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Page No 3.18:

Question 1:

The following numbers are not perfect squares. Give reason.
(i) 1547
(ii) 45743
(iii) 8948
(iv) 333333

Answer:

A number ending with 2, 3, 7 or 8 cannot be a perfect square.
(i) Its last digit is 7. Hence, 1547 cannot be a perfect square.
(ii) Its last digit is 3. Hence, 45743 cannot be a perfect square.
(iii) Its last digit is 8. Hence, 8948 cannot be a perfect square.
(iv) Its last digit is 3. Hence, 333333 cannot be a perfect square.

Page No 3.18:

Question 2:

Show that the following numbers are not perfect squares:
(i) 9327
(ii) 4058
(iii) 22453
(iv) 743522

Answer:

A number ending with 2, 3, 7 or 8 cannot be a perfect square.
(i) Its last digit is 7. Hence, 9327 is not a perfect square.
(ii) Its last digit is 8. Hence, 4058 is not a perfect square.
(iii) Its last digit is 3. Hence, 22453 is not a perfect square.
(iv) Its last digit is 2. Hence, 743522 is not a perfect square.

Page No 3.18:

Question 3:

The square of which of the following numbers would be an odd number?
(i) 731
(ii) 3456
(iii) 5559
(iv) 42008

Answer:

The square of an odd number is always odd.
(i) 731 is an odd number. Hence, its square will be an odd number.
(ii) 3456 is an even number. Hence, its square will not be an odd number.
(iii) 5559 is an odd number. Hence, its square will not be an odd number.
(iv) 42008 is an even number. Hence, its square will not be an odd number.

Hence, only the squares of 731 and 5559 will be odd numbers.



Page No 3.19:

Question 4:

What will be the units digit of the squares of the following numbers?
(i) 52
(ii) 977
(iii) 4583
(iv) 78367
(v) 52698
(vi) 99880
(vii) 12796
(viii) 55555
(ix) 53924

Answer:

The units digit is affected only by the last digit of the number. Hence, for each question, we only need to examine the square of its last digit.
(i) Its last digit is 2. Hence, the units digit is 22, which is equal to 4.
(ii) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 72), which is 9.
(iii) Its last digit is 3. Hence, the units digit is 32, which is equal to 9.
(iv) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 72), which is 9.
(v) Its last digit is 8. Hence, the units digit is the last digit of 64 (64 = 82), which is 4.
(vi) Its last digit is 0. Hence, the units digit is 02, which is equal to 0.
(vii) Its last digit is 6. Hence, the units digit is the last digit of 36 (36 = 62), which is 6.
(viii) Its last digit is 5. Hence, the units digit is the last digit of 25 (25 = 52), which is 5.
(ix) Its last digit is 4. Hence, the units digit is the last digit of 16 (16 = 42), which is 6.

Page No 3.19:

Question 5:

Observe the following pattern
              1 + 3 = 22
       1 + 3 + 5 = 32
1 + 3 × 5 + 7 = 42
and write the value of 1 + 3 + 5 + 7 + 9 + ... upto n terms.

Answer:

From the pattern, we can say that the sum of the first n positive odd numbers is equal to the square of the n-th positive number. Putting that into formula:
1 + 3 + 5 + 7 + ...  n =  n2, where the left hand side consists of n terms.

Page No 3.19:

Question 6:

Observe the following pattern
22 − 12 = 2 + 1
32 − 22 = 3 + 2
42 − 32 = 4 + 3
52 − 42 = 5 + 4
and find the value of
(i) 1002 − 992
(ii) 1112 − 1092
(iii) 992 − 962

Answer:

From the pattern, we can say that the difference between the squares of two consecutive numbers is the sum of the numbers itself.
In a formula:
 
n+12 - n2 = n+1 + n

Using this formula, we get:
(i) 1002 − 992   = (99 + 1) + 99
                        = 199
(ii) 1112 − 1092 = 1112 − 1102 + 1102 − 1092
                        = (111 + 110) + (110 + 109)
                        = 440
(iii) 992 − 962 = 992 − 982 + 982 − 972 + 972 − 962
                         = 99 + 98 + 98 + 97 + 97 + 96
                         = 585

Page No 3.19:

Question 7:

Which of the following triplets are pythagorean?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(v) (16, 63, 65)
(vi) (12, 35, 38)

Answer:

Only (i), (ii), (iv) and (v) are Pythagorean triplets.
A triplet (a, b, c) is called Pythagorean if the sum of the squares of the two smallest numbers is equal to the square of the biggest number.

(i) The two smallest numbers are 8 and 15. The sum of their squares is:
82 + 152 = 289 = 172
Hence, (8, 15, 17) is a Pythagorean triplet.

(ii) The two smallest numbers are 18 and 80. The sum of their squares is:
182 + 802 = 6724 = 822
Hence, (18, 80, 82) is a Pythagorean triplet.

(iii) The two smallest numbers are 14 and 48. The sum of their squares is:
142 + 482 = 2500, which is not equal to 512 = 2601
Hence, (14, 48, 51) is not a Pythagorean triplet.

(iv) The two smallest numbers are 10 and 24. The sum of their squares is:
102 + 242 = 676 = 262
Hence, (10, 24, 26) is a Pythagorean triplet.

(v) The two smallest numbers are 16 and 63. The sum of their squares is:
162 + 632 = 4225 = 652
Hence, (16, 63, 65) is a Pythagorean triplet.

(vi) The two smallest numbers are 12 and 35. The sum of their squares is:
122 + 352 = 1369, which is not equal to 382 = 1444
Hence, (12, 35, 38) is not a Pythagorean triplet.

Page No 3.19:

Question 8:

Observe the following pattern
1×2+2×3=2×3×43
1×2+2×3+3×4=3×4×53
1×2+2×3+3×4+4×5=4×5×63
and find the value of
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6)

Answer:

The RHS of the three equalities is a fraction whose numerator is the multiplication of three consecutive numbers and whose denominator is 3.
If the biggest number (factor) on the LHS is 3, the multiplication of the three numbers on the RHS begins with 2.
If the biggest number (factor) on the LHS is 4, the multiplication of the three numbers on the RHS begins with 3.
If the biggest number (factor) on the LHS is 5, the multiplication of the three numbers on the RHS begins with 4.
Using this pattern, (1 x 2) + (2 x 3) + (3 x 4) + (4 x 5) + (5 x 6) has 6 as the biggest number. Hence, the multiplication of the three numbers on the RHS will begin with 5. Finally, we have:

1 × 2 + 2 ×3 + 3 × 4 + 4 × 5 + 5 × 6 = 5 × 6 × 73 = 70

Page No 3.19:

Question 9:

Observe the following pattern
               1=121×1+1          1+2=122×2+1     1+2+3=123×3+11+2+3+4=124×4+1
and find the values of each of the following:
(i) 1 + 2 + 3 + 4 + 5 + ... + 50
(ii) 31 + 32 + ... + 50

Answer:

Observing the three numbers for right hand side of the equalities:
The first equality, whose biggest number on the LHS is 1, has 1, 1 and 1 as the three numbers.
The second equality, whose biggest number on the LHS is 2, has 2, 2 and 1 as the three numbers.
The third equality, whose biggest number on the LHS is 3, has 3, 3 and 1 as the three numbers.
The fourth equality, whose biggest number on the LHS is 4, has 4, 4 and 1 as the three numbers.
Hence, if the biggest number on the LHS is n, the three numbers on the RHS will be n, n and 1.
Using this property, we can calculate the sums for (i) and (ii) as follows:

(i) 1 + 2 + 3 +........ + 50 = 12 × 50 × (50 + 1) = 1275 

(ii) The sum can be expressed as the difference of the two sums as follows:

31 + 32 + .....+ 50 = (1 + 2 + 3 + ......+ 50) - ( 1 + 2 + 3 + ......+30)

The result of the first bracket is exactly the same as in part (i).
     
1 + 2 + ....+ 50 = 1275

Then, the second bracket:
      1+2+......+30 = 1230×30+1 = 465
Finally, we have:
31 + 32 + .... + 50 = 1275 - 465  = 810



Page No 3.20:

Question 10:

Observe the following pattern
                    12=161×1+1×2×1+1             12+22=162×2+1×2×2+1      12+22+32=163×3+1×2×3+112+22+32+42=164×4+1×2×4+1
and find the values of each of the following:
(i) 12 + 22 + 32 + 42 + ... + 102
(ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122

Answer:

Observing the six numbers on the RHS of the equalities:
The first equality, whose biggest number on the LHS is 1, has 1, 1, 1, 2, 1 and 1 as the six numbers.
The second equality, whose biggest number on the LHS is 2, has 2, 2, 1, 2, 2 and 1 as the six numbers.
The third equality, whose biggest number on the LHS is 3, has 3, 3, 1, 2, 3 and 1 as the six numbers.
The fourth equality, whose biggest number on the LHS is 4, has numbers 4, 4, 1, 2, 4 and 1 as the six numbers.
Note that the fourth number on the RHS is always 2 and the sixth number is always 1. The remaining numbers are equal to the biggest number on the LHS.
Hence, if the biggest number on the LHS is n, the six numbers on the RHS would be n, n, 1, 2, n and 1.
Using this property, we can calculate the sums for (i) and (ii) as follows:

(i) 12 + 22 + .......+ 102 = 16×10×10+1×2×10+1
                                           
                                            = 16 × 10 × 11 × 12  = 385.

(ii) The sum can be expressed as the difference of the two sums as follows:

52 + 62 +.......+ 122 =  12 + 22 +......+122 -  12 + 22 +......+42

The sum of the first bracket on the RHS:

                                  12 + 22 +.....+122 = 1612×(12+1)×(2×12+1)                                    = 650    
                                     
The second bracket is:

                                     12 + 22 + ......+ 42  = 16×4×4+1×2×4+1 =  16 × 4 × 5 × 9 = 30
                                    
Finally, the wanted sum is:
 
                                     52 + 62 + ...... + 122  = (12 + 22 +..... + 122) - ( 12 + 22 +..... + 122)  =650 - 30 = 620

Page No 3.20:

Question 11:

Which of the following numbers are squares of even numbers?
121, 225, 256, 324, 1296, 6561, 5476, 4489, 373758

Answer:

The numbers whose last digit is odd can never be the square of even numbers. So, we have to leave out 121, 225, 6561 and 4489, leaving only 256, 324, 1296, 5476 and 373758. For each number, use prime factorisation method and make pairs of equal factors.
(i) 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
             = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2)
There are no factors that are not paired. Hence, 256 is a perfect square. The square of an even number is always even. Hence, 256 is the square of an even number.

