Haryana State Board HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots InText Questions and Answers.

## Haryana Board 8th Class Maths Solutions 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. [∵ 6^{2} = 36]

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

Try These (Page 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 which you can decide by looking at their one’s digit that they are not square numbers.

Solution:

(i) 1057 can not be a perfect square because the digit in the one’s place is 7.

(ii) 23453 is not a perfect square.

[∵ The digit end with 3]

(iii) 7928 is not a perfect square.

[∵ The digit end with 8]

(iv) 222222 is not a perfect square.

[∵ The digit end with 2]

(v) 1069 may be a perfect square.

[For perfect square the digits in the one’s place must be 0, 1, 4, 5, 6 or 9]

(vi) 2061 may be a perfect square.

Question 2.

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

Solution:

Five numbers which is not a perfect square are 223, 237, 25968, 7727 and 8888.

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 has 1 or 9 in the unit’s place, the it’s square ends in 1.

∴ 161^{2} = 161 × 161 and 109

= 109 × 109

are the two numbers whose unit’s place ends in 1 and 9 respectively.

Question 2.

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

(ii) 19^{2}

(ii) 24^{2}

(iii) 26^{2}

(iv) 36^{2}

(v) 34^{2}

Solution:

(ii) 24^{2}

(iii) 26^{2}

(iv) 36^{2} and

(v) 34^{2} would have digit 6 at unit place.

[∵ One’s place should be either 4 or 6]

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:

(i) 6 [∵ 4^{2} = 16]

(ii) 9 [∵ 7^{2} = 49]

(iii) 4 [∵ 8^{2} = 64]

(iv) 0 [∵ 0^{2} = 0]

(v) 4 [∵ 2^{2} = 4]

(vi) 6 [∵ 6^{2} = 6]

Question 2.

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:

(i) 727, odd number.

(ii) 158, even number.

(iii) 269, odd number.

(iv) 1980, even number.

[∵ Squares of even numbers are always even and squares of odd numbers are.always odd]

Question 3.

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

(i) 60

(ii) 400.

Solution:

(i) Two zeros

(ii) 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:

Total numbers of natural number between 9^{2} and 10^{2}.

2n = 2 × 9 = 18

Here, n = 9

Similarly, total number of natural numbers between

11^{2} and 12^{2} = 2n

= 2 × 11 = 22.

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 between 1002 and 1012

= 2n, where n smallest number.

= 2 × 100 = 200

(ii) Non square numbers between 90^{2} and 91^{2}

= 2n = 2 × 90= 180 [∵ = 90]

(iii) Non square numbers between 1000^{2} and 1001^{2}

= 2n = 2 × 1000 [∵ n = 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:

(i) 121 – 1 = 120

120 – 3 = 117

117 – 5 = 112

……………………

……………………

and so on.

We will get the result is zero in the end after 11 steps.

So, 121 is a perfect square.

[Do the same calculations as explained in part (i)]

(ii) 55 is not a perfect square,

(iii) 81 is 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:

(i) 21^{2} = 441

= [\(\frac{21^{2}-1}{2}\) + \(\frac{21^{2}+1}{2}\)]

= 220 + 221

(ii) 13^{2} = 169

= [\(\frac{13^{2}-1}{2}\) + \(\frac{13^{2}+1}{2}\)]

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

= \(\frac{169-1}{2}\) + \(\frac{169+1}{2}\) = 81 + 82

(iii) Similarly, 11^{2} = 121 = 60 + 61.

192 = 361 = 180 + 181.

Question 2.

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

Solution:

No, the reverse is not true

4 + 5 = 9 = 3^{2}

but 6 + 7 = 13

is not a perfect square.

Try These (Page 95)

Question 1.

Write the square, making use of the above pattern.

(i) 111111^{2}

(ii) 1111111^{2}

Solution:

(i) 111111^{2} = 12345654321

[∵ 1^{2} = 1, 11^{2} = 121, 111^{2} = 12321 and so on]

(ii) 1111111^{2} = 1234567654321

[Follow the same pattern]

Question 2.

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

6667^{2} = 44448889

………………………….

………………………….

Try These (Page 97)

Question 1.

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

(i) 15

(ii) 95

(iii) 105

(iv) 205.

