = 1024

Hence, 1024 is closer to 1000, than 961

∴ √1000 = 32 (app.)

(iii) 350

100 < 350 < 400

√100 = 10

√400 = 20

10 < √350 < 20

But we are not very close to square number.

then 18^{2} = 324

19^{2} = 361

So, 350 is closer to 361, then 324

Hence √350 = 19 (app.)

(iv) 500

100 < 500 < 625

√100 = 10

√625 = 25

10 < √500 < 25

But we are not very close to the square number. So,

22^{2} = 484 and 23^{2} = 529

22 < √500 < 23

But 500 is closer to 484 than 529

Hence, √500 = 22 (app.)