(ii) 324 = 2 x 2 x 3 x 3 x 3 x 3
              = (2 x 2) x (3 x 3) x (3 x 3)
There are no factors that are not paired. Hence, 324 is a perfect square. The square of an even number is always even. Hence, 324 is the square of an even number.

(iii)1296 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3
                 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3)
There are no factors that are not paired. Hence, 1296 is a perfect square. The square of an even number is always even. Hence, 1296 is the square of an even number.

(iv) 5476 = 2 x 2 x 37 x 37
                 = (2 x 2) x (37 x 37)
There are no factors that are not paired. Hence, 5476 is a perfect square. The square of an even number is always even. Hence, 5476 is the square of an even number.

(v) 373758 = 2 x 3 x 7 x 11 x 809
Here, each factor appears only once, so grouping them into pairs of equal factors is not possible. It means that 373758 is not the square of an even number.

Hence, the numbers that are the squares of even numbers are 256, 324, 1296 and 5476.

Page No 3.20:

Question 12:

By just examining the units digits, can you tell which of the following cannot be whole squares?
(i) 1026
(ii) 1028
(iii) 1024
(iv) 1022
(v) 1023
(vi) 1027

Answer:

If the units digit of a number is 2, 3, 7 or 8, the number cannot be a whole square.
(i) 1026 has 6 as the units digit, so it is possibly a perfect square.
(ii) 1028 has 8 as the units digit, so it cannot be a perfect square.
(iii) 1024 has 4 as the units digit, so it is possibly a perfect square.
(iv) 1022 has 2 as the units digit, so it cannot be a perfect square.
(v) 1023 has 3 as the units digit, so it cannot be a perfect square.
(vi) 1027 has 7 as the unit digit, so it cannot be a perfect square.
Hence, by examining the units digits, we can be certain that 1028, 1022, 1023 and 1027 cannot be whole squares.

Page No 3.20:

Question 13:

Write five numbers for which you cannot decide whether they are squares.

Answer:

A number whose unit digit is 2, 3, 7 or 8 cannot be a perfect square.
On the other hand, a number whose unit digit is 1, 4, 5, 6, 9 or 0 might be a perfect square (although we will have to verify whether it is a perfect square or not).
Applying the above two conditions, we cannot quickly decide whether the following numbers are squares of any numbers:
1111, 1444, 1555, 1666, 1999

Page No 3.20:

Question 14:

Write five numbers which you cannot decide whether they are square just by looking at the unit's digit.

Answer:

A number whose unit digit is 2, 3, 7 or 8 cannot be a perfect square.
On the other hand, a number whose unit digit is 1, 4, 5, 6, 9 or 0 might be a perfect square although we have to verify that.
Applying these two conditions, we cannot determine whether the following numbers are squares just by looking at their unit digits:
1111, 1001, 1555, 1666 and 1999

Page No 3.20:

Question 15:

Write true (T) or false (F) for the following statements.
(i) The number of digits in a square number is even.
(ii) The square of a prime number is prime.
(iii)  The sum of two square numbers is a square number.
(iv) The difference of two square numbers is a square number.
(v) The product of two square numbers is a square number.
(vi) No square number is negative.
(vii) There is no square number between 50 and 60.
(viii) There are fourteen square number upto 200.

Answer:

(i) False
Example: 100 is the square of a number but its number of digits is three, which is not an even number.

(ii) False
If p is a prime number, its square is p2, which has at least three factors: 1, p and p2. Since it has more than two factors, it is not a prime number.

(iii) False
1 is the square of a number (1 = 12). But  1 + 1 = 2, which is not the square of any number.

(iv) False
4 and 1 are squares (4  = 22, 1 = 12). But  4 − 1 = 3, which is not the square of any number.

(v) True
If a2 and b2 are two squares, their product is a2 x b2 = (a x b)2, which is a square.

(vi) True
The square of a negative number will be positive because negative times negative is positive.

(vii) True
72 = 49 and 82 = 64. 7 and 8 are consecutive numbers and hence there are no square numbers between 50 and 60.

(viii) True
142 is equal to 196, which is below 200. There are 14 square numbers below 200.



Page No 3.32:

Question 1:

Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication:
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96

Answer:


(i) Here, a = 2, b = 5
Step 1. Make 3 columns and write the values of a2, 2 x a x b, and b2 in these columns.

Column I Column II Column III
a2 2 x a x b b2
4 20 25
 
Step 2. Underline the unit digit of b2 (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II).
Column I Column II Column III
a2 2 x a x b b2
4 20 + 2 25
  22  
 
Step 3. Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a2 in Column I.
Column I Column II Column III
a2 2 x a x b b2
4 + 2 20 + 2 25
6 22  
 
Step 4. Underline the number in Column I.
Column I Column II Column III
a2 2 x a x b b2
4 + 2 20 + 2 25
6 22  
 
Step 5. Write the underlined digits at the bottom of each column to obtain the square of the given number.
In this case, we have:
252 = 625
Using multiplication:
    25
    25
  125
  50  
 625
This matches with the result obtained by the column method.

(ii) Here, a = 3, b = 7
Step 1. Make 3 columns and write the values of a2, 2 x a x b, and b2 in these columns.
Column I Column II Column III
a2 2 x a x b b2
9 42 49
 
Step 2. Underline the unit digit of b2 (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II).
Column I Column II Column III
a2 2 x a x b b2
9 42 + 4 49
  46  
 
Step 3. Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a2 in Column I.
Column I Column II Column III
a2 2 x a x b b2
9 + 4 42 + 4 49
13 46  
 
Step 4. Underline the number in Column I.
Column I Column II Column III
a2 2 x a x b b2
9 + 4 42 + 4 49
13 46  
 
Step 5. Write the underlined digits at the bottom of each column to obtain the square of the given number.
In this case, we have:
372 = 1369
Using multiplication:
    37
    37
  259
111 
1369
This matches with the result obtained using the column method.

(iii) Here, a = 5, b = 4
Step 1. Make 3 columns and write the values of a2, 2 x a x b and b2 in these columns.
Column I Column II Column III
a2 2 x a x b b2
25 40 16
 
Step 2. Underline the unit digit of b2 (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II).
Column I Column II Column III
a2 2 x a x b b2
25 40 + 1 16
  41  
 
Step 3. Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a2 in Column I.
Column I Column II Column III
a2 2 x a x b b2
25 + 4 40 + 1 16
29 41  
 
Step 4. Underline the number in Column I.
Column I Column II Column III
a2 2 x a x b b2
25 + 4 40 + 1 16
29 41  
 
Step 5. Write the underlined digits at the bottom of each column to obtain the square of the given number.
In this case, we have:
542 = 2916
Using multiplication:
    54
    54
  216
270 
2916
This matches with the result obtained using the column method.

(iv) Here, a = 7, b = 1
Step 1. Make 3 columns and write the values of a2, 2 x a x b and b2 in these columns.
Column I Column II Column III
a2 2 x a x b b2
49 14 1
 
Step 2. Underline the unit digit of b2 (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II).
Column I Column II Column III
a2 2 x a x b b2
49 14 + 0 1
  14  
 
Step 3. Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a2 in Column I.
Column I Column II Column III
a2 2 x a x b b2
49 + 1 14 + 0 1
50 14  
 
Step 4. Underline the number in Column I.
Column I Column II Column III
a2 2 x a x b b2
49 + 1 14 + 0 1
50 14  
 
Step 5. Write the underlined digits at the bottom of each column to obtain the square of the given number.
In this case, we have:
712 = 5041
Using multiplication:
    71
    71
    71
497  
5041
This matches with the result obtained using the column method.

(v) Here, a = 9, b = 6
Step 1. Make 3 columns and write the values of a2, 2 x a x b and b2 in these columns.
Column I Column II Column III
a2 2 x a x b b2
81 108 36
 
Step 2. Underline the unit digit of b2 (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II).
Column I Column II Column III
a2 2 x a x b b2
81 108 + 3 36
  111  
 
Step 3. Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a2 in Column I.
Column I Column II Column III
a2 2 x a x b b2
81 + 11 108 + 3 36
92 111  
 
Step 4. Underline the number in Column I.
Column I Column II Column III
a2 2 x a x b b2
81 + 11 108 + 3 36
92 111  
 
Step 5. Write the underlined digits at the bottom of each column to obtain the square of the given number.
In this case, we have:
962 = 9216
Using multiplication:
    96
    96
  576
864  
9216
This matches with the result obtained using the column method.

Page No 3.32:

Question 2:

Find the squares of the following numbers using diagonal method:
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171

Answer:

(i)

982 = 9604
(ii)

2732 = 74529

(iii)

3482 = 121104

(iv)

2952 = 87025

(v)

1712 = 29241

Page No 3.32:

Question 3:

Find the squares of the following numbers:
(i) 127
(ii) 503
(iii) 451
(iv) 862
(v) 265

Answer:

We will use visual method as it is the most efficient method to solve this problem.

(i) We have:
127 = 120 + 7
Hence, let us draw a square having side 127 units. Let us split it into 120 units and 7 units.

Hence, the square of 127 is 16129.

(ii) We have:
503 = 500 + 3
Hence, let us draw a square having side 503 units. Let us split it into 500 units and 3 units.

Hence, the square of 503 is 253009.

(iii) We have:
451 = 450 + 1
Hence, let us draw a square having side 451 units. Let us split it into 450 units and 1 units.

Hence, the square of 451 is 203401.

(iv) We have:
862 = 860 + 2
Hence, let us draw a square having side 862 units. Let us split it into 860 units and 2 units.

Hence, the square of 862 is 743044.

(v) We have:
265 = 260 + 5
Hence, let us draw a square having side 265 units. Let us split it into 260 units and 5 units.

Hence, the square of 265 is 70225.

Page No 3.32:

Question 4:

Find the squares of the following numbers:
(i) 425
(ii) 575
(iii) 405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995

Answer:

Notice that all numbers except the one in question (vii) has 5 as their respective unit digits. We know that the square of a number with the form n5 is a number ending with 25 and has the number n(n + 1) before 25.

