# NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Root | Class 8th Maths

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# NCERT Solutions for Class 8 Maths chapter 6 Square and Square Root

class 8th maths

**NCERT Solutions for Class 8 Maths chapter 6 Square and Square Root**

class 8th maths

Ex 6.1

class 8th maths

Ex 6.1

**Exercise 6.1**

**1. What will be the unit digit of the squares of the following numbers?**

(i) 81 | (ii) 272 | (iii) 799 | (iv) 3853 |

(v) 1234 | (vi) 26387 | (vii) 52698 | (viii) 99880 |

(ix) 12796 | (x) 55555 |

**Solutions :**

**(i) 81**

One’s in number 81 is = 1

∴ 1 × 1 = 1

So, unit digit of the square of 81 is = 1

**(ii) 272**

One’s in number 81 is = 2

∴ 2 × 2 = 4

So, unit digit of the square of 272 is = 4

**(iii) 799**

One’s in number 799 is = 9

∴ 9 × 9 = 81

So, unit digit of the square of 799 is = 1

**(vi) 3853**

One’s in number 3853 is = 3

∴ 3 × 3 = 9

So, unit digit of the square of 3853 is = 9

**(v) 1234**

One’s in number 1234 is = 4

∴ 4 × 4 = 16

One’s in number 1234 is = 6

**(vi) 26387**

One’s in number 26387 is = 7

∴ 7 × 7 = 49

So, unit digit of the square of 26387 is = 9

**(vii) 52698**

One’s in number 52698 is = 8

∴ 8 × 8 = 64

So, unit digit of the square of 52698 is = 4

**(viii) 99880**

One’s in number 99880 is = 0

∴ 0 × 0 = 0

So, unit digit of the square of 99880 is = 0

**(xi) 12796**

One’s in number 12796 is = 6

∴ 6 × 6 = 36

So, unit digit of the square of 12796 is = 6

**(x) 55555**

One’s in number 55555 is = 5

∴ 5 × 5 = 25

So, unit digit of the square of 55555 is = 5

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

(i) 1057 | (ii) 23453 | (iii) 7928 | (iv) 222222 |

(v) 64000 | (vi) 89722 | (vii) 222000 | (viii) 505050 |

**Solutions :**

Numbers whose units digit is 2, 3, 7 or 8 are not perfect square numbers.

Even if there is an odd numbers of zero at the end of the number, that number will not be a perfect square number.

**(i) 1057**

The unit’s digit of the number 1057 is 7, so it is not a perfect square number.

**(ii) 23453**

The unit’s digit of the number 23453 is 3, so it is not a perfect square number.

**(iii) 7928**

The unit’s digit of the number 7928 is 8, so it is not a perfect square number.

**(iv) 222222**

The unit’s digit of the number 222222 is 2, so it is not a perfect square number.

**(v) 64000**

The number 64000 has odd numbers of zeros at the end, so it is not a perfect square.

**(vi) 89722**

The unit’s digit of the number 89722 is 2, so it is not a perfect square number.

**(vii) 222000**

The number 222000 has odd numbers of zeros at the end, so it is not a perfect square.

**(viii) 505050**

The number 505050 has odd numbers of zeros at the end, so it is not a perfect square.

**3. 3. The squares of which of the following would be odd numbers?**

(i) 431 | (ii) 2826 | (iii) 7779 | (vi) 82004 |

**Solutions :**

The square of odd numbers is an odd number.

**(i) 431**

Number 431 is an odd number, so its square will also be an odd number.

**(ii) 2826**

The number 2826 is an even number, so its square will also be an even number.

**(iii) 7779**

The number 7779 is an even number, so its square will also be an even number.

**(iv) 82004**

The number 82004 is an even number, so its square will also be an even number.

**4. Observe the following pattern and find the missing digits.**

11² | = | 121 |

101² | = | 10201 |

1001² | = | 1002001 |

100001² | = | 1………..2……….1 |

10000001² | = | ……………………………… |

**Solutions :**

11² | = | 121 |

101² | = | 10201 |

1001² | = | 1002001 |

100001² | = | 10000200001 |

10000001² | = | 100000020000001 |

**5. Observe the following pattern and supply the missing numbers.**

11² | = | 121 |

101² | = | 10201 |

10101² | = | 102030201 |

1010101² | = | ……………………….. |

…………………² | = | 10203040504030201 |

**Solutions :**

11² | = | 121 |

101² | = | 10201 |

10101² | = | 102030201 |

1010101² | = | 1020304030201 |

101010101² |
= | 10203040504030201 |

**6. Using the given pattern, find the missing numbers.**

1^{²} + 2^{²} + 2^{²} = 3^{²}

2^{²} + 3^{²} + 6^{²} = 7^{²}

3^{²} + 4^{²} + 12^{²} = 13^{²}

4^{²} + 5^{²} + __^{2} = 21²

5^{²} + __^{2} + 30^{²} = 31^{²}

6^{²} + 7^{²} + __^{2} = __^{2}

**Solutions :**

The second number in the given pattern is obtained by adding 1 to the first number. When the first number is multiplied by the second number, the third number is obtained and the number after the equal sign is obtained by adding one to the third number.

1^{²} + 2^{²} + 2^{²} = 3^{²}

2^{²} + 3^{²} + 6^{²} = 7^{²}

3^{²} + 4^{²} + 12^{²} = 13^{²}

4^{²} + 5^{²} + **20**^{2} = 21²

5^{²} + **6**^{2} + 30^{²} = 31^{²}

6^{²} + 7^{²} + **42**^{2} = **43**^{2}

**7. Without adding, find the sum.**

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

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

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

**Solutions :**

The sum of the first n odd natural numbers is n².

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

Sum of first 5 odd natural numbers = 5^{2}

= 25

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

Sum of first 10 odd natural numbers = 10^{2}

= 100

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

Sum of first 12 odd natural numbers = 12^{2}

= 144

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

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

**हल :**

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

The sum of the first n odd natural numbers is n².

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

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

The sum of the first n odd natural numbers is n².

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

**9. How many numbers lie between squares of the following numbers?**

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

**Solutions :**

Numbers between squares of any two numbers = difference of square of two numbers − 1

**(i) 12 and 13**

12² = 144

13² = 169

difference of square of both = 169 − 144

= 25

Now, one less than 25 = 25 − 1 = 24

So, There are 24 numbers between the squares of 12 and 13.