Solution:

(i) 15^{2} = (1 × 2) hundreds + 25,

= 2 × 100 + 25 = 225

(ii) 95^{2} [∵ (a_{5})^{2} = a(a + 1) hundred + 25,

where (a_{5}) is a number with unit digit as 5]

= (9 × 10) hundreds + 25

= 9000 + 25 = 9025

(iii) (105)^{2} = (10 × 11) × 100 + 25

= 11025

(iv) (205)^{2} = (20 × 21) × 100 + 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) Since 11^{2} = 121

∴ Square root of 121 = 11

(ii) Since 14^{2} = 196

∴ Square root of 196 = 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) 121 – 1 = 120 120 – 3 = 117

117 – 5 = 112 112 – 7 = 105

105 – 9 = 96 96 – 11 = 85

85 – 13 = 72 72 – 15 = 57

57 – 17 = 40 40 – 19 = 21

21 – 21 = 0

We observe that the number 121 reduced to zero after subtracting 11 consecutive odd numbers starting from 1. Thus, 121 is a perfect square.

∴ 121 = 11^{2} ⇒ \(\sqrt {121}\) = 11

(ii) 55 – 1 = 54 54 – 3 = 51

51 – 5 = 46 46 – 7 = 39

39 – 9 = 30 30 – 11 = 19

19 – 13 = 6

Here the no. do not reduced to zero. So, 55 is not a perfect square.

(iii) 36 – 1 = 35 35 – 3 = 32

32 – 5 = 27 27 – 7 = 20

20 – 9 = 11 11 – 11 = 0

We obtained 0 at 6th step

∴ 36 = 6^{2} (It is a perfect square)

or \(\sqrt {36}\) = 6

(iv) 49 – 1 = 48 48 – 3 = 45

45 – 5 = 40 40 – 7 = 33

33 – 9 = 24 24 – 11 = 13

13 – 13 = 0

We have subtracted successive odd numbers starting from 1 and obtained 0 at 7th step.

∴ 49 = 7^{2} ⇒ \(\sqrt {49}\) = 7 .

This is a perfect square.

(v) 90 – 1 = 89 89 – 3 = 86

86 – 5 = 81 81 – 7 = 74

74 – 9 = 65 65 – 11 = 54

54 – 13 = 41 41 – 17 = 24

24 – 19 = 5

∵ We do not obtain zero in the end.

∴ 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

Here, the number of digits (n) = 5 (odd)

∴ The number of digits in the square roots of 25600

= \(\frac{(n+1)}{2}\) = \(\frac{(5+1)}{2}\) = 3

(ii) 100000000

Here, the number of digits (n) = 9 (odd)

∴ The number of digits in the square roots of the number

= \(\frac{(n+1)}{2}\) = \(\frac{(9+1)}{2}\) = 5

∴ The number of digits in the square of the number

= \(\frac{(n+1)}{2}\) = \(\frac{(5+1)}{2}\) = \(\frac{(6)}{2}\) = 3

Try These (Page 107)

Question 1.

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

(i) \(\sqrt {80}\)

(ii) \(\sqrt {1000}\)

(iii) \(\sqrt {350}\)

(iv) \(\sqrt {500}\)

Solution:

(i) We know that

80 < 81 < 100

and \(\sqrt {81}\) = 9 and \(\sqrt {100}\) = 10

But 9 < \(\sqrt {80}\) < 10

Now, \(\sqrt {80}\) is much closer to 9^{2} = 81

then 10^{2} = 100

So, \(\sqrt {80}\) is approximately 9.

(ii) We know that

100 < 1000 < 10000

and \(\sqrt {100}\) = 10

\(\sqrt {1000}\) = 100

So, 10 < \(\sqrt {1000}\) < 10

But still we are not very close to square number.

We know that 31^{2} = 961

and 32^{2} = 1024

But 1024 is much closer to 1000 than 961.

So, \(\sqrt {1000}\) = 32 (approximately).

(iii) We know that

100 < 350 < 400

and \(\sqrt {100}\) = 10

\(\sqrt {400}\) = 20

So, 10 < \(\sqrt {350}\) < 20

But still you are not very close to the square number.

We know that 18^{2} = 324 and 19^{2} = 361

∴ 18 < \(\sqrt {350}\) < 19

and 350 is much closer to 361 than to 324

So, \(\sqrt {350}\) is approximately equal to 19.

(iv) We know that

100 < 500 < 625

and < \(\sqrt {100}\) =10

< \(\sqrt {625}\) = 25

So, 10 < \(\sqrt {500}\) < 25

But we are still not very close to the square number.

We know that

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

∴ 2 < \(\sqrt {500}\) < 23

But 500 is closer to 484 than to 529.

Hence, < \(\sqrt {500}\) is approximately equal to 22.