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Class 8 NCERT Solutions- Chapter 6 Squares and Square Roots – Exercise 6.4

  • Last Updated : 09 Mar, 2021

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

(i) 2304 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar. 

Thus we have, \overline{23} \overline{04}

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Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend. 



Here, we have 23

Divide and get the remainder.

Here, we get 7

Step 3: Bring down the remaining number under the next bar to the right of the remainder. 

Here, its 04

So now the new dividend is 704.

Step 4: Double the quotient and enter it with a blank on its right. 

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 88 × 8 = 704.

So we choose the new digit as 8. 

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number. 

Hence, √2304 = 48

(ii) 4489 

Solution:



Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{44} \overline{89}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 44

Divide and get the remainder.

Here, we get 8

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 89

So now the new dividend is 889.



Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 127 × 7 = 889.

So we choose the new digit as 7.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √4489 = 67

(iii) 3481 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{34} \overline{81}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 34

Divide and get the remainder.

Here, we get 9

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 81



So now the new dividend is 981.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 109 × 9 = 981.

So we choose the new digit as 9.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √3481 = 59

(iv) 529

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{5} \overline{29}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 5

Divide and get the remainder.

Here, we get 1

Step 3: Bring down the remaining number under the next bar to the right of the remainder.



Here, its 29

So now the new dividend is 129.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 23 × 3 = 129.

So we choose the new digit as 3.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √529 = 23

(v) 3249 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{32} \overline{49}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 32

Divide and get the remainder.

Here, we get 7



Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 49

So now the new dividend is 749.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 107 × 7 = 749.

So we choose the new digit as 7.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √3249 = 57

(vi) 1369 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{13} \overline{69}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 13

Divide and get the remainder.

Here, we get 4

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 69

So now the new dividend is 469.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 37 × 7 = 469.

So we choose the new digit as 7.



Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √1369 = 37

(vii) 5776 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{57} \overline{76}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 57

Divide and get the remainder.

Here, we get 8

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 76

So now the new dividend is 876.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 146 × 6 = 876.

So we choose the new digit as 6.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √5776 = 76

(viii) 7921

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{79} \overline{21}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 79



Divide and get the remainder.

Here, we get 15

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 21

So now the new dividend is 1521.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 169 × 9 = 1521.

So we choose the new digit as 9.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √7921 = 89

(ix) 576 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{5} \overline{76}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 5

Divide and get the remainder.

Here, we get 1

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 76

So now the new dividend is 176.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 44 × 4 = 176.

So we choose the new digit as 4.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √576 = 24

(x) 1024 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{10} \overline{24}



Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 10

Divide and get the remainder.

Here, we get 1

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 24

So now the new dividend is 124.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 62 × 2 = 124.

So we choose the new digit as 2.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √1024 = 32

(xi) 3136 

Solution:

Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{31} \overline{36}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 31

Divide and get the remainder.

Here, we get 6

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 36

So now the new dividend is 636.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 106 × 6 = 636.

So we choose the new digit as 6.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √3136 = 56

(xii) 900

Solution:



Step 1: Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, \overline{9} \overline{00}

Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend.

Here, we have 9

Divide and get the remainder.

Here, we get 0

Step 3: Bring down the remaining number under the next bar to the right of the remainder.

Here, its 0

So now the new dividend is 000.

Step 4: Double the quotient and enter it with a blank on its right.

Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 60 × 0 = 000.

So we choose the new digit as 0.

Get the remainder.

Step 6: Since the remainder is 0 and no digits are left in the given number.

Hence, √900 = 30

Question 2. Find the number of digits in the square root of each of the following numbers (without any calculation).

If n is number of digits in a square number then

Number of digits in the square root = \frac {n}{2}   if n is even 

and, \frac{n+1}{2}    if n is odd.

(i) 64 

Solution:

Here, n = 2, which is even

So, number of digits in square root is = \frac {n}{2}

\frac {2}{2}

= 1

(ii) 144 

Solution:

Here, n = 3, which is odd

So, number of digits in square root is = \frac {n+1}{2}

\frac {4}{2}

= 2

(iii) 4489 

Solution:

Here, n = 4, which is even

So, number of digits in square root is = \frac {n}{2}

\frac {4}{2}

= 2

(iv) 27225

Solution:

Here, n = 5, which is odd



So, number of digits in square root is = \frac {n+1}{2}

\frac {6}{2}

= 3

(v) 390625 

Solution:

Here, n = 6, which is even

So, number of digits in square root is = \frac {n}{2}

\frac {6}{2}

= 3

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

(i) 2.56 

Solution:

To find the square root of a decimal number we put bars on the integral part of the number in the usual manner. And place bars on the decimal part on every pair of digits beginning with the first decimal place. 

We get \overline{2}.\overline{56}

Since the remainder is 0 and no digits are left in the given number.

Hence, √2.56 = 1.6

(ii) 7.29 

Solution:

To find the square root of a decimal number we put bars on the integral part of the number in the usual manner. And place bars on the decimal part on every pair of digits beginning with the first decimal place.

We get \overline{7}.\overline{29}

Since the remainder is 0 and no digits are left in the given number.

