CHAPTER-3-SQUARES AND SQUARE ROOTS

EXERCISE 3(A)

1. (i)
Solution:
59

( 59 )=
2
59 × 59
= 3481

(ii)
Solution:
6.3

( 63=
)
2
6.3 × 63
= 39.69

(iii)
Solution:
15

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 (15)=
2
15 × 15
= 225

2. (i)
Solution:
11025

11025 = 3× 3× 5× 5× 7 × 7
= 32 × 52 × 7 2
= 3× 5× 7
= 105

(ii)
Solution:
396900

39600 = 3× 3× 3× 3× 7 × 7 × 2 × 5× 2 × 5
= 22 × 32 × 32 × 52 × 7 2
= 2 × 3× 3× 5× 7
= 630

Square root of 194481 = 441

3. (i)

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Solution:
Given number = 2592
Factor of 2592
2592 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2 × 2 × 2 × 2 × 2 × 3× 3× 3× 3
Here prime factor 2 is not paired so we need to pair it.
Required smallest number = 2

(ii)
Solution:
Given number = 12708
Factor of 12748

12748 = 2 × 2 × 3187
Here prime factor 3187 is not paired so we need to pair it.
Required smallest number = 3187

4.
Solution:
Given number = 10368
Factor of 10368
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10368 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 3× 3
Here prime factor 2 is not paired so we will divide it by by 2 .
Smallest required number = 2

Then factor become: 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 ×3

Square root of remaining number

= 2 × 2 × 2 × 3× 3
= 72
Square root of remaining number
= 72

5. (i)
Solution:
0.1764
1764
0.1764 =
10000
Now write factor of both numerator and denominator

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 1764 2 × 2 × 3 × 3 × 7 × 7
=
10000 10 × 10 × 10 × 10
2 × 2 × 3× 3× 7 × 7
=
10 × 10 × 10 × 10
2 × 2 × 3× 3× 7 × 7
0.1764 =
10 × 10 × 10 × 10
2 × 3× 7
=
10 × 10
= 0.42
Square root of 0.1764 = 0.42

(ii)
Solution:
1
96
25
1 2401
96 =
25 25

Now write the factor of both numerator and denominator.

2401 7 × 7 × 7 × 7
=
25 5× 5
7×7
=
5
= 9.8
1
Square root of 96 = 9.8
25
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 (iii)
Solution:
0.0169

169
0.0169 =
10000
Now,
Write the factor of both numerator and denominator.
169 13 × 13
=
10000 10 × 10 × 10 × 10
13
=
10 × 10
= 0.13

Square root of 0.0169 = 0.13

6. (i)
Solution:

14.4
22.5

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 144 × 10 144
=
225 × 10 225
144 12 × 12
=
225 15 × 15
12 × 12 12 × 12
=
15 × 15 15 × 15
12
= 0.8
15
= 0.8

14.4
= 0.8
22.5

(ii)
Solution:
0.225
28.9

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 0.225 × 10
289 × 100
225 × 10 225
=
289 × 1000 289 × 100
225 15 × 15
=
289 × 100 17 × 17 × 10 × 10
15
= 0.1071
17 × 10
0.225
= 0.1071
28.9

(iii)
Solution:

25 13
× 2 × 0.25
32 18

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 25 49 25
= × ×
32 18 100
5× 5 7×7 5× 5
= × ×
2 × 2 × 2 × 2 × 2 2 × 3 × 3 10 × 10
5× 5× 7 × 7 × 5× 5
=
2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 10 × 10
5× 7 × 5
=
2 × 2 × 2 × 3 × 10
175
=
240
= 0.723
25 13
× 2 × 0.25 = 0.723
32 18

(iv)
Solution:

4 4 7
1 × 14 × 2
5 44 55

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 9 637 117
× ×
5 44 55
3 × 3 7 × 7 × 13 3 × 3 × 13
× ×
5 2 × 2 × 11 5 × 11
3 × 3 × 7 × 7 × 3 × 3 × 13 × 13
5 × 5 × 2 × 2 × 11× 11
3 × 7 × 3 × 13
=
5 × 2 × 11
819
=
110
= 7.445