(i) Here, n = 42
n(n + 1) = (42)(43) = 1806
4252 = 180625

(ii) Here, n = 57
n(n + 1) = (57)(58) = 3306
5752 = 330625

(iii) Here n = 40
n(n + 1) = (40)(41) = 1640
4052 = 164025

(iv) Here n = 20
n(n + 1) = (20)(21) = 420
2052 =  42025

(v) Here n = 9
n(n + 1) = (9)(10) = 90
952 = 9025

(vi) Here n = 74
n(n + 1) = (74)(75) = 5550
7452 = 555025

(vii) We know:
The square of a three-digit number of the form 5ab = (250 + ab)1000 + (ab)2
5122 = (250+12)1000 + (12)2 = 262000 + 144 = 262144

(viii) Here, n = 99
n(n + 1) = (99)(100) = 9900
9952 = 990025

Page No 3.32:

Question 5:

Find the squares of the following numbers using the identity (a + b)2 = a2 + 2ab + b2:
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605

Answer:

(i) On decomposing:
405 = 400 + 5
Here, a = 400 and b = 5
Using the identity (a + b)2 = a2 + 2ab + b2:
4052 = (400 + 5)2 = 4002 + 2(400)(5) + 52 = 160000 + 4000 + 25 = 164025

(ii) On decomposing:
510 = 500 + 10
Here, a = 500 and b = 10
Using the identity (a + b)2 = a2 + 2ab + b2:
5102 = (500 + 10)2 = 5002 + 2(500)(10) + 102 = 250000 + 10000 + 100 = 260100

(iii) On decomposing:
1001 = 1000 + 1
Here, a = 1000 and b = 1
Using the identity (a + b)2 = a2 + 2ab + b2:
10012 = (1000 + 1)2 = 10002 + 2(1000)(1) + 12 = 1000000 + 2000 + 1 = 1002001

(iv) On decomposing:
209 = 200 + 9
Here, a = 200 and b = 9
Using the identity (a + b)2 = a2 + 2ab + b2:
2092 = (200 + 9)2 = 2002 + 2(200)(9) + 92 = 40000 + 3600 + 81 = 43681

(v) On decomposing:
605 = 600 + 5
Here, a = 600 and b = 5
Using the identity (a + b)2 = a2 + 2ab + b2:
6052 = (600 + 5)2 = 6002 + 2(600)(5) + 52 = 360000 + 6000 + 25 = 366025

Page No 3.32:

Question 6:

Find the squares of the following numbers using the identity (ab)2 = a2 − 2ab + b2:
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599

Answer:

(i) Decomposing: 395 = 400 − 5
Here, a = 400 and b = 5
Using the identity (ab)2 = a2 − 2ab + b2:

3952 = (400 5)2 = 4002 2(400)(5) + 52 = 160000 4000 + 25 = 156025

(ii) Decomposing: 995 = 1000 − 5
Here, a = 1000 and b = 5
Using the identity (ab)2 = a2 − 2ab + b2:

9952 = (1000 5)2 = 10002 2(1000)(5) + 52 = 1000000 10000 + 25 = 990025

(iii) Decomposing: 495 = 500 − 5
Here, a = 500 and b = 5
Using the identity (ab)2 = a2 − 2ab + b2:

4952 = (500 5)2 = 5002 2(500)(5) + 52 = 250000 5000 + 25 = 245025

(iv) Decomposing: 498 = 500 − 2
Here, a = 500 and b = 2
Using the identity (ab)2 = a2 − 2ab + b2

4982 = (500 2)2 = 5002 2(500)(2) + 22 = 250000 2000 + 4 = 248004

(v) Decomposing: 99 = 100 − 1
Here, a = 100 and b = 1
Using the identity (ab)2 = a2 − 2ab + b2:

992 = (100 1)2 = 1002 2(100)(1) + 12 = 10000 200 + 1 = 9801

(vi) Decomposing: 999 = 1000 - 1
Here, a = 1000 and b = 1
Using the identity (ab)2 = a2 − 2ab + b2:

9992 = (1000 1)2 = 10002 2(1000)(1) + 12 = 1000000 2000 + 1 = 998001

(vii) Decomposing: 599 = 600 − 1
Here, a = 600 and b = 1
Using the identity (ab)2 = a2 − 2ab + b2:

5992 = (600 1)2 = 6002 2(600)(1) + 12 = 360000 1200 + 1 = 358801

Page No 3.32:

Question 7:

Find the squares of the following numbers by visual method:
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99

Answer:

(i) We have:
52 = 50 + 2
Let us draw a square having side 52 units. Let us split it into 50 units and 2 units.

The sum of the areas of these four parts is the square of 52. Thus, the square of 52 is 2704.

(ii) We have:
95 = 90 + 5
Let us draw a square having side 95 units. Let us split it into 90 units and 5 units.

The sum of the areas of these four parts is the square of 95. Thus, the square of 95 is 9025.

(iii) We have:
505 = 500 + 5
Let us draw a square having side 505 units. Let us split it into 500 units and 5 units.

The sum of the areas of these four parts is the square of 505. Thus, the square of 505 is 255025.

(iv) We have:
702 = 700 + 2
Let us draw a square having side 702 units. Let us split it into 700 units and 2 units.

The sum of the areas of these four parts is the square of 702. Thus, the square of 702 is 492804.

(v) We have:
99 = 90 + 9
Let us draw a square having side 99 units. Let us split it into 90 units and 9 units.

The sum of the areas of these four parts is the square of 99. Thus, the square of 99 is 9801.



Page No 3.38:

Question 1:

Write the possible unit's digits of the square root of the following numbers. Which of these numbers are odd square roots?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025

Answer:

(i) The unit digit of the number 9801 is 1. So, the possible unit digits are 1 or 9 (Table 3.4). Note that 9801 is equal to 992. Hence, the square root is an odd number.

(ii) The unit digit of the number 99856 is 6. So, the possible unit digits are 4 or 6 (Table 3.4). Since its last digit is 6 (an even number), it cannot have an odd number as its square root.

(iii) The unit digit of the number 998001 is 1. So, the possible unit digits are 1 or 9. Note that 998001 is equal to (33 x 37)2. Hence, the square root is an odd number.

(iv) The unit digit of the number 657666025 is 5. So, the only possible unit digit is 5. Note that 657666025 is equal to (5 x 23 x 223)2. Hence, the square root is an odd number.

Hence, among th given numbers, (i), (iii) and (iv) have odd numbers as their square roots.

Page No 3.38:

Question 2:

Find the square root of each of the following by prime factorization.
(i) 441
(ii) 196
(iii) 529
(iv) 1764
(v) 1156
(vi) 4096
(vii) 7056
(viii) 8281
(ix) 11664
(x) 47089
(xi) 24336
(xii) 190969
(xiii) 586756
(xiv) 27225
(xv) 3013696

Answer:

(i) Resolving 441 into prime factors:
441 = 3 x 3 x 7 x 7

Grouping the factors into pairs of equal factors:
441 = (3 x 3) x (7 x 7)
Taking one factor for each pair, we get the square root of 441:
3 x 7 = 21

(ii) Resolving 196 into prime factors:
196 = 2 x 2 x 7 x 7

Grouping the factors into pairs of equal factors:
196 = (2 x 2) x (7 x 7)
Taking one factor for each pair, we get the square root of 196:
2 x 7 = 14

(iii) Resolving 529 into prime factors:
529 = 23 x 23

Grouping the factors into pairs of equal factors:
529= (23 x 23)
Taking one factor for each pair, we get the square root of 529 as 23.

(iv) Resolving 1764 into prime factors:
1764 = 2 x 2 x 3 x 3 x 7 x 7

Grouping the factors into pairs of equal factors:
1764 = (2 x 2) x (3 x 3) x (7 x 7)
Taking one factor for each pair, we get the square root of 1764:
2 x 3 x 7 = 42

(v) Resolving 1156 into prime factors:
1156 = 2 x 2 x 17 x 17

Grouping the factors into pairs of equal factors:
1156 = (2 x 2) x (17 x 17)
Taking one factor for each pair, we get the square root of 1156:
2 x 17 = 34

(vi) Resolving 4096 into prime factors:
4096 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the factors into pairs of equal factors:
4096 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2)
Taking one factor for each pair, we get the square root of 4096:
(2 x 2) x (2 x 2) x (2 x 2) = 64

(vii) Resolving 7056 into prime factors:
7056 = 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7

Grouping the factors into pairs of equal factors:
7056 = (2 x 2) x (2 x 2) x (3 x 3) x (7 x 7)
Taking one factor for each pair, we get the square root of 705:
2 x 2 x 3 x 7 = 84

(viii) Resolving 8281 into prime factors:
8281 = 7 x 7 x 13 x 13

Grouping the factors into pairs of equal factors:
8281 = (7 x 7) x (13 x 13)
Taking one factor for each pair, we get the square root of 8281:
7 x 13 = 91

(ix) Resolving 11664 into prime factors:
11664 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

Grouping the factors into pairs of equal factors:
11664 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3)
Taking one factor for each pair, we get the square root of 11664:
2 x 2 x 3 x 3 x 3 = 108

(x) Resolving 47089 into prime factors:
47089 = 7 x 7 x 31 x 31

Grouping the factors into pairs of equal factors:
47089 = (7 x 7) x (31 x 31)
Taking one factor for each pair, we get the square root of 47089:
7 x 31 = 217

(xi) Resolving 24336 into prime factors:
24336 = 2 x 2 x 2 x 2 x 3 x 3 x 13 x 13

Grouping the factors into pairs of equal factors:
24336 = (2 x 2) x (2 x 2) x (3 x 3) x (13 x 13)
Taking one factor for each pair, we get the square root of 24336:
2 x 2 x 3 x 13 = 156

(xii) Resolving 190969 into prime factors:
190969 = 19 x 19 x 23 x 23

Grouping the factors into pairs of equal factors:
190969 = (19 x 19) x (23 x 23)
Taking one factor for each pair, we get the square root of 190969:
19 x 23 = 437

(xiii) Resolving 586756 into prime factors:
586756 = 2 x 2 x 383 x 383

Grouping the factors into pairs of equal factors:
586756 = (2 x 2) x (383 x 383)
Taking one factor for each pair, we get the square root of 586756:
2 x 383 = 766

(xiv) Resolving 27225 into prime factors:
27225 = 3 x 3 x 5 x 5 x 11 x 11

Grouping the factors into pairs of equal factors:
27225 = (3 x 3) x (5 x 5) x (11 x 11)
Taking one factor for each pair, we get the square root of 27225:
3 x 5 x 11 = 165

(xv) Resolving 3013696 into prime factors:
3013696 = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 31 x 31

Grouping the factors into pairs of equal factors:
3013696 = (2 x 2) x (2 x 2) x (2 x 2) x (7 x 7) x (31 x 31)
Taking one factor for each pair, we get the square root of 3013696:
2 x 2 x 2 x 7 x 31 = 1736

Page No 3.38:

Question 3:

Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.

