Squares And Square Roots

(i) 484

Resolving 484 into prime factors we get,

484 = 2 × 2 × 11 × 11

Now,

Grouping the factors into pairs of equal factors, we get:

484 = (2 × 2) × (11 × 11)

We observe that all are paired so,

484 is a perfect square

(ii) 625

Resolving 625 into prime factors we get,

625 = 5 × 5 × 5 × 5

Now,

Grouping the factors into pairs of equal factors, we get:

625 = (5 × 5) × (5 × 5)

We observe that all are paired so,

625 is a perfect square

(iii) 576

Resolving 576 into prime factors we get,

576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

Now,

Grouping the factors into pairs of equal factors, we get:

576 = (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)

We observe that all are paired so,

576 is a perfect square

(iv) 941

Resolving 941 into prime factors we get,

941 = 941 × 1

Now,

As 941 itself is a prime number

Hence,

It do not have a perfect square

(v) 961

Resolving 961 into prime factors we get,

961 = 31 × 31

Now,

Grouping the factors into pairs of equal factors, we get:

961 = (31 × 31)

We observe that all are paired so,

961 is a perfect square

(vi) 2500

Resolving 2500 into prime factors we get,

2500 = 2 × 2 × 5 × 5 × 5 × 5

Now,

Grouping the factors into pairs of equal factors, we get:

2500 = (2 × 2) × (5 × 5) × (5 × 5)

We observe that all are paired so,

2500 is a perfect square

(i) 1156

Resolving 1156 into prime factors we get,

1156 = 2 × 2 × 17 × 17

Now, grouping the factors into pairs of equal factors

We get,

1156 = (2 × 2) × (17 × 17)

As all factors are paired

Hence, 1156 is a perfect square

Again,

1156 = (2 × 17) × (2 × 17)

= 34 × 34

= (34)²

Thus, 1156 is a square of 34

(ii) 2025

Resolving 2025 into prime factors we get,

2025 = 3 × 3 × 3 × 3 × 5 × 5

Now, grouping the factors into pairs of equal factors

We get,

2025 = (3 × 3) × (3 × 3) × (5 × 5)

As all factors are paired

Hence, 2025 is a perfect square

Again,

2025 = (3 × 3 × 5) × (3 × 3 × 5)

= 45 × 45

= (45)²

Thus, 2025 is a square of 45

(iii)14641

Resolving 14641 into prime factors we get,

14641 = 11 × 11 × 11 × 11

Now, grouping the factors into pairs of equal factors

We get,

14641 = (11 × 11) × (11 × 11)

As all factors are paired

Hence, 14641 is a perfect square

Again,

14641 = (11 × 11) × (11 × 11)

= 121 × 121

= (121)²

Thus, 14641 is a square of 121

(iv) 4761

Resolving 4761 into prime factors we get,

4761 = 3 × 3 × 23 × 23

Now, grouping the factors into pairs of equal factors

We get,

4761 = (3 × 3) × (23 × 23)

As all factors are paired

Hence, 4761 is a perfect square

Again,

4761 = (3 × 23) × (3 × 23)

= 69 × 69

= (69)²

Thus, 4761 is a square of 69

(i) 23805

Resolving 23805 into prime factors, we get

23805 = 3 × 3 × 23 × 23 × 5

Obtained factors can be paired into equal factors except for 5

To pair it equally multiply with 5

23805 × 5 = 3 × 3 × 5 × 5 × 23 × 23

Again,

23805 × 5 = (3× 5 × 23) × (3 × 5 × 23)

= 345 × 345

= (345)²

Therefore, product is the square of 345

(ii) 12150

Resolving 12150 into prime factors, we get

12150 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 × 2

Obtained factors can be paired into equal factors except for 2

To pair it equally multiply with 2

12150 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3

Again,

12150 × 2 = (5 × 3 × 2 × 2 × 2) × (5 × 3 × 2 × 2 × 2)

= 120 × 120

= (120)²

Therefore, product is the square of 120

(iii) 7688

Resolving 7688 into prime factors, we get

7688 = 2 × 2 × 31 × 31 × 2

Obtained factors can be paired into equal factors except for 2

To pair it equally multiply with 2

7688 × 2 = 2 × 2 × 2 × 2 × 31 × 31

Again,

7688 × 2 = (2× 2 × 31) × (2 × 2 × 31)

= 124 × 124

= (124)²

Therefore, product is the square of 124

(i) 12283

Resolving 14283 into prime factors, we get

14283 = 3 × 3 × 3 × 23 × 23

Obtained factors can be paired into equal factors except for 3

So, eliminate 3 by diving the dividing the number with 3

= (3 × 3) × (23 × 23)

Again,

= (3 × 23) × (3 × 23)

= 69 × 69

= (69)²

Therefore,

The resultant is the square of 69

(ii) 1800

Resolving 1800 into prime factors, we get

1800 = 2 × 2 × 5 × 5 × 3 × 3 × 2

Obtained factors can be paired into equal factors except for 2

So, eliminate 2 by diving the dividing the number with 2

= (2 × 2) × (3 × 3) × (5 × 5)

