Rd Sharma 2019 2020 Solutions for Class 8 Math Chapter 3 Squares And Square Roots are provided here with simple step-by-step explanations. These solutions for Squares And Square Roots are extremely popular among Class 8 students for Math Squares And Square Roots Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 2020 Book of Class 8 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2019 2020 Solutions. All Rd Sharma 2019 2020 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### 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 2^{2}, which is equal to 4.

(ii) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 7^{2}), which is 9.

(iii) Its last digit is 3. Hence, the units digit is 3^{2}, which is equal to 9.

(iv) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 7^{2}), which is 9.

(v) Its last digit is 8. Hence, the units digit is the last digit of 64 (64 = 8^{2}), which is 4.

(vi) Its last digit is 0. Hence, the units digit is 0^{2}, which is equal to 0.

(vii) Its last digit is 6. Hence, the units digit is the last digit of 36 (36 = 6^{2}), which is 6.

(viii) Its last digit is 5. Hence, the units digit is the last digit of 25 (25 = 5^{2}), which is 5.

(ix) Its last digit is 4. Hence, the units digit is the last digit of 16 (16 = 4^{2}), which is 6.

#### Page No 3.19:

#### Question 5:

Observe the following pattern

1 + 3 = 2^{2}

1 + 3 + 5 = 3^{2}

1 + 3 × 5 + 7 = 4^{2}

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 =* *n*^{2}, where the left hand side consists of *n* terms.

#### Page No 3.19:

#### Question 6:

Observe the following pattern

2^{2} − 1^{2} = 2 + 1

3^{2} − 2^{2} = 3 + 2

4^{2} − 3^{2} = 4 + 3

5^{2} − 4^{2} = 5 + 4

and find the value of

(i) 100^{2} − 99^{2}

(ii) 111^{2} − 109^{2}

(iii) 99^{2} − 96^{2}

#### 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:

${\left(n+1\right)}^{2}-{\left(n\right)}^{2}=\left(n+1\right)+n$

Using this formula, we get:

(i) 100^{2} − 99^{2} = (99 + 1) + 99

= 199

(ii) 111^{2} − 109^{2} = 111^{2} − 110^{2 }+ 110^{2} − 109^{2}

= (111 + 110) + (110 + 109)

= 440

(iii) 99^{2} − 96^{2} = 99^{2} − 98^{2 }+ 98^{2 }− 97^{2} + 97^{2 }− 96^{2 }

= 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:

8^{2} + 15^{2} = 289 = 17^{2}

Hence, (8, 15, 17) is a Pythagorean triplet.

(ii) The two smallest numbers are 18 and 80. The sum of their squares is:

18^{2} + 80^{2} = 6724 = 82^{2}

Hence, (18, 80, 82) is a Pythagorean triplet.

(iii) The two smallest numbers are 14 and 48. The sum of their squares is:

14^{2} + 48^{2} = 2500, which is not equal to 51^{2} = 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:

10^{2} + 24^{2} = 676 = 26^{2}

Hence, (10, 24, 26) is a Pythagorean triplet.

(v) The two smallest numbers are 16 and 63. The sum of their squares is:

16^{2} + 63^{2} = 4225 = 65^{2}

Hence, (16, 63, 65) is a Pythagorean triplet.

(vi) The two smallest numbers are 12 and 35. The sum of their squares is:

12^{2} + 35^{2} = 1369, which is not equal to 38^{2} = 1444

Hence, (12, 35, 38) is not a Pythagorean triplet.

#### Page No 3.19:

#### Question 8:

Observe the following pattern

$\left(1\times 2\right)+\left(2\times 3\right)=\frac{2\times 3\times 4}{3}$

$\left(1\times 2\right)+\left(2\times 3\right)+\left(3\times 4\right)=\frac{3\times 4\times 5}{3}$

$\left(1\times 2\right)+\left(2\times 3\right)+\left(3\times 4\right)+\left(4\times 5\right)=\frac{4\times 5\times 6}{3}$

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:

$1\times 2+2\times 3+3\times 4+4\times 5+5\times 6=\frac{5\times 6\times 7}{3}=70$

#### Page No 3.19:

#### Question 9:

Observe the following pattern

$1=\frac{1}{2}\left\{1\times \left(1+1\right)\right\}\phantom{\rule{0ex}{0ex}}1+2=\frac{1}{2}\left\{2\times \left(2+1\right)\right\}\phantom{\rule{0ex}{0ex}}1+2+3=\frac{1}{2}\left\{3\times \left(3+1\right)\right\}\phantom{\rule{0ex}{0ex}}1+2+3+4=\frac{1}{2}\left\{4\times \left(4+1\right)\right\}$

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:

$\left(\mathrm{i}\right)1+2+3+........+50=\frac{1}{2}\times 50\times (50+1)=1275$

(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:

$1+2+......+30=\frac{1}{2}\left(30\times \left(30+1\right)\right)$ = 465

Finally, we have:

$31+32+....+50=1275-465=810$

#### Page No 3.20:

#### Question 10:

Observe the following pattern

${1}^{2}=\frac{1}{6}\left[1\times \left(1+1\right)\times \left(2\times 1+1\right)\right]\phantom{\rule{0ex}{0ex}}{1}^{2}+{2}^{2}=\frac{1}{6}\left[2\times \left(2+1\right)\times \left(2\times 2+1\right)\right]\phantom{\rule{0ex}{0ex}}{1}^{2}+{2}^{2}+{3}^{2}=\frac{1}{6}\left[3\times \left(3+1\right)\times \left(2\times 3+1\right)\right]\phantom{\rule{0ex}{0ex}}{1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}=\frac{1}{6}\left[4\times \left(4+1\right)\times \left(2\times 4+1\right)\right]$

and find the values of each of the following:

(i) 1^{2} + 2^{2} + 3^{2} + 4^{2 }+ ... + 10^{2}

(ii) 5^{2} + 6^{2} + 7^{2} + 8^{2} + 9^{2} + 10^{2} + 11^{2} + 12^{2}

#### 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:

$\left(\mathrm{i}\right){1}^{2}+{2}^{2}+.......+{10}^{2}=\frac{1}{6}\times \left[10\times \left(10+1\right)\times \left(2\times 10+1\right)\right]$

$=\frac{1}{6}\times \left[10\times 11\times 12\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=385$.

