#### Page No 96:

#### Question 1:

What will be the unit digit of the squares of the following numbers?

(i) 81 (ii) 272

(iii) 799 (iv) 3853

(v) 1234 (vi) 26387

(vii) 52698 (viii) 99880

(ix) 12796 (x) 55555

#### Answer:

We know that if a number has its unit’s place digit as *a*, then its square will end with the unit digit of the multiplication *a* × *a*.

(i) 81

Since the given number has its unit’s place digit as 1, its square will end with the unit digit of the multiplication (1 ×1 = 1) i.e., 1.

(ii) 272

Since the given number has its unit’s place digit as 2, its square will end with the unit digit of the multiplication (2 × 2 = 4) i.e., 4.

(iii) 799

Since the given number has its unit’s place digit as 9, its square will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.

(iv) 3853

Since the given number has its unit’s place digit as 3, its square will end with the unit digit of the multiplication (3 × 3 = 9) i.e., 9.

(v) 1234

Since the given number has its unit’s place digit as 4, its square will end with the unit digit of the multiplication (4 × 4 = 16) i.e., 6.

(vi) 26387

Since the given number has its unit’s place digit as 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.

(vii) 52698

Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.

(viii) 99880

Since the given number has its unit’s place digit as 0, its square will have two zeroes at the end. Therefore, the unit digit of the square of the given number is 0.

(xi) 12796

Since the given number has its unit’s place digit as 6, its square will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.

(x) 55555

Since the given number has its unit’s place digit as 5, its square will end with the unit digit of the multiplication (5 × 5 = 25) i.e., 5.

#### Question 2:

The following numbers are obviously not perfect squares. Give reason.

(i) 1057 (ii) 23453

(iii) 7928 (iv) 222222

(v) 64000 (vi) 89722

(vii) 222000 (viii) 505050

#### Answer:

The square of numbers may end with any one of the digits 0, 1, 5, 6, or 9. Also, a perfect square has even number of zeroes at the end of it.

(i) 1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.

(ii) 23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.

(iii) 7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.

(iv) 222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(v) 64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

(vi) 89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(vii) 222000 has three zeroes at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

(viii) 505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

#### Question 3:

The squares of which of the following would be odd numbers?

(i) 431 (ii) 2826

(iii) 7779 (iv) 82004

#### Answer:

The square of an odd number is odd and the square of an even number is even. Here, 431 and 7779 are odd numbers.

Thus, the square of 431 and 7779 will be an odd number.

#### Question 4:

Observe the following pattern and find the missing digits.

11^{2} = 121

101^{2} = 10201

1001^{2} = 1002001

100001^{2} = 1…2…1

10000001^{2} = …

#### Answer:

In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,

100001^{2} = 10000200001

10000001^{2} = 100000020000001

#### Question 5:

Observe the following pattern and supply the missing number.

11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} = …

…^{2} = 10203040504030201

#### Answer:

By following the given pattern, we obtain

1010101^{2 }= 1020304030201

101010101^{2} = 10203040504030201

#### Question 6:

Using the given pattern, find the missing numbers.

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} + _ ^{2} = 21^{2}

5^{2} + _ ^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + _ ^{2} = __^{2}

#### Answer:

From the given pattern, it can be observed that,

(i) The third number is the product of the first two numbers.

(ii) The fourth number can be obtained by adding 1 to the third number.

Thus, the missing numbers in the pattern will be as follows.

4^{2} + 5^{2} + = 21^{2}

5^{2} + + 30^{2} = 31^{2}

6^{2} + 7^{2} + =

#### Question 7:

Without adding find the sum

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

#### Answer:

We know that the sum of first *n* odd natural numbers is *n*^{2}.

(i) Here, we have to find the sum of first five odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 = (5)^{2} = 25

(ii) Here, we have to find the sum of first ten odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)^{2} = 100

(iii) Here, we have to find the sum of first twelve odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)^{2} = 144

#### Question 8:

(i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11odd numbers.

#### Answer:

We know that the sum of first *n* odd natural numbers is *n*^{2}.

(i) 49 = (7)^{2}

Therefore, 49 is the sum of first 7 odd natural numbers.

49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 121 = (11)^{2}

Therefore, 121 is the sum of first 11 odd natural numbers.

