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Question 1

Which of the following numbers are perfect squares? 11, 12, 16, 32, 36

Solution:
11 is not a perfect square because it is a prime number.
12 is not a perfect square because its units digit is 2.
16 is a perfect square because 16 = 4 x 4. 
32 is not a perfect square because its units digit is 2.
36 is a perfect square because 36 = 6 x 6.

Solution:
256 and 1296 are squares of even numbers because they are even numbers.
121, 225 and 6561 are not squares of even numbers because they are odd numbers.

Question 3

Which of the following numbers are perfect squares? 100, 1000, 330550, 12345600000

Solution:
100 is a perfect square because the number of zeros in the end is even.
1000, 330550 and 12345600000 are not perfect squares because the number of zeros in the end is odd.

Question 4

Write down the correct number in the box:   [For every natural number n, we have 
       (n + 1)2 - n2 = (n + 1 - n)(n + 1 + n) = (n + 1) + n.]
        (a) 232 - 222     = ________
        (b) 1012 -1002  = _________
        (c) 5512 - 5502 = __________

Solution:
(a) 232 - 222     =23 + 22 = 45 
(b) 1012 -1002  =101 + 100 = 201
(c) 5512 - 5502  =551 + 550 = 1101

Question 5

Write down the correct number in each box:
        (a) 652 = 6 x 7(100) + 52 = 4200 + 25 =_________ 
        (b) 752 = 7 x 8(100) + 52 = 5600 + 25 =__________ 

Solution:
(a) 652 = 6x 7(100) + 52 = 4200 + 25 = 4225
(b) 752 = 7 x 8(100) + 52 = 5600 + 25 = 5625

Question 6

Which of the following triplets are Pythagorean?
        (i) (1, 2, 3)    (ii) (3, 4, 5)     (iii) (6, 8, 10)      
        [We know that the three natural numbers m, n, p are said to be Pythagorean triplets if m2 + n2 = p2.] 

Solution:
(i) 12 + 22 = 1 + 4 = 5 ¹ 32
(ii) 32 + 42 = 9 + 16 = 25 = 52
(iii) 62 + 82 = 36 + 64 = 100 = 102
Therefore, (3, 4, 5) and (6, 8, 10) are Pythagorean triplets.

Question 7

Which of the following triplets are Pythagorean?
        (i) (1, 1, 1)          (ii) (2, 2, 3)
        [We know that the three natural numbers m, n, p are said to be Pythagorean triplets if m2 + n2 = p2.] 

Solution:
(i) 12 + 12 = 1 + 1 = 2 ¹  12
(ii) 22 + 22 = 4 + 4 = 8 ¹  32
Therefore there are no Pythagorean triplets.

Question 8

Observe a pattern in the following and find the missing numbers:
                 222
       121 = _______
                1+2+1
        
                          (333)2
       12321 =_________
                    1+2+3+2+1

        1234321                =__________ 
        123454321            =__________
        12345654321        =__________
        1234567654321    =__________ 
        123456787654321=___________
                                                                  (999999999)2
        
12345678987654321 =
___________________________________
                                              1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1

Solution:
                  222
121    =   ______
              1+2+1

                 (333)2
12321 =   __________
            1+2+3+2+1

                      (4444)2
1234321 =   ______________
               1+2+3+4+3+2+1

                            (55555)2
123454321 = ___________________
                   1+2+3+4+5+4+3+2+1

                                      (666666)2
12345654321 =  ________________________
                       1+2+3+4+5+6+5+4+3+2+1
 
                                        (7777777)2
1234567654321 =  _____________________________
                            1+2+3+4+5+6+7+6+5+4+3+2+1

                                               (88888888)2
123456787654321 = _________________________________
                              1+2+3+4+5+6+7+8+7+6+5+4+3+2+1



Solution:
The squares of all natural between 80 and 90 are as follows:
812 = 6561
822 = 6724
832 = 6889
842 = 7056 
852 = 7225
862 = 7396
872 = 7569
882 = 7744
892 = 7921

         a) 25             b) 64       c) 81          d) 100            e) 169  
         f) 225            g) 400     h) 4900      i). 39 × 39       j) 320 × 320

Solution:
a) 25 = 52
... The square root of 25 is 5.

b) 64 = 82
... The square root of 64 is 8.

c) 81 = 92
... The square root of 81 is 9.

d)100 = 102
... The square root of 100 is 10.

e)169 = 132
... The square root of 169 is 13.

f) 225 = 152
... The square root of 225 is 15.

g)400 = 4 × 100 = 22 × 102 = 202
... The square root of 400 is 20.

h) 4900 = 49
× 100 = 72 × 102 = 702
... The square root of 4900 is 70.

i) The square root of 39 x 39 is 39.

j) The square root of 320 x 320 is 320.