Answer:

The prime factorisation of 180:
180 = 2 x 2 x 3 x 3 x 5
Grouping the factors into pairs of equal factors, we get:
180 = (2 x 2) x (3 x 3) x 5
The factor, 5 does not have a pair. Therefore, we must multiply 180 by 5 to make a perfect square. The new number is:
(2 x 2) x (3 x 3) x (5 x 5) = 900
Taking one factor from each pair on the LHS, the square root of the new number is 2 x 3 x 5, which is equal to 30.

Page No 3.38:

Question 4:

Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.

Answer:

The prime factorisation of 147:
147 = 3 x 7 x 7
Grouping the factors into pairs of equal factors, we get:
147 = 3 x (7 x 7)
The factor, 3 does not have a pair. Therefore, we must multiply 147 by 3 to make a perfect square. The new number is:
(3 x 3) x (7 x 7) = 441
Taking one factor from each pair on the LHS, the square root of the new number is 3 x 7, which is equal to 21.

Page No 3.38:

Question 5:

Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.

Answer:

The prime factorisation of 3645:
3645 = 3 x 3 x 3 x 3 x 3 x 3 x 5
Grouping the factors into pairs of equal factors, we get:
3645 = (3 x 3) x (3 x 3) x (3 x 3) x 5
The factor, 5 does not have a pair. Therefore, we must divide 3645 by 5 to make a perfect square. The new number is:
(3 x 3) x (3 x 3) x (3 x 3) = 729
Taking one factor from each pair on the LHS, the square root of the new number is 3 x 3 x 3, which is equal to 27.

Page No 3.38:

Question 6:

Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained.

Answer:

The prime factorisation of 1152:
1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Grouping the factors into pairs of equal factors, we get:
1152 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 2
The factor, 2, at the end, does not have a pair. Therefore, we must divide 1152 by 2 to make a perfect square. The new number is:
(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) = 576
Taking one factor from each pair on the LHS, the square root of the new number is 2 x 2 x 2 x 3, which is equal to 24.

Page No 3.38:

Question 7:

The product of two numbers is 1296. If one number is 16 times the other, find the numbers.

Answer:

Let the two numbers be a and b.
From the first statement, we have:
a x b = 1296
If one number is 16 times the other, then we have:
b = 16 x a.
Substituting this value in the first equation, we get:
a x (16 x a) = 1296
By simplifying both sides, we get:
a2 = 1296/16 = 81
Hence, a is the square root of 81, which is 9.
To find b, use equation b = 16 x a.
Since a = 9:
b = 16 x 9 = 144
So, the two numbers satisfying the question are 9 and 144.

Page No 3.38:

Question 8:

A welfare association collected Rs 202500 as donation from the residents. If each paid as many rupees as there were residents, find the number of residents.

Answer:

Let R be the number of residents.
Let r be the money in rupees donated by each resident.
Total donation = R x r = 202500
Since the money received as donation is the same as the number of residents:
r = R.
Substituting this in the first equation, we get:
R x R = 202500
R2 = 202500
R2 = (2 x 2) x (5 x 5) x (5 x 5) x (3 x 3)2
R = 2 x 5 x 5 x 3 x 3 = 450
So, the number of residents is 450.

Page No 3.38:

Question 9:

A society collected Rs 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?

Answer:

Let M be the number of members.
Let r be the amount in paise donated by each member.
The total contribution can be expressed as follows:
M x r = Rs 92.16 = 9216 paise
Since the amount received as donation is the same as the number of members:
r = M
Substituting this in the first equation, we get:
M x M = 9216
M2 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
M2 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3)
M = 2 x 2 x 2 x 2 x 2 x 3 = 96
To find r, we can use the relation r = M.
Let M be the number of members.
Let r be the amount in paise donated by each member.
The total contribution can be expressed as follows:
M x r = Rs 92.16 = 9216 paise
Since the amount received as donation is the same as the number of members:
r = 96
So, there are 96 members and each paid 96 paise.

Page No 3.38:

Question 10:

A school collected Rs 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school?

Answer:

Let S be the number of students.
Let r be the money donated by each student.
The total contribution can be expressed by (S)(r) = Rs 2304
Since each student paid as many paise as the number of students, then r = S. Substituting this in the first equation, we get:
S x S = 2304
S2 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
S2 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3)
S = 2 x 2 x 2 x 2 x 3 = 48
So, there are 48 students in total in the school.

Page No 3.38:

Question 11:

The area of a square field is 5184 cm2. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.

Answer:

First, we have to find the perimeter of the square.
The area of the square is r2, where r is the side of the square.
Then, we have the equation as follows:
r2 = 5184 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3)
Taking the square root, we get r = 2 x 2 x 2 x 3 x 3 = 72
Hence the perimeter of the square is 4 x r = 288 m

Now let L be the length of the rectangular field.
Let W be the width of the rectangular field.
The perimeter is equal to the perimeter of square. 
Hence, we have:
2(L + W) = 288
Moreover, since the length is twice the width:
L = 2 x W.
Substituting this in the previous equation, we get:
2 x (2 x W + W) = 288
               3 x W = 144
                     W = 48
To find L:
L = 2 x W = 2 x 48 = 96
∴ Area of the rectangular field = L x W = 96 x 48 = 4608 m2

Page No 3.38:

Question 12:

Find the least square number, exactly divisible by each one of the numbers:
(i) 6, 9, 15 and 20
(ii) 8, 12, 15 and 20

Answer:

(i) The smallest number divisible by 6, 9, 15 and 20 is their L.C.M., which is equal to 60.
Factorising 60 into its prime factors:
60 = 2 x 2 x 3 x 5
Grouping them into pairs of equal factors:
60 = (2 x 2) x 3 x 5
The factors 3 and 5 are not paired. To make 60 a perfect square, we have to multiply it by 3 x 5, i.e . by15.
The perfect square is 60 x 15, which is equal to 900.

(i) The smallest number divisible by 8, 12, 15 and 20 is their L.C.M., which is equal to 120.
Factorising 120 into its prime factors:
120 = 2 x 2 x 2 x 3 x 5
Grouping them into pairs of equal factors:
120 = (2 x 2) x 2 x 3 x 5
The factors 2, 3 and 5 are not paired. To make 120 into a perfect square, we have to multiply it by 2 x 3 x 5, i.e. by 30.
The perfect square is 120 x 30, which is equal to 3600.

Page No 3.38:

Question 13:

Find the square roots of 121 and 169 by the method of repeated subtraction.

Answer:

To find the square root of 121:
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
In total, there are 11 numbers to subtract from 121. Hence, the square root of 121 is 11.

To find the square root of 169:
169 − 1 = 168
168 − 3 = 165
165 − 5 = 160
160 − 7 = 153
153 − 9 = 144
144 − 11 = 133
133 − 13 = 120
120 − 15 = 105
105 − 17 = 88
88 − 19 = 69
69 − 21 = 48
48 − 23 = 25
25 − 25 = 0
In total, there are 13 numbers to subtract from 169. Hence, the square root of 169 is 13.

Page No 3.38:

Question 14:

Write the prime factorization of the following numbers and hence find their square roots.
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056

Answer:

(i) The prime factorisation of 7744:
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
Grouping them into pairs of equal factors, we get:
7744 = (2 × 2) × (2 × 2) × (2 × 2) × (11 × 11)
Taking one factor from each pair, we get :
sqrt{7744}=2times2times2times2times11=88

(ii) The prime factorisation of 9604:
9604 = 2 × 2 × 7 × 7 × 7 × 7
Grouping them into pairs of equal factors, we get:
9604 = (2 × 2) × (7 × 7) × (7 × 7)
Taking one factor from each pair, we get:
sqrt{9604}=2times7times7=98

(iii) The prime factorisation of 5929:
5929 = 7 x 7 x 11 x 11
Grouping them into pairs of equal factors, we get:
5929 = (7 × 7) × (11 × 11)
Taking one factor from each pair, we get:
sqrt{5929}=7times11=77

(iv) The prime factorisation of 7056:
7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Grouping them into pairs of equal factors, we get:
7056 = (2 × 2) × (2 × 2) × (3 × 3) × (7 × 7)
Taking one factor from each pair, we get:
sqrt{7056}=2times2times3times7=84

Page No 3.38:

Question 15:

The students of class VIII of a school donated Rs 2401 for PM's National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Answer:

Let S be the number of students.
Let r be the amount in rupees donated by each student.

The total donation can be expressed by:
S × r = Rs. 2401
Since the total amount in rupees is equal to the number of students, r is equal to S.
Substituting this in the first equation:
S × S = 2401
S2 = (7 × 7) × (7 × 7)
S = 7 × 7 = 49
So, there are 49 students in the class.

Page No 3.38:

Question 16:

A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.

Answer:

Since 71 students were left out, there are only 5929 (6000 - 71) students remaining.
Hence, the number of rows or columns is simply the square root of 5929.
Factorising 5929 into its prime factors:
5929 = 7 × 7 × 11 × 11
Grouping them into pairs of equal factors:
5929 = (7 x 7) x (11 x 11)
The square root of 5929
= sqrt{5929}=7times11=77
Hence, in the arrangement, there were 77 rows of students.



Page No 3.4:

Question 1:

Which of the following numbers are perfect squares?
(i) 484
(ii) 625
(iii) 576
(iv) 941
(v) 961
(vi) 2500

Answer:

(i) 484 = 222
(ii) 625 = 252
(iii) 576 = 242
(iv) Perfect squares closest to 941 are 900 (302) and 961 (312). Since 30 and 31 are consecutive numbers, there are no perfect squares between 900 and 961. Hence, 941 is not a perfect square.
(v) 961 = 312
(vi) 2500 = 502

Hence, all numbers except that in (iv), i.e. 941, are perfect squares.

Page No 3.4:

Question 2:

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number in each case:
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761

Answer:

In each problem, factorise the number into its prime factors.

(i) 1156 = 2 x 2 x 17 x 17
Grouping the factors into pairs of equal factors, we obtain:
1156 = (2 x 2) x (17 x 17)
No factors are left over. Hence, 1156 is a perfect square. Moreover, by grouping 1156 into equal factors:
1156 = (2 x 17) x (2 x 17)
         = (2 x 17)2
Hence, 1156 is the square of 34, which is equal to 2 x 17.

(ii) 2025 = 3 x 3 x 3 x 3 x 5 x 5
Grouping the factors into pairs of equal factors, we obtain:
2025 = (3 x 3) x (3 x 3) x (5 x 5)
No factors are left over. Hence, 2025 is a perfect square. Moreover, by grouping 2025 into equal factors:
2025 = (3 x 3 x 5) x (3 x 3 x 5)
          = (3 x 3 x 5)2
Hence, 2025 is the square of 45, which is equal to 3 x 3 x 5.