121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

#### Question 9:

How many numbers lie between squares of the following numbers?

(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

#### Answer:

We know that there will be 2*n* numbers in between the squares of the numbers *n* and (*n* + 1).

(i) Between 12^{2} and 13^{2}, there will be 2 × 12 = 24 numbers

(ii) Between 25^{2} and 26^{2}, there will be 2 × 25 = 50 numbers

(iii) Between 99^{2} and 100^{2}, there will be 2 × 99 = 198 numbers

#### Page No 98:

#### Question 1:

Find the square of the following numbers

(i) 32 (ii) 35

(iii) 86 (iv) 93

(v) 71 (vi) 46

#### Answer:

(i) 32^{2} = (30 + 2)^{2}

= 30 (30 + 2) + 2 (30 + 2)

= 30^{2} + 30 × 2 + 2 × 30 + 2^{2}

= 900 + 60 + 60 + 4

= 1024

(ii) The number 35 has 5 in its unit’s place. Therefore,

35^{2} = (3) (3 + 1) hundreds + 25

= (3 × 4) hundreds + 25

= 1200 + 25 = 1225

(iii) 86^{2 }= (80 + 6)^{2}

= 80 (80 + 6) + 6 (80 + 6)

= 80^{2} + 80 × 6 + 6 × 80 + 6^{2}

= 6400 + 480 + 480 + 36

= 7396

(iv) 93^{2 }= (90 + 3)^{2}

= 90 (90 + 3) + 3 (90 + 3)

= 90^{2} + 90 × 3 + 3 × 90 + 3^{2}

= 8100 + 270 + 270 + 9

= 8649

(v) 71^{2 }= (70 + 1)^{2}

= 70 (70 + 1) + 1 (70 + 1)

= 70^{2} + 70 × 1 + 1 × 70 + 1^{2}

= 4900 + 70 + 70 + 1

= 5041

(vi) 46^{2 }= (40 + 6)^{2}

= 40 (40 + 6) + 6 (40 + 6)

= 40^{2} + 40 × 6 + 6 × 40 + 6^{2}

= 1600 + 240 + 240 + 36

= 2116

#### Question 2:

Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14

(iii) 16 (iv) 18

#### Answer:

For any natural number *m* > 1, 2*m*, *m*^{2} − 1, *m*^{2} + 1 forms a Pythagorean triplet.

(i) If we take *m*^{2} + 1 = 6, then *m*^{2} = 5

The value of *m* will not be an integer.

If we take *m*^{2} − 1 = 6, then *m*^{2} = 7

Again the value of *m* is not an integer.

Let 2*m* = 6

*m* = 3

Therefore, the Pythagorean triplets are 2 × 3, 3^{2} − 1, 3^{2} + 1 or 6, 8, and 10.

(ii) If we take *m*^{2} + 1 = 14, then *m*^{2} = 13

The value of *m* will not be an integer.

If we take *m*^{2} − 1 = 14, then *m*^{2} = 15

Again the value of *m* is not an integer.

Let 2*m *= 14

*m* = 7

Thus, *m*^{2} − 1 = 49 − 1 = 48 and *m*^{2} + 1 = 49 + 1 = 50

Therefore, the required triplet is 14, 48, and 50.

(iii) If we take *m*^{2} + 1 = 16, then *m*^{2} = 15

The value of *m* will not be an integer.

If we take *m*^{2} − 1= 16, then *m*^{2} = 17

Again the value of *m* is not an integer.

Let 2*m* = 16

*m *= 8

Thus, *m*^{2} − 1 = 64 − 1 = 63 and *m*^{2} + 1 = 64 + 1 = 65

Therefore, the Pythagorean triplet is 16, 63, and 65.

(iv) If we take *m*^{2} + 1 = 18,

*m*^{2} = 17

The value of *m* will not be an integer.

If we take *m*^{2} − 1 = 18, then *m*^{2} = 19

Again the value of *m* is not an integer.

Let 2*m *=18

*m *= 9

Thus, *m*^{2} − 1 = 81 − 1 = 80 and *m*^{2} + 1 = 81 + 1 = 82

Therefore, the Pythagorean triplet is 18, 80, and 82.

#### Page No 102:

#### Question 1:

What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801 (ii) 99856

(iii) 998001 (iv) 657666025

#### Answer:

(i) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.