**(ii) 25 and 26 **

25² = 625

26² = 676

difference of square of both = 676 − 625

= 51

Now, one less than 51 = 51 − 1 = 50

So, There are 50 numbers between the squares of 25 and 26.

**(iii) 99 and 100**

99² = 9801

100² = 10000

difference of square of both = 10000 − 9801

= 199

Now, one less than 199 = 199 − 1 = 198

So, There are 198 numbers between the squares of 99 and 100.

**NCERT Solutions for Class 8 Maths chapter 6 **

class 8th math

Ex 6.2

class 8th math

Ex 6.2

Exercise 6.2

**1. Find the square of the following numbers.**

(i) 32 | (ii) 35 | (iii) 86 | (iv) 93 |

(v) 71 | (vi) 46 |

**Solutions : **

**(i) 32**

identity (a + b)^{2} = a^{2} + 2ab + b^{2}

(30 + 2)^{2} = 30^{2} + 2 × 30 × 2 + 2^{2}

32^{2} = 900 + 120 + 4

32^{2} = 1024

So the square of 32 is 1024.

**(ii) 35**

identity (a + b)^{2} = a^{2} + 2ab + b^{2}

(35 + 4)^{2} = 30^{2} + 2 × 30 × 5 + 5^{2}

35^{2} = 900 + 300 + 25

35^{2} = 1225

So the square of 35 is 1225.

**(iii) 86**

identity (a + b)^{2} = a^{2} + 2ab + b^{2}

(80 + 6)^{2} = 80^{2} + 2 × 80 × 6 + 6^{2}

86^{2} = 6400 + 960 + 36

86^{2} = 7396

So the square of 86 is 7396.

**(iv) 93**

identity (a + b)^{2} = a^{2} + 2ab + b^{2}

(90 + 3)^{2} = 90^{2} + 2 × 90 × 3 + 3^{2}

93^{2} = 8100 + 540 + 9

93^{2} = 8649

So the square of 93 is 8649.

**(v) 71**

identity (a + b)^{2} = a^{2} + 2ab + b^{2}

(70 + 1)^{2} = 70^{2} + 2 × 70 × 1 + 1^{2}

71^{2} = 4900 + 140 + 1

71^{2} = 5041

So the square of 71 is 5041.

**(vi) 46**

identity (a + b)^{2} = a^{2} + 2ab + b^{2}

(40 + 6)^{2} = 40^{2} + 2 × 40 × 6 + 6^{2}

40^{2} = 1600 + 480 + 36

40^{2} = 5041

So the square of 46 is 5041.

**2. Write a Pythagorean triplet whose one member is.**

(i) 6 | (ii) 14 | (iii) 16 | (iv) 18 |

**Solutions :**

From the simple form 2m, m² – 1, m² + 1 we can find the Pythagorean triplet.

**(i) 6 **

If we take 2*m* = 6 then –

\(\displaystyle m=\frac{{6}}{2}\)

*m* = 3

which is the integer value for *m*.

So, *m*² – 1

put the value of *m*

= 3² – 1

= 9 – 1

= 8

same as, put the value of *m* in *m*² + 1

3² + 1

= 9 + 1

= 10

So the Pythagorean triples are 6, 8 and 10.

**(ii) 14 **

If we take 2*m* = 14 then

\(\displaystyle m=\frac{{14}}{2}\)

*m* = 7

which is the integer value for *m*.

So, *m*² – 1

put the value of *m*

= 7² – 1

= 49 – 1

= 48

same as, put the value of *m* in *m*² + 1

7² + 1

= 49 + 1

= 50

So the Pythagorean triples are 14, 48 and 50.

**(iii) 16 **

If we take 2*m* = 16 then

\(\displaystyle m=\frac{{16}}{2}\)

*m* = 8

which is the integer value for *m*.

So, *m*² – 1

put the value of *m*

= 8² – 1

= 64 – 1

= 63

same as, put the value of *m* in *m*² + 1

8² + 1

= 64 + 1

= 65

So the Pythagorean triples are 16, 63 and 65.

**(iv) 18 **

If we take 2*m* = 18 then

\(\displaystyle m=\frac{{18}}{2}\)

*m* = 9

which is the integer value for *m*.

So, *m*² – 1

put the value of *m*

= 9² – 1

= 81 – 1

= 80

same as, put the value of *m* in *m*² + 1

9² + 1

= 81 + 1

= 82

So the Pythagorean triples are 9, 80 and 82.

**NCERT Solutions for Class 8 Maths chapter 6 **

class 8th math

Ex 6.3

class 8th math

Ex 6.3

Exercise 6.3

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

(i) 9801 | (ii) 99856 | (iii) 998001 | (iv) 657666025 |

**Solutions : **

**(i) 9801**

The unit’s digit in the number 9801 is 1.

So the units digit 1 comes only when 1 is multiplied by 1 or 9 by 9.

So the unit’s digit in the square root of the number 9801 will be 1 or 9.

**(ii) 99856**

The unit’s digit in the number 99856 is 6.

So the units digit 6 comes only when 4 is multiplied by 4 or 6 by 6.

So the unit’s digit in the square root of the number 99856 will be 4 or 6.

**(iii) 998001**

The unit’s digit in the number 998001 is 1.

So the units digit 1 comes only when 1 is multiplied by 1 or 9 by 9.

So the unit’s digit in the square root of the number 998001 will be 1 or 9.

**(iv) 657666025**

The unit’s digit in the number 657666025 is 5.

So the unit’s digit 5 comes only when 5 is multiplied by 5.

Hence the unit’s digit in the square root of the number 657666025 will be 5.

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

**Solutions : **

**100 square root by the method of repeated subtraction – **

As we know that the sum of the first *n* odd natural numbers is *n*².

100 – 1 = 99 | 99 – 3 = 96 | 96 – 5 = 91 | 91 – 7 = 84 |

85 – 9 = 75 | 76 – 11 = 64 | 66 – 13 = 51 | 53 – 15 = 36 |

38 – 17 = 19 | 19 – 19 = 0 |

Subtracting consecutive odd numbers from the number 100, 0 comes in the 10th term.

So the square root of 100 will be 10.

\(\displaystyle \sqrt{{100}}=10\)

**169 square root by the method of repeated subtraction – **

As we know that the sum of the first *n* odd natural numbers is *n*².

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 |

Subtracting consecutive odd numbers from the number 169, 0 comes in the 13th term.

So the square root of 169 will be 13.