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

${5}^{2}+{6}^{2}+.......+{12}^{2}=\left({1}^{2}+{2}^{2}+......+{12}^{2}\right)-\left({1}^{2}+{2}^{2}+......+{4}^{2}\right)$

The sum of the first bracket on the RHS:

${1}^{2}+{2}^{2}+.....+{12}^{2}=\frac{1}{6}\left[12\times (12+1)\times (2\times 12+1)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=650$

The second bracket is:

${1}^{2}+{2}^{2}+......+{4}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\times \left[4\times \left(4+1\right)\times \left(2\times 4+1\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\times 4\times 5\times 9=30$

Finally, the wanted sum is:

${5}^{2}+{6}^{2}+......+{12}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=({1}^{2}+{2}^{2}+.....+{12}^{2})-({1}^{2}+{2}^{2}+.....+{12}^{2})\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=650-30=620$

#### 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 *p*^{2}, which has at least three factors: 1, *p* and *p*^{2}. Since it has more than two factors, it is not a prime number.

(iii) False

1 is the square of a number (1 = 1^{2}). But 1 + 1^{ }= 2, which is not the square of any number.

(iv) False

4 and 1 are squares (4 = 2^{2}, 1 = 1^{2}). But 4 − 1^{ }= 3, which is not the square of any number.

(v) True

If *a*^{2} and *b*^{2} are two squares, their product is *a*^{2}^{ }x* **b*^{2} = (*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

7^{2} = 49 and 8^{2} = 64. 7 and 8 are consecutive numbers and hence there are no square numbers between 50 and 60.

(viii) True

14^{2} 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 *a*^{2}, 2 x *a* x *b*, and *b*^{2} in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

4 | 20 | 25 |

Step 2. Underline the unit digit of

*b*

^{2}(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 |

a^{2} |
2 x a x b |
b^{2} |

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

*a*

^{2}in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

4 + 2 | 20 + 2 | 25 |

6 | 22 |

Step 4. Underline the number in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

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:

25

^{2}= 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

*a*

^{2}, 2 x

*a*x

*b*, and

*b*

^{2}in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

9 | 42 | 49 |

Step 2. Underline the unit digit of

*b*

^{2}(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 |

a^{2} |
2 x a x b |
b^{2} |

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

*a*

^{2}in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

9 + 4 | 42 + 4 | 49 |

13 | 46 |

Step 4. Underline the number in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

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:

37

^{2}= 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

*a*

^{2}, 2 x

*a*x

*b*and

*b*

^{2}in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

25 | 40 | 16 |

Step 2. Underline the unit digit of

*b*

^{2}(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 |

a^{2} |
2 x a x b |
b^{2} |

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

*a*

^{2}in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

25 + 4 | 40 + 1 | 16 |

29 | 41 |

Step 4. Underline the number in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

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:

54

^{2}= 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

*a*

^{2}, 2 x

*a*x

*b*and

*b*

^{2}in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

49 | 14 | 1 |

Step 2. Underline the unit digit of

*b*

^{2}(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 |

a^{2} |
2 x a x b |
b^{2} |

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

*a*

^{2}in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

49 + 1 | 14 + 0 | 1 |

50 | 14 |

Step 4. Underline the number in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

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:

71

^{2}= 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

*a*

^{2}, 2 x

*a*x

*b*and

*b*

^{2}in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

81 | 108 | 36 |

Step 2. Underline the unit digit of

*b*

^{2}(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 |

a^{2} |
2 x a x b |
b^{2} |

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

*a*

^{2}in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

81 + 11 | 108 + 3 | 36 |

92 | 111 |

Step 4. Underline the number in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b |
b^{2} |

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:

96

^{2}= 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)

$\therefore $ 98^{2} = 9604

(ii)

$\therefore $ 273^{2} = 74529

(iii)

$\therefore $ 348^{2} = 121104

(iv)

$\therefore $ 295^{2} = 87025

(v)

$\therefore $ 171^{2} = 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 *n*5 is a number ending with 25 and has the number *n*(*n *+ 1) before 25.