Again,

= (2 × 3 × 5) × (2 × 3 × 5)

= 30 × 30

= (30)²

Therefore,

The resultant is the square of 30

(iii) 2904

Resolving 2904 into prime factors, we get

2904 = 2 × 2 × 11 × 11 × 2 × 3

Obtained factors can be paired into equal factors except for 2 and 3

So, eliminate 6 by diving the dividing the number with 6

= (2 × 2) × (11 × 11)

Again,

= (2 × 11) × (2 × 11)

= 22 × 22

= (22)²

Therefore,

The resultant is the square of 22

11: Since 11 is a prime number,

Hence, it is not a perfect square

12: Since, 12 is ending with 2,

Hence, not a perfect square

16: Since, 16 = 4 × 4

= (16)²

Therefore, it is a perfect square

32: Since, 32 is ending with 2,

Hence, not a perfect square

36: Since, 36 = 6²

Hence, it is a perfect square

50: Since, 50 = 5² × 2

Hence, it is not a perfect square

64: Since, 64 = 8²

Hence, it is a perfect square

79: Since it is a prime number so it cannot be a perfect square

81: Since, 81 = 9²

Hence, it is a perfect square

111: Since, 111 is a prime number so it cannot be a perfect square

121: Since, 121 = 11²

Hence, it is perfect square

Since,

189 = 3² × 3 × 7

It cannot be written as pair of two equal factors, so 189 is not a perfect square

Since,

225 = (5 × 5) × (3 × 3)

It can be written as pair of two equal factors, so 22 is a perfect square

Since,

2048 = (2 × 2) × (2 × 2) × (2 × 2) (2 × 2) × (2 × 2) × 2

All the factors cannot be written as pair of two equal factors, so 189 is not a perfect square

Since,

343 = (7 × 7) × 7

It cannot be written as pair of two equal factors, so 343 is not a perfect square

Since,

441 = (7 × 7) × (3 × 3)

It can be written as pair of two equal factors, so 441 is a perfect square

Since,

2916 = (3 × 3) × (3 × 3) × (3 × 3) × (2 × 2)

It can be written as pair of two equal factors, so 2916 is a perfect square

Since,

11025 = (5 × 5) × (3 × 3) × (7 × 7)

It can be written as pair of two equal factors, so 11025 is a perfect square

Since,

3549 = (13 × 13) × 3 × 7

It cannot be written as pair of two equal factors, so

3549 is not a perfect square

(i) 8820

8820 = (2 × 2) × (3 × 3) × (7 × 7) × 5

In the above factors only 5 is unpaired

So, multiply the number with 5 to make it paired

Again,

8820 × 5 = 2 × 2 × 3 × 3 × 7 × 7 × 5 × 5

= (2 × 2) × (3 × 3) × (7 × 7) (5 × 5)

= (2 × 3 × 7 × 5) × (2 × 3 × 7 × 5)

= 210 × 210

= (210)²

So, the product is the square of 210

(ii) 3675

3675 = (5 × 5) × (7 × 7) × 3

In the above factors only 3 is unpaired

So, multiply the number with 3 to make it paired

Again,

3675 × 3 = 5 × 5 × 7 × 7 × 3 × 3

= (5 × 5) × (7 × 7) × (3 × 3)

= (3 × 5 × 7) × (3 × 5 × 7)

= 105 × 105

= (105)²

So, the product is the square of 105

(iii) 605

605 = 5 × (11 × 11)

In the above factors only 5 is unpaired

So, multiply the number with 5 to make it paired

Again,

605 × 5 = 5 × 5 × 11 × 11

= (5 × 5) × (11 × 11)

= (5 × 11) × (5 × 11)

= 55 × 55

= (55)²

So, the product is the square of 55

(iv) 2880

2880 = 5 × (3 × 3) × (2 × 2) × (2 × 2) × (2 × 2)

In the above factors only 5 is unpaired

So, multiply the number with 5 to make it paired

Again,

2880 × 5 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= (2 × 2) × (2 × 2) × (2 × 2) (3 × 3) × (5 × 5)

= (2 × 2 × 2 × 3 × 5) × (2 × 2 × 2 × 3 × 5)

= 120 × 120

= (120)²

So, the product is the square of 120

(v) 4056

4056 = (2 × 2) × (13 × 13) × 2 × 3

In the above factors only 2 and 3 are unpaired

So, multiply the number with 6 to make it paired

Again,

4056 × 6 = 2 × 2 × 13 × 13 × 2 × 2 × 3 × 3

= (2 × 2) × (13 × 13) × (2 × 2) (3 × 3)

= (2 × 2 × 3 × 13) × (2 × 2 × 3 × 13)

= 156 × 156

= (156)²

So, the product is the square of 156

(vi) 3468

3468 = (2 × 2) × 3 × (17 × 17)

In the above factors only 3 are unpaired

So, mulityply the number with 3 to make it paired

3468 × 3 = (2 × 2) × (3 × 3) × (17 × 17)