Hence, √7.29 = 2.7

(iii) 51.84 

Solution:

To find the square root of a decimal number we put bars on the integral part of the number in the usual manner. And place bars on the decimal part on every pair of digits beginning with the first decimal place.

We get \overline{51}.\overline{84}

Since the remainder is 0 and no digits are left in the given number.

Hence, √51.84 = 7.2

(iv) 42.25

Solution:

To find the square root of a decimal number we put bars on the integral part of the number in the usual manner. And place bars on the decimal part on every pair of digits beginning with the first decimal place.

We get \overline{42}.\overline{25}



Since the remainder is 0 and no digits are left in the given number.

Hence, √42.25 = 6.5

(v) 31.36

Solution:

To find the square root of a decimal number we put bars on the integral part of the number in the usual manner. And place bars on the decimal part on every pair of digits beginning with the first decimal place.

We get \overline{31}.\overline{36}

Since the remainder is 0 and no digits are left in the given number.

Hence, √31.36 = 5.6

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

Here, remainder is the least required number to be subtracted from the given number to get a perfect square.

(i) 402

Solution: 

By following all the steps for obtaining square root, we get

Here remainder is 2

2 is the least required number to be subtracted from 402 to get a perfect square

New number = 402 – 2 = 400

Thus, √400 = 20

(ii) 1989 

Solution:

By following all the steps for obtaining square root, we get

Here remainder is 53

53 is the least required number to be subtracted from 1989.

New number = 1989 – 53 = 1936

Thus, √1936 = 44

(iii) 3250 

Solution:

By following all the steps for obtaining square root, we get

Here remainder is 1

1 is the least required number to be subtracted from 3250 to get a perfect square.

New number = 3250 – 1 = 3249

Thus, √3249 = 57

(iv) 825

Solution:

By following all the steps for obtaining square root, we get

Here, the remainder is 41

41 is the least required number which can be subtracted from 825 to get a perfect square.

New number = 825 – 41 = 784

Thus, √784 = 28

(v) 4000

Solution:

By following all the steps for obtaining square root, we get

Here, the remainder is 31

31 is the least required number which should be subtracted from 4000 to get a perfect square.

New number = 4000 – 31 = 3969

Thus, √3969 = 63

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

(i) 525 

Solution:

By following all the steps for obtaining square root, we get

Here remainder is 41

It represents that 222 is less than 525.

Next number is 23, 

Where, 232 = 529

Hence, the number to be added = 529 – 525 = 4

New number = 525+4 = 529

Thus, √529 = 23

(ii) 1750 

Solution:

By following all the steps for obtaining square root, we get

Here the remainder is 69

It represents that 412 is less than in 1750.

The next number is 42 

Where, 422 = 1764

Hence, number to be added to 1750 = 1764 – 1750 = 14

New number = 1750 + 14 = 1764

√1764 = 42

(iii) 252 

Solution:

By following all the steps for obtaining square root, we get

Here the remainder is 27.

It represents that 152 is less than 252.

The next number is 16 

Where,162 = 256

Hence, number to be added to 252 = 256 – 252 = 4

New number = 252 + 4 = 256

and √256 = 16

(iv) 1825

Solution:

By following all the steps for obtaining square root, we get

The remainder is 61.

It represents that 422 is less than in 1825.

Next number is 43 



Where, 432 = 1849

Hence, number to be added to 1825 = 1849 – 1825 = 24

New number = 1825 + 24 = 1849

and √1849 = 43

(v) 6412

Solution:

By following all the steps for obtaining square root, we get

Here, the remainder is 12.

It represents that 802 is less than in 6412.

The next number is 81 

Where, 812 = 6561

Hence, the number to be added = 6561 – 6412 = 149

New number = 6412 + 149 = 6561

and √6561 = 81

Question 6. Find the length of the side of a square whose area is 441 m2.

Solution:

Let the side of square be x m.

Area of square = x2

 According to the given question,

x2 = 441

x = √441

Hence, the side of square is 21 m.

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

(a) If AB = 6 cm, BC = 8 cm, find AC 

Solution:

In right triangle ABC

AC2 = AB2 + BC2  [By Pythagoras Theorem]

AC2 = 62 + 82

AC2 = 100

AC = √100

AC = 10 cm

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

Solution:

In right triangle ABC

AC2 = AB2 + BC2  [By Pythagoras Theorem]

132 = AB2 + 52

AB2 = 132 – 52

AB2 = (13+5) (13-5)

AB2 = 18 × 8

AB2 = 144

AB = √144

AB = 12 cm

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

Solution:

Let the number of rows and columns be x.

Total number of plants = x2

x2 = 1000 

x = √1000

Here the remainder is 39

So the 312 is less than 1000.

Next number is 32 

Where, 322 = 1024

Hence the number to be added = 1024 – 1000 = 24

Hence, the minimum number of plants required by him = 24.

Question 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 and columns be x.

Total number of plants = x2

x2 = 500

x = √500

Here the remainder is 16

New Number 500 – 16 = 484

and, √484 = 22

Thus, 16 students will be left out in this arrangement.




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