(iii) 14641 = 11 x 11 x 11 x 11
Grouping the factors into pairs of equal factors, we obtain:
14641 = (11 x 11) x (11 x 11)
No factors are left over. Hence, 14641 is a perfect square. The above expression is already grouped into equal factors:
14641 = (11 x 11) x (11 x 11)
            = (11 x 11)2
Hence, 14641 is the square of 121, which is equal to 11 x 11.

(iv) 4761 = 3 x 3 x 23 x 23
Grouping the factors into pairs of equal factors, we obtain:
4761 = (3 x 3) x (23 x 23)
No factors are left over. Hence, 4761 is a perfect square. The above expression is already grouped into equal factors:
4761 = (3 x 23) x (3 x 23)
            = (3 x 23)2
Hence, 4761 is the square of 69, which is equal to 3 x 23.

Page No 3.4:

Question 3:

Find the smallest number by which the given number must bew multiplied so that the product is a perfect square:
(i) 23805
(ii) 12150
(iii) 7688

Answer:

Factorise each number into its prime factors.

(i) 23805 = 3 x 3 x 5 x 23 x 23



Grouping 23805 into pairs of equal factors:
23805 = (3 x 3) x (23 x 23) x 5
Here, the factor 5 does not occur in pairs. To be a perfect square, every prime factor has to be in pairs. Hence, the smallest number by which 23805 must be multiplied is 5.

(ii) 12150 = 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5


Grouping 12150 into pairs of equal factors:
12150  = (3 x 3 x 3 x 3) x (5 x 5) x 2 x 3
Here, 2 and 3 do not occur in pairs. To be a perfect square, every prime factor has to be in pairs. Hence. the smallest number by which 12150 must be multiplied is 2 x 3, i.e. by 6.

(iii) 7688 = 2 x 2 x 2 x 31 x 31


Grouping 7688 into pairs of equal factors:
7688 = (2 x 2) x (31 x 31) x 2
Here, 2 does not occur in pairs. To be a perfect square, every prime factor has to be in pairs. Hence, the smallest number by which 7688 must be multiplied is 2.

Page No 3.4:

Question 4:

Find the smallest number by which the given number must be divided so that the resulting number is a perfect square:
(i) 14283
(ii) 1800
(iii) 2904

Answer:


For each question, factorise the number into its prime factors.

(i) 14283 = 3 x 3 x 3 x 23 x 23


Grouping the factors into pairs:
14283 = (3 x 3) x (23 x 23) x 3
Here, the factor 3 does not occur in pairs. To be a perfect square, all the factors have to be in pairs. Hence, the smallest number by which 14283 must be divided for it to be a perfect square is 3.

(ii) 1800= 2 x 2 x 2 x 3 x 3 x 5 x 5


Grouping the factors into pairs:
1800 = (2 x 2) x (3 x 3) x (5 x 5) x 2
Here, the factor 2 does not occur in pairs. To be a perfect square, all the factors have to be in pairs. Hence, the smallest number by which 1800 must be divided for it to be a perfect square is 2.

(iii) 2904 = 2 x 2 x 2 x 3 x 11 x 11



Grouping the factors into pairs:
2904 = (2 x 2) x (11 x 11) x 2 x 3
Here, the factors 2 and 3 do not occur in pairs. To be a perfect square, all the factors have to be in pairs. Hence, the smallest number by which 2904 must be divided for it to be a perfect square is 2 x 3, i.e. 6.

Page No 3.4:

Question 5:

Which of the following numbers are perfect squares?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121

Answer:

11: The perfect squares closest to 11 are 9 (9 = 32) and 16 (16 = 42). Since 3 and 4 are consecutive numbers, there are no perfect squares between 9 and 16, which means that 11 is not a perfect square.
12: The perfect squares closest to 12 are 9 (9 =32) and 16 (16 = 42). Since 3 and 4 are consecutive numbers, there are no perfect squares between 9 and 16, which means that 12 is not a perfect square.
16 = 42
32: The perfect squares closest to 32 are 25 (25 = 52) and 36 (36 = 62). Since 5 and 6 are consecutive numbers, there are no perfect squares between 25 and 36, which means that 32 is not a perfect square.
36 = 62
50: The perfect squares closest to 50 are 49 (49 = 72) and 64 (64 = 82). Since 7 and 8 are consecutive numbers, there are no perfect squares between 49 and 64, which means that 50 is not a perfect square.
64 = 82
79: The perfect squares closest to 79 are 64 (64 = 82) and 81 (81 = 92). Since 8 and 9 are consecutive numbers, there are no perfect squares between 64 and 81, which means that 79 is not a perfect square.
81 = 92
111: The perfect squares closest to 111 are 100 (100 = 102) and 121 (121 = 112). Since 10 and 11 are consecutive numbers, there are no perfect squares between 100 and 121, which means that 111 is not a perfect square.
121 = 112

Hence, the perfect squares are 16, 36, 64, 81 and 121.

Page No 3.4:

Question 6:

Using prime factorization method, find which of the following numbers are perfect squares?
189, 225, 2048, 343, 441, 2916, 11025, 3549

Answer:


(i) 189 = 3 x 3 x 3 x 7

Grouping them into pairs of equal factors:
189 = (3 x 3) x 3 x 7
The factors 3 and 7 cannot be paired. Hence, 189 is not a perfect square.

(ii) 225 = 3 x 3 x 5 x 5


Grouping them into pairs of equal factors:
225 = (3 x 3) x (5 x 5)
There are no left out of pairs. Hence, 225 is a perfect square.

(iii) 2048 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2


Grouping them into pairs of equal factors:
2048 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x 2
The last factor, 2 cannot be paired. Hence, 2048 is not a perfect square.

(iv) 343 = 7 x 7 x 7



Grouping them into pairs of equal factors:
343 = (7 x 7) x 7
The last factor, 7 cannot be paired. Hence, 343 is not a perfect square.

(v) 441 = 3 x 3 x 7 x 7


Grouping them into pairs of equal factors:
441 = (3 x 3) x (7 x 7)
There are no left out of pairs. Hence, 441 is a perfect square.

(vi) 2916 = 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3


Grouping them into pairs of equal factors:
2916 = (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3)
There are no left out of pairs. Hence, 2916 is a perfect square.

(vii) 11025 = 3 x 3 x 5 x 5 x 7 x 7


Grouping them into pairs of equal factors:
11025 = (3 x 3) x (5 x 5) x (7 x 7)
There are no left out of pairs. Hence, 11025 is a perfect square.

(viii) 3549 = 3 x 7 x 13 x 13


Grouping them into pairs of equal factors:
3549 = (13 x 13) x 3 x 7
The last factors, 3 and 7 cannot be paired. Hence, 3549 is not a perfect square.

Hence, the perfect squares are 225, 441, 2916 and 11025.

Page No 3.4:

Question 7:

By what number should each of the following numbers be multiplied to get a perfect square in each case? Also, find the number whose square is the new number.
(i) 8820
(ii) 3675
(iii) 605
(iv) 2880
(v) 4056
(vi) 3468
(vii) 7776

Answer:

Factorising each number.

(i) 8820 = 2 x 2 x 3 x 3 x 5 x 7 x 7

Grouping them into pairs of equal factors:
8820 = (2 x 2) x (3 x 3) x (7 x 7) x 5
The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 8820 must be multiplied by 5 for it to be a perfect square.
The new number would be (2 x 2) x (3 x 3) x (7 x 7) x (5 x 5).
Furthermore, we have:
(2 x 2) x (3 x 3) x (7 x 7) x (5 x 5) = (2 x 3 x 5 x 7) x (2 x 3 x 5 x 7)
Hence, the number whose square is the new number is:
2 x 3 x 5 x 7 = 210

(ii) 3675 = 3 x 5 x 5 x 7 x 7

Grouping them into pairs of equal factors:
3675 = (5 x 5) x (7 x 7) x 3
The factor, 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3675 must be multiplied by 3 for it to be a perfect square.
The new number would be (5 x 5) x (7 x 7) x (3 x 3).
Furthermore, we have:
(5 x 5) x (7 x 7) x (3 x 3) = (3 x 5 x 7) x (3 x 5 x 7)
Hence, the number whose square is the new number is:
3 x 5 x 7 = 105

(iii) 605 = 5 x 11 x 11


Grouping them into pairs of equal factors:
605 = 5 x (11 x 11)
The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 605 must be multiplied by 5 for it to be a perfect square.
The new number would be (5 x 5) x (11 x 11).
Furthermore, we have:
(5 x 5) x (11 x 11) = (5 x 11) x (5 x 11)
Hence, the number whose square is the new number is:
5 x 11 = 55

(iv) 2880 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5


Grouping them into pairs of equal factors:
2880 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 5
There is a 5 as the leftover. For a number to be a perfect square, each prime factor has to be paired. Hence, 2880 must be multiplied by 5 to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5).
Furthermore, we have:
(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5) = (2 x 2 x 2 x 3 x 5) x (2 x 2 x 2 x 3 x 5)
Hence, the number whose square is the new number is:
2 x 2 x 2 x 3 x 5 = 120

(v) 4056 = 2 x 2 x 2 x 3 x 13 x 13


Grouping them into pairs of equal factors:
4056 = (2 x 2) x (13 x 13) x 2 x 3
The factors at the end, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 4056 must be multiplied by 6 (2 x 3) for it to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (3 x 3) x (13 x 13).
Furthermore, we have:
(2 x 2) x (2 x 2) x (3 x 3) x (13 x 13) = (2 x 2 x 3 x 13) x (2 x 2 x 3 x 13)
Hence, the number whose square is the new number is:
2 x 2 x 3 x 13 = 156

(vi) 3468 = 2 x 2 x 3 x 17 x 17


Grouping them into pairs of equal factors:
3468 = (2 x 2) x (17 x 17) x 3
The factor 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3468 must be multiplied by 3 for it to be a perfect square.
The new number would be (2 x 2) x (17 x 17) x (3 x 3).
Furthermore, we have:
(2 x 2) x (17 x 17) x (3 x 3) = (2 x 3 x 17) x (2 x 3 x 17)
Hence, the number whose square is the new number is:
2 x 3 x 17 = 102

(vii) 7776 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3


Grouping them into pairs of equal factors:
7776 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x 2 x 3
The factors, 2 and 3 at the end are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 7776 must be multiplied by 6 (2 x 3) for it to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3).
Furthermore, we have:
(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3) = (2 x 2 x 2 x 3 x 3 x 3) x (2 x 2 x 2 x 3 x 3 x 3)
Hence, the number whose square is the new number is:
2 x 2 x 2 x 3 x 3 x 3 = 216

Page No 3.4:

Question 8:

By what numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575

Answer:

Factorising each number.