= (2 × 3 × 17) × (2 × 3 × 17)

= 102 × 102

= (102)²

So, the product is the square of 102

(vii) 7776

7776 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2 × 3

In the above factors only 2 and 3 are unpaired

So, multiply the number with 6 to make it paired

Again,

7776 × 6 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= (2 × 2) × (2 × 2) × (2 × 2) (3 × 3) × (3 × 3) × (3 × 3)

= (2 × 2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 3 × 3 × 3)

= 216 × 216

= (216)²

So, the product is the square of 216

(i) 16562

16562 = (7 × 7) × (13 × 13) × 2

= (7 × 7) × (13 × 13)

= (7 × 13) × (7 × 13)

= 91 × 91

= 91²

Therefore, the resultant is the square of 91

(ii) 3698

3698 = 2 × (43 × 43)

= 43 × 43

= 43²

Therefore, the numbers must be divided by 2 and resultant is square of 43

(iii) 5103

5103 = (3 × 3) × (3 × 3) × (3 × 3) × 7

= (3 × 3 × 3) × (3 × 3 × 3)

= 27 × 27

= 27²

Therefore, the number must be divided by 7 and resultant is square of 27

(iv) 3174

3174 = 2 × 3 × (23 × 23)

= 23 × 23

= 23²

Therefore, the number must be divided by 6 and the resultant is square of 23

(v) 1575

1575 = 3 × 3 × 5 × 5 × 7

= 3 × 3 × 5 × 5

= (3 × 5) × (3 × 5)

= 15 × 15

= 15²

Therefore, the number must be divided by 7 and the resultant is square of 15

Greatest 2 digit number = 99

Hence, greatest 2 digit perfect square number is:

99 – 18 = 81

Smallest 3 digit number = 100

At first we will find the square root of 100

Hence, the least number that is a perfect square is 100 itself

Factors of 4851 are:

4851 = 3 × 3 × 7 × 7 × 11

Pairs = 3² × 7²

Hence, 4851 should be multiplied by 11 in order to get a perfect square when smallest number multiplied to 4851

Factors of 28812 are:

28812 = 2 × 2 × 3 × 3 × 3 × 17 × 17

Pairs = 2² × 3² × 17²

Hence, 28812 should be divided by 3 in order to get a perfect square when divided by the least number

The square root will be:

2 × 3 × 17 = 102

Factors of 1152 are:

1152 = 2⁷ × 3²

Pairs = 2⁶ × 3²

Hence, 1152 should be divided by 2 in order to get the perfect square.

Hence the number after division by 2 = 1152/2 = 576

Factors of 576 are = 2⁶ × 3² = 24²

Hence, resulting number is the square of 24.

Numbers ending with 2, 3, 7 or 8 are not perfect squares. So,

(i) 1547

(ii) 45743

(iii) 8948

(iv) 333333 are not perfect squares

Hence, 7, 8, 3, 2 as ending numbers respectively. As mentioned above ending with 2, 3, 7, 8 are not perfect square. So, these given numbers are not perfect squares

Square of an odd number is an odd number

Square of an even number is an even number

(i) 731: It is an odd number so its square is also odd number

(ii) 3456: It is an even number so its square is also even number

(iii) 5559: It is an odd number so its square is also odd number

(iv) 42008: It is an even number so its square is also even number

(i) 52

Unit digit of (52)² = unit digit of (2)² = 4

(ii) 977

Unit digit of (977)² = unit digit of (7)² = 9

(iii) 4583

Unit digit of (4583)² = unit digit of (3)² = 9

(iv) 78367

Unit digit of (78367)² = unit digit of (7)² = 9

(v) 52698

Unit digit of (52698)² = unit digit of (8)² = 4

(vi) 99880

Unit digit of (99880)² = unit digit of (0)² = 0

(vii) 12796

Unit digit of (12796)² = unit digit of (6)² = 6

(viii) 55555

Unit digit of (55555)² = unit digit of (5)² = 5

(ix) 53924

Unit digit of (53924)² = unit digit of (4)² = 6

The pattern here is the square of the number on the Right-hand side is equal to the sum of all the numbers on the left-hand side.

Thus, for n terms,

1 + 3 + 5 +…..n terms = n² [As there are n terms]

(i) 100² – 99²

= 100 + 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 + 92) + (97 + 96)

= 585

(i) (8, 15, 17)

L.H.S = 8² + 15² = 289

R.H.S = 17² = 289

L.H.S = R.H.S

So, it is Pythagoras

(ii) (18, 80, 82)

L.H.S = 18² + 80² = 6724

R.H.S = 82² = 6724

L.H.S = R.H.S

So, it is Pythagoras

(iii) (14, 48, 51)

L.H.S = 14² + 48² = 2500

R.H.S = 51² = 2601

L.H.S ≠ R.H.S

So, it is not Pythagoras

(iv) (10, 24, 26)

L.H.S = 10² + 24² = 676

R.H.S = 26² = 676

L.H.S = R.H.S

So, it is Pythagoras

(v) (16, 63, 65)

