Mathematics Chapter 6 Squares And Square Roots

Since, 1 x 1 = 1
         2 x 2 = 4
         3 x 3 = 9
         4 x 4 = 16
         5 x 5 = 25
         6 x 6 = 36
         7 x 7 = 49
Thus, 36 is a perfact square number between 30 and 40.
Since, 7 x 7 = 49 and 8 x 8 = 64. it mean there is no perfect number between 49 and 64, and thus there is not perfect number between 50 and 60.

(i) 1057

∵ The ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)

∴ 1057 cannot be a square number.

(ii) 23453

∵ The ending digit is 3 (which is not one of 0, 1, 4, 5, 6 and 9)

∴ 23453 cannot be a square number.

(iii) 7928

∵ The ending digit is 8 (which is not one of 0, 1, 4, 5, 6 and 9)

∴ 7928 cannot be a square number.

(iv) 222222

∵ The ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9)

∴ 222222 cannot be a square number,

(v) 1069

∵ The ending digit is 9.

∴ It may or may not be a sqaure number.
Also,   30 x 30 = 900
         31 x 31 = 691
         32 x 32 = 1024
         33 x 33 = 1089

i.e. No natural number between 1024 and 1089 is a square number.

∴ 1069 cannot be a square number.

(vi) 2061

∵ The ending digit is 1

∴ It may or may not be a square number.

∵ 45 × 45 = 2025

and 46 × 46 = 2116

i.e. No natural number between 2025 and 2116 is a square number.

∴ 2061 is not a square number.

We can write many numbers which do not end with 0, 1, 4, 5, 6 or 9. (i.e. which are not square number). Five such numbers can be:

1234, 4312, 5678, 87543, 1002007.

Any natural number ending in 0, 1, 4, 5, 6 or 9 can be or cannot be a square-number. Five such numbers are:

56790, 3671, 2454, 76555, 69209

Property 2. If a number has 1 or 9 in the unit’s place, then its square ends in 1.

For example:

(1)² = 1, (9)² = 81, (11)² = 121, (19)² = 361, (21)² = 441.

The squares of those numbers end in 1 which end in either 1 or 9. The squares of 161 and 109 would end in 1.

Property 3. When a square number ends in 6, then the number whose square it is, will have 4 or 6 in its unit place.

(i) 19²: Unit’s place digit = 9

∴ 19² would not have unit’s digit as 6.

(ii) 24²: Unit’s place digit = 4

∴ 24² would have unit’s digit as 6.

(iii) 26²: Unit’s place digit = 6

∴ 26² would have 6 as unit’s place.

(iv) 36²: Unit place digit = 6

∴ 36² would end in 6.

(v) 34²: Since, the unit place digit is 4

∴ 342 would have unit place digit as 6.

(i) ∵ Ending digit = 4 and 4² = 16

∴ (1234) will have 6 as the one’s digit.

(ii) ∵ Ending digit is 7 and 7² = 49

∴ (26387)² will have 9 as the one’s digit.

(iii) ∵ Ending digit is 8, and 8² = 64

∴ (52692)² will end in 4.

(iv) ∵ Ending digit is 0.

∴ (99880)² will end in 0.

(v) ∵ 2² = 4

 ∴ Ending digit of (21222)² is 4.

(vi) ∵ 6² = 36

 ∴ Ending digit of (9106)² is 6.

(i) 727

Since 727 is an odd number.

 ∴ It square is also an odd number.

(ii) 158

Since 158 is an even number.

 ∴ Its square is also an even number.

(iii) 269

Since 269 is an odd number.

 ∴ Its square is also an odd number.

(iv) 1980

Since 1980 is an even number.

 ∴ Its square is also an even number.

(i) In 60, number of zero is 1

 ∴ Its square will have 2 zeros.

(ii) ∵ There are 2 zeros in 400.

 ∴ Its square will have 4 zeros.

Property 6. The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.

Property 7. There are 2n non-perfect square numbers between the squares of the numbers n and n + 1.

(a) Between 9² and 10²

Here, n = 9 and n + 1 = 10

 ∴ Natural number between 9² and 10² are (2 × n) or 2 x 9, i.e. 18.

(b) Between 11² and 12²

Here, n = 11 and n + 1 = 12

 ∴ Natural numbers between 11² and 12² are (2 × n) or 2 × 11, i.e. 22.

(i) Between 100² and 101²
Here, n=100 ∴  n x2 = 100 x 2 = 200

∴ 200 non-square numbers lie between 100² and 101² .
(ii)  Between 90² and 91²
Here, n=90  ∴  2 x n = 2 x 90 or 180

∴  180 non-square numbers lie between 90 and 91.