#### Question 2:

Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257

(iii) 408 (iv) 441

#### Answer:

The perfect squares of a number can end with any of the digits 0, 1, 4, 5, 6, or 9 at unit’s place. Also, a perfect square will end with even number of zeroes, if any.

(i) Since the number 153 has its unit’s place digit as 3, it is not a perfect square.

(ii) Since the number 257 has its unit’s place digit as 7, it is not a perfect square.

(iii) Since the number 408 has its unit’s place digit as 8, it is not a perfect square.

(iv) Since the number 441 has its unit’s place digit as 1, it is a perfect square.

#### Question 3:

Find the square roots of 100 and 169 by the method of repeated subtraction.

#### Answer:

We know that the sum of the first *n* odd natural numbers is *n*^{2}.

Consider.

(i) 100 − 1 = 99 (ii) 99 − 3 = 96 (iii) 96 − 5 = 91

(iv) 91 − 7 = 84 (v) 84 − 9 = 75 (vi) 75 − 11= 64

(vii) 64 − 13 = 51 (viii) 51 − 15 = 36 (ix) 36 − 17 = 19

(x) 19 − 19 = 0

We have subtracted successive odd numbers starting from 1 to 100, and obtained 0 at 10^{th} step.

Therefore,

The square root of 169 can be obtained by the method of repeated subtraction as follows.

(i) 169 − 1 = 168 (ii) 168 − 3 = 165 (iii) 165 − 5 = 160

(iv) 160 − 7 = 153 (v) 153 − 9 = 144 (vi) 144 − 11 = 133

(vii) 133 − 13 = 120 (viii) 120 − 15 = 105 (ix) 105 − 17 = 88

(x) 88 − 19 = 69 (xi) 69 − 21 = 48 (xii) 48 − 23 = 25

(xiii)25 − 25 = 0

We have subtracted successive odd numbers starting from 1 to 169, and obtained 0 at 13^{th} step.

Therefore,

#### Question 4:

Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729 (ii) 400

(iii) 1764 (iv) 4096

(v) 7744 (vi) 9604

(vii) 5929 (viii) 9216

(ix) 529 (x) 8100

#### Answer:

(i) 729 can be factorised as follows.

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

729 = 3 × 3 × 3 × 3 × 3 × 3

∴ = 27

(ii) 400 can be factorised as follows.

2 | 400 |

2 | 200 |

2 | 100 |

2 | 50 |

5 | 25 |

5 | 5 |

1 |

400 = 2 × 2 × 2 × 2 × 5 × 5

∴ = 20

(iii) 1764 can be factorised as follows.

2 | 1764 |

2 | 882 |

3 | 441 |

3 | 147 |

7 | 49 |

7 | 7 |

1 |

1764 = 2 × 2 × 3 × 3 × 7 × 7

∴ = 42

(iv) 4096 can be factorised as follows.

2 | 4096 |

2 | 2048 |

2 | 1024 |

2 | 512 |

2 | 256 |

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ = 64

(v) 7744 can be factorised as follows.

2 | 7744 |

2 | 3872 |

2 | 1936 |

2 | 968 |

2 | 484 |

2 | 242 |

11 | 121 |

11 | 11 |

1 |

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

∴ = 88

(vi) 9604 can be factorised as follows.

2 | 9604 |

2 | 4802 |

7 | 2401 |

7 | 343 |

7 | 49 |

7 | 7 |

1 |

9604 = 2 × 2 × 7 × 7 × 7 × 7

∴ = 98

(vii) 5929 can be factorised as follows.

7 | 5929 |

7 | 847 |

11 | 121 |

11 | 11 |

1 |

5929 = 7 × 7 × 11 × 11

∴ = 77

(viii) 9216 can be factorised as follows.

2 | 9216 |

2 | 4608 |

2 | 2304 |

2 | 1152 |

2 | 576 |

2 | 288 |

2 | 144 |

2 | 72 |

2 | 36 |

2 | 18 |

3 | 9 |

3 | 3 |

1 |

9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3× 3

∴ = 96

(ix) 529 can be factorised as follows.

23 | 529 |

23 | 23 |

1 |

529 = 23 × 23

(x) 8100 can be factorised as follows.