L.H.S = 16² + 63² = 4225

R.H.S = 65² = 4225

L.H.S = R.H.S

So, it is Pythagoras

(vi) (12, 35, 38)

L.H.S = 12² + 35² = 1369

R.H.S = 38² = 1444

L.H.S ≠ R.H.S

So, it is Pythagoras

From observation:

(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6) =

= 70

R.H.S = [No. of terms in L.H.S × (No. of terms + 1)] (Therefore, only when L.H.S starts with 1)

Therefore,

(i) 1 + 2 + 3 +…..50 = [50 × (50 + 1)]

= 25 × 51

= 1275

(ii) 31 + 32 +…..+50 = (1 + 2 + 3 + …. + 50) – (1 + 2 + ….. 30)

= 1275 – [ (30 × 30 +1)]

= 1275 – 465

= 810

R.H.S = [(No. of terms in L.H.S) × (No. + 1) × (2 × No. + 1)]

(i) 1² + 2² + 3² + 4² + …… + 10² = [10 (10 + 1) × (2 × 10 + 1)]

= [2310]

= 385

(ii) 5² + 6² +….. + 12² = 1² + 2² + ….. 12² – (1² + 2² + 3³ + 4²)

= [12 × (12 + 1) × (2 × 12 + 1)] - [4 ×(4 + 1) × (2 × 4 + 1)]

= 650 – 30

= 620

Only even numbers be the square of even numbers

So, 256, 324, 1296, 5476, 373758 can be square of even numbers but 373758 is not a perfect square

So, 256, 324, 1296, 5476 are numbers

Numbers ending with 2, 3, 7, 8 cannot be perfect square. So,

1028 (iv) 1022 (v) 1023 (vi) 1027 cannot be whole squares.

All the natural numbers whose unit digit is 0, 1, 4, 5, 6 or 9 can not be said surely if they are square numbers or not

Any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number.

Hence,

The five examples are:

(i) 2061

The ending digit is 1. Hence, it may or may not be a square number

(ii) 1069

The ending digit is 9. Hence, it may or may not be a square number

(iii) 1234

The ending digit is 4. Hence, it may or may not be a square number

(iv) 56790

The ending digit is 0. Hence, it may or may not be a square number

(v) 76555

The ending digit is 5. Hence, it may or may not be a square number

(i) False: Because 169 is square number with odd digit

(ii) False: Square of 3 (Prime) is 9 (not prime)

(iii) False: Sum of 2² and 3² is 13 which is not square number

(iv) False: Difference of 3² and 2² is 5, which is not square number

(v) True: Because the square of 2² and 3² is 36 which is square of 6

(vi) True: As (-2)² is 4, i.e. not negative

(vii) True: As there is no square number between them

(viii) True: The fourteen numbers upto 200 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196

(i) 25

Here, a = 2, b = 5

25² = 625

And,

25² = 25× 25 = 625

(ii) 37

Here, a = 3, b = 7

37² = 1369

And,

37² = 37× 37 = 1369

(iii) 54

Here, a = 5, b = 4

54² = 2916

And,

54² = 54× 54 = 2916

(iv) 71

Here, a = 7, b = 1

71² = 4941

And,

71² = 71× 71 = 4941

(v) 96

Here, a = 9, b = 6

96² = 9216

And,

96² = 96× 96 = 9216

(i) 98

Step I: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step II: Draw square and divide it into n² sub-squares of the same size by drawing (n - 1) horizontal and (n - 1) vertical lines.

Step III: Draw the diagonals of each sub-square.

Step IV: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares as shown below.

Step V: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step VI: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step VII: Obtain the required square by writing the digits from the left-most side.

(98)² = 9604

(ii) 273

Step I: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step II: Draw square and divide it into n² sub-squares of the same size by drawing (n - 1) horizontal and (n - 1) vertical lines.

Step III: Draw the diagonals of each sub-square.

Step IV: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares as shown below.

Step V: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step VI: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step VII: Obtain the required square by writing the digits from the left-most side.

(273)²= 74529

(iii)348

Step I: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step II: Draw square and divide it into n² sub-squares of the same size by drawing (n - 1) horizontal and (n - 1) vertical lines.

Step III: Draw the diagonals of each sub-square.

Step IV: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares as shown below.

Step V: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step VI: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step VII: Obtain the required square by writing the digits from the left-most side.

348² = 121104

(iv) 295

Step I: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step II: Draw square and divide it into n² sub-squares of the same size by drawing (n - 1) horizontal and (n - 1) vertical lines.

Step III: Draw the diagonals of each sub-square.

Step IV: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares as shown below.

Step V: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step VI: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step VII: Obtain the required square by writing the digits from the left-most side.

(295)² = 87025

(v) 171

Step I: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step II: Draw square and divide it into n² sub-squares of the same size by drawing (n - 1) horizontal and (n - 1) vertical lines.

Step III: Draw the diagonals of each sub-square.

Step IV: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares as shown below.

Step V: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step VI: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step VII: Obtain the required square by writing the digits from the left-most side.