4 21 7
1 × 14 × 2 = 7.445
5 44 55

7. (i)
Solution:

32 × 63 × 24

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 32 × ( 2 × 3) × 2 × 2 × 2 × 3
3

3× 3× 2 × 2 × 2 × 3× 3× 3× 2 × 2 × 2 × 3
2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 3× 3× 3× 3
= 2 × 2 × 2 × 3× 3× 3
= 216
32 × 63 × 24 =
216

Value of 32 × 63 × 24 =
216

(ii)
Solution:

( 0.5)
3
× 6 × 35

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 3
 5
  × 2 × 3 × 35
 10 
53
× 2 × 3 × 35
103
5× 5× 5
× 2 × 3× 3× 3× 3× 3× 3
10 × 10 × 10
5× 5× 5
× 3× 3× 3× 3× 3× 3
10 × 10 × 10
5× 5× 3× 3× 3× 3× 3× 3
10 × 10
5× 3× 3× 3
=
10
135
=
10
= 13.5

(iii)
Solution:

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  71  0.169 10
 5 + × ×
 25  1.6 1000
196 169 10
× ×
25 16 1000
14 × 14 13 × 13 1
× ×
5× 5 4 × 4 10 × 10
14 × 13
=
5 × 4 × 10
182
=
200
= 0.91

(iv)
Solution:

 3 −3 
5 2 
 4 10 

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  11 −3 
= 5 
 4 10 
 11× 5 − 3 × 2 
= 5 
 20 

= 5
( 55 − 6 )
20
5 × 49
=
20
5× 7 × 7
=
5× 2× 2
7
=
2
= 3.5

(v)
Solution:

248 + 52 + 144

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= 248 + 52 + 12 × 12 { 144 = 12

= 248 + 52 + 12
= 248 + 64
= 248 + 8 × 8
248 + 8
= { 64 =
8

= 256
= 16 × 16
= 16

8.
Solution:
Let number of days in tour be = x so according to question.
Rupees he spend every day = x
Total amount he spend= x × x
= x2
And given total amount = 1296 so both are equal.
x 2 = 1296
x = 1296
= 36
He spend total 36 days in tour.
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 9.
Solution:
Let total rows = x
So total column will be also x
Total student in x rows and x columns = x 2
16 student left out of total 745 student
Student in respected row, columns
= 745 − 16
= 729
According to question
x 2 = 729
x = 729
= 27
Required number of rows = 27

10. (i)
Solution:
Mirror image of 12 is 21
Square of 12 = 144
And square of 21 = 441
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Their square 144 and 441 is also mirror image of each other

(ii)
Solution:
Mirror image of 112 is 211
Square of 112 = 12544
And square of 221 = 44521
So 12544 and 44521 are mirror image of each other.

11.
Solution:
The required smallest perfect square number divisible 3,4,5 and 6 is
divisible by LCM of 3,4,5 and 6
LCM of 3,4,5 and 6 = 60
Now,
Factor of 60

60 = 2 × 2 × 3 × 5
To make perfect square number divisible by 3,4,5 and 6 it must be
multiply by 3 × 5
Required perfect square number

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= 2 × 2 × 3× 5× 3× 3× 5
= 900
So required perfect square number = 900

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 CHAPTER-3-SQUARES AND SQUARE ROOTS

EXERCISE 3(B)

1. (i)
Solution:
4761

Square root of 4761 = 69

(ii)
Solution:
7744

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Square root of 7744 = 88
Find square root:

(iii)
Solution:
15129

Square root of 15129 = 123

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(iv)
Solution:

Square root of 0.2916 = 0.54

(v)
Solution:
0.035

Square root of 0.001225 = 0.035

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(vi)
Solution:
0.023104

Square root of 0.023104 = 0.152

(vii)
Solution:
27.3529

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Square root of 27.3529 = 5.23

2. (i)
Solution:
4.2025

Square root of 4.2025 = 2.05

(ii)
Solution:
531.7636

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Square root of 531.7636 = 23.06