2 | 8100 |

2 | 4050 |

3 | 2025 |

3 | 675 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

∴ = 90

#### Question 5:

For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252 (ii) 180

(iii) 1008 (iv) 2028

(v) 1458 (vi) 768

#### Answer:

(i)252 can be factorised as follows.

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

1 |

252 = 2 × 2 × 3 × 3 × 7

Here, prime factor 7 does not have its pair.

If 7 gets a pair, then the number will become a perfect square. Therefore, 252 has to be multiplied with 7 to obtain a perfect square.

252 × 7 = 2 × 2 × 3 × 3 × 7 × 7

Therefore, 252 × 7 = 1764 is a perfect square.

∴

(ii)180 can be factorised as follows.

2 | 180 |

2 | 90 |

3 | 45 |

3 | 15 |

5 | 5 |

1 |

180 = 2 × 2 × 3 × 3 × 5

Here, prime factor 5 does not have its pair. If 5 gets a pair, then the number will become a perfect square. Therefore, 180 has to be multiplied with 5 to obtain a perfect square.

180 × 5 = 900 = 2 × 2 × 3 × 3 × 5 × 5

Therefore, 180 × 5 = 900 is a perfect square.

∴ = 30

(iii)1008 can be factorised as follows.

2 | 1008 |

2 | 504 |

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

1 |

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here, prime factor 7 does not have its pair. If 7 gets a pair, then the number will become a perfect square. Therefore, 1008 can be multiplied with 7 to obtain a perfect square.

1008 × 7 = 7056 = 2 × 2 ×2 × 2 × 3 × 3 × 7 × 7

Therefore, 1008 × 7 = 7056 is a perfect square.

∴ = 84

(iv) 2028 can be factorised as follows.

2 | 2028 |

2 | 1014 |

3 | 507 |

13 | 169 |

13 | 13 |

1 |

2028 = 2 × 2 × 3 × 13 × 13

Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a perfect square. Therefore, 2028 has to be multiplied with 3 to obtain a perfect square.

Therefore, 2028 × 3 = 6084 is a perfect square.

2028 × 3 = 6084 = 2 × 2 × 3 × 3 × 13 × 13

∴ = 78

(v) 1458 can be factorised as follows.

2 | 1458 |

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, prime factor 2 does not have its pair. If 2 gets a pair, then the number will become a perfect square. Therefore, 1458 has to be multiplied with 2 to obtain a perfect square.

Therefore, 1458 × 2 = 2916 is a perfect square.

1458 × 2 = 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

∴ = 54

(vi) 768 can be factorised as follows.

2 | 768 |

2 | 384 |

2 | 192 |

2 | 96 |

2 | 48 |

2 | 24 |

2 | 12 |

2 | 6 |

3 | 3 |

1 |

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a perfect square. Therefore, 768 has to be multiplied with 3 to obtain a perfect square.

Therefore, 768 × 3 = 2304 is a perfect square.

768 × 3 = 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

∴ = 48

#### Question 6:

For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252 (ii) 2925

(iii) 396 (iv) 2645

(v) 2800 (vi) 1620

#### Answer:

(i) 252 can be factorised as follows.

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

1 |

Here, prime factor 7 does not have its pair.

If we divide this number by 7, then the number will become a perfect square. Therefore, 252 has to be divided by 7 to obtain a perfect square.

252 ÷7 = 36 is a perfect square.

(ii) 2925 can be factorised as follows.

3 | 2925 |

3 | 975 |

5 | 325 |

5 | 65 |

13 | 13 |

1 |

Here, prime factor 13 does not have its pair.

If we divide this number by 13, then the number will become a perfect square. Therefore, 2925 has to be divided by 13 to obtain a perfect square.

2925 ÷13 = 225 is a perfect square.

(iii)396 can be factorised as follows.

2 | 396 |

2 | 198 |

3 | 99 |

3 | 33 |

11 | 11 |

1 |

Here, prime factor 11 does not have its pair.

If we divide this number by 11, then the number will become a perfect square. Therefore, 396 has to be divided by 11 to obtain a perfect square.

396 ÷11 = 36 is a perfect square.

(iv) 2645 can be factorised as follows.

5 | 2645 |

23 | 529 |

23 | 23 |

1 |

Here, prime factor 5 does not have its pair.