\(\displaystyle \sqrt{{169}}=13\)

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

(i) 729 | (ii) 400 | (iii) 1764 | (iv) 4096 |

(v) 7744 | (vi) 9604 | (vii) 5929 | (viii) 9216 |

(ix) 529 | (x) 8100 |

**Solutions : **

**(i) 729**

\(\displaystyle \begin{array}{l}\underline{{3\left| {729} \right.}}\\\underline{{3\left| {243} \right.}}\\\underline{{3\left| {81} \right.}}\\\underline{{\left. 3 \right|27}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\left. {} \right|1\end{array}\)

prime factors of the number 729 = 3 × 3 × 3 × 3 × 3 × 3

On creating a pair of prime factors = 3 × 3 × 3 × 3 × 3 × 3

= 3 × 3 × 3

= 27

So, \(\displaystyle \sqrt{{729}}=27\)

**(ii) 400 **

\(\displaystyle \begin{array}{l}\underline{{2\left| {400} \right.}}\\\underline{{2\left| {200} \right.}}\\\underline{{2\left| {100} \right.}}\\\underline{{\left. 2 \right|50}}\\\underline{{5\left| {25} \right.}}\\\underline{{5\left| 5 \right.}}\\\left. {} \right|1\end{array}\)

prime factors of the number 400 = 2 × 2 × 2 × 2 × 5 × 5

On creating a pair of prime factors = 2 × 2 × 2 × 2 × 5 × 5

= 2 × 2 × 5

= 20

So, \(\displaystyle \sqrt{{400}}=20\)

**(iii) 1764**

\(\displaystyle \begin{array}{l}\underline{{2\left| {1764} \right.}}\\\underline{{2\left| {882} \right.}}\\\underline{{3\left| {441} \right.}}\\\underline{{\left. 3 \right|147}}\\\underline{{7\left| {49} \right.}}\\\underline{{7\left| 7 \right.}}\\\left. {} \right|1\end{array}\)

prime factors of the number 1764 = 2 × 2 × 3 × 3 × 7 × 7

On creating a pair of prime factors = 2 × 2 × 3 × 3 × 7 × 7

= 2 × 3 × 7

= 42

So, \(\displaystyle \sqrt{{1764}}=42\)

**(iv) 4096**

\(\displaystyle \begin{array}{l}\underline{{2\left| {4096} \right.}}\\\underline{{2\left| {2048} \right.}}\\\underline{{2\left| {1024} \right.}}\\\underline{{2\left| {512} \right.}}\\\underline{{2\left| {256} \right.}}\\\underline{{2\left| {128} \right.}}\\\underline{{2\left| {64} \right.}}\\\underline{{2\left| {32} \right.}}\\\underline{{2\left| {16} \right.}}\\\underline{{2\left| 8 \right.}}\\\underline{{2\left| 4 \right.}}\\\underline{{2\left| 2 \right.}}\\\left. {} \right|1\end{array}\)

prime factors of the number 4096= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2 × 2 × 2 × 2 × 2 × 2

= 64

So, \(\displaystyle \sqrt{{4096}}=64\)

**(v) 7744**

\(\displaystyle \begin{array}{l}\,\,\underline{{2\left| {7744} \right.}}\\\,\,\underline{{2\left| {3872} \right.}}\\\,\,\underline{{2\left| {1936} \right.}}\\\,\,\underline{{2\left| {968} \right.}}\\\underline{{\,\,2\left| {484} \right.}}\\\underline{{\,\,2\left| {242} \right.}}\\\underline{{11\left| {121} \right.}}\\\underline{{11\left| {11} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)

prime factors of the number 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

= 2 × 2 × 2 × 11

= 88

So, \(\displaystyle \sqrt{{7744}}=88\)

**(vi) 9604**

\(\displaystyle \begin{array}{l}\underline{{2\left| {9604} \right.}}\\\underline{{2\left| {4802} \right.}}\\\underline{{7\left| {2401} \right.}}\\\underline{{7\left| {343} \right.}}\\\underline{{7\left| {49} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factors of the number 9604 = 2 × 2 × 7 × 7 × 7 × 7

On creating a pair of prime factors = 2 × 2 × 7 × 7 × 7 × 7

= 2 × 7 × 7

= 98

So, \(\displaystyle \sqrt{{9604}}=98\)

**(vii) 5929**

\(\displaystyle \begin{array}{l}\underline{{\,\,7\left| {5929} \right.}}\\\underline{{\,\,7\left| {847} \right.}}\\\underline{{11\left| {121} \right.}}\\\underline{{11\left| {11} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)

prime factors of the number 5929 = 7 × 7 × 11 × 11

On creating a pair of prime factors = 7 × 7 × 11 × 11

= 7 × 11

= 77

So, \(\displaystyle \sqrt{{5929}}=77\)

**(viii) 9216**

\(\displaystyle \begin{array}{l}\underline{{2\left| {9216} \right.}}\\\underline{{2\left| {4608} \right.}}\\\underline{{2\left| {2304} \right.}}\\\underline{{2\left| {1152} \right.}}\\\underline{{2\left| {576} \right.}}\\\underline{{2\left| {288} \right.}}\\\underline{{2\left| {144} \right.}}\\\underline{{2\left| {72} \right.}}\\\underline{{2\left| {36} \right.}}\\\underline{{2\left| {18} \right.}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\left. {} \right|1\end{array}\)

prime factors of the number 9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

= 2 × 2 × 2 × 2 × 2 × 3

= 96

So, \(\displaystyle \sqrt{{9216}}=96\)

**(ix) 529**

\(\displaystyle \begin{array}{l}\underline{{23\left| {529} \right.}}\\\underline{{23\left| {23} \right.}}\\\left. {\,\,\,\,\,} \right|1\end{array}\)

prime factors of the number 529 = 23 × 23

On creating a pair of prime factors = 23 × 23

= 23

So, \(\displaystyle \sqrt{{529}}=23\)

**(x) 8100**

\(\displaystyle \begin{array}{l}\underline{{2\left| {8100} \right.}}\\\underline{{2\left| {4050} \right.}}\\\underline{{3\left| {2025} \right.}}\\\underline{{3\left| {675} \right.}}\\\underline{{3\left| {225} \right.}}\\\underline{{3\left| {75} \right.}}\\\underline{{5\left| {25} \right.}}\\\underline{{5\left| 5 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factors of the number 8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

On creating a pair of prime factors = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

= 2 × 3 × 3 × 5

So, \(\displaystyle \sqrt{{8100}}=90\)

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

(i) 252 | (ii) 180 | (iii) 1008 | (iv) 2028 |

(v) 1458 | (vi) 768 |

**Solutions : **

**(i) 252**

\(\displaystyle \begin{array}{l}\underline{{2\left| {252} \right.}}\\\underline{{2\left| {126} \right.}}\\\underline{{3\left| {63} \right.}}\\\underline{{3\left| {21} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number 252 = 2 × 2 × 3 × 3 × 7

On creating a pair of prime factors = 2 × 2 × 3 × 3 × 7

As the prime factor 7 has no pair, 252 is not a perfect square.