(iii)  Between 1000² and 1001²
Here, n = 1000         ∴  2 x n = 2 x 1000 or 2000
∴  2000 non-square lie between 1000² and 10012.

(i) 121

∵            121 - 1 = 120                                       85-13 = 72
             120 - 3 = 117                                       72 - 15 = 57
             117 - 5 = 112                                       54 - 17 = 40
             112 - 7 = 105                                       40 - 19 = 21
             105 - 9 = 96                                         21 - 21 = 0
              96 - 11 = 85 
i.e. 121 = 1+3+5+7+9+11+13+15+17+19+21. Thus , 121 is a perfact square.
(ii) 55

∵               55 - 1 = 54                                    30 - 11 = 19
             54 - 3 = 51                                    19 - 13 = 6
             51 - 5 = 46                                     6 - 15 = -9
             46 - 7 = 39
             39 - 9 = 30
Since, 55 cannot be expressed as the sum of successive old numbers starting from 1.

∴  55 is not a perfact square.
(iii)    81
      Since,           81 - 1 = 80                            56 - 11 = 45
                         80 - 3 = 77                            45 - 13 = 32
                         77 - 5 = 72                            32 - 15 = 17
                         72 - 7 = 65                            17 - 17 = 0
                          65 - 9 =56

∴ 81 = 1+3+5+7+9+11+13+15+17
Thus,  81 is a perfact square.

(iv)      49
       Since,             49 - 1 = 48
                            48 - 3 = 45
                            45 - 5 = 40
                            40 - 7 = 33
                            33 - 9 = 24
                            24 - 11 = 13
                            13 -  13 = 0
      

∴ 49 = 1+3+5+7+9+11+13
Thus, 69  is not a perfact square.

(v) 69 
Since,         69 - 1 = 68                                 44 - 11 = 33
                 68 - 3 = 65                                 33 - 13 = 20
                 65 - 5 = 60                                 20 - 15 = 5
                 60 - 7 = 53                                 5 - 17 = 12
                 53 - 9 = 44

∴   69 cannot be expressed as the sum of consecutive odd num numbers starting fron 1. Thus, 69 is not a perfect square.

(i)  n = 21
∴   
      

∴  
or    21²  = 220 + 221 = 441

(ii)  n = 13
      

∴            13² = 85+84 = 169

           
Similarly,  
  (iii)            11² = 60 + 61 = 1216
          
  (iv)          19² = 180 + 181 = 361
            

No, it is not always true.
for example  (i)  5 + 6 = 11, 11 is not a perfect square.
                 (ii) 21 + 22 = 43, 43 is not a perfect square.

(i)   9² - 8² = 81-64 = 17 = 9 + 8
      10² - 9² = 100-81 = 19 = 10 + 9
      15² - 14² = 225-196 = 29 = 15 + 14
      101² - 100² = 10201-10000 = 201 = 101 + 100
(ii)   If (n+1) and (n-1) are two consecutive even or odd natural numbers, then (n+1) x (n-1) = n² -1
For example       
                10 x 12 = (11 - 1)  x (11+1) = 11² - 1
                11 x 13 = (12 - 1) x (12 + 1) = 12² -1
                 25 x 27 = (22 -1) x (26 - 1) x (26 + 1) = 26² - 1

Using above pattern we can write,

(i) (111111)² = 12345654321

(ii) (1111111)² = 1234567654321

Using the above pattern, we can have,

(i) 6666667² = 44444448888889

(ii) 66666667² = 4444444488888889

(i) ∵ 1 × 1 = 1

 ∴ The unit’s digit of (81 )² will be 1.

(ii) ∵ 2 × 2 = 4

 ∴ The unit’s digits of (272)² will be 4.

(iii) Since, 9 × 9 = 81

 ∴ The unit’s digit of (799)² will be 1.

(iv) Since, 3 × 3 = 9

 ∴ The unit’s digit of (3853)² will be 9.

(v) Since, 4 × 4= 16

 ∴ The unit’s digit of (1234)² will be 6.

(vi) Since 7 × 7 = 49

 ∴ The unit’s digit of (26387)² will be 9.

(vii) Since, 8 × 8 = 64

 ∴ The unit’s digit of (52698)² will be 4.

(viii) Since 0 × 0 = 0

 ∴ The unit’s digit of (99880)² will be 0.

(ix) Since 6 x 6 = 36

 ∴ The unit’s digit of (12796)² will be 6.