         (a) 16      (b) 196     (c) 529   (d) 400      (e) 1764       (f) 4096  
         (g) 7744  (h) 11664 (i) 4900  (j) 47089  (k) 298116

Solution:
a)
   

16 = 2 x 2 x 2 x 2 = (22 x 22) = (2 2)2
... The square root of 16 = 2 ´ 2= 4
    
b)

   

196 = 2 x 2 x 7 x 7 = (22 x 72) = (2 x 7)2
... The square root of 196 = 2
× 7= 14
    
c)

   

529 = 23 x 23 = 232
... The square root of 529 = 23 
     
d)

  
 

400 = 2×2×2×2×5×5 = (22 x 22 x 52) = (2 x 2 x 5)2
... The square root of 400 = 2
×2×5= 20

e) 1764 = 2
×2×3×3×7×7
    
  

1764 =  2×2×3×3×7×7 = (22 x 32 x 72) = (2 x 3 x 7)2
... The square root of 1764 = 2
×3×7= 42

f)
  


... The square root of 4096=2×2×2×2×2×2= 64

g)
  


h)
   


(i)
   
   



j)

 


k)

   




Solution:

         

9408 = 2×2×2×2×2×3×7×7.

The prime factors 2 and 7 occur in pairs. 
But prime factor 3 doesn't have a pair.
Therefore, 3 is the smallest number by which 9408 
must be divided so that it becomes a perfect square.

... Perfect square = 9408¸3= 3136
                          = 2
×2×2×2×2×2×7×7

... Square root = 2
×2×2×7 = 56


Solution:
There are as many students in a row as there are rows in the auditorium means 



5929 = 7×11×11 


...77 rows are there in the auditorium.



Solution:
The amount paid by each student = The total number of students in the school.

  
... The amount paid by each student = (The total number of students in the school) 
                                                                      = Rs.2304 = 230400 paise

... The total number of students in the school = 230400 = 102304 
 

               
There are 480 students in the school.


Question 15

By just examining the units, can you tell which of the following cannot be perfect squares?

           (i) 1026
          (ii) 1028
         (iii) 1024
         (iv) 1022
          (v) 1023
         (vi) 1027

Solution:
(i), (ii), (iv), (v) and (vi) cannot be whole perfect squares. 
This is because square numbers do not end in the digits 2,3,7 or 8.

Question 16

Find the square roots of the perfect square 44944 by factorising.

Solution:

The prime factorization of 44944 is 2 x 2 x 2 x 2 x 53 x 53 = (22 x 22 x 532) = (2 x 2 x 53)2
Thus the square root of 44944 is 2 x 2 x 53 = 212.


            
Solution:
        
The prime factorization of 28900 is 2 × 2 × 5 × 5 × 17 × 17                 
Thus the square root of 28900 is 2
× 5 × 17 = 170

Question 18

If =24, find 2m + 1.

Solution:
Given =24

m = 24 x 24

m =  576

2m + 1 = 2(576) + 1 = 1152 + 1 = 1153

Question 19

Find the square root of the following rational number:


Solution:
           

Question 20

Find the square root of the following rational number:

Solution:


Question 21

Find the square root of the rational number:
                
Solution:


Question 22

If =2, find a.                              

Solution:
=2

a = × = 2

Question 23

Find the squares of the following rational numbers

         (i) (ii) (iii)                              

Solution:
(i)
(ii)
(iii)

Question 24

Find the smallest number that should be multiplied by 5408 to make it a perfect square.

Solution:

The prime factorization of 5408 is 2 × 2 × 2 × 2 × 2 × 13 × 13.
The number 2 does not have a pair and hence the smallest number that should be multiplied by 5408 to make it a perfect square is 2.



Solution:

             00
The square root of 0.04 is 0.2




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