(i) Here, *n* = 42

$\therefore $ *n*(*n *+ 1) = (42)(43) = 1806

$\therefore $ 425^{2} = 180625

(ii) Here, *n* = 57

$\therefore $ *n*(*n *+ 1) = (57)(58) = 3306

$\therefore $ 575^{2} = 330625

(iii) Here *n* = 40

$\therefore $ *n*(*n *+ 1) = (40)(41) = 1640

$\therefore $ 405^{2} = 164025

(iv) Here *n* = 20

$\therefore $ *n*(*n *+ 1) = (20)(21) = 420

$\therefore $ 205^{2} = 42025

(v) Here *n* = 9

$\therefore $ *n*(*n *+ 1) = (9)(10) = 90

$\therefore $ 95^{2} = 9025

(vi) Here *n* = 74

$\therefore $ *n*(*n *+ 1) = (74)(75) = 5550

$\therefore $ 745^{2} = 555025

(vii) We know:

The square of a three-digit number of the form 5*ab** =* (250 + *ab*)1000 + (*ab*)^{2}

$\therefore $ 512^{2} = (250+12)1000 + (12)^{2} = 262000 + 144 = 262144

(viii) Here, *n* = 99

$\therefore $ *n*(*n *+ 1) = (99)(100) = 9900

$\therefore $ 995^{2} = 990025

#### Page No 3.32:

#### Question 5:

Find the squares of the following numbers using the identity (*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}:

(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} = *a*^{2} + 2*ab* + *b*^{2}:

405^{2} = (400 + 5)^{2} = 400^{2} + 2(400)(5) + 5^{2} = 160000 + 4000 + 25 = 164025

(ii) On decomposing:

510 = 500 + 10

Here, *a* = 500 and *b* = 10

Using the identity (*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}:

510^{2} = (500 + 10)^{2} = 500^{2} + 2(500)(10) + 10^{2} = 250000 + 10000 + 100 = 260100

(iii) On decomposing:

1001 = 1000 + 1

Here, *a* = 1000 and *b* = 1

Using the identity (*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}:

1001^{2} = (1000 + 1)^{2} = 1000^{2} + 2(1000)(1) + 1^{2} = 1000000 + 2000 + 1 = 1002001

(iv) On decomposing:

209 = 200 + 9

Here, *a* = 200 and *b* = 9

Using the identity (*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}:

209^{2} = (200 + 9)^{2} = 200^{2} + 2(200)(9) + 9^{2} = 40000 + 3600 + 81 = 43681

(v) On decomposing:

605 = 600 + 5

Here, *a* = 600 and *b* = 5

Using the identity (*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}:

605^{2} = (600 + 5)^{2} = 600^{2} + 2(600)(5) + 5^{2} = 360000 + 6000 + 25 = 366025

#### Page No 3.32:

#### Question 6:

Find the squares of the following numbers using the identity (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

(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 (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

395^{2} = (400 − 5)^{2} = 400^{2 }− 2(400)(5) + 5^{2} = 160000 − 4000 + 25 = 156025

(ii) Decomposing: 995 = 1000 − 5

Here, *a* = 1000 and *b* = 5

Using the identity (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

995^{2} = (1000 − 5)^{2} = 1000^{2 }− 2(1000)(5) + 5^{2} = 1000000 − 10000 + 25 = 990025

(iii) Decomposing: 495 = 500 − 5

Here, *a* = 500 and *b* = 5

Using the identity (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

495^{2} = (500 − 5)^{2} = 500^{2} − 2(500)(5) + 5^{2} = 250000 − 5000 + 25 = 245025

(iv) Decomposing: 498 = 500 − 2

Here, *a* = 500 and *b* = 2

Using the identity (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

498^{2} = (500 − 2)^{2} = 500^{2} − 2(500)(2) + 2^{2} = 250000 − 2000 + 4 = 248004

(v) Decomposing: 99 = 100 − 1

Here, *a* = 100 and *b* = 1

Using the identity (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

99^{2} = (100 − 1)^{2} = 100^{2} − 2(100)(1) + 1^{2} = 10000 − 200 + 1 = 9801

(vi) Decomposing: 999 = 1000 - 1

Here, *a* = 1000 and *b* = 1

Using the identity (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

999^{2} = (1000 − 1)^{2} = 1000^{2} − 2(1000)(1) + 1^{2} = 1000000 − 2000 + 1 = 998001

(vii) Decomposing: 599 = 600 − 1

Here, *a* = 600 and *b* = 1

Using the identity (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}:

599^{2} = (600 − 1)^{2} = 600^{2 }− 2(600)(1) + 1^{2} = 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 99^{2}. 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 (3^{3} 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:

*a*^{2} = 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:

$\therefore $ *r = R*.

Substituting this in the first equation, we get:

*R x R* = 202500

*R*^{2} = 202500

*R*^{2} = (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:

$\therefore $ *r = M*

Substituting this in the first equation, we get:

*M x M* = 9216

*M*^{2} = 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} = (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:

$\therefore $ *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

*S*^{2} = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

*S*^{2} = (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 cm^{2}. 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* **r*^{2}, where *r* is the side of the square.

Then, we have the equation as follows:

*r ^{2}* = 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 m

^{2}

#### 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 $\times $ 2 $\times $ 2 $\times $ 2 $\times $ 2 $\times $ 2 $\times $ 11 $\times $ 11

Grouping them into pairs of equal factors, we get:

7744 = (2 $\times $ 2) $\times $ (2 $\times $ 2) $\times $ (2 $\times $ 2) $\times $ (11 $\times $ 11)

Taking one factor from each pair, we get :

(ii) The prime factorisation of 9604:

9604 = 2 $\times $ 2 $\times $ 7 $\times $ 7 $\times $ 7 $\times $ 7

Grouping them into pairs of equal factors, we get:

9604 = (2 $\times $ 2) $\times $ (7 $\times $ 7) $\times $ (7 $\times $ 7)

Taking one factor from each pair, we get:

(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 $\times $ 7) $\times $ (11 $\times $ 11)

Taking one factor from each pair, we get:

(iv) The prime factorisation of 7056:

7056 = 2 $\times $ 2 $\times $ 2 $\times $ 2 $\times $ 3 $\times $ 3 $\times $ 7 $\times $ 7

Grouping them into pairs of equal factors, we get:

7056 = (2 $\times $ 2) $\times $ (2 $\times $ 2) $\times $ (3 $\times $ 3) $\times $ (7 $\times $ 7)

Taking one factor from each pair, we get:

#### 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* $\times $ *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* $\times $ *S* = 2401

*S*^{2} = (7 $\times $ 7) $\times $ (7 $\times $ 7)

*S* = 7 $\times $ 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 $\times $ 7 $\times $ 11 $\times $ 11

Grouping them into pairs of equal factors:

5929 = (7 x 7) x (11 x 11)

The square root of 5929

=

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 = 22^{2}

(ii) 625 = 25^{2}

(iii) 576 = 24^{2}

(iv) Perfect squares closest to 941 are 900 (30^{2}) and 961 (31^{2}). 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 = 31^{2}

(vi) 2500 = 50^{2}

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 = 3^{2}) and 16 (16 = 4^{2}). 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 =3^{2}) and 16 (16 = 4^{2}). 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 = 4^{2}

32: The perfect squares closest to 32 are 25 (25 = 5^{2}) and 36 (36 = 6^{2}). 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 = 6^{2}

50: The perfect squares closest to 50 are 49 (49 = 7^{2}) and 64 (64 = 8^{2}). 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 = 8^{2}

79: The perfect squares closest to 79 are 64 (64 = 8^{2}) and 81 (81 = 9^{2}). 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 = 9^{2}

111: The perfect squares closest to 111 are 100 (100 = 10^{2}) and 121 (121 = 11^{2}). 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 = 11^{2}

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 10^{2} is equal to 100 and 9^{2} 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.

1^{2} = 1

2^{2} = 4

3^{2} = 9

4^{2} = 16

5^{2} = 25

6^{2} = 36

7^{2} = 49

8^{2} = 64

9^{2} = 81

10^{2} = 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 47^{2}. 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 441^{2}. 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 162^{2}. 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 127^{2}. 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 2098^{2}. 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 75^{2}. 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 71^{2}. 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 2125^{2}. 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 194^{2}. 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 712^{2}. 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 32^{2}. 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 317^{2}. 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.

$\therefore $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 m^{2}. 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 m^{2}

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 $\times $ 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:

$t=\frac{s}{v}$

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 m^{2} 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:

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:

$\therefore $ Length of one side of the square = 73 m

The circumference of the square is 73 $\times $ 4 = 292 m

$\therefore $ Total cost of fencing the lawn at Rs. 5 per metre = 292 $\times $ 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 = 31^{2}

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 48^{2}. Hence, 4 must be added to 2300 to get a perfect square.

#### Page No 3.48:

#### Question 1:

Find the square root of:

(i) $\frac{441}{961}$

(ii) $\frac{324}{841}$

(iii) $4\frac{29}{29}$

(iv) $2\frac{14}{25}$

(v) $2\frac{137}{196}$

(vi) $23\frac{26}{121}$

(vii) $25\frac{544}{729}$

(viii) $75\frac{46}{49}$

(ix) $3\frac{942}{2209}$

(x) $3\frac{334}{3025}$

(xi) $21\frac{2797}{3364}$

(xii) $38\frac{11}{25}$

(xiii) $23\frac{394}{729}$

(xiv) $21\frac{51}{169}$

(xv) $10\frac{151}{225}$

#### Answer:

(i) We know:

$\sqrt{\frac{441}{961}}=\frac{\sqrt{441}}{\sqrt{961}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{441}=\sqrt{\left(3\times 3\right)\times \left(7\times 7\right)}=3\times 7=21\phantom{\rule{0ex}{0ex}}\sqrt{961}=\sqrt{31\times 31}=31\phantom{\rule{0ex}{0ex}}\therefore \sqrt{\frac{441}{961}}=\frac{21}{31}$

(ii)We know:

$\sqrt{\frac{324}{841}}=\frac{\sqrt{324}}{\sqrt{841}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{324}=\sqrt{2\times 2\times 3\times 3\times 3\times 3}=2\times 3\times 3=18\phantom{\rule{0ex}{0ex}}\sqrt{841}=\sqrt{29\times 29}=29\phantom{\rule{0ex}{0ex}}\therefore \sqrt{\frac{324}{841}}=\frac{18}{29}$

(iii) By looking at the book's answer key, the fraction should be $\sqrt{4\frac{29}{49}},\mathrm{not}\sqrt{4\frac{29}{29}}$.

We know:

$\sqrt{4\frac{29}{49}}=\sqrt{\frac{225}{49}}=\frac{\sqrt{225}}{\sqrt{49}}\phantom{\rule{0ex}{0ex}}\sqrt{225}=15\phantom{\rule{0ex}{0ex}}\sqrt{49}=7\phantom{\rule{0ex}{0ex}}\therefore \sqrt{4\frac{29}{49}}=\frac{15}{7}\phantom{\rule{0ex}{0ex}}$

(iv) We know:

$\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{\sqrt{64}}{\sqrt{25}}=\frac{8}{5}$

(v) We know:

$\sqrt{2\frac{137}{196}}=\sqrt{\frac{529}{196}}=\frac{\sqrt{529}}{\sqrt{196}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{529}=\sqrt{23\times 23}=23\phantom{\rule{0ex}{0ex}}\sqrt{196}=\sqrt{2\times 2\times 7\times 7}=2\times 7=14\phantom{\rule{0ex}{0ex}}\therefore \sqrt{2\frac{137}{196}}=\frac{23}{14}$