If 7 gets a pair then the number will become perfect square. So, we multiply 252 by 7 to get, **
**= 252 × 7

= 1764 is a perfect square.

252 = 2 × 2 × 3 × 3 × 7 ×

**7**

Now each prime factor is in a pair. Therefore,

= 2 × 3 × 7

Thus the required smallest multiple of 252 is 1764 which is a perfect square.

\(\displaystyle \sqrt{{1764}}=42\)

**(ii) 180**

\(\displaystyle \begin{array}{l}\underline{{2\left| {180} \right.}}\\\underline{{2\left| {90} \right.}}\\\underline{{3\left| {45} \right.}}\\\underline{{3\left| {15} \right.}}\\\underline{{5\left| 5 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number = 2 × 2 × 3 × 3 × 5

On creating a pair of prime factors = 2 × 2 × 3 × 3 × 5

As the prime factor 5 has no pair, 180 is not a perfect square.

If 5 gets a pair then the number will become perfect square. So, we multiply 180 by 7 to get, **
**= 180 × 5

= 900 is a perfect square.

= 2 × 2 × 3 × 3 × 5 ×

**5**

Now each prime factor is in a pair. Therefore,

= 2 × 3 × 5

Thus the required smallest multiple of 180 is 900 which is a perfect square.

\(\displaystyle \sqrt{{900}}=30\)

**(iii) 1008
**\(\displaystyle \begin{array}{l}\underline{{2\left| {1008} \right.}}\\\underline{{2\left| {504} \right.}}\\\underline{{2\left| {252} \right.}}\\\underline{{2\left| {126} \right.}}\\\underline{{3\left| {63} \right.}}\\\underline{{3\left| {21} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number = 2 × 2 × 2 × 2 × 3 × 3 × 7

On creating a pair of prime factors = 2 × 2 × 2 × 2 × 3 × 3 × 7

As the prime factor 7 has no pair, 1008 is not a perfect square.

If 7 gets a pair then the number will become perfect square. So, we multiply 1008 by 7 to get,

**= 1008 × 7**

= 7056 is a perfect square.

= 2 × 2 × 2 × 2 × 3 × 3 × 7 ×

**7**

Now each prime factor is in a pair. Therefore,

= 2 × 2 × 3 × 7

Thus the required smallest multiple of 1008 is 7056 which is a perfect square. \(\displaystyle \sqrt{{7056}}=84\)

**(iv) 2028**

\(\displaystyle \begin{array}{l}\,\,\underline{{2\left| {2028} \right.}}\\\,\,\underline{{2\left| {1014} \right.}}\\\,\,\underline{{3\left| {507} \right.}}\\\underline{{13\left| {169} \right.}}\\\underline{{13\left| {13} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number = 2 × 2 × 3 × 13 × 13

On creating a pair of prime factors = 2 × 2 × 3 × 13 × 13

If 3 gets a pair then the number will become perfect square. So, we multiply 2028 by 3 to get **
**= 2028 × 3

= 6084 is a perfect square.

= 2 × 2 × 3 ×

**3**× 13 × 13

Now each prime factor is in a pair. Therefore,

= 2 × 3 × 13

Thus the required smallest multiple of 2028 is 6084 which is a perfect square.

\(\displaystyle \sqrt{{6084}}=78\)

**(v) 1458**

\(\displaystyle \begin{array}{l}\underline{{2\left| {1458} \right.}}\\\underline{{3\left| {729} \right.}}\\\underline{{3\left| {243} \right.}}\\\underline{{3\left| {81} \right.}}\\\underline{{3\left| {27} \right.}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number = 2 × 3 × 3 × 3 × 3 × 3 × 3

On creating a pair of prime factors = 2 × 3 × 3 × 3 × 3 × 3 × 3

If 2 gets a pair then the number will become perfect square. So, we multiply 1458 by 2 to get **
**= 1458 × 2

= 2916 is a perfect square.

= 2 ×

**2**× 3 × 3 × 3 × 3 × 3 × 3

Now each prime factor is in a pair. Therefore,

= 2 × 3 × 3 × 3

Thus the required smallest multiple of 1458 is 2916 which is a perfect square.

\(\displaystyle \sqrt{{2916}}=54\)

**(vi) 768**

\(\displaystyle \begin{array}{l}\underline{{2\left| {768} \right.}}\\\underline{{2\left| {384} \right.}}\\\underline{{2\left| {192} \right.}}\\\underline{{2\left| {96} \right.}}\\\underline{{2\left| {48} \right.}}\\\underline{{2\left| {24} \right.}}\\\underline{{2\left| {12} \right.}}\\\underline{{2\left| 6 \right.}}\\\underline{{3\left| 3 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

If 3 gets a pair then the number will become perfect square. So, we multiply 768 by 3 to get **
**= 768 × 3

= 2304 is a perfect square.

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 ×

**3**

Now each prime factor is in a pair. Therefore,

= 2 × 2 × 2 × 2 × 3

Thus the required smallest multiple of 768 is 2304 which is a perfect square.

\(\displaystyle \sqrt{{768}}=48\)

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

(i) 252 | (ii) 2925 | (iii) 396 | (iv) 2645 |

(v) 2800 | (vi) 1620 |

**Solutions : **

**(i) 252**

\(\displaystyle \begin{array}{l}\underline{{2\left| {252} \right.}}\\\underline{{2\left| {126} \right.}}\\\underline{{3\left| {63} \right.}}\\\underline{{3\left| {21} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number 252 = 2 × 2 × 3 × 3 × 7

On creating a pair of prime factors = 2 × 2 × 3 × 3 × 7

If we divide 252 by the factor 7, then

= 252 ÷ 7

= 36 is a perfect square.