(i) 16562 = 2 x 7 x 7 x 13 x 13


Grouping them into pairs of equal factors:
16562 = 2 x (7 x 7) x (13 x 13)
The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 16562 must be divided by 2 for it to be a perfect square.
The new number would be (7 x 7) x (13 x 13).
Furthermore, we have:
(7 x 7) x (13 x 13) = (7 x 13) x (7 x 13)
Hence, the number whose square is the new number is:
7 x 13 = 91

(ii) 3698 = 2 x 43 x 43


Grouping them into pairs of equal factors:
3698 = 2 x (43 x 43)
The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3698 must be divided by 2 for it to be a perfect square.
The new number would be (43 x 43).
Hence, the number whose square is the new number is 43.

(iii) 5103 = 3 x 3 x 3 x 3 x 3 x 3 x 7


Grouping them into pairs of equal factors:
5103 = (3 x 3) x (3 x 3) x (3 x 3) x 7
The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 5103 must be divided by 7 for it to be a perfect square.
The new number would be (3 x 3) x (3 x 3) x (3 x 3).
Furthermore, we have:
(3 x 3) x (3 x 3) x (3 x 3) = (3 x 3 x 3) x (3 x 3 x 3)
Hence, the number whose square is the new number is:
3 x 3 x 3 = 27

(iv) 3174 = 2 x 3 x 23 x 23


Grouping them into pairs of equal factors:
3174 = 2 x 3 x (23 x 23)
The factors, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3174 must be divided by 6 (2 x 3) for it to be a perfect square.
The new number would be (23 x 23).

Hence, the number whose square is the new number is 23.    

(v) 1575 = 3 x 3 x 5 x 5 x 7



Grouping them into pairs of equal factors:
1575 = (3 x 3) x (5 x 5) x 7
The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 1575 must be divided by 7 for it to be a perfect square.
The new number would be (3 x 3) x (5 x 5).
Furthermore, we have:
(3 x 3) x (5 x 5) = (3 x 5) x (3 x 5)
Hence, the number whose square is the new number is:
3 x 5 = 15

Page No 3.4:

Question 9:

Find the greatest number of two digits which is a perfect square.

Answer:

We know that 102 is equal to 100 and 92 is equal to 81.
Since 10 and 9 are consecutive numbers, there is no perfect square between 100 and 81. Since 100 is the first perfect square that has more than two digits, 81 is the greatest two-digit perfect square.

Page No 3.4:

Question 10:

Find the least number of three digits which is perfect square.

Answer:

Let us make a list of the squares starting from 1.
12 = 1
22 = 4
32 = 9
42 = 16
52 = 25
62 = 36
72 = 49
82 = 64
92 = 81
102 = 100
The square of 10 has three digits. Hence, the least three-digit perfect square is 100.



Page No 3.43:

Question 1:

Find the square root of each of the following by long division method:
(i) 12544
(ii) 97344
(iii) 286225
(iv) 390625
(v) 363609
(vi) 974169
(vii) 120409
(viii) 1471369
(ix) 291600
(x) 9653449
(xi) 1745041
(xii) 4008004
(xiii) 20657025
(xiv) 152547201
(xv) 20421361
(xvi) 62504836
(xvii) 82264900
(xviii) 3226694416
(xix) 6407522209
(xx) 3915380329

Answer:

 (i)


Hence, the square root of 12544 is 112.

(ii)


Hence, the square root of 97344 is 312.

(iii)

Hence, the square root of 286225 is 535.

(iv)

Hence, the square root of 390625 is 625.

(v)


Hence, the square root of 363609 is 603.

(vi)


Hence, the square root of 974169 is 987.


Hence, the square root of 120409 is 347.

(viii)

Hence, the square root of 1471369 is 1213.

(ix)

Hence, the square root of 291600 is 540.


Hence, the square root of 9653449 is 3107.

(xi)

Hence, the square root of 1745041 is 1321.

(xii)

Hence, the square root of 4008004 is 2002.

(xiii)

Hence, the square root of 20657025 is 4545.

(xiv)

Hence, the square root of 152547201 is 12351.

(xv)

Hence, the square root of 20421361 is 4519.

(xvi)

Hence, the square root of 6250486 is 7906.

(xvii)

Hence, the square root of 82264900 is 9070.

(xviii)

Hence, the square root of 3226694416 is 56804.

(xix)

Hence, the square root of 6407522209 is 80047.

(xx)

Hence, the square root of 3915380329 is 625763.

Page No 3.43:

Question 2:

Find the least number which must be subtracted from the following numbers to make them a perfect square:
(i) 2361
(ii) 194491
(iii) 26535
(iv) 16160
(v) 4401624

Answer:

(i) Using the long division method:

We can see that 2361 is 57 more than 472. Hence, 57 must be subtracted from 2361 to get a perfect square.

(ii) Using the long division method:

We can see that 194491 is 10 more than 4412. Hence, 10 must be subtracted from 194491 to get a perfect square.

(iii) Using the long division method:

We can see that 26535 is 291 more than 1622. Hence, 291 must be subtracted from 26535 to get a perfect square.

(iv) Using the long division method:

We can see that 16160 is 31 more than 1272. Hence, 31 must be subtracted from 16160 to get a perfect square.

(v) Using the long division method:

We can be see that 4401624 is 20 more than 20982. Hence, 20 must be subtracted from 4401624 to get a perfect square.

Page No 3.43:

Question 3:

Find the least number which must be added to the following numbers to make them a perfect square:
(i) 5607
(ii) 4931
(iii) 4515600
(iv) 37460
(v) 506900

Answer:


(i) Using the long division method:

We can see that 5607 is 18 more than 752. Hence, we have to add 18 to 5607 to get a perfect square.

(ii) Using the long division method:

We can see that 4931 is 110 more than 712. Hence, we have to add 110 to 4931 to get a perfect square.

(iii) Using the long division method:

We can see that 4515600 is 25 more than 21252. Hence, we have to add 25 to 4515600 to get a perfect square.

(iv) Using the long division method:

We can see that 37460 is 176 more than 1942. Hence, we have to add 176 to 37460 to get a perfect square.

(v) Using the long division method:

We can see that 506900 is 44 more than 7122. Hence, we have to add 44 to 506900 to get a perfect square.

Page No 3.43:

Question 4:

Find the greatest number of 5 digits which is a perfect square.

Answer:

The greatest number with five digits is 99999. To find the greatest square number with five digits, we must find the smallest number that must be subtracted from 99999 in order to make a perfect square. For that, we have to find the square root of 99999 by the long division method as follows:

Hence, we must subtract 143 from 99999 to get a perfect square:
99999 − 143 = 99856

Page No 3.43:

Question 5:

Find the least number of 4 digits which is a perfect square.

Answer:

The least number with four digits is 1000. To find the least square number with four digits, we must find the smallest number that must be added to 1000 in order to make a perfect square. For that, we have to find the square root of 1000 by the long division method as shown below:

1000 is 24 (124 − 100) less than the nearest square number 322. Thus, 24 must be added to 1000 to be a perfect square.
1000 + 24 = 1024
Hence, the smallest perfect square number with four digits is 1024.

Page No 3.43:

Question 6:

Find the least number of six digits which is a perfect square.

Answer:

The least number with six digits is 100000. To find the least square number with six digits, we must find the smallest number that must be added to 100000 in order to make a perfect square. For that, we have to find the square root of 100000 by the long division method as follows:

100000 is 489 (4389 − 3900) less than 3172. Hence, to be a perfect square, 489 should be added to 100000.
100000 + 489 = 100489
Hence, the least number of six digits that is a perfect square is 100489.



Page No 3.44:

Question 7:

Find the greatest number of 4 digits which is a perfect square.

Answer:

The greatest number with four digits is 9999. To find the greatest perfect square with four digits, we must find the smallest number that must be subtracted from 9999 in order to make a perfect square. For that, we have to find the square root of 9999 by the long division method as shown below:

We must subtract 198 from 9999 to make a perfect square:
9999 − 198 = 9801

Hence, the greatest perfect square with four digits is 9801.

Page No 3.44:

Question 8:

A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row.

Answer:

60 soldiers are left out. 
Remainaing soldiers = 8160 − 60 = 8100
The number of soldiers in each row to form a perfect square would be the square root of 8100. We have to find the square root of 8100 by the long division method as shown below:


Hence, the number of soldiers in each row to form a perfect square is 90.

Page No 3.44:

Question 9:

The area of a square field is 60025 m2. A man cycles along its boundary at 18 km/hr. In how much time will he return at the starting point?

Answer:

Area of the square field = 60025 m2
The length of the square field would be the square root of 60025.
Using the long division method:

Hence, the length of the square field is 245 m.
The square has four sides, so the number of boundaries of the field is 4.
The distance s covered by the man = 245 m × 4 = 980 m = 0.98 km
If the velocity v is 18 km/hr, the required time t can be calculated using the following formula:
t=sv
t = {frac{0.98}{18}} = 0.054 hr=3minutes,16seconds
So, the man will return to the starting point after 3 minutes and 16 seconds.

Page No 3.44:

Question 10:

The cost of levelling and turfing a square lawn at Rs 2.50 per m2 is Rs 13322.50. Find the cost of fencing it at Rs 5 per metre.

Answer:

First, we have to find the area of the square lawn, which the total cost divided by the cost of levelling and turfing per square metre:

 textrm{Area of a square} = frac{13322.5}{2.5} = 5329textrm{ m}^{2}
The length of one side of the square is equal to the square root of the area. We will use the long division method to find it as shown below:

Length of one side of the square = 73 m
The circumference of the square is 73 × 4 = 292 m
Total cost of fencing the lawn at Rs. 5 per metre = 292 × 5 = Rs. 1460

Page No 3.44:

Question 11:

Find the greatest number of three digits which is a perfect square.

Answer:

The greatest number with three digits is 999. To find the greatest perfect square with three digits, we must find the smallest number that must be subtracted from 999 in order to get a perfect square. For that, we have to find the square root by the long division method as shown below:
 
So, 38 must be subtracted from 999 to get a perfect square.
999 − 38 = 961
961 = 312
Hence, the greatest perfect square with three digits is 961.

Page No 3.44:

Question 12:

Find the smallest number which must be added to 2300 so that it becomes a perfect square.

Answer:

To find the square root of 2300, we use the long division method:

23000 is 4 (704 − 700) less than 482. Hence, 4 must be added to 2300 to get a perfect square.