(vi) We know:

$\sqrt{23\frac{26}{121}}=\sqrt{\frac{2809}{121}}=\frac{\sqrt{2809}}{\sqrt{121}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{121}=11\phantom{\rule{0ex}{0ex}}\therefore \sqrt{23\frac{26}{121}}=\frac{53}{11}$

(vii) We know:

$\sqrt{25\frac{544}{729}}=\sqrt{\frac{18769}{729}}=\frac{\sqrt{18769}}{\sqrt{729}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{729}=27\phantom{\rule{0ex}{0ex}}\therefore \sqrt{25\frac{544}{729}}=\frac{137}{27}$

(viii) We know:

$\sqrt{75\frac{46}{49}}=\sqrt{\frac{3721}{49}}=\frac{\sqrt{3721}}{\sqrt{49}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{49}=7\phantom{\rule{0ex}{0ex}}\therefore \sqrt{75\frac{46}{49}}=\frac{61}{7}$

(ix) We know:

$\sqrt{3\frac{942}{2209}}=\sqrt{\frac{7569}{2209}}=\frac{\sqrt{7569}}{\sqrt{2209}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\therefore \sqrt{3\frac{942}{2209}}=\frac{87}{47}$

(x) We know:

$\sqrt{3\frac{334}{3025}}=\sqrt{\frac{9409}{3025}}=\frac{\sqrt{9409}}{\sqrt{3025}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\therefore \sqrt{3\frac{334}{3025}}=\frac{97}{55}$

(xi) We know:

$\sqrt{21\frac{2797}{3364}}=\sqrt{\frac{73441}{3364}}=\frac{\sqrt{73441}}{\sqrt{3364}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\therefore \sqrt{21\frac{2797}{3364}}=\frac{271}{58}$

(xii) We know:

$\sqrt{38\frac{11}{25}}=\sqrt{\frac{961}{25}}=\frac{\sqrt{961}}{\sqrt{25}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{961}=31\phantom{\rule{0ex}{0ex}}\sqrt{25}=5\phantom{\rule{0ex}{0ex}}\therefore \sqrt{38\frac{11}{25}}=\frac{31}{5}$

(xiii) We know:

$\sqrt{23\frac{394}{729}}=\sqrt{\frac{17161}{729}}=\frac{\sqrt{17161}}{\sqrt{729}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{729}=27\phantom{\rule{0ex}{0ex}}\therefore \sqrt{23\frac{394}{729}}=\frac{131}{27}=4\frac{23}{27}$

(xiv) We know:

$\sqrt{21\frac{51}{169}}=\sqrt{\frac{3600}{169}}=\frac{\sqrt{3600}}{169}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{3600}=\sqrt{60\times 60}=60\phantom{\rule{0ex}{0ex}}\sqrt{169}=\sqrt{13\times 13}=13\phantom{\rule{0ex}{0ex}}\therefore \sqrt{21\frac{51}{169}}=\frac{60}{13}=4\frac{8}{13}\phantom{\rule{0ex}{0ex}}$

(xv) We know:

$\sqrt{10\frac{151}{225}}=\sqrt{\frac{2401}{225}}=\frac{\sqrt{2401}}{\sqrt{225}}$

Now let us compute the square roots of the numerator and the denominator separately.

$\sqrt{2401}=\sqrt{7\times 7\times 7\times 7}=7\times 7=49\phantom{\rule{0ex}{0ex}}\sqrt{225}=\sqrt{3\times 3\times 5\times 5}=3\times 5=15\phantom{\rule{0ex}{0ex}}\therefore \sqrt{10\frac{151}{225}}=\frac{49}{15}=3\frac{4}{15}$

#### Page No 3.48:

#### Question 2:

Find the value of:

(i) $\frac{\sqrt{80}}{\sqrt{405}}$

(ii) $\frac{\sqrt{441}}{\sqrt{625}}$

(iii) $\frac{\sqrt{1587}}{\sqrt{1728}}$

(iv) $\sqrt{72}\times \sqrt{338}$

(v) $\sqrt{45}\times \sqrt{20}$

#### Answer:

(i) We have:

(ii) Computing the square roots:

$\therefore $

(iii) We have:

(by dividing both numbers by 3)

Computing the square roots of the numerator and the denominator:

$\therefore $

(iv) We have:

=

(v) We have:

= 30

#### Page No 3.48:

#### Question 3:

The area of a square field is $80\frac{244}{729}$ 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

We have

Now, to calculate the square root of the numerator and the denominator:

We know that:

$\mathrm{Therefore},\mathrm{length}\mathrm{of}\mathrm{one}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{field}=\frac{242}{27}=8\frac{26}{27}m$

#### Page No 3.49:

#### Question 4:

The area of a square field is $30\frac{1}{4}{\mathrm{m}}^{2}.$ 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 .