(171)² = 29241

(i) (127)² = 127 × 127

= 16129

(ii) (503)² = 503 × 503

= 253009

(iii) (451)² = 451 × 451

= 203401

(iv) (862)² = 862 × 862

= 743044

(v) (265)² = 265 × 265

= 70225

(i) 425

We know that,

The square of 425 is:

(425)² = 425 × 425

= 180625

Hence, the square of 425 is 180625

(ii) 575

We know that,

The square of 575 is:

(575)² = 575 × 575

= 330625

Hence, the square of 575 is 330625

(iii) 405

We know that,

The square of 405 is:

(405)² = 405 × 405

= 164025

Hence, the square of 405 is 164025

(iv) 205

We know that,

The square of 205 is:

(205)² = 205 × 205

= 42025

Hence, the square of 205 is 42025

(v) 95

We know that,

The square of 95 is:

(95)² = 95 × 95

= 9025

Hence, the square of 95 is 9025

(vi) 745

We know that,

The square of 745 is:

(745)² = 745 × 745

= 555025

Hence, the square of 745 is 555025

(vii) 512

We know that,

The square of 512 is:

(512)² = 512 × 512

= 262144

Hence, the square of 512 is 262144

(viii) 995

We know that,

The square of 995 is:

(995)² = 995 × 995

= 990025

Hence, the square of 995 is 990025

(i) 405

We have,

(405)² = (400 + 5)²

= (400)² + 5² + 2 (400) (5)

= 160000 + 25 + 4000

= 164025

(ii) 510

We have,

(510)² = (500 + 10)²

= 250000 + 100 + 10000

= 260100

(iii) 1001

We have,

(1001)² = (1000 + 1)²

= (1000)² + 1 + 2 (1000)

= 1000000 + 1 + 2000

= 1002001

(iv) 209

We have,

(209)² = (200 + 9)²

= (200)² + 9² + 2 (200) (9)

= 40000 + 81 + 3600

= 43681

(v) 605

We have,

(605)² = (600 + 5)²

= (600)² + 5² + 2 (600) (5)

= 360000 + 25 + 6000

= 366025

(i) 395

395 = (400 – 5)²

= (400)² + 5² – 2 (400) (5)

= 160000 + 25 – 4000

= 156025

(ii) 995

995 = (1000 – 5)²

= (1000)² + 5² – 2 (1000) (5)

= 1000000 + 25 – 10000

= 990025

(iii)495

495 = (500 – 5)²

= (500)² + 5² – 2 (500) (5)

= 250000 + 25 – 5000

= 245025

(iv) 498

498 = (500 – 2)²

= (500)² + 2² – 2 (500) (2)

= 250000 + 4 – 2000

= 248004

(v) 99

99 = (100 – 1)²

= (100)² + 1² – 2 (100) (1)

= 10000 + 1 – 200

= 9799

(vi) 999

999 = (1000 – 1)²

= (1000)² + 1² – 2 (1000) (1)

= 1000000 + 1 – 2000

= 998001

(vii)599

(600 – 1)²

= (600)² + 1² – 2 (600) (1)

= 360000 + 1 – 1200

= 358801

(i) 52, (52)² = (50 + 2)²

= 50² + 2² + (2 × 50 × 2)

= 2500 + 4 + 200

= 2704

(ii) 95, (95)² = (100 - 5)²

= 100² + 5² - (2 × 5 × 100)

= 10000 + 25 - 1000

= 9025

(iii) 505, (505)² = (505 + 5)²

= 500² + 5² + (2 × 500 × 5)

= 250000 + 25 + 5000

= 255025

(iv) 702, (702)² = (700 + 2)²

= 700² + 2² + (2 × 700 × 2)

= 140000 + 4 + 2800

= 142804

(v) 99, (99)² = (100 - 1)²

= 100² + 1² - (2 × 100 × 1)