Page No 3.48:

Question 1:

Find the square root of:
(i) 441961
(ii) 324841
(iii) 42929
(iv) 21425
(v) 2137196
(vi) 2326121
(vii) 25544729
(viii) 754649
(ix) 39422209
(x) 33343025
(xi) 2127973364
(xii) 381125
(xiii) 23394729
(xiv) 2151169
(xv) 10151225

Answer:

(i) We know:
441961=441961
Now, let us compute the square roots of the numerator and the denominator separately.
441=3×3×7×7=3×7=21961=31×31=31441961=2131

(ii)We know:
324841=324841
Now, let us compute the square roots of the numerator and the denominator separately.
324=2×2×3×3×3×3=2×3×3=18841=29×29=29324841=1829
(iii) By looking at the book's answer key, the fraction should be 42949, not 42929.
We know:
42949=22549=22549225=1549=742949=157

(iv) We know:
21425=6425=6425=85

(v) We know:
2137196=529196=529196
Now, let us compute the square roots of the numerator and the denominator separately.
529=23×23=23196=2×2×7×7=2×7=142137196=2314

(vi) We know:
2326121=2809121=2809121
Now, let us compute the square roots of the numerator and the denominator separately.

121=112326121=5311

(vii) We know:
25544729=18769729=18769729
Now, let us compute the square roots of the numerator and the denominator separately.

729=2725544729=13727

(viii) We know:
754649=372149=372149
Now, let us compute the square roots of the numerator and the denominator separately.

49=7754649=617

(ix) We know:
39422209=75692209=75692209
Now, let us compute the square roots of the numerator and the denominator separately.

39422209=8747

(x) We know:
33343025=94093025=94093025
Now, let us compute the square roots of the numerator and the denominator separately.

33343025=9755

(xi) We know:
2127973364=734413364=734413364
Now, let us compute the square roots of the numerator and the denominator separately.

2127973364=27158

(xii) We know:
381125=96125=96125
Now, let us compute the square roots of the numerator and the denominator separately.
961=3125=5381125=315

(xiii) We know:
23394729=17161729=17161729
Now, let us compute the square roots of the numerator and the denominator separately.

729=2723394729=13127=42327

(xiv) We know:
2151169=3600169=3600169
Now, let us compute the square roots of the numerator and the denominator separately.
3600=60×60=60169=13×13=132151169=6013=4813

(xv) We know:
10151225=2401225=2401225
Now let us compute the square roots of the numerator and the denominator separately.
2401=7×7×7×7=7×7=49225=3×3×5×5=3×5=1510151225=4915=3415

Page No 3.48:

Question 2:

Find the value of:
(i) 80405
(ii) 441625
(iii) 15871728
(iv) 72×338
(v) 45×20

Answer:

(i) We have:
frac{sqrt{80}}{sqrt{405}}=sqrt{frac{80}{405}}=sqrt{frac{16}{81}}=frac{sqrt{16}}{sqrt{81}}=frac{4}{9}

(ii) Computing the square roots:
sqrt{441}=sqrt{left (3times3 right )timesleft (7times7 right )}=3times7=21
sqrt{625}=sqrt{left (5times5 right )timesleft (5times5 right )}=5times5=25
frac{sqrt{441}}{sqrt{625}}=frac{21}{25}

(iii) We have:

frac{sqrt{1587}}{sqrt{1728}}=sqrt{frac{529}{576}}      (by dividing both numbers by 3)
Computing the square roots of the numerator and the denominator:
sqrt{529}=sqrt{23times23}=23
sqrt{576}=sqrt{24times24}=24

frac{sqrt{1587}}{sqrt{1728}}=frac{23}{24}

(iv) We have:
sqrt{72}timessqrt{338} =sqrt{72times338} =sqrt{2times2times2times3times3times2times13times13}
==sqrt{2times2times2times2times3times3times13times13}  =2times2times3times13
                                                       =156
(v) We have:
sqrt{45}timessqrt{20} = sqrt{3times3times5times2times2times5}
                          = sqrt{3times3times2times2times5times5}
                          = 30

Page No 3.48:

Question 3:

The area of a square field is 80244729 square metres. Find the length of each side of the field.

Answer:

The length of one side is the square root of the area of the field. Hence, we need to calculate the value of sqrt{80frac{244}{729}}
We have
sqrt{80frac{244}{729}}=sqrt{frac{58564}{729}}=frac{sqrt{58564}}{sqrt{729}}
Now, to calculate the square root of the numerator and the denominator:

We know that:
sqrt{729}=27

Therefore, length of one side of the field = 24227 = 82627 m



Page No 3.49:

Question 4:

The area of a square field is 3014m2. Calculate the length of the side of the square.

Answer:

The length of one side is equal to the square root of the area of the field. Hence, we just need to calculate the value of sqrt{30frac{1}{4}}.
We have:
sqrt{30frac{1}{4}}=frac{sqrt{121}}{sqrt{4}}

Now, calculating the square root of the numerator and the denominator:
sqrt{121}=sqrt{11times11}=11
sqrt{4}=2

Therefore, the length of the side of the square = 3014= 112 = 512 m

Page No 3.49:

Question 5:

Find the length of a side of a square playground whose area is equal to the area of a rectangular field of diamensions 72 m and 338 m.

Answer:

The area of the playground = 72 × 338 = 24336 m2
The length of one side of a square is equal to the square root of its area. Hence, we just need to find the square root of 24336.

Hence, the length of one side of the playground is 156 metres.



Page No 3.5:

Question 11:

Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect suqare.

Answer:

Prime factorisation of 4851:
4851 = 3 x 3 x 7 x 7 x 11


Grouping them into pairs of equal factors:
4851 = (3 x 3) x (7 x 7) x 11
The factor, 11 is not paired. The smallest number by which 4851 must be multiplied such that the resulting number is a perfect square is 11.

Page No 3.5:

Question 12:

Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.

Answer:

Prime factorisation of 28812:
28812 = 2 x 2 x 3 x 7 x 7 x 7 x 7


Grouping them into pairs of equal factors:
28812 = (2 x 2) x (7 x 7) x (7 x 7) x 3
The factor, 3 is not paired. Hence, the smallest number by which 28812 must be divided such that the resulting number is a perfect square is 3.

Page No 3.5:

Question 13:

Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the number whose square is the resulting number.

Answer:

Prime factorisation of 1152:
1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3


Grouping them into pairs of equal factors:
1152 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 2
The factor, 2 at the end is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 1152 must be divided by 2 for it to be a perfect square.

The resulting number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3).
Furthermore, we have:

(2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) = (2 x 2 x 2 x 3) x (2 x 2 x 2 x 3)
Hence, the number whose square is the resulting number is:
2 x 2 x 2 x 3 = 24



Page No 3.52:

Question 1:

Find the square root in decimal form:
84.8241

Answer:


Hence, the square root of 84.8241 is 9.21.

Page No 3.52:

Question 2:

Find the square root in decimal form:
0.7225

Answer:


Hence, the square root of 0.7225 is 0.85.

Page No 3.52:

Question 3:

Find the square root in decimal form:
0.813604

Answer:


Hence, the square root of 0.813604 is 0.902.

Page No 3.52:

Question 4:

Find the square root in decimal form:
0.00002025

Answer:


Hence, the square root of 0.00002025 is 0.0045.

Page No 3.52:

Question 5:

Find the square root in decimal form:
150.0625

Answer:


Hence, the square root of 150.0625 is 12.25.

Page No 3.52:

Question 6:

Find the square root in decimal form:
225.6004

Answer:


Hence, the square root of 225.6004 is 15.02

Page No 3.52:

Question 7:

Find the square root in decimal form:
3600.720036

Answer:


Hence, the square root of 3600.720036 is 60.006.

Page No 3.52:

Question 8:

Find the square root in decimal form:
236.144689

Answer:


Hence, the square root of 236.144689 is 15.367.

Page No 3.52:

Question 9:

Find the square root in decimal form:
0.00059049

Answer:


Hence, the square root of 0.00059049 is 0.0243.

Page No 3.52:

Question 10:

Find the square root in decimal form:
176.252176

Answer:

Hence, the square root of 176.252176 is 13.276.

 

Page No 3.52:

Question 11:

Find the square root in decimal form:
9998.0001

Answer:


Hence, the square root of 9998.0001 is 99.99.

Page No 3.52:

Question 12:

Find the square root in decimal form:
0.00038809

Answer:


Hence, the square root of 0.00038809 is 0.0197.

Page No 3.52:

Question 13:

What is that fraction which when multiplied by itself gives 227.798649?

Answer:

We have to find the square root of the given number.



Hence, the fraction, which when multiplied by itself, gives 227.798649 is 15.093.

Page No 3.52:

Question 14:

The area of a square playground is 256.6404 square metres. Find the length of one side of the playground.

Answer:

The length of one side of the playground is the square root of its area.


So, the length of one side of the playground is 16.02 metres.

Page No 3.52:

Question 15:

What is the fraction which when multiplied by itself gives 0.00053361?

Answer:

We have to find the square root of the given number.


Hence, the fraction, which when multiplied by itself, gives 0.00053361 is 0.0231.

Page No 3.52:

Question 16:

Simplify:
(i) 59.29-5.2959.29+5.29
(ii) 0.2304+0.17640.2304-0.1764

Answer:

(i) We have:
59.29=5929100=7×7×11×1110=7×1110=7.75.29=529100=529100=2310=2.359.29-5.2959.29+5.29=7.7-2.37.7+2.3=5.410=0.54


(ii) We have:
0.2304=230410000             =2×2×2×2×2×2×2×2×3×310000             =2×2×2×2×3100             =0.480.1764=176410000             =2×2×3×3×7×710000             =2×3×7100             =0.42 0.2304+0.17640.2304-0.1764=0.48+0.420.48-0.42=0.90.06=15          
                  

Page No 3.52:

Question 17:

Evaluate 50625 and hence find the value of 506.25+5.0625

Answer:

We have:
50625=3×3×3×3×5×5×5×5=3×3×5×5=225

Next, we will calculate 506.25 and 5.0625

506.25=50625100=50625100=22510=22.55.0625=5062510000=5062510000=225100=2.25506.25+5.0625=22.5+2.25=24.75

Page No 3.52:

Question 18:

Find the value of 103.0225 and hence find the value of
(i) 10302.25
(ii) 1.030225

Answer:

The value of 103.0225 is:


Hence, the square root of 103.0225 is 10.15.