(iii)
Solution:
0.007225

Square root of 0.007225 = 0.085

3.(i)

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Solution:
245 correct to two places of decimal

245.0000

Square root of 245 up to two place of decimal = 15.65

(ii)
Solution:
496

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Square root of 496 up to two places decimal = 22.27

(iii)
Solution:
82.6

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Square root of 82.6 up to two decimal = 9.08

(iv)
Solution:
0.065

Square root of 0.065 up to three decimal = 0.254

(v)
Solution:
5.2005

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Square root of 5.2005 up to two decimal = 2.28

(vi)
Solution:
0.602

Square root of 0.602 up to two decimal 0.77

4.(i)
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Solution:
4 19
3= = 3.8
5 5

4
Square root of 3 = 1.94
5

(ii)
Solution:
7 55
6= = 6.875
8 8

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 7
Square root of 6 up to two decimal = 2.62
8
5.(i)
Solution:
796

So 12 must be subtracted from 796 so that if will be perfect square.

(ii)
Solution:
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1886

37 must be subtracted from 1886 so that if will be perfect square.

(iii)
Solution:
23497

88 must be subtracted from given number to get perfect square

number .

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6.(i)
Solution:
511

Clearly 511 is greater than 222 .

On adding the enquired number to 511 we shall be getting 232 or
= 529 .
= 529 − 511
Required number
= 18

(ii)
Solution:

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Clearly 7172 is greater than square of 84.
On adding required number we shall get 852 or = 7225 .
Required number
= 7225 − 7172
= 53

(iii)
Solution:

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Cleary 55078 is greater than ( 234 )
2

On adding required number we shall get square of ( 235 ) or

2

( 235)
2
= 55225 .

Required number
= 55225 − 55078
= 147
Required is 147

7.
Solution:
Square root of 7

Square root of 7 up to two decimal = 2.67

Now,

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 4+ 7
4− 7

Rationalization
=
(4 + 7 ) × (4 + 7 )
(4 − 7 ) (4 + 7 )
(4 + 7 )
2

=
16 − 7
4+ 7
=
3
Now,

4+ 7
Value of
3
4 + 2.67
=
3
{
7 2.62

= 2.223

8.
Solution:

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 5

Square root of 5 up to two decimal = 2.23

Now,

3− 5
=
(3 − 5 ) × (3 − 5 )
3+ 5 (3 + 5 ) (3 − 5 )
( 5)
2
3−
=
9−5
( 5)
2
9+ − 2×3 5
=
4
14 − 6 5
=
4

Now, put the value of 5 = 2.23

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 14 − 6 × 2.23
=
4
0.62
=
4
= 0.155

9.(i)
Solution:
1764
2809
1764
Square root of
2809

1764 1764
=
2809 2809
Find square root of numerator and denominator and divide then
square root of 1764 .

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 1764 = 42

(ii)
Solution:

2809

2809 = 53
1764 42
=
2809 53
= 0.792

(iii)
Solution:
507
4107

507 =

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 507 = 22.5
Similarly finding square root of 4107

4107 = 22.08

507 22.5
Now =
4107 64.08
= 0.3511

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(iv)
Solution:

108 × 2028

219024

108 × 2028 =
468

(v)
Solution:

Square root of 0.01 + 0.0064

= 0.01 + 0.0064

First find square root of 0.0064

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 0.0064 = 0.08

0.01 + 0.0064 =0.01 + 0.08

= 0.09
Now again find square root of 0.009

= 0.3

10. (i)
Solution:

7.832

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We have to find square root up to two decimal place.

Required value of square root of 7.832 up to 2 decimal = 2.79

(ii)
Solution:
7.832
We have find root two significant digits

Required value up to two significant of digit = 2.7

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 CHAPTER-3-SQUARES AND SQUARE ROOTS

EXERCISE 3(C)

1.
(i)
Solution:
3051
For square of a number unit place digit should be 0, 1, 4, 5, 6 or 9.
So,
This can be square of a number.

(ii)
Solution:
2332
Unit place digit = 2
For square of a number unit place digit should be 0, 1, 4, 5, 6 or 9.
So,
This can’t be square of a number.