= 10000 + 1 - 200

= 9301

(i) 9801

Unit digit = 1

Unit digit of square root = 1 or 9

As number is odd, square root is also odd

(ii) 99856

Unit digit = 6

Unit digit of square root = 4 or 6

As number is even, square root is also even

(iii) 998001

Unit digit = 1

Unit digit of square root = 1 or 9

As number is odd, square root is also odd

(iv) 657666025

Unit digit = 5

Unit digit of square root = 5

As number is odd, square root is also odd

(i) 441

441 = 3² × 7²

= 3 × 7

= 21

(ii) 196

196 = 2² × 7²

= 2 × 7

= 14

(iii) 529

529 = 23²

= 23

(iv) 1764

1764 = 2² × 3² × 7²

= 2 × 3 × 7

= 42

(v) 1156

1156 = 2² × 17²

= 2 × 17

= 34

(vi) 4096

4096 = 2¹²

= 2⁶

= 64

(vii) 7056

7056 = 2² × 2² × 21²

= 2 × 2 × 21

= 84

(viii) 8281

8281 = 91²

= 91

(ix) 11664

11664 = 2² × 2² × 3² × 3² × 3²

= 2 × 2 × 3 × 3× 3

= 108

(x) 47089

47089 = 217²

= 217

(xi) 24336

24336 = 2² × 2² × 3² × 13²

= 2 × 2 × 3 × 13

= 156

(xii) 190969

190969 = 23² × 19²

= 23 × 19

= 437

(xiii) 586756

586756 = 2² × 383²

= 2 × 383

= 766

(xiv) 27225

27225 = 5² × 3² × 11²

= 5 × 3 × 11

= 165

(xv) 3013696

3013696 = 2⁶ × 217²

= 2³ × 217

= 1736

180 = 2² × 3² × 5

= (2 × 2) × (3 × 3) × 5

To make the unpaired 5 into paired, multiply the number with 5

Therefore,

180 × 5 = 2² × 3² × 5²

Hence, square root of number = × = 2 × 3 × 5

= 30

147 = 7² × 3

To make the unpaired 3 into paired, multiply the number with 3

Therefore,

147 × 3 = 7² × 3²

Hence, square root of number = √147 × √3 = 7 × 3

= 21

3645 = 5 × (3 × 3) × (3 × 3) × 3

Here 5 and 3 are unpaired so we have to divide 3645 with 5 × 3 = 15

Therefore,

= 3² × 3²

Hence,

Square root of numbers = = 3 × 3

= 9

1152 = (2 × 2) × (2 × 2) × 2 × (3 × 3)

Here 2 is unpaired so we have to divide 1152 with 2

Therefore,

= 2² × 2² × 2² × 3²

Hence,

Square root of numbers = = 2 × 2 × 2 × 3

= 24

Let a and b be two numbers

a × b = 1296

a = 16b

= 16 b × b

= 1296

b² = 81

b = 9

Therefore,

a = 144 and b = 9

Let total residents be a

Therefore, each paid Rs. a

Total collection = a (a) = a²

given, Total Collection = 202500

a =
a = √(2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 5)
a = 2 × 3 × 3 × 5 × 5
a = 450

Therefore,

Total residents = 450

Let there were a members

Therefore, each attributed a paise

Therefore,

a (a), i.e. total cost collected = 9216 paise

a² = 9216

a =

= 2 × 2 × 2 × 12

= 96

Therefore, there were 96 members and each contributed 96 paise

Let, a be number of school students

Therefore, each student contributed a paise

Total money obtained = a²paise

= 230400 paise

a =

=

= 10

a = 10 × 2 × 2 × 12

a = 480

Therefore, there were 480 students

Let ‘a’ be the side of square field

Therefore,

a² = 5184 m²

a = m

a = 2 × 2 ×2 × 9

= 72 m

Perimeter of square = 4a

= 288 m

Perimeter of rectangle = 2 (l + b)