Therefore, the smallest number be divide is 7.

prime factor of new number 36 = 2 × 2 × 3 × 3

= 2 × 3

= 6

Thus, square root of new number is \(\displaystyle \sqrt{{36}}=6\)

**(ii) 2925**

\(\displaystyle \begin{array}{l}\underline{{\,\,3\left| {2925} \right.}}\\\underline{{\,\,3\left| {975} \right.}}\\\underline{{\,\,5\left| {325} \right.}}\\\underline{{\,\,5\left| {65} \right.}}\\\underline{{13\left| {13} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number 2925 = 3 × 3 × 5 × 5 × 13

On creating a pair of prime factors = 3 × 3 × 5 × 5 × 13

If we divide 2925 by the factor 13, then

= 2925 ÷ 13

= 225 is a perfect square.

Therefore, the smallest number be divide is 13.

prime factor of new number 225 = 3 × 3 × 5 × 5

= 3 × 5

= 15

Thus, square root of new number is \(\displaystyle \sqrt{{225}}=15\)

**(iii) 396**

\(\displaystyle \begin{array}{l}\underline{{\,\,2\left| {396} \right.}}\\\underline{{\,\,2\left| {198} \right.}}\\\underline{{\,\,3\left| {99} \right.}}\\\underline{{\,\,3\left| {33} \right.}}\\\underline{{11\left| {11} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number 396 = 2 × 2 × 3 × 3 × 11

On creating a pair of prime factors = 2 × 2 × 3 × 3 × 11

If we divide 396 by the factor 11, then

= 396 ÷ 11

= 36 is a perfect square.

Therefore, the smallest number be divide is 11.

prime factor of new number 36 = 2 × 2 × 3 × 3

= 2 × 3

= 6

Thus, square root of new number is \(\displaystyle \sqrt{{36}}=6\)

**(iv) 2645**

\(\displaystyle \begin{array}{l}\underline{{\,\,\,5\left| {2645} \right.}}\\\underline{{23\left| {529} \right.}}\\\underline{{23\left| {23} \right.}}\\\,\,\,\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number 2645 = 5 × 23 × 23

On creating a pair of prime factors = 5 × 23 × 23

If we divide 2645 by the factor 5, then

= 2645 ÷ 5

= 529 is a perfect square.

Therefore, the smallest number be divide is 5.

prime factor of new number 529 = 23 × 23

= 23

Thus, square root of new number is \(\displaystyle \sqrt{{529}}=23\)

**(v) 2800**

\(\displaystyle \begin{array}{l}\underline{{2\left| {2800} \right.}}\\\underline{{2\left| {1400} \right.}}\\\underline{{2\left| {700} \right.}}\\\underline{{2\left| {350} \right.}}\\\underline{{5\left| {175} \right.}}\\\underline{{5\left| {35} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number 2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

On creating a pair of prime factors = 2 × 2 × 2 × 2 × 5 × 5 × 7

If we divide 2800 by the factor 7, then

= 2800 ÷ 7

= 400 is a perfect square.

Therefore, the smallest number be divide is 13.

prime factor of new number 400 = 2 × 2 × 2 × 2 × 5 × 5

= 2 × 2 × 5

Thus, square root of new number is \(\displaystyle \sqrt{{400}}=20\)

**(vi) 1620**

\(\displaystyle \begin{array}{l}\underline{{2\left| {1620} \right.}}\\\underline{{2\left| {810} \right.}}\\\underline{{3\left| {405} \right.}}\\\underline{{3\left| {135} \right.}}\\\underline{{3\left| {45} \right.}}\\\underline{{3\left| {15} \right.}}\\\underline{{5\left| 5 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

prime factor of the number 1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

On creating a pair of prime factors = 2 × 2 × 3 × 3 × 3 × 3 × 5

If we divide 1620 by the factor 5, then

= 1620 ÷ 5

= 324 is a perfect square.

Therefore, the smallest number be divide is 13.

prime factor of new number 324 = 2 × 2 × 3 × 3 × 3 × 3

= 2 × 3 × 3

Thus, square root of new number is \(\displaystyle \sqrt{{324}}=18\)

**7. The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.**

**Solutions :**

Let the number of students in the class = *x*

And each student deposited in the Prime Minister’s National Fund = ₹ *x*

The total amount donated to the Prime Minister’s National Fund = 2401

According to Question –

*x* × *x* = 2401

*x*² = 2401

\(\displaystyle x=\sqrt{{2401}}\)

\(\displaystyle \begin{array}{l}\underline{{7\left| {2401} \right.}}\\\underline{{7\left| {343} \right.}}\\\underline{{7\left| {49} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

\(\displaystyle x=\sqrt{{\underline{{7\times 7}}\times \underline{{7\times 7}}}}\)

*x* = 7 × 7

*x* = ₹ 49

So there were 49 students in the school and each student donated ₹ 49.

**8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
**

**Solutions :**

Let the number of rows =

*x*

Number of plants in each row =

*x*

Total number of plants = 2025

According to Question –

*x*×

*x*= 2025

*x*² = 2025

\(\displaystyle x=\sqrt{{2025}}\)

\(\displaystyle \begin{array}{l}\underline{{5\left| {2025} \right.}}\\\underline{{5\left| {405} \right.}}\\\underline{{3\left| {81} \right.}}\\\underline{{3\left| {27} \right.}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)

\(\displaystyle x=\sqrt{{\underline{{5\times 5}}\times \underline{{3\times 3}}\times \underline{{3\times 3}}}}\)

*x*= 5 × 3 × 3

*x*= 45

So the number of rows will be 45 and the number of plants in each row will be 45.

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

**Solution :**

This can be solved in two steps. First we will find the smallest common multiple. And after that find the required class number.

The smallest number into which 4, 9 and 10 will be divided is its L.C.M Is.

\(\displaystyle \begin{array}{l}\underline{{2\left| {4,9,10} \right.}}\\\underline{{2\left| {2,9,5} \right.}}\\\underline{{3\left| {1,9,5} \right.}}\\\underline{{3\left| {1,3,5} \right.}}\\\underline{{5\left| {1,1,5} \right.}}\\\,\,\,\left| {1,1,1} \right.\end{array}\)

L.C.M of numbers 4, 9 and 10. S. = 180 है।

prime factors of LCM = 2 × 2 × 3 × 3 × 5

In prime factorization, we see that pairs of 5 are not formed. Hence 180 is not a perfect square number.

If 5 gets a pair then the number will become perfect square. So, we multiply 180 by 5 to get,

So, if we multiply 180 by 5 then we get 900, which is a perfect square number.