(x) Since, 5 x 5 = 25

 ∴ The unit’s digit of (55555)² will be 5.

(i) 1057

Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)

 ∴ 1057 is not a perfect square.

(ii) 23453

Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9).

 ∴ 23453 is not a perfect square.

(iii) 7928

Since, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9).

 ∴ 7928 is not a perfect square.

(iv) 222222

Since, the ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).

 ∴ 222222 is not a perfect square.

(v) 64000

Since, the number of zeros is odd.

 ∴ 64000 is not a perfect square.

(vi) 89722

Since, the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9).

 ∴ 89722 is not a perfect square.

(vii) 222000

Since, the number of zeros is odd.

 ∴ 222000 is not a perfect square.

(viii) 505050

The unit’s digit is odd zero.

 ∴ 505050 can not be a perfect square.

Since the square of an odd natural number is odd and that of an even number is an even number.

∴    (i)    The square of 431 is an odd number
            (∵  431 is an odd number)
     (ii)   The square of 2826 is an even nnumber.
            (∵ 2826 is an odd number)
     (iii)     The square of 7779 is an odd number
            (∵ 7779 is an odd number)
     (iv)  The square of 82004 is an even nnumber.
            (∵ 82004 is an odd number)

Observing the above pattern, we have

(i) (100001)² = 10000200001

(ii) (10000001)² = 100000020000001

Observing the above, we have

(i)(1010101)² = 1020304030201

(ii) 10203040504030201 = (101010101 )²

The missing numbers are

(i) 4² + 5² + 20² = 21² (ii) 5² + 6² + 30² = 31²

(iii) 6² + 7² + 42² = 43

(i)   The sum of first 5 odd numbers = 5²
                                               = 25
(ii)   The sum of first 10 odd numbers = 10²
                                                 = 100
(iii)   The sum of first 12 odd numbers = 12²
                                                 = 144

49 = 7² = Sum of first 7 odd numbers
         =  1 + 3+ 5 +7 +9 + 11 + 13

121 = 11² = Sum of first 11 odd numbers
         =  1 + 3+ 5 +7 +9 + 11 + 13 + 15 + 17 + 19 + 21

Since between n² and (n + l)², there are 2n non-square numbers.

 ∴ (i) Between 12² and 13², there are 2 × 12, i.e. 24 numbers

(ii) Between 25² and 26^:, there are 2 × 25, i.e. 50 numbers

(iii) Between 99² and 100², there are 2 × 99, i.e. 198 numbers

(27)² = (20 + 7)²
Using the formula (a + b)² = a² + 2ab + b² , we have
                         (20 + 7)² = (20)² + 2 x (20) x (7) + (7)²
                         = 400 + 280 + 49
                         = 729
Thus,
                     (27)² =729

Since, a Pythagorean triplet is given by 2n, n² - 1 and n² + 1.

∴                                   2n = 15 or n = is not an integer.
So, let us assume that
                                
                            n² - 1 = 15
or                             n²  = 15 + 1 = 16
or                              n² = 4² , i. e. n = 4
Now, the required Pythagorean triplet is
                            2n, n² - 1 and n² + 1
or,                         2(4), 4² - 1 and 4² + 1
or                                   8, 15 and 17

(i)    (15)² = 1 x (1 + 1) x 100 + 25
                = 1 x 2 x 100 + 25
                = 200 + 25 = 225
(ii)    (95)² = 9 x (9 + 1) x 100 + 25
                = 9 x 10 x 100 + 25
                = 9000 + 25 = 9025
(iii)   (105)² = 10 x (10 + 1) x 100 + 25
                = 10 x 11 x 100 + 25
                = 11000 + 25 = 11025
(iv)    (205)² = 20 x (20 + 1) x 100 + 25
                = 20 x 21 x 100 + 25
                = 42000 + 25 = 42025  

(i)     (32)² = (30 + 2)² 
                = 30² + 2(30)(2)+(2)²
                = 900 + 120 + 4
                = 1024
(i)     (35)² = (30 + 5)² 
                = 30² + 2(30)(5)+(5)²
                = 900 + 300 + 25
                = 1225
Second method
               (35)² = 3 x (3 + 1) x 100 + 25 
                = 3 x 4 x 100 + 25
                = 1200 + 25
                = 1225\
(iii)       (86)² = (80 + 6)² 
                = (80)² + 2(80)(6)+(6)²
                = 6400 + 960 + 9
                = 8649
(iv)    (93)² = (90 + 3)² 
                = (90)² + 2(90)(3)+(3)²
                = 8100 + 540 + 9
                = 8649
(v)    (71)² = (70 + 1)² 
                = (70)² + 2(70)(1)+(1)²
                = 4900 + 140 + 1
                = 5041
(vi)   (46)² = (40 + 6)² 
                = (40)² + 2(40)(6)+(6)²
                = 1600 + 480 + 36
                = 2116