Now, we can solve the following questions as shown below:

(i) 10302.25=103.0225×100=103.0225×100=10.15×10=101.5(ii) 1.030225=103.0225100=103.0225100=10.1510=1.015



Page No 3.56:

Question 1:

Find the square root of each of the following correct to three places of decimal.
(i) 5
(ii) 7
(iii) 17
(iv) 20
(v) 66
(vi) 427
(vii) 1.7
(viii) 23.1
(ix) 2.5
(x) 237.615
(xi) 15.3215
(xii) 0.9
(xiii) 0.1
(xiv) 0.016
(xv) 0.00064
(xvi) 0.019
(xvii) 78
(xviii) 512
(xix) 212
(xx) 28758

Answer:

(i) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 5 up to three decimal places is 2.236.

(ii) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 7 up to three decimal places is 2.646.

(iii) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 17 up to three decimal places is 4.123.

(iv) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 20 up to three decimal places is 4.472.

(v) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 66 up to three decimal places is 8.124.

(vi) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 427 up to three decimal places is 20.664.

(vii) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 1.7 up to three decimal places is 1.304.

(viii) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 23.1 up to three decimal places is 4.806.

(ix) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 2.5 up to three decimal places is 1.581.

(x) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 237.615 up to three decimal places is 15.415.

(xi) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 15.3215 up to three decimal places is 3.914.

(xii) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 0.9 up to three decimal places is 0.949.

(xiii) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 0.1 up to three decimal places is 0.316.

(xiv) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 0.016 up to three decimal places is 0.126.

(xv) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 0.00064 up to three decimal places is 0.025.

(xvi) We can find the square root up to three decimal places by using long division until we get four decimal places and then rounding it to three decimal places.

Hence, the square root of 0.019 up to three decimal places is 0.138.

(xvii) We can find the square root up to four decimal places by expanding 7/8 to decimal form up to eight digits to the right of the decimal point as shown below:
78=0.875
Hence, we have:

So, the square root of 7/8 up to three decimal places is 0.935.

(xviii) We can find the square root up to four decimal places by expanding 5/12 to decimal form up to eight digits to the right of the decimal point as shown below:
52=0.41666666
Hence, we have:

So, the square root of 5/12 up to three decimal places is 0.645.

(xix) We can find the square root up to four decimal places by expanding  212 into decimal form up to eight digits to the right of the decimal point as shown below:
212=2.50000000
But, this is the same with the value 2.5 in problem (ix). Hence, the square root of   212 is 1.581.

(xx) We can find the square root up to four decimal places by expanding 28758 into decimal form up to eight digits to the right of the decimal point as shown below:
28758=287.62500000
Hence, we have:

So, the square root of  28758 up to three decimal places is 16.960.



Page No 3.57:

Question 2:

Find the square root of 12.0068 correct to four decimal places.

Answer:


 12.0068=3.46508 
We can round it off to four decimal places, i.e. 3.4651.

Page No 3.57:

Question 3:

Find the square root of 11 correct to five decimal places.

Answer:

Using the long division method:

 11=3.31662

Page No 3.57:

Question 4:

Given that: 2=1.414, 3=1.732, 5=2.236 and 7=2.646, evaluate each of the following:

(i) 1447
(ii) 25003

Answer:

Given:  7 = 2.646
(i) 1447=1447=122.646=4.536

Given: 3 = 1.732
(ii) 25003=25003=501.732=28.867

Page No 3.57:

Question 5:

Given that: 2=1.414, 3=1.732, 5=2.236 and 7=2.646, find the square roots of the following:
(i) 19675
(ii) 40063
(iii) 1507
(iv) 2565
(v) 2550

Answer:

From the given values, we can simplify the expressions in the following manner:

(i) 19675=1453=145×1.732=1.617(ii) 40063=2037=203×2.646=2.520(iii) 1507=52×37=5×1.414×1.7322.646=4.628(iv) 2565=165=162.236=7.155(v) 2750=3352=3×1.7325×1.414=0.735

  



Page No 3.61:

Question 1:

Using square root table, find the square root
7

Answer:

From the table, we directly find that the square root of 7 is 2.646.

Page No 3.61:

Question 2:

Using square root table, find the square root
15

Answer:

Using the table to find 3 and 5
15=3×5       =1.732×2.236       =3.873
              
       

Page No 3.61:

Question 3:

Using square root table, find the square root
74

Answer:

  Using the table to find 2 and 37

74=2×37        =1.414×6.083        =8.602
               
         

Page No 3.61:

Question 4:

Using square root table, find the square root
82

Answer:

Using the table to find 2 and 41

82=2×41       =1.414×6.403       =9.055
                  
         

Page No 3.61:

Question 5:

Using square root table, find the square root
198

Answer:

Using the table to find 2 and 11

198=2×9×11        =1.414×3×3.317        = 14.070
                   
           

Page No 3.61:

Question 6:

Using square root table, find the square root
540

Answer:

Using the table to find 3 and 5

540=54×10         =2×33×5         =2×3×1.732×2.2361         =23.24
            
                

Page No 3.61:

Question 7:

Using square root table, find the square root
8700

Answer:

Using the table to find 3 and 29

8700=3×29×100          =1.7321×5.385×10          =93.27
                      
              

Page No 3.61:

Question 8:

Using square root table, find the square root
3509

Answer:

Using the table to find 29

3509=121×29          =11×5.3851          =59.235
                         
              

Page No 3.61:

Question 9:

Using square root table, find the square root
6929

Answer:

Using the table to find 41

6929=169×41          =13 ×6.4031          =83.239
                       
              

Page No 3.61:

Question 10:

Using square root table, find the square root
25725

Answer:

Using the table to find 3 and 7

25725=3×5×5×7×7×7            =3×5×7×7            =1.732 ×5×7×2.646            =160.41
                          
                          
                 

Page No 3.61:

Question 11:

Using square root table, find the square root
1312

Answer:

Using the table to find 2 and 41

1312=2×2×2×2×2×41          =2×22×41          =2×2×1.414×6.4031          =36.222
              
                       
              

Page No 3.61:

Question 12:

Using square root table, find the square root
4192

Answer:

4192=2×2×2×2×2×131          =2×22×131
              
The square root of 131 is not listed in the table. Hence, we have to apply long division to find it.

Substituting the values:
            =   2×2×11.4455          (using the table to find 2)
              = 64.75

Page No 3.61:

Question 13:

Using square root table, find the square root
4955

Answer:

On prime factorisation:
4955 is equal to 5 × 991, which means that 4955=5×11.
The square root of 991 is not listed in the table; it lists the square roots of all the numbers below 100.
Hence, we have to manipulate the number such that we get the square root of a number less than 100. This can be done in the following manner:
4955=49.55×100=49.55×10
Now, we have to find the square root of 49.55.
We have: 49=7  and 50=7.071 .
Their difference is 0.071.
Thus, for the difference of 1 (50 - 49), the difference in the values of the square roots is 0.071.
For the difference of 0.55, the difference in the values of the square roots is:
0.55 × 0.0701 = 0.03905
49.55=7+0.03905=7.03905

Finally, we have:
4955=49.55×10=7.03905×10=70.3905

Page No 3.61:

Question 14:

Using square root table, find the square root
99144

Answer:

99144=3×3×11144
             = 31112             
           
            =3×3.316612                    (using the square root table to find 11)
             =0.829
       
             

Page No 3.61:

Question 15:

Using square root table, find the square root
57169

Answer:

57169=3×19169          
             1.732×4.358913          (using the square root table to find 3 and 19)
              0.581

Page No 3.61:

Question 16:

Using square root table, find the square root
101169

Answer:

101169=101169
The square root of 101 is not listed in the table. This is because the table lists the square roots of all the numbers below 100.
Hence, we have to manipulate the number such that we get the square root of a number less than 100. This can be done in the following manner:
101=1.01×100=1.01×10
Now, we have to find the square root of 1.01.

We have:
 1=1 and 2=1.414
Their difference is 0.414.
Thus, for the difference of 1 (2 - 1), the difference in the values of the square roots is 0.414.
For the difference of 0.01, the difference in the values of the square roots is:
0.01 × 0.414 = 0.00414
1.01=1+0.00414=1.00414101=1.01×10=1.00414×10=10.0414

Finally, 101169=1011313=10.041413=0.772
This value is really close to the one from the key answer.

Page No 3.61:

Question 17:

Using square root table, find the square root
13.21

Answer:

From the square root table, we have:
 13=3.606 and 14=2×7=3.742
Their difference is 0.136.
Thus, for the difference of 1 (14 - 13), the difference in the values of the square roots is 0.136.
For the difference of 0.21, the difference in the values of their square roots is:
0.136×0.21=0.02856
13.21=3.606+0.028563.635

Page No 3.61:

Question 18:

Using square root table, find the square root

Answer:

We have to find 21.97.
From the square root table, we have:
21=3×7=4.583 and 22=2×11=4.690
Their difference is 0.107.
Thus, for the difference of 1 (22 - 21), the difference in the values of the square roots is 0.107.
For the difference of 0.97, the difference in the values of their square roots is:
0.107×0.97=0.104
21.97=4.583+0.1044.687

Page No 3.61:

Question 19:

Using square root table, find the square root
110

Answer:

110=2×5×11        =1.414×2.236×3.317        (Using the square root table to find all the square roots)        =10.488
                  
           

Page No 3.61:

Question 20:

Using square root table, find the square root
1110

Answer:

1110=2×3×5×37          =1.414×1.732×2.236×6.083       (Using the table to find all the square roots )          =33.312
                          

Page No 3.61:

Question 21:

Using square root table, find the square root
11.11

Answer:

We have:
11=3.317 and 12=3.464
Their difference is 0.1474.
Thus, for the difference of 1 (12 - 11), the difference in the values of the square roots is 0.1474.
For the difference of 0.11, the difference in the values of the square roots is:
0.11 × 0.1474 = 0.0162
11.11=3.3166+0.0162=3.3283.333

Page No 3.61:

Question 22:

The area of a square field is 325 m2. Find the approximate length of one side of the field.

Answer:

The length of one side of the square field will be the square root of 325.
325=5×5×13         =5×13         =5×3.605         =18.030 
           
Hence, the length of one side of the field is 18.030 m.

Page No 3.61:

Question 23:

Find the length of a side of a sqiare, whose area is equal to the area of a rectangle with sides 240 m and 70 m.

Answer:

The area of the rectangle = 240 m × 70 m = 16800 m2
Given that the area of the square is equal to the area of the rectangle.
Hence, the area of the square will also be 16800 m2.
The length of one side of a square is the square root of its area.
16800=2×2×2×2×2×3×5×5×7            =2×2×52×3×7            =2042m=129.60 m
                 
                 
  Hence, the length of one side of the square is 129.60 m



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