We have:

Now, calculating the square root of the numerator and the denominator:

$\mathrm{Therefore},\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{square}=\sqrt{30\frac{1}{4}}=\frac{11}{2}=5\frac{1}{2}\mathrm{m}$

#### 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 $\times $ 338 = 24336 m^{2}

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) $\frac{\sqrt{59.29}-\sqrt{5.29}}{\sqrt{59.29}+\sqrt{5.29}}$

(ii) $\frac{\sqrt{0.2304}+\sqrt{0.1764}}{\sqrt{0.2304}-\sqrt{0.1764}}$

#### Answer:

(i) We have:

$\sqrt{59.29}=\sqrt{\frac{5929}{100}}=\frac{\sqrt{7\times 7\times 11\times 11}}{10}=\frac{7\times 11}{10}=7.7\phantom{\rule{0ex}{0ex}}\sqrt{5.29}=\sqrt{\frac{529}{100}}=\frac{\sqrt{529}}{\sqrt{100}}=\frac{23}{10}=2.3\phantom{\rule{0ex}{0ex}}\frac{\sqrt{59.29}-\sqrt{5.29}}{\sqrt{59.29}+\sqrt{5.29}}=\frac{7.7-2.3}{7.7+2.3}=\frac{5.4}{10}=0.54$

(ii) We have:

$\sqrt{0.2304}=\sqrt{\frac{2304}{10000}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3}}{\sqrt{10000}}\phantom{\rule{0ex}{0ex}}=\frac{2\times 2\times 2\times 2\times 3}{100}\phantom{\rule{0ex}{0ex}}=0.48\phantom{\rule{0ex}{0ex}}\sqrt{0.1764}=\sqrt{\frac{1764}{10000}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2\times 2\times 3\times 3\times 7\times 7}}{\sqrt{10000}}\phantom{\rule{0ex}{0ex}}=\frac{2\times 3\times 7}{100}\phantom{\rule{0ex}{0ex}}=0.42\phantom{\rule{0ex}{0ex}}\frac{\sqrt{0.2304}+\sqrt{0.1764}}{\sqrt{0.2304}-\sqrt{0.1764}}=\frac{0.48+0.42}{0.48-0.42}=\frac{0.9}{0.06}=15\phantom{\rule{0ex}{0ex}}$

#### Page No 3.52:

#### Question 17:

Evaluate $\sqrt{50625}$ and hence find the value of $\sqrt{506.25}+\sqrt{5.0625}$

#### Answer:

We have:

$\sqrt{50625}=\sqrt{3\times 3\times 3\times 3\times 5\times 5\times 5\times 5}=3\times 3\times 5\times 5=225\phantom{\rule{0ex}{0ex}}$

Next, we will calculate $\sqrt{506.25}\mathrm{and}\sqrt{5.0625}$

$\sqrt{506.25}=\sqrt{\frac{50625}{100}}=\frac{\sqrt{50625}}{\sqrt{100}}=\frac{225}{10}=22.5\phantom{\rule{0ex}{0ex}}\sqrt{5.0625}=\sqrt{\frac{50625}{10000}}=\frac{\sqrt{50625}}{\sqrt{10000}}=\frac{225}{100}=2.25\phantom{\rule{0ex}{0ex}}\sqrt{506.25}+\sqrt{5.0625}=22.5+2.25=24.75$

#### Page No 3.52:

#### Question 18:

Find the value of $\sqrt{103.0225}$ and hence find the value of

(i) $\sqrt{10302.25}$

(ii) $\sqrt{1.030225}$

#### 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:

$\left(\mathrm{i}\right)\sqrt{10302.25}=\sqrt{103.0225\times 100}=\sqrt{103.0225}\times \sqrt{100}=10.15\times 10=101.5\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\sqrt{1.030225}=\sqrt{\frac{103.0225}{100}}=\frac{\sqrt{103.0225}}{\sqrt{100}}=\frac{10.15}{10}=1.015$

#### 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) $\frac{7}{8}$

(xviii) $\frac{5}{12}$

(xix) $2\frac{1}{2}$

(xx) $287\frac{5}{8}$

#### 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:

$\frac{7}{8}=0.875$

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:

$\frac{5}{2}=0.41666666$

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 $2\frac{1}{2}$ into decimal form up to eight digits to the right of the decimal point as shown below:

$2\frac{1}{2}=2.50000000$

But, this is the same with the value 2.5 in problem (ix). Hence, the square root of $2\frac{1}{2}$ is 1.581.

(xx) We can find the square root up to four decimal places by expanding $287\frac{5}{8}$ into decimal form up to eight digits to the right of the decimal point as shown below:

$287\frac{5}{8}=287.62500000$

Hence, we have:

So, the square root of $287\frac{5}{8}$ 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:

$\therefore $ $\sqrt{12.0068}=3.46508$

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:

$\therefore $ $\sqrt{11}=3.31662$

#### Page No 3.57:

#### Question 4:

Given that: $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236\mathrm{and}\sqrt{7}=2.646,$ evaluate each of the following:

(i) $\sqrt{\frac{144}{7}}$

(ii) $\sqrt{\frac{2500}{3}}$

#### Answer:

Given: $\sqrt{7}=2.646$

$\left(i\right)\sqrt{\frac{144}{7}}=\frac{\sqrt{144}}{\sqrt{7}}=\frac{12}{2.646}=4.536$

Given: $\sqrt{3}=1.732$

$\left(\mathrm{ii}\right)\sqrt{\frac{2500}{3}}=\frac{\sqrt{2500}}{\sqrt{3}}=\frac{50}{1.732}=28.867$

#### Page No 3.57:

#### Question 5:

Given that: $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236\mathrm{and}\sqrt{7}=2.646,$ find the square roots of the following:

(i) $\frac{196}{75}$

(ii) $\frac{400}{63}$

(iii) $\frac{150}{7}$

(iv) $\frac{256}{5}$

(v) $\frac{25}{50}$

#### Answer:

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

$\left(\mathrm{i}\right)\sqrt{\frac{196}{75}}=\frac{14}{5\sqrt{3}}=\frac{14}{5\times 1.732}=1.617\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\sqrt{\frac{400}{63}}=\frac{20}{3\sqrt{7}}=\frac{20}{3\times 2.646}=2.520\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\sqrt{\frac{150}{7}}=\frac{5\sqrt{2}\times \sqrt{3}}{\sqrt{7}}=\frac{5\times 1.414\times 1.732}{2.646}=4.628\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\sqrt{\frac{256}{5}}=\frac{16}{\sqrt{5}}=\frac{16}{2.236}=7.155\phantom{\rule{0ex}{0ex}}\left(\mathrm{v}\right)\sqrt{\frac{27}{50}}=\frac{3\sqrt{3}}{5\sqrt{2}}=\frac{3\times 1.732}{5\times 1.414}=0.735$

#### 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 $\sqrt{3}\mathrm{and}\sqrt{5}$

$\sqrt{15}=\sqrt{3}\times \sqrt{5}\phantom{\rule{0ex}{0ex}}=1.732\times 2.236\phantom{\rule{0ex}{0ex}}=3.873\phantom{\rule{0ex}{0ex}}$

#### Page No 3.61:

#### Question 3:

Using square root table, find the square root

74

#### Answer:

Using the table to find $\sqrt{2}\mathrm{and}\sqrt{37}$

$\sqrt{74}=\sqrt{2}\times \sqrt{37}\phantom{\rule{0ex}{0ex}}=1.414\times 6.083\phantom{\rule{0ex}{0ex}}=8.602$

#### Page No 3.61:

#### Question 4:

Using square root table, find the square root

82

#### Answer:

Using the table to find $\sqrt{2}\mathrm{and}\sqrt{41}$

$\sqrt{82}=\sqrt{2}\times \sqrt{41}\phantom{\rule{0ex}{0ex}}=1.414\times 6.403\phantom{\rule{0ex}{0ex}}=9.055$

#### Page No 3.61:

#### Question 5:

Using square root table, find the square root

198

#### Answer:

Using the table to find $\sqrt{2}\mathrm{and}\sqrt{11}$

$\sqrt{198}=\sqrt{2}\times \sqrt{9}\times \sqrt{11}\phantom{\rule{0ex}{0ex}}=1.414\times 3\times 3.317\phantom{\rule{0ex}{0ex}}=14.070$

#### Page No 3.61:

#### Question 6:

Using square root table, find the square root

540

#### Answer:

Using the table to find $\sqrt{3}\mathrm{and}\sqrt{5}$

$\sqrt{540}=\sqrt{54}\times \sqrt{10}\phantom{\rule{0ex}{0ex}}=2\times 3\sqrt{3}\times \sqrt{5}\phantom{\rule{0ex}{0ex}}=2\times 3\times 1.732\times 2.2361\phantom{\rule{0ex}{0ex}}=23.24$

#### Page No 3.61:

#### Question 7:

Using square root table, find the square root

8700

#### Answer:

Using the table to find $\sqrt{3}\mathrm{and}\sqrt{29}$

$\sqrt{8700}=\sqrt{3}\times \sqrt{29}\times \sqrt{100}\phantom{\rule{0ex}{0ex}}=1.7321\times 5.385\times 10\phantom{\rule{0ex}{0ex}}=93.27$

#### Page No 3.61:

#### Question 8:

Using square root table, find the square root

3509

#### Answer:

Using the table to find $\sqrt{29}$

$\sqrt{3509}=\sqrt{121}\times \sqrt{29}\phantom{\rule{0ex}{0ex}}=11\times 5.3851\phantom{\rule{0ex}{0ex}}=59.235$

#### Page No 3.61:

#### Question 9:

Using square root table, find the square root

6929

#### Answer:

Using the table to find $\sqrt{41}$

$\sqrt{6929}=\sqrt{169}\times \sqrt{41}\phantom{\rule{0ex}{0ex}}=13\times 6.4031\phantom{\rule{0ex}{0ex}}=83.239$

#### Page No 3.61:

#### Question 10:

Using square root table, find the square root

25725

#### Answer:

Using the table to find $\sqrt{3}\mathrm{and}\sqrt{7}$

$\sqrt{25725}=\sqrt{3\times 5\times 5\times 7\times 7\times 7}\phantom{\rule{0ex}{0ex}}=\sqrt{3}\times 5\times 7\times \sqrt{7}\phantom{\rule{0ex}{0ex}}=1.732\times 5\times 7\times 2.646\phantom{\rule{0ex}{0ex}}=160.41$

#### Page No 3.61:

#### Question 11:

Using square root table, find the square root

1312

#### Answer:

Using the table to find $\sqrt{2}\mathrm{and}\sqrt{41}$

$\sqrt{1312}=\sqrt{2\times 2\times 2\times 2\times 2\times 41}\phantom{\rule{0ex}{0ex}}=2\times 2\sqrt{2}\times \sqrt{41}\phantom{\rule{0ex}{0ex}}=2\times 2\times 1.414\times 6.4031\phantom{\rule{0ex}{0ex}}=36.222$

#### Page No 3.61:

#### Question 12:

Using square root table, find the square root

4192

#### Answer:

$\sqrt{4192}=\sqrt{2\times 2\times 2\times 2\times 2\times 131}\phantom{\rule{0ex}{0ex}}=2\times 2\sqrt{2}\times \sqrt{131}$

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

Substituting the values:

= $2\times 2\times 11.4455$ (using the table to find $\sqrt{2}$)

= 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 $\times $ 991, which means that $\sqrt{4955}=\sqrt{5}\times \sqrt{11}$.