(i)  Let         2n = 6                  ∴  n = 3
     Now,      n² -1 = 3² - 1 = 8
     and        n² +1 = 3² + 1 = 10
Thus the required Pythagorean triplet is 6,8,10.
(ii)  Let         2n = 14                  ∴  n = 7
      Now,      n² -1 = 7² - 1 = 48
      and        n² +1 = 7² + 1 = 50
Thus the required Pythagorean triplet is 14,48,50.
(iii)  Let         2n = 16                  ∴  n = 8
       Now,      n² -1 = 8² - 1 = 63
       and        n² +1 = 8² + 1 = 65
∴ The required Pythagorean triplet is 16,63,65.
(iv)  Let         2n = 18                  ∴  n = 9
     Now,      n² -1 = 9² - 1 = 80
     and        n² +1 = 9² + 1 = 82
∴ The required Pythagorean triplet is 18,80,82.

121, 144, 169, 196, 225, 256 and 289

17, 18, 19, 20, 21, 22, 23 and 24

11

60 + 61

We have P = Rs 2,10,000   R=5% p.a.,  T = 2 Years

∵ Reduction is there, we use the following formula:

∴                       
                            = Rs
                            = Rs 210000
                            = Rs (525 ) = Rs 1,89,525

∴      The value of the car after 2 years = Rs 1,89,525

(15)²

8, 15, 17

12, 35, 37

(i) The square root of 121 is 11.

(ii) The square root of 196 is 14.

(i) Since (–1) × (–1) = 1

i.e. (–1)² = 1

 ∴ Square root of 1 can also be –1.

Similarly,

(ii) Yes (–2) is a square root of 4.

(iii) Yes (–9) is a square root of 81.
Since we have to consider the positive square roots only and the symbol for a  positive square root is .

∴  = 4 (and not (-4)

    Similarly,     means, the positive square root of 25 i.e. 5.
We can also find the square root by subtracting successive odd numbers starting from 1.

(i) Subtracting the successive odd numbers from 121, we have
                  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

∴    = 11               ( ∵   We had subtract the first 11 odd numbers)

(ii)   ∵                          55 - 1 = 54                       54 - 3 = 51
                            51 - 5 = 46                        46 - 7 = 39
                            39 - 9 = 30                      30 - 11 = 19
                             19 - 13 = 6                     6 - 15 = -9

and we  do not reach to o

∴ 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

and we have obtained 0 after subtracting 6 successive odd numbers.

∴  36 is a perfect square.
Thus, = 6

∵  We have obtained o after successive subtraction of 7 odd numbers

∴   49 is a perfect square.
Thus, 

∴  90 is not a perfect square.

The possible digit at one’s place of the square root of:
(i) 9801 can be 1 or 9.
( ∵ 1 x 1 = 1 and 9 x 9 = 81)
(ii) 99856 can be 4 or 6.
( ∵ 4 x 4 = 16 and 6 x 6 = 36)
(iii) 99001 can be 1 or 9
( ∵ 1 x 1 = 1 and 9 x 9 = 81)
(iv) 657666025 can be 5.
( ∵  5 x 5 = 25)

We know that the ending digit of perfect square is 0, 1, 4, 5, 6, and 9.

 ∴ A number ending in 2, 3, 7 or 8 can never be a perfect square.

 ∴ (i) 153, cannot be a perfect square.

(ii) 257, cannot be a perfect square.

(iii) 408, cannot be a perfect square.

(iv) 441, can be a perfect square.

Thus, (i) 153, (ii) 257 and (iii) 408 are surely not perfect squares.

(i)
          
      We have      100 - 1 = 99                 99 - 3 = 96            96 - 5 = 91
                        91 - 7 = 84                  84 - 9 = 75            75 - 11 = 64
                        64 - 13 = 51                 51 - 15 = 36           36 - 17 = 19
                         19 - 19 = 0

∵ We reach at 0 by successive subtraction of 10 odd numbers.

∴  = 10

∵ We reach at 0 by successive subtraction of 13 odd numbers.

∴  = 13

Yes it is true that,
Number of digit of the perfect square     n
Number of digits of the square root  (when 'n' is even)
Number of digit of the perfect square   n
Number of digits of the square root  (when 'n' is odd)
Example    529 (is perfect square) and  n = 3 (even number)

∴  Number of digits if its square root  =  

                                                 = = 2
Also, square root of 529 = 23 (2-digits).