**10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
वह सबसे छोटी वर्ग संख्या ज्ञात कीजिए जो प्रत्येक 8, 15 और 20 से विभाजित हो जाए।**

**Solution :**

This can be solved in two steps. First we will find the smallest common multiple. And after that find the required class number.

The smallest number into which 8, 15 and 20 will be divided is its L.C.M’s.

\(\displaystyle \begin{array}{l}\underline{{2\left| {8,15,20} \right.}}\\\underline{{2\left| {4,9,10} \right.}}\\\underline{{2\left| {2,9,5} \right.}}\\\underline{{3\left| {1,9,5} \right.}}\\\underline{{3\left| {1,3,5} \right.}}\\\underline{{5\left| {1,1,5} \right.}}\\\,\,\,\left| {1,1,1} \right.\end{array}\)

The L.C.M of numbers 8, 15 and 20. = 360 है।

prime factor of L.C.M 360 are = 2 × 2 × 2 × 3 × 3 × 5

In prime factorization, we see that pairs of 2, 5 are not formed. Hence 360 is not a perfect square number.

If 2 and 5 gets a pair then the number will become perfect square. So, we multiply 360 by 10 to get,

So, if we multiply 360 by 10 then we get 3600, which is a perfect square number.

**NCERT Solutions for Class 8 Maths chapter 6 **

class 8th math

Ex 6.4

class 8th math

Ex 6.4

**1. Find the square root of each of the following numbers by Division method.**

**(i) 2304**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,} \right|\,\,\,\,\,\,48\,\,\,}}\\\left. {\,\,\,4} \right|\,\,\,\,\,\underline{{23}}\underline{{04}}\\\underline{{\left. {+4} \right|-16\,\downarrow \,}}\\\left. {\,88} \right|\,\,\,\,\,\,\,\,704\\\underline{{\left. {\,\,\,\,8} \right|\,\,\,-\,\,704}}\\\left. {\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
**Step 1** Starting with the unit digit, put a bar on each pair. Thus the number is written as 5284. \(\displaystyle \overline{{23}}\overline{{04}}\)

**Step 2** The first pair on the left is 23. Now we have to find a number which when multiplied by it will either give 23 or less than 23.

as, 4² = 4 × 4 = 16

5² = 5 × 5 = 25

So we take the number 4.

**Step 3** Take this number as the divisor and quotient with the dividend 23 at the bottom of the leftmost bar. Divide and find the remainder.

**Step 4** Write the number in the middle of the next bar to the right of the remainder. So the next composite number will be 704.

**Step 5** Double the denominator and write it with a space on the right.

**Step 6** To fill in the blanks guess the largest possible digit which will be the new digit in the quotient and multiplying the new divisor by the new quotient, the product will be less than or equal to the quotient.

In this situation-

89 × 9 = 801

88 × 8 = 704

Since 88 × 8 = 704, So to get the remainder, the new digit 8 is chosen.

**Step 7** Because the remainder is 0, and there are no digits left in the given number.

So, \(\displaystyle \sqrt{{2304}}=48\)

**(ii) 4489**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,67\,\,\,\,\,}}\\\left. {\,\,\,\,\,\,6} \right|\,\,\,\,\,\overline{{44}}\overline{{89}}\\\underline{{\left. {+\,\,\,6} \right|-36\,\,\downarrow \,}}\\\left. {127} \right|\,\,\,\,\,\,\,\,\,889\\\underline{{\left. {\,\,\,\,\,7} \right|\,\,\,\,\,-\,889}}\\\left. {\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{4489}}=67\)

**(iii) 3481**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,59\,\,\,\,\,}}\\\left. {\,\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{34}}\overline{{81}}\\\underline{{\left. {+\,\,\,5} \right|-25\,\,\downarrow \,}}\\\left. {109} \right|\,\,\,\,\,\,\,\,981\\\underline{{\left. {\,\,\,\,\,9} \right|\,\,\,\,\,-\,981}}\\\left. {\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{3481}}=59\)

**(iv) 529**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,23\,\,\,\,\,}}\\\left. {\,\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{29}}\\\underline{{\left. {+\,\,\,2} \right|-4\,\,\downarrow \,}}\\\left. {\,\,\,43} \right|\,\,\,\,\,\,129\\\underline{{\left. {\,\,\,\,\,\,3} \right|\,\,-\,129}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{529}}=23\)

**(v) 3249**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,57\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\overline{{32}}\overline{{49}}\\\underline{{\left. {+\,\,5} \right|-25\,\,\downarrow \,}}\\\left. {107} \right|\,\,\,\,\,\,749\\\underline{{\left. {\,\,\,\,\,7} \right|\,\,-\,749}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{3249}}=57\)

**(vi) 1369**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,37\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\overline{{13}}\overline{{69}}\\\underline{{\left. {+\,\,3} \right|-9\,\,\downarrow \,}}\\\left. {\,\,67} \right|\,\,\,\,\,469\\\underline{{\left. {\,\,\,\,\,7} \right|-469}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{1369}}=37\)

**(vii) 5776**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,76\,\,\,\,\,}}\\\left. {\,\,\,\,\,7} \right|\,\,\,\,\,\overline{{57}}\overline{{76}}\\\underline{{\left. {+\,\,7} \right|-49\,\,\downarrow \,}}\\\left. {146} \right|\,\,\,\,\,\,\,\,876\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,\,\,-876}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{5776}}=76\)

**(viii) 7921**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,89\,\,\,\,\,}}\\\left. {\,\,\,\,\,8} \right|\,\,\,\,\,\overline{{79}}\overline{{21}}\\\underline{{\left. {+\,\,8} \right|-64\,\,\downarrow \,}}\\\left. {169} \right|\,\,\,\,\,1521\\\underline{{\left. {\,\,\,\,\,9} \right|\,-1521}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{7921}}=89\)

**(ix) 576**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,24\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{76}}\\\underline{{\left. {+\,\,2} \right|-4\,\,\downarrow \,}}\\\left. {\,\,44} \right|\,\,\,\,\,176\\\underline{{\left. {\,\,\,\,\,4} \right|\,-176}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{576}}=24\)

**(x) 1024**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,32\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\,\,\overline{{10}}\overline{{24}}\\\underline{{\left. {+\,\,3} \right|\,\,\,\,-9\,\,\downarrow \,}}\\\left. {\,\,62} \right|\,\,\,\,\,\,\,\,124\\\underline{{\left. {\,\,\,\,\,2} \right|\,\,\,\,-124}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{1024}}=32\)