= 288 m

2 (2b + b) = 288

b = 48 and l = 96

Area of rectangle = 96 × 48 m²

= 4608 m²

(i) 6, 9, 15 and 20

L.C.M of given 4 numbers is 180

180 = 2² × 3² × 5

To make it a perfect square, we have to multiply the number with 5

Therefore,

180 × 5 = 2² × 3² × 5²

900 is the least square number divisible by 6, 9, 15 and 20

3600 is the least square number divisible by 8, 12, 15 and 20

(ii) 8, 2, 15 and 20

L.C.M of given 4 numbers is 360

360 = 2² × 3² ×2 × 5

To make it a perfect square, we have to multiply the number with 2 × 5 = 10

Therefore,

360 × 10 = 2² × 3² × 5² × 2²

121 – 1 = 120

120 – 3 = 117

117 – 5 = 112

112 – 7 = 115

115 – 9 = 106

106 – 11 = 95

95 – 13 = 82

82 – 15 = 67

67 – 17 = 50

50 – 19 = 31

31 – 21 = 10

Clearly, we have performed operation 11 times

Therefore,

= 11

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

Clearly, we have performed subtraction 13 times

Therefore,

= 13

(i) 7744

7744 = 2² × 2² × 2² × 11²

= 2 × 2× 2 × 11

=88

(ii) 9604

9604 = 2² × 7² × 7²

= 2 × 7× 7

=

(iii) 5929

5929 = 11² × 7²

= 11 × 7

=77

(iv) 7056

7056 = 2² × 2² × 7² × 3²

= 2 × 2× 7 × 3

=84

Let a be the number of students

Therefore,

Each student denoted a rupee

So,

Total amount collected = a × a rupees

= 2401

a² = 2401

a = 49

Therefore,

There are 49 students in the class

Let a be number of rows

Therefore,

No. of columns = a

Total number of students who sat in field = a²

Total students = a² + 71

= 6000

a² = 5929

a =

a = 11 × 7

= 77

Therefore, total number of rows = 77

(i) 12544

Therefore,

= 112

(ii) 97344

Therefore,

= 312

(iii) 286225

Therefore,

= 535

(iv) 390625

= 625

(v) 363609

Therefore,

= 603

(vi) 974169

Therefore,

= 987

(vii) 120409

Therefore,

= 347

(viiii) 1471369

Therefore,

= 1213

(ix) 291600

Therefore,

= 540

(x) 9653449

Therefore,

= 3107

(xi) 1745041

Therefore,

= 1321

(xii) 4008004

Therefore,

= 2002

(xiii) 20657025

= 4545

(xiv) 152547201

Therefore,

= 12351

(xv) 20421361

Therefore,

= 4519

(xvi) 62504836

Therefore,

= 7906

(xvii) 82264900

Therefore,

= 9070

(xviii) 3226694416

= 56804

(xix) 6407522209

Therefore,

= 80047

(xx) 3915380329

(i) 2361

Hence,

57 must be subtracted from 2361 in order to get a perfect square

(ii) 194491

Hence,

10 must be subtracted from 194491 in order to get a perfect square

(iii) 26535

Hence,

291 must be subtracted from 26535 in order to get a perfect square

(iv) 161605

Hence,

1 must be subtracted from 161605 in order to get a perfect square

(v) 4401624

Hence,

20 must be subtracted from 4401624 in order to get a perfect square number

(i) 5607

The remainder is 131

Hence, (74)² < 5607

The next perfect square number is:

(75)² = 5625 > 5607

Hence, the number to be added = 5625 – 5607

= 18

(ii)4931

The remainder is 31

Hence, (70)² < 4931

The next perfect square number is:

(71)² = 5041 > 4931

Hence, the number to be added = 5041 – 4931

= 110

(iii) 4515600

The remainder is 4224

Hence, (2124)² < 4515600

The next perfect square number is:

(2125)² = 4515625 > 4515600

Hence, the number to be added = 4515625 – 4515600

= 25

(iv) 37460

The remainder is 211

Hence, (193)² < 37460

The next perfect square number is:

(194)² = 37636 > 37460

Hence, the number to be added = 37636 – 37460

= 176

(v) 506900

The remainder is 1379

Hence, (711)² < 506900

The next perfect square number is:

(712)² = 506944 > 506900

Hence, the number to be added = 506944 – 506900

= 44

We know that,

Greatest 5 digit number = 99999

The remainder is 143

Therefore,

The greatest 5 digit perfect square number is:

99999 – 143

= 99856

Hence, 99856 is the required greatest 5 digit perfect square number

We know that,

Least 4 digit number = 1000

The remainder is 39

Therefore,

(31)² < 1000

Hence,

The next perfect square number is:

(32)² = 1024 > 1000

Hence, 1024 is the required number

We know that,

Least 6 digit number = 100000

The remainder is 144

Therefore,

(316)² < 100000

Hence, the next perfect square

(317)² = 100489 > 100000

Hence, 100489 is the required number

We know that,

Greatest 4 digit number = 9999

The remainder is 198

Hence,

The greatest 4 digit perfect square number = 9999 – 198

= 9801

Total number of soldiers = 8160

Number of soldiers left out = 60

Number of soldiers arranged in rows to form a perfect square = 8160 – 60

= 8100

Hence, number of soldiers in each row =

=

= 90

Area of square field = 60025 m²

Speed of cyclist = 18 km/h

= 18 ×

= 5 m/s²

Area = 60025 m²

Side² = 60025

Side =

= 245

Therefore,

Total length of boundary = 4 × Side

= 4 × 245

= 980 m

Hence,

Time taken =

= 196 seconds

= 3 minutes and 16 seconds

Rate of leveling and turning a square lawn = 2.50 per m²

Total cost of leveling and turning = Rs. 13322.50

Total area of square lawn =

= 5329 m²

Side of square lawn =

= 73 m

Total length of lawn = 4 × 73

= 292 m

Cost of fencing the lawn at Rs 5 per metre = 292 × 5

= Rs. 1460

We know that,

Largest 3 digit number = 999

The remainder is 38

Hence,

The greatest 3-digit perfect square number = 999 – 38

= 961

At first we have to find,

The square root of 2300

So, the square root of 2300 is:

The remainder is 91

Hence,

(47)² < 2300

Now, the next perfect square number is (48)² = 2304 > 2300

Hence,

The smallest number that must be added to 2300 to get a perfect square is:

2304 – 2300

= 4

(i)

(ii)

(iii)

=

(iv)

=

(v)

=

(vi)

=

(vii)

=

(viii)

=

(ix)

=

(x)

=

(xi)

=

(xii)

=

(xiii)

=

(xiv)

=

(xv)

=

(i) = (Cancelling numerator and denominator with 5)

= (Therefore, = 4, = 9)

(ii)

= = (Therefore, = 21, = 25)

(iii) = (Cancelling numerator and denominator with 3)

= (Therefore, = 23, = 24)

(iv)

= ×

We know that,

× =

= 2² × 3 × 13

= 156

(v)

= ×

We know that,

× =

= 5 × 9 × 2

= 90

Given area = 80 × m²

= m²

If L is length of each side

Therefore,

L² =

L = (Therefore, = )

=

Given, area = 30 × m²

= m²

If L is length of each side then,

L² =

L = =

= (Therefore, )

Therefore, length is

Area of rectangular field = l × b

= 72 × 338 m²

= 24336 m²

Area of square = L² = 24336 m²

L =

= 156 m

Therefore, 156 m is the length of side of square playground.