If we divide this number by 5, then the number will become a perfect square.

Therefore, 2645 has to be divided by 5 to obtain a perfect square.

2645 ÷5 = 529 is a perfect square.

(v)2800 can be factorised as follows.

2 | 2800 |

2 | 1400 |

2 | 700 |

2 | 350 |

5 | 175 |

5 | 35 |

7 | 7 |

1 |

Here, prime factor 7 does not have its pair.

If we divide this number by 7, then the number will become a perfect square.

Therefore, 2800 has to be divided by 7 to obtain a perfect square.

2800 ÷7 = 400 is a perfect square.

∴

(vi)1620 can be factorised as follows.

2 | 1620 |

2 | 810 |

3 | 405 |

3 | 135 |

3 | 45 |

3 | 15 |

5 | 5 |

1 |

Here, prime factor 5 does not have its pair.

If we divide this number by 5, then the number will become a perfect square.

Therefore, 1620 has to be divided by 5 to obtain a perfect square.

1620 ÷5 = 324 is a perfect square.

∴

#### Question 7:

The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’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:

It is given that each student donated as many rupees as the number of students of the class. Number of students in the class will be the square root of the amount donated by the students of the class.

The total amount of donation is Rs 2401.

Number of students in the class =

∴

Hence, the number of students in the class is 49.

#### Page No 103:

#### Question 8:

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

#### Answer:

It is given that in the garden, each row contains as many plants as the number of rows.

Hence,

Number of rows = Number of plants in each row

Total number of plants = Number of rows × Number of plants in each row

Number of rows × Number of plants in each row = 2025

(Number of rows)^{2} = 2025

∴

Thus, the number of rows and the number of plants in each row is 45.

#### Question 9:

Find the smallest square number that is divisible by each of the numbers 4, 9, and 10.

#### Answer:

The number that will be perfectly divisible by each one of 4, 9, and 10 is their LCM. The LCM of these numbers is as follows.

2 | 4, 9, 10 |

2 | 2, 9, 5 |

3 | 1, 9, 5 |

3 | 1, 3, 5 |

5 | 1, 1, 5 |

1, 1, 1 |

LCM of 4, 9, 10 = 2 × 2 × 3 × 3 × 5 =180

Here, prime factor 5 does not have its pair. Therefore, 180 is not a perfect square. If we multiply 180 with 5, then the number will become a perfect square. Therefore, 180 should be multiplied with 5 to obtain a perfect square.

Hence, the required square number is 180 × 5 = 900

#### Question 10:

Find the smallest square number that is divisible by each of the numbers 8, 15,

and 20.

#### Answer:

The number that is perfectly divisible by each of the numbers 8, 15, and 20 is their LCM.

2 | 8, 15, 20 |

2 | 4, 15, 10 |

2 | 2, 15, 5 |

3 | 1, 15, 5 |

5 | 1, 5, 5 |

1, 1, 1 |

LCM of 8, 15, and 20 = 2 × 2 × 2 × 3 × 5 =120

Here, prime factors 2, 3, and 5 do not have their respective pairs. Therefore, 120 is not a perfect square.

Therefore, 120 should be multiplied by 2 × 3 × 5, i.e. 30, to obtain a perfect square.

Hence, the required square number is 120 × 2 × 3 × 5 = 3600

#### Page No 107:

#### Question 1:

Find the square root of each of the following numbers by division method.

(i) 2304 (ii) 4489

(iii) 3481 (iv) 529

(v) 3249 (vi) 1369

(vii) 5776 (viii) 7921

(ix) 576 (x) 1024

(xi) 3136 (xii) 900

#### Answer:

(i) The square root of 2304 can be calculated as follows.

48 | |

4 | |

88 | 704704 |

0 |

∴

(ii) The square root of 4489 can be calculated as follows.

67 | |

6 | |

127 | 889889 |

0 |

∴

(iii) The square root of 3481 can be calculated as follows.

59 | |

5 | |

109 | 981981 |

0 |

Therefore,

(iv) The square root of 529 can be calculated as follows.

23 | |

2 | |

43 | 129129 |

0 |

∴

(v) The square root of 3249 can be calculated as follows.