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:

$\sqrt{4955}=\sqrt{49.55\times 100}=\sqrt{49.55}\times 10$

Now, we have to find the square root of 49.55.

We have: $\sqrt{49}=7\mathrm{and}\sqrt{50}=7.071$ .

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 $\times $ 0.0701 = 0.03905

$\therefore $ $\sqrt{49.55}=7+0.03905=7.03905$

Finally, we have:

$\sqrt{4955}=\sqrt{49.55}\times 10=7.03905\times 10=70.3905$

#### Page No 3.61:

#### Question 14:

Using square root table, find the square root

$\frac{99}{144}$

#### Answer:

$\sqrt{\frac{99}{144}}=\frac{\sqrt{3\times 3\times 11}}{\sqrt{144}}$

= $\frac{3\sqrt{11}}{12}$

=$\frac{3\times 3.3166}{12}$ (using the square root table to find $\sqrt{11}$)

=0.829

#### Page No 3.61:

#### Question 15:

Using square root table, find the square root

$\frac{57}{169}$

#### Answer:

$\sqrt{\frac{57}{169}}=\frac{\sqrt{3}\times \sqrt{19}}{\sqrt{169}}\phantom{\rule{0ex}{0ex}}$

$\frac{1.732\times 4.3589}{13}$ (using the square root table to find $\sqrt{3}$ and $\sqrt{19}$)

$0.581$

#### Page No 3.61:

#### Question 16:

Using square root table, find the square root

$\frac{101}{169}$

#### Answer:

$\sqrt{\frac{101}{169}}=\frac{\sqrt{101}}{\sqrt{169}}$

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:

$\sqrt{101}=\sqrt{1.01\times 100}=\sqrt{1.01}\times 10$

Now, we have to find the square root of 1.01.

We have:

$\sqrt{1}=1\mathrm{and}\sqrt{2}=1.414$

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 $\times $ 0.414 = 0.00414

$\therefore $ $\sqrt{1.01}=1+0.00414=1.00414\phantom{\rule{0ex}{0ex}}\sqrt{101}=\sqrt{1.01}\times 10=1.00414\times 10=10.0414$

Finally, $\sqrt{\frac{101}{169}}=\frac{\sqrt{101}}{1313}=\frac{10.0414}{13}=0.772$

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:

$\sqrt{13}=3.606\mathrm{and}\sqrt{14}=\sqrt{2}\times \sqrt{7}=3.742$

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:

$0.136\times 0.21=0.02856$

$\therefore $ $\sqrt{13.21}=3.606+0.02856\approx 3.635$

#### Page No 3.61:

#### Question 18:

Using square root table, find the square root

#### Answer:

We have to find $\sqrt{21.97}$.

From the square root table, we have:

$\sqrt{21}=\sqrt{3}\times \sqrt{7}=4.583\mathrm{and}\sqrt{22}=\sqrt{2}\times \sqrt{11}$=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:

$0.107\times 0.97=0.104$

$\therefore $ $\sqrt{21.97}=4.583+0.104\approx 4.687$

#### Page No 3.61:

#### Question 19:

Using square root table, find the square root

110

#### Answer:

$\sqrt{110}=\sqrt{2}\times \sqrt{5}\times \sqrt{11}\phantom{\rule{0ex}{0ex}}=1.414\times 2.236\times 3.317(\mathrm{Using}\mathrm{the}\mathrm{square}\mathrm{root}\mathrm{table}\mathrm{to}\mathrm{find}\mathrm{all}\mathrm{the}\mathrm{square}\mathrm{roots})\phantom{\rule{0ex}{0ex}}=10.488$

#### Page No 3.61:

#### Question 20:

Using square root table, find the square root

1110

#### Answer:

$\sqrt{1110}=\sqrt{2}\times \sqrt{3}\times \sqrt{5}\times \sqrt{37}\phantom{\rule{0ex}{0ex}}=1.414\times 1.732\times 2.236\times 6.083(\mathrm{Using}\mathrm{the}\mathrm{table}\mathrm{to}\mathrm{find}\mathrm{all}\mathrm{the}\mathrm{square}\mathrm{roots})\phantom{\rule{0ex}{0ex}}=33.312$

#### Page No 3.61:

#### Question 21:

Using square root table, find the square root

11.11

#### Answer:

We have:

$\sqrt{11}=3.317\mathrm{and}\sqrt{12}=3.464$

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 $\times $ 0.1474 = 0.0162

$\therefore $ $\sqrt{11.11}=3.3166+0.0162=3.328\approx 3.333$

#### Page No 3.61:

#### Question 22:

The area of a square field is 325 m^{2}. 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.

$\therefore $ $\sqrt{325}=\sqrt{5\times 5\times 13}\phantom{\rule{0ex}{0ex}}=5\times \sqrt{13}\phantom{\rule{0ex}{0ex}}=5\times 3.605\phantom{\rule{0ex}{0ex}}=18.030$

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 $\times $ 70 m = 16800 m^{2}

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 m^{2}.

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

$\therefore $ $\sqrt{16800}=\sqrt{2\times 2\times 2\times 2\times 2\times 3\times 5\times 5\times 7}\phantom{\rule{0ex}{0ex}}=2\times 2\times 5\sqrt{2\times 3\times 7}\phantom{\rule{0ex}{0ex}}=20\sqrt{42}\mathrm{m}=129.60\mathrm{m}$

Hence, the length of one side of the square is 129.60 m

View NCERT Solutions for all chapters of Class 8