(iii)

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Solution:
5684
Unit place digit = 4
For square of a number unit place digit should be 0, 1, 4, 5, 6 or 9.
So, This can’t be a square of a number.

(iv)
Solution:
6908
Unit place digit = 8
For square of a number unit place digit should be 0, 1, 4, 5, 6 or 9.
So, This can’t be a square of a number.

(v)
Solution:
50699
Unit place digit = 9
For square of a number the unit place digit should be 0, 1, 4, 5, 6 or
9.
So, This can’t be a square of a number.

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2.
(i)
Solution:
57
Unit place digit = 7
If number has 1 or 9 at its unit place.
Then square of the number will have 1 at unit place.
So,
57 will not have 1 at unit place.

(ii)
Solution:
81
Unit place digit = 1
A number has 1 or 9 at unit place will have 1 at its unit place of its
square.
So,
81’s square will have 1 at unit place.

(iii)
Solution:

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139
Unit place digit = 9
A number with 1 or 9 at unit place will have 1 at its unit place of its
square.
So,
Square of 139 will have 1 at unit place.

(iv)
Solution:
73
Unit place digit = 3
A number with 1 or 9 at unit place will have 1 at its unit place of its
square.
So,
73 will have 1 at its square’s unit place.

(v)
Solution:
64
Unit place digit = 4
A number with 1 or 9 at unit place will have 1 at its unit place of its
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square.
So,
64 will have 1 at unit place of its square.

3.
(i)
Solution:
322
Unit place digit of 32 = 2
Square of 2 = 4
So, 322 will not have 1 at its unit place.

(ii)
Solution:
57 2
Unit place digit of 57 = 7
Square of 7 = 49 with 9 at its unit place
So, 57 2 will not have 1 at its unit place.

(iii)
Solution:

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692
Unit place digit of 69 = 9
Square of 9 = 81 have 1 at unit place

So, 692 will have 1 at its unit place.

(iv)
Solution:
3212
Unit place digit of 321 = 1
Square of 1 = 1
So, 3212 will have 1 at its unit place.

(v)
Solution:
2652
Unit place digit of 265 = 5
Square of 5 = 25 with 5 at unit place
So, 2652 will not have 1 at its unit place.

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4.
(i)
Solution:
35
Unit place digit = 5
If unit place digit of a number is 4 or 6, then square will always
have 6 at its unit’s place.
So,
35 will not have 6 at its unit place of its square.

(ii)
Solution:
23
Unit place digit = 3
If unit place digit of a number is 4 or 6, then square will always
have 6 at its unit’s place.
So,
Square 23 will not have 6 at its unit place.

(iii)
Solution:

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64
Unit place digit = 4
If unit place digit of a number is 4 or 6, then square will always
have 6 at its unit’s place.
So,
Square of 64 will have 6 at its unit place.

(iv)
Solution:
76
Unit place digit = 6
If unit place digit of a number is 4 or 6, then square will always
have 6 at its unit’s place.
So,
Square of 76 will have 6 at its unit place.

(v)
Solution:
98
Unit place digit = 8
If a number have 4 or 6 at its unit’s place, then square of the number

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will have 6 at its unit’s place.
So,
98 will not have 6 at its unit’s place of its unit’s place of its square.

5.
(i)
Solution:
262
Unit of place digit = 6
If a number have 4 or 6 at its unit’s place, then its square will have 6
at its unit’s place.
So,
262 will have 6 at its unit’s place.

(ii)
Solution:
492
Unit of place digit = 9
If a number have 4 or 6 at its unit’s place then its square will have 6
at its unit’s place.

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So,
492 will not have 6 at its unit’s place.

(iii)
Solution:
342
Unit of place digit = 4
If a number have 4 or 6 at its unit’s place then its square will have 6
at its unit’s place.
So,
342 will have 6 at its unit’s place.

(iv)
Solution:
432
Unit of place digit = 3
If a number have 4 or 6 at its unit’s place then its square will have 6
at its unit’s place.
So,
432 will not have at its unit’s place.