**(xi) 3136**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,56\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{31}}\overline{{36}}\\\underline{{\left. {+\,\,5} \right|\,-25\,\,\downarrow \,}}\\\left. {106} \right|\,\,\,\,\,\,636\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,-636}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{3136}}=56\)

**(xii) 900**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,30\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\,\,\overline{9}\overline{{00}}\\\underline{{\left. {+\,\,3} \right|\,\,-9\,\,\downarrow \,}}\\\left. {\,\,60} \right|\,\,\,\,\,\,000\\\underline{{\left. {\,\,\,\,\,0} \right|\,\,-000}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{900}}=30\)

**3. Find the square root of the following decimal numbers.**

(i) 2.56 | (ii) 7.29 | (iii) 51.84 | (iv) 42.25 |

(v) 31.36 |

**Solutions : **

**(i) 2.56**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,1.6\,\,\,\,\,}}\\\left. {\,\,\,\,\,1} \right|\,\,\,\,\,\overline{2}.\overline{{56}}\\\underline{{\left. {+\,\,1} \right|\,\,-1\,\,\downarrow \,}}\\\,\,\left. {26} \right|\,\,\,\,\,\,\,\,156\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,-156}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{2.56}}=1.6\)

**(ii) 7.29**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,2.7\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{7}.\overline{{29}}\\\underline{{\left. {+\,\,2} \right|\,\,-4\,\,\downarrow \,}}\\\,\,\left. {47} \right|\,\,\,\,\,\,329\\\underline{{\left. {\,\,\,\,\,7} \right|\,\,-329}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{7.29}}=2.7\)

**(iii) 51.84**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,7.2\,\,\,\,\,}}\\\left. {\,\,\,\,\,7} \right|\,\,\,\,\,\overline{{51}}.\overline{{84}}\\\underline{{\left. {+\,\,7} \right|\,-49\,\,\downarrow \,}}\\\left. {142} \right|\,\,\,\,\,\,\,284\\\underline{{\left. {\,\,\,\,\,2} \right|\,\,\,-284}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{51.84}}=7.2\)

**(iv) 42.25**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,6.5\,\,\,\,\,}}\\\left. {\,\,\,\,\,6} \right|\,\,\,\,\,\overline{{42}}.\overline{{25}}\\\underline{{\left. {+\,\,6} \right|\,-36\,\,\downarrow \,}}\\\left. {125} \right|\,\,\,\,\,\,\,\,\,625\\\underline{{\left. {\,\,\,\,\,5} \right|\,\,\,\,\,-625}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{42.25}}=6.5\)

**(v) 31.36**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,5.6\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{31}}.\overline{{36}}\\\underline{{\left. {+\,\,5} \right|\,\,-25\,\,\downarrow \,}}\\\left. {106} \right|\,\,\,\,\,\,\,\,636\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,\,\,\,-636}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{31.36}}=5.6\)

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

(i) 402 | (ii) 1989 | (iii) 3250 | (iv) 825 |

(v) 4000 |

**Solutions : **

**(i) 402**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,2\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{4}\overline{{02}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,4\,\,\,} \right|\,\,\,\,\,\,\,\,02\end{array}\)

The remainder is 2, so subtracting 2 from 402 gives the number 400.

which is a perfect square number. Whose square root will be 20.

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,20\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{4}\overline{{00}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,40} \right|\,\,\,\,\,\,\,\,00\\\underline{{\,\,\,\,\left. {\,\,\,} \right|\,\,\,\,\,\,\,\,00}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)

\(\displaystyle \sqrt{{400}}=20\)

**(ii) 1989**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,4\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{19}}\overline{{89}}\\\underline{{\left. {+\,\,4} \right|\,-16\,\,\downarrow \,}}\\\left. {\,\,\,84} \right|\,\,\,\,\,\,\,389\\\underline{{\,\,\,\,\left. {\,\,4} \right|\,\,\,\,\,\,\,336}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,53\end{array}\)

The remainder is 53, so subtracting 53 from 1989 gives the number 1936.

which is a perfect square number. Whose square root will be 44.

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,44\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{19}}\overline{{36}}\\\underline{{\left. {+\,\,4} \right|\,-16\,\,\downarrow \,}}\\\left. {\,\,\,84} \right|\,\,\,\,\,\,\,336\\\underline{{\,\,\,\,\left. {\,\,4} \right|\,\,\,\,\,\,\,336}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,\times \end{array}\)

\(\displaystyle \sqrt{{1936}}=44\)

**(iii) 3250**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,57\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{32}}\overline{{50}}\\\underline{{\left. {+\,\,5} \right|\,-25\,\,\downarrow \,}}\\\left. {107} \right|\,\,\,\,\,\,\,750\\\underline{{\,\,\,\,\left. {\,7} \right|\,\,\,\,\,\,\,749}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,1\end{array}\)

The remainder is 1, so subtracting 1 from 3250 gives the number 3249.

which is a perfect square number. whose square root will be 57.

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,57\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{32}}\overline{{49}}\\\underline{{\left. {+\,\,5} \right|\,-25\,\,\downarrow \,}}\\\left. {107} \right|\,\,\,\,\,\,\,749\\\underline{{\,\,\,\,\left. {\,7} \right|\,\,\,\,\,\,\,749}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,\times \end{array}\)

\(\displaystyle \sqrt{{3249}}=57\)

**(iv) 825**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,28\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{8}\overline{{25}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,48} \right|\,\,\,\,\,425\\\underline{{\,\,\,\,\left. {\,8} \right|\,\,\,\,\,384}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,41\end{array}\)

The remainder is 41, so subtracting 41 from 825 gives the number 784.

which is a perfect square number. Whose square root will be 28.

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,28\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{7}\overline{{84}}\\\underline{{\left. {+\,\,2} \right|\,-3\,\,\downarrow \,}}\\\left. {\,\,48} \right|\,\,\,\,\,384\\\underline{{\,\,\,\,\left. {\,8} \right|\,\,\,\,\,384}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \end{array}\)

\(\displaystyle \sqrt{{784}}=28\)

**(v) 4000**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,63\,\,\,\,\,}}\\\left. {\,\,\,\,\,6} \right|\,\,\,\,\,\overline{{40}}\overline{{00}}\\\underline{{\left. {+\,\,6} \right|\,-36\,\,\downarrow \,}}\\\left. {123} \right|\,\,\,\,\,\,\,400\\\underline{{\,\,\,\,\left. {\,3} \right|\,\,\,\,\,\,\,369}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,31\end{array}\)

The remainder is 31, so subtracting 31 from 4000 gives the number 3969.

which is a perfect square number. Whose square root will be 63.