84.8241

Therefore,

√84.8241 = 9.21

0.7225

√0.7225 = 0.85

0.81304

= 0.902

0.00002025

150.0625

= 12.25

225.6004

= 15.02

3600.720036

= 60.006

236.144689

= 15.367

0.00059049

= 0.0243

176.252176

= 13.276

9998.0001

= 99.99

0.00038809

= 0.0197

a =

Given: area = L² = 256.6 m²

L =

a² = 0.00053361

Therefore,

a = 0.0231

(i)

At first, we find

Therefore,

=

= = 7.7

And,

=

= = 2.3

Now,

= 0.54

(ii)

At first, we find

Therefore,

=

= = 0.44

And,

=

= = 0.42

Now,

= 15

=

Now,

=

= = 22.5

=

= = 2.25

+

= 22.5 + 2.25

= 24.75

=

Now,

(i)

= 10 × 10.15

(ii) =

= 1.015

(i) 5 = 2.236

= 2.236

(ii) 7 = 2.647

= 2.646

(iii) 17 = 4.123

= 4.123

(iv) 20 = 4.472

= 4.472

(v) 66 = 8.124

= 8.124

(vi) 427 = 20.664

= 20.664

(vii) 1.7 = 1.304

= 1.304

(viii) 23.1 = 4.806

= 4.806

(ix) 2.5 = 1.581

= 1.581

(x) 237.615 = 15.415

= 15.415

(xi) 15.3215 = 3.914

= 3.914

(xii) 0.9 = 0.949

= 0.949

(xiii) 0.1 = 0.316

= 0.316

(xiv) 0.016 = 0.126

(xv) 0.00064 = 0.025

(xvi) 0.019 = 0.138

= 0.138

(xvii) = 0.875

= 0.875

(xviii) = 0.416

= 0.645

(xix) = 2.500000

(xx) = 287.62

Hence,

The square root of 12.0068 is:

Hence,

Hence,

= 3.4651 approx

The square root of 11 is:

Hence,

= 3.31662

(i) =

=

= 4.535

(ii) =

=

=

= 28.867

(i)

=

=

=

=

=

= 1.50

(ii)

=

=

=

=

=

= 2.519

(iii)

=

=

=

=

=

= 4.627

(iv)

=

=

=

=

= 7.155

(v)

=

=

=

=

= 0.735

From square root table,

Square root of 7 is:

= 2.645

Therefore,

The square root of 7 is 2.645

From square root table,

Square root of 15 is:

= 3.872

Therefore,

The square root of 15 is 3.872

From square root table,

Square root of 74 is:

= 8.602

Therefore,

The square root of 74 is 8.602

From square root table,

Square root of 82 is:

= 9.055

Therefore,

The square root of 82 is 9.055

From square root table,

Square root of 198 is:

= 14.071

Therefore,

The square root of 198 is 14.071

From square root table,

Square root of 540 is:

= 23.237

Therefore,

The square root of 540 is 23.237

From square root table,

Square root of 8700 is:

= 93.237

Therefore,

The square root of 8700 is 93.237

From square root table,

Square root of 3509 is:

= 59.236

Therefore,

The square root of 3509 is 59.236

From square root table,

Square root of 6929 is:

= 83.240

Therefore,

The square root of 6929 is 83.240

From square root table,

Square root of 25720 is:

= 160.374

Therefore,

The square root of 25720 is 160.374

From square root table,

Square root of 1312 is:

= 36.221

Therefore,

The square root of 1312 is 36.221

From square root table,

Square root of 4192 is:

= 64.745

Therefore,

The square root of 4192 is 64.745

From square root table,

Square root of 49555 is:

= 222.609

Therefore,

The square root of 49555 is 222.609

From square root table,

Square root of is:

= 0.829

Therefore,

The square root of is 0.829

From square root table,

Square root of is:

= 0.580

Therefore,

The square root of is 0.580

From square root table,

Square root of is:

= 0.773

Therefore,

The square root of is 0.773

From square root table,

Square root of 13.21 is:

= 3.634

Therefore,

The square root of 13.21 is 3.634

From square root table,

Square root of 21.97 is:

= 4.687

Therefore,

The square root of 21.97 is 4.687

From square root table,

Square root of 110 is:

= 10.488

Therefore,

The square root of 110 is 10.488

From square root table,

Square root of 1110 is:

= 33.316

Therefore,

The square root of 1110 is 33.316

From square root table,

Square root of 11.11 is:

= 3.333

Therefore,

The square root of 11.11 is 3.333

Area of the field = 325 m²

In order to find approximate length of the side of the field we will have to calculate the square root of 325

= 18.027 m

Hence,

The approximate length of one side of the field is 18.027 m

According to the question,

Area of square = Area of rectangle

Side² = 240 × 70

Side =

=

= 20

= 20 × 6.48

= 129.60 m