1296 (is perfect square) and n = 4 (even number)

∴ Number of digits of its square root  = 

                                                 = = 2
Now                                   = 36 (2-digits)

(i) 25600

∵ n = 5 ( an odd number

∴ Its square root will have  
I.e.    digits

(ii)   100000000

∵ n = 9 odd number

∴  Number of digits of its square root =

                                                 =
(iii)   36864

∵  n = 5 odd number

∴  Number of digits in its square root =
                                   
                                                 =

(i)  

∵ 10² = 100, 9² = 81, 8² = 64
and 80 is between 64 and 81.
i.e. 64 < 80 < 81
or 8² < 80 < 9²
or
Thus, lies between 8 and 9.

We know that  30² = 900, 31² = 961, 32² = 1024

∴  1000 lies between 961 and 1024.
i.e. 916 < 1000 < 1024
or 31² < 1000 < 32²
or
Thus, lies between 31 and 32.

(iii)  

We have  18² = 324, 19² = 361

Since, 350 lies between 324 and 316.
i.e. 324 < 350 < 361
or 18² < 350 < 19²
or
Thus, lies between 18 and 19.

(iv)  

∵  22² = 484  and 23² = 529
Since. 500 lies between 484 and 529
i.e. 484 < 500 < 529
or 22² < 500 < 23²
or
Thus, lies between 22 and 23.

If ‘n’ stands for number of digits in the given number, then

(i) For 64, n = 2 [even number]

∴ Number of digit is its square root =

(ii) For 144, n=3 (odd number)
∴ Number of digit is its square root =

(iii)  For 4489, n=4 (even number)
∴ Number of digit is its square root =

(iv) For 27225, n=5 (odd number)
∴ Number of digit is its square root =

(v) For 390625, n=6 (even number)
∴ Number of digit is its square root =

1

5

64

100

(a) 9, (b) 12, (c) 15, (d) 18, (e) 84

80

No, 3 : 84

3 : 56

900

(23) 23  (ii) 64 (iii) 120 (iii) 120 (iv) 27 (v) 36

198

Initial cost (value) of the scooter (P) = Rs 42000
                          Depreciation rate = 8% p.a.
                                
                                Time = 1 year n = 1
                        Using A = , we have
                                A = Rs
                                   = Rs 42000
                                   = Rs 420 92 = Rs 38640
Thus, the value of the scooter after 1 year will be Rs 38640.

9801

69 : 37

(a) 4.2   (b) 3.5

48 m

49

Since,
         
          1 x 1 = 1                       2 x 2 = 4
          3 x 3 = 1                       4 x 4 = 16
          5 x 5 = 25                     6 x 6 = 36
          7 x 7 = 49                    8 x 8 = 64

∴ Perfect squares between 30 adn 50 are 36 adn 49.

(i) in 2346, the ending digit = 5

∵ 5² = 25

∴ (2345)² will have as 5 as the one's digit

(ii) in 3456, the ending digit = 6

∵ (6)² = 36

∴ (3456)² will have as 6 as the one's digit

(i) We have n = 23

∴
        =

∴      
  23² = 264 + 265

(ii)    We have n = 17

∴ 
         =

∴    17² = 144 + 145

(iii)   We have n = 27

∴     

          

∴         (27)² = 364 + 365

We know that, between n² and (n + l)², there are 2n non-square numbers.
(i) Between 27 and 28, there are 2 × 27, i.e 54 numbers.

(ii) Between 16 and 17, there are 2 × 16, i.e. 32 numbers.

(iii) Between 97 and 98, there are 2 × 97, i.e. 194 numbers.

We have   169 = 13² 
               = Sum of first odd numbers
               = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25
               = (1 + 25) + (3 + 23) + (5 + 21) + (7+ 19) + (9 + 17) + (11 + 15) +13
               = 26 + 26 + 26 + 26 +26 + 26 + 13
               =  (6 x 26) + 13 = 156 + 13 = 169

When ‘n’ is a member of a Pythagorean triplet, then the triplet is
n² -1,    2n,    n² +1
or      (13² - 1),     (2 x 13),    (13² + 1)
or      (169 - 1),     (26),         (169 + 1)
or       168,            26           and 170     

(i) 2          (ii) 4

180 + 180 

(i) 52                                (ii) 62

841

20, 24, 26

A.

1

A.

an even number

C.

4

B.

n²

C.

4

B.

2n, n² – 1 and n² + 1

B.

C.