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

(i) 525 | (ii) 1750 | (iii) 252 | (iv) 1825 |

(v) 6412 |

**Solutions : **

**(i) 525**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,22\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{25}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,42} \right|\,\,\,\,\,125\\\underline{{\,\,\,\,\left. {\,2} \right|\,\,\,\,\,\,\,84}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,41\end{array}\)

On doing the square root of the number 525 by the division method, a remainder of 41 is left. which shows that

22² < 525

next whole number 23² = 529

so the required number = 23² – 525

529 – 525 = 4

So adding 4 to the number 525 will make a perfect square number whose square root will be 23.

**(ii) 1750**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,41\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{17}}\overline{{50}}\\\underline{{\left. {+\,\,4} \right|-16\,\,\downarrow \,}}\\\left. {\,\,\,81} \right|\,\,\,\,\,\,\,\,150\\\underline{{\,\,\,\,\,\left. 1 \right|\,\,\,\,\,\,\,\,\,\,\,81}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,69\end{array}\)

On doing the square root of the number 1750 by the division method, a remainder of 69 is left. which shows that

41² < 1750

next whole number 42² = 1764

so the required number = 42² – 1750

1764 – 1750 = 14

So adding 14 to the number 1750 will make a perfect square number whose square root will be 42.

**(iii) 252**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,15\,\,\,\,\,}}\\\left. {\,\,\,\,\,1} \right|\,\,\,\,\,\overline{2}\overline{{52}}\\\underline{{\left. {+\,\,1} \right|\,\,-1\,\,\downarrow \,}}\\\left. {\,\,25} \right|\,\,\,\,\,\,152\\\underline{{\,\,\,\,\,\left. 5 \right|\,\,\,\,\,\,125}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,27\end{array}\)

On doing the square root of the number 252 by the division method, a remainder of 27 is left. which shows that

15² < 252

next whole number 16² = 256

so the required number = 16² – 252

256 – 252 = 4

So adding 4 to the number 252 will make a perfect square number whose square root will be 16.

**(iv) 1825**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,42\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{18}}\overline{{25}}\\\underline{{\left. {+\,\,4} \right|\,\,-16\,\,\downarrow \,}}\\\left. {\,\,82} \right|\,\,\,\,\,\,\,\,225\\\underline{{\,\,\,\,\,\left. 2 \right|\,\,\,\,\,\,\,\,164}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,61\end{array}\)

On doing the square root of the number 1825 by the division method, a remainder of 61 is left. which shows that

42² < 1825

next whole number 43² = 1849

so the required number = 43² – 1825

1849 – 1825 = 24

Therefore, adding 24 to the number 1825 will become a perfect square number whose square root will be 43.

**(v) 6412**

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,80\,\,\,\,\,}}\\\left. {\,\,\,\,\,8} \right|\,\,\,\,\,\overline{{64}}\overline{{12}}\\\underline{{\left. {+\,\,8} \right|\,\,-64\,\,\downarrow \,}}\\\left. {160} \right|\,\,\,\,\,\,\,\,012\\\underline{{\,\,\,\,\,\left. 0 \right|\,\,\,\,\,\,\,\,000}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,12\end{array}\)

On doing the square root of the number 6412 by the division method, a remainder of 12 is left. which shows that

80² < 6412

next whole number 81² = 6561

so the required number = 81² – 6412

6561 – 6412 = 149

So adding 149 to the number 6412 will make a perfect square number whose square root will be 81.

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

**Solutions :**

Let the side of the square be

*x*.

area of square = 441

*m*²

side × side = 441

*x*×

*x*= 441

*x*² = 441

\(\displaystyle x=\sqrt{{441}}\)

On finding the square root by division method –

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,21\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{4}\overline{{41}}\\\underline{{\left. {+\,\,2} \right|\,\,-4\,\,\downarrow \,}}\\\left. {\,\,41} \right|\,\,\,\,\,\,041\\\underline{{\,\,\,\,\,\left. 1 \right|\,\,\,\,\,\,\,\,\,\,41}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \end{array}\)

So the side of the square will be

*x*= 21

*m*

**7. In a right triangle ABC, ∠B = 90°.**

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

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

**Solutions : **

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

we know that

(hypotenuse)^{2} = (base)^{2} + (perpendicular)^{2}

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

Putting the value in the formula –

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

(AC)^{2} = 36 + 64

(AC)^{2} = 100

\(\displaystyle AC=\sqrt{{100}}\)

AC = 10 *cm*

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

we know that

(hypotenuse)^{2} = (base)^{2} + (perpendicular)^{2}

(AB)^{2} = (AC)^{2} – (BC)^{2}

(AB)^{2} = (13)^{2} – (5)^{2}

(AB)^{2} = 169 – 25

(AB)^{2} = 144

\(\displaystyle AB=\sqrt{{144}}\)

AB = 12 cm

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

**Solution :**

Let the number of rows =

*x*

and the number of plants in each row =

*x*

According to Question –

*x*×

*x*= 1000

*x*² = 1000

\(\displaystyle x=\sqrt{{1000}}\)

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,31\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\,\,\overline{{10}}\overline{{00}}\\\underline{{\left. {+\,\,3} \right|\,\,\,\,-9\,\,\downarrow \,}}\\\left. {\,\,91} \right|\,\,\,\,\,\,\,\,100\\\underline{{\,\,\,\,\,\left. 1 \right|\,\,\,\,\,\,\,\,\,\,91}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,9\end{array}\)

On doing the square root of the number 1000 by the division method, a remainder of 9 is left. which shows that

31² < 1000

next whole number 32² = 1024

so the required number = 32² – 1000

1024 – 1000 = 24

So adding 24 to the number 1024 will become a perfect square number.

The gardener will need 24 plants.

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

**Solution :**

Let the number of rows =

*x*

and number of columns =

*x*

According to Question –

*x*×

*x*= 500

*x*² = 500

\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,22\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{00}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,42} \right|\,\,\,\,\,\,100\\\underline{{\,\,\,\,\,\left. 2 \right|\,\,\,\,\,\,\,84}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,16\end{array}\)

The remainder is 16, so subtracting 16 from 500 gives the number 484.

which is a perfect square number. Whose square root will be 22.

So 16 students have to go out to make arrangements.