57 | |

5 | |

107 | 749749 |

0 |

∴

(vi) The square root of 1369 can be calculated as follows.

37 | |

3 | |

67 | 469469 |

0 |

∴

(vii) The square root of 5776 can be calculated as follows.

76 | |

7 | |

146 | 876876 |

0 |

∴

(viii) The square root of 7921 can be calculated as follows.

89 | |

8 | |

169 | 15211521 |

0 |

∴

(ix) The square root of 576 can be calculated as follows.

24 | |

2 | |

44 | 176176 |

0 |

∴

(x) The square root of 1024 can be calculated as follows.

32 | |

3 | |

62 | 124124 |

0 |

∴

(xi) The square root of 3136 can be calculated as follows.

56 | |

5 | |

106 | 636636 |

0 |

∴

(xii) The square root of 900 can be calculated as follows.

30 | |

3 | |

60 | 0000 |

0 |

∴

#### Question 2:

Find the number of digits in the square root of each of the following numbers (without any calculation).

(i) 64 (ii) 144

(iii) 4489 (iv) 27225

(v) 390625

#### Answer:

(i) By placing bars, we obtain

Since there is only one bar, the square root of 64 will have only one digit in it.

(ii) By placing bars, we obtain

Since there are two bars, the square root of 144 will have 2 digits in it.

(iii) By placing bars, we obtain

Since there are two bars, the square root of 4489 will have 2 digits in it.

(iv) By placing bars, we obtain

Since there are three bars, the square root of 27225 will have three digits in it.

(v) By placing the bars, we obtain

Since there are three bars, the square root of 390625 will have 3 digits in it.

#### Question 3:

Find the square root of the following decimal numbers.

(i) 2.56 (ii) 7.29

(iii) 51.84 (iv) 42.25

(v) 31.36

#### Answer:

(i) The square root of 2.56 can be calculated as follows.

1. 6 | |

1 | |

26 | 156156 |

0 |

∴

(ii) The square root of 7.29 can be calculated as follows.

2. 7 | |

2 | |

47 | 329329 |

0 |

∴

(iii) The square root of 51.84 can be calculated as follows.

7.2 | |

7 | |

142 | 284284 |

0 |

∴

(iv) The square root of 42.25 can be calculated as follows.

6.5 | |

6 | |

125 | 625625 |

0 |

∴

(v) The square root of 31.36 can be calculated as follows.

5.6 | |

5 | |

106 | 636636 |

0 |

∴

#### Question 4:

Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 402 (ii) 1989

(iii) 3250 (iv) 825

(v) 4000

#### Answer:

(i) The square root of 402 can be calculated by long division method as follows.

20 | |

2 | |

40 | 0200 |

2 |

The remainder is 2. It represents that the square of 20 is less than 402 by 2. Therefore, a perfect square will be obtained by subtracting 2 from the given number 402.

Therefore, required perfect square = 402 − 2 = 400

And,

(ii) The square root of 1989 can be calculated by long division method as follows.

44 | |

4 | |

84 | 389336 |

53 |

The remainder is 53. It represents that the square of 44 is less than 1989 by 53. Therefore, a perfect square will be obtained by subtracting 53 from the given number 1989.

Therefore, required perfect square = 1989 − 53 = 1936

And,

(iii) The square root of 3250 can be calculated by long division method as follows.

57 | |

5 | |

107 | 750749 |

1 |

The remainder is 1. It represents that the square of 57 is less than 3250 by 1. Therefore, a perfect square can be obtained by subtracting 1 from the given number 3250.

Therefore, required perfect square = 3250 − 1 = 3249

And,

(iv) The square root of 825 can be calculated by long division method as follows.

28 | |

2 | |

48 | 425384 |

41 |

The remainder is 41. It represents that the square of 28 is less than 825 by 41. Therefore, a perfect square can be calculated by subtracting 41 from the given number 825.

Therefore, required perfect square = 825 − 41 = 784

And,

(v) The square root of 4000 can be calculated by long division method as follows.

63 | |

6 | |

123 | 400369 |

31 |

The remainder is 31. It represents that the square of 63 is less than 4000 by 31. Therefore, a perfect square can be obtained by subtracting 31 from the given number 4000.