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(v)
Solution:
2442
Unit of place digit = 4
If a number have 4 or 6 at its unit’s place then its square will have 6
at its unit’s place.
So,
2442 will have 6 at its unit’s place.

6.
Solution:
If a number ends with ‘n’ zeroes; its square ends with ‘2n’ zeroes.
Here, n = 3
So,
Zero in its square = 2n = 2 × 3 = 6

7.
Solution:
If a number ends with ‘n’ zeroes then its square will have ‘2n’
zeroes.
Here, 2n = 10

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So,
10
n
= = 5
2
So,
Number have 5 zeroes.

8.
Solution:
If is not possible for the square of a number to end with 5 zeroes,
because the number of zeroes in square of a number is even number
(i.e. 2n), where n is number of zeroes in number.

9. (i)
Solution:
2162
Unit place digit = 2
A perfect square number have 0, 1, 4, 5, 6 or 9 at its unit’s place.
So,
2162 is not a perfect square.

(ii)

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Solution:
6843
Unit place digit = 3
A perfect square number have 0, 1, 4, 5, 6 or 9 at its unit’s place.
So,
6843 is not a perfect square.

(iii)
Solution:
9637
Unit place digit = 7
A perfect square number have 0, 1, 4, 5, 6 or 9 at its unit’s place.
So,
9137 is not a perfect square.

(iv)
Solution:
6598
Unit place digit = 8
A perfect square number have 0, 1, 4, 5, 6 or 9 at its unit’s place.

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So,
6598 is not a perfect square.

10. (i)
Solution:
23
23 is an odd number.
Square of an odd number is an odd number and square of an even
number is an even number.
So,
Square of 23 is an odd number.

(ii)
Solution:
54
54 is an even number.
Square of an odd number is an odd number and square of an even
number is an even number.
So,
Square of 54 is an even number.

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 (iii)
Solution:
76
76 is an even number.
Square of an odd number is an odd number and square of an even
number is an even number.
So,
Square of 76 is an even number.

(iv)
Solution:
75
75 is an odd number.
Square of an odd number is an odd number and square of an even
number is an even number.
So,
Square of 75 is an odd number.

11.
Solution:

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Number of zeroes in a square of a number is multiple of 2 (i.e. 2n).
Number of zeroes in 640 = 1
Number of zeroes in 81000 = 3
Number of zeroes in 3600000 = 5
As number of zeroes in each given numbers are not multiple of 2 so
these can’t be a perfect square.

12. (i)
Solution:
37 2 − 362
For any natural number n,

( n + 1) − n 2 = ( n + 1) + n
2

Here,
n = 36
n +1 = 37
So,
37 2 − 362 = ( 36 + 1) + 36
= 37 + 36
= 73

(ii)

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Solution:
852 − 842
For any natural number n,

( n + 1) − n 2 = ( n + 1) + n
2

Here,
n = 84
n +1 = 85
So,
852 − 842 = 85 + 84 = 169

(iii)
Solution:
1012 − 1002
For any natural number n,

( n + 1) − n 2 = ( n + 1) + n
2

Here,
n = 100
n +1 =101
So,
1012 − 1002 =101 + 100 = 201

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13. (i)
Solution:
Then sum of first n odd natural numbers = n 2
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
n = 12
2 2
Sum
= n= 12= 144

(ii)
Solution:
1 + 3 + 5 + 7 + 9 + ............36 + 41
The sum of first 20 odd natural numbers
= ( 20 )
2

= 400

(iii)
Solution:
1 + 3 + 5 + 7 + 9 + ............51 + 53
The sum of first 26 odd natural numbers
= ( 26 )
2

= 676

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14.
Solution:
For any three natural numbers p, q, r
if p 2 + q 2 =
r2
Then p, q, r are known as Pythagorean triplets.
Following are three sets of Pythagorean triplets
a) 3, 4, 5
32 + 42 =
52
b) 6, 8, 10

( 6 ) + (8)
2 2
102
=

c) 5, 12, 13

( 5) + (12 )
2 2
132
=

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