Therefore, required perfect square = 4000 − 31 = 3969

And,

#### Question 5:

Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 525 (ii) 1750

(iii) 252 (iv) 1825

(v) 6412

#### Answer:

(i) The square root of 525 can be calculated by long division method as follows.

22 | |

2 | |

42 | 12584 |

41 |

The remainder is 41.

It represents that the square of 22 is less than 525.

Next number is 23 and 23^{2} = 529

Hence, number to be added to 525 = 23^{2} − 525 = 529 − 525 = 4

The required perfect square is 529 and

(ii) The square root of 1750 can be calculated by long division method as follows.

41 | |

4 | |

81 | 15081 |

69 |

The remainder is 69.

It represents that the square of 41 is less than 1750.

The next number is 42 and 42^{2} = 1764

Hence, number to be added to 1750 = 42^{2} − 1750 = 1764 − 1750 = 14

The required perfect square is 1764 and

(iii) The square root of 252 can be calculated by long division method as follows.

15 | |

1 | |

25 | 152125 |

27 |

The remainder is 27. It represents that the square of 15 is less than 252.

The next number is 16 and 16^{2} = 256

Hence, number to be added to 252 = 16^{2} − 252 = 256 − 252 = 4

The required perfect square is 256 and

(iv) The square root of 1825 can be calculated by long division method as follows.

42 | |

4 | |

82 | 225164 |

61 |

The remainder is 61. It represents that the square of 42 is less than 1825.

The next number is 43 and 43^{2} = 1849

Hence, number to be added to 1825 = 43^{2} − 1825 = 1849 − 1825 = 24

The required perfect square is 1849 and

(v) The square root of 6412 can be calculated by long division method as follows.

80 | |

8 | |

160 | 0120 |

12 |

The remainder is 12.

It represents that the square of 80 is less than 6412.

The next number is 81 and 81^{2} = 6561

Hence, number to be added to 6412 = 81^{2} − 6412 = 6561 − 6412 = 149

The required perfect square is 6561 and

#### Question 6:

Find the length of the side of a square whose area is 441 m^{2}.

#### Answer:

Let the length of the side of the square be* x* m.

Area of square = (*x*)^{2} = 441 *m*^{2}

The square root of 441 can be calculated as follows.

21 | |

2 | |

41 | 04141 |

0 |

∴

Hence, the length of the side of the square is 21 m.

#### Question 7:

In a right triangle ABC, ∠B = 90°.

(a) If AB = 6 cm, BC = 8 cm, find AC

(b) If AC = 13 cm, BC = 5 cm, find AB

#### Answer:

(a) ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (6 cm)^{2} + (8 cm)^{2}

AC^{2} = (36 + 64) cm^{2} =100 cm^{2}

AC=

AC = 10 cm

(b) ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

AC^{2} = AB^{2} + BC^{2}

(13 cm)^{2} = (AB)^{2} + (5 cm)^{2}

AB^{2} = (13 cm)^{2} − (5 cm)^{2} = (169 − 25) cm^{2} = 144 cm^{2}

AB=

AB = 12 cm

#### Question 8:

A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.

#### Answer:

It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same.

We have to find the number of more plants that should be there, so that when the gardener plants them, the number of rows and columns are same.

That is, the number which should be added to 1000 to make it a perfect square has to be calculated.

The square root of 1000 can be calculated by long division method as follows.

31 | |

3 | |

61 | 10061 |

39 |

The remainder is 39. It represents that the square of 31 is less than 1000.

The next number is 32 and 32^{2} = 1024

Hence, number to be added to 1000 to make it a perfect square

= 32^{2} − 1000 = 1024 − 1000 = 24

Thus, the required number of plants is 24.

#### Question 9:

These are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?

#### Answer:

It is given that there are 500 children in the school. They have to stand for a P.T. drill such that the number of rows is equal to the number of columns.

The number of children who will be left out in this arrangement has to be calculated. That is, the number which should be subtracted from 500 to make it a perfect square has to be calculated.

The square root of 500 can be calculated by long division method as follows.

22 | |

2 | |

42 | 10084 |

16 |

The remainder is 16.

It shows that the square of 22 is less than 500 by 16. Therefore, if we subtract 16 from 500, we will obtain a perfect square.

Required perfect square = 500 − 16 = 484

Thus, the number of children who will be left out is 16.