These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.4

Question 1.

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

(i) 2304

(ii) 4489

(iii) 3481

(iv) 529

(v) 3249

(vi) 1369

(vii) 5776

(viii) 7921

(ix) 576

(x) 1024

(xi) 3136

(xii) 900

Solution:

(i) 2304

∴ √2304 = 48

(ii) 4489

∴ √4489 = 67

(iii) 3481

∴ √3481 = 59

(iv) 529

∴ √529 = 23

(v) 3249

∴ √3249 = 57

(vi) 1369

∴ √1369 = 37

(vii) 5776

∴ √5776 = 76

(viii) 7921

∴ √7921 = 89

(ix) 576

∴ √576 = 24

(x) 1024

∴ √1024 = 32

(xi) 3136

∴ √3136 = 56

(xii) 900

∴ √900 = 30

Question 2.

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

(i) 64

(ii) 144

(iii) 4489

(iv) 27225

(v) 390625

Solution:

If n stands fnr nnmher of dibits in the given number then

(i) For 64

Here, n = 2 (even number)

Number of digits in the square root = \(\frac{n}{2}=\frac{2}{2}=1\)

(ii) For 144

Here, n = 3 (odd number)

Number of digits in the, square root = \(\frac{n+1}{2}=\frac{3+4}{2}=\frac{4}{2}=2\)

(iii) For 4489

Here, n = 4 (even number)

Number of digits in the square root = \(\frac{\mathrm{n}}{2}=\frac{4}{2}=2\)

(iv) For 27225

Here, n = 5 (odd number)

Number of digits in the square root = \(\frac{\mathrm{n}+1}{2}=\frac{5+1}{2}=\frac{6}{2}=3\)

(v) For 390625

Here, n = 6 (even number)

Number of digits in the square root = \(\frac{n}{2}=\frac{6}{2}=3\)

Question 3.

Find the square root of the following decimal numbers.

(i) 2.56

(ii) 7.29

(iii) 51.84

(iv) 42.25

(v) 31.36

Solution:

(i) 2.56

∴ √2.56 = 1.6

(ii) 7.29

∴ √7.29 = 2.7

(iii) 51.84

∴ √51.84 = 7.2

(iv) 42.25

∴ √42.25 = 6.5

(v) 31.36

∴ √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.

(i) 402

(ii) 1989

(iii) 3250

(iv) 825

(v) 4000

Solution:

(i) 402

The required least number to be subtracted from 402 is 2.

∴ 402 – 2 = 400

∴ √400 = 20

(ii) 1989

The remainder is 53

∴ The required least numbers to be subtracted from the given number is 53.

1989 – 53 = 1936

∴ √1936 = 44

(iii) 3250

The remainder is 1

∴ The required least number to be subtracted from the given number is 1.

3250 – 1 = 3249

∴ √3249 = 57

(iv) 825

The remainder is 41

The required least number to be subtracted from the given number is 41.

825 – 41 = 784

∴ √784 = 28

(v) 4000

The remainder is 31

The required least number to be subtracted from the given number is 31.

4000 – 31 = 3969

∴ √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

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

Solution:

(i) 525

The remainder is 41.

i.e. 525 > 22^{2} and the next perfect square number is 23^{2} = 529

The required number to be added = 23^{2} – 525

= 529 – 525

= 4

Hence, 4 should be added to 525 to get a perfect square

number.

525 + 4 = 529

∴ √529 = 23

(ii) 1750

The remainder is 69.

1750 > 41^{2} and the next perfect square number is 42^{2} = 1764

The required number to be added = 42^{2} – 1750

= 1764 – 1750

= 14

The next perfect square number = 1750 + 14 = 1764

∴ √1764 = 42

(iii) 252

The remainder is 27.

This shows that 15^{2} < 252

The next perfect square number is 16^{2} = 256.

The required number to be added = 16^{2} – 252

= 256 – 252

= 4

The perfect square number = 252 + 4 = 256

∴ √256 = 16

(iv) 1825

The remainder is 61.

This shows that 42^{2} < 1825

The next perfect square is 43^{2} = 1849

Hence, the required number to be added = 43^{2} – 1825

= 1849 – 1825

= 24

The perfect square number so obtained 1825 + 24 = 1849

Hence, √1849 = 43

(v) 6412

The remainder is 12.

This shows that 80^{2} < 6412

The next perfect square number is 81^{2} = 6561

The required number to be added = 6561 – 6412 = 149

The perfect square number = 6412 + 149 = 6561

Hence, √6561 = 81

Question 6.

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

Solution:

Let x be the side of a square.

∴ Area of square = x^{2}

Given, Area of square = 441

x^{2} = 441

⇒ x = √441 = 21

So, length of the side of the square = 21 m.

Question 7.

In a right triangle ABC, ∠B = 90°.

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

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

Solution:

(a) In the right triangle ABC, ∠B = 90°

By Pythagoras Theorem,

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

⇒ AC^{2} = 6^{2} + 8^{2}

⇒ AC^{2} = 36 + 64

⇒ AC^{2} = 100

⇒ AC = √100 = 10

∴ Length of AC = 10 cm

(b) In the right triangle ABC, ∠B = 90°

By Pythagoras theorem,

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

⇒ AB^{2} = AC^{2} – BC^{2}

⇒ AB^{2} = 13^{2} – 5^{2}

⇒ AB^{2} = 169 – 25

⇒ AB^{2} = 144

⇒ AB = √144 = 12

∴ Length of 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 be ‘x’

The number of columns is x

The number of plants = x × x = x^{2}

which is a perfect square

Let us find out the square root of 1000 by division method

The remainder is 39

This shows that 31^{2} < 1000

The next perfect square number is 32^{2} = 1024

Minimum number of plants he needs besides 100 plants = 32^{2} – 1000

= 1024 – 1000

= 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 be x and the number of columns is x

∴ Number of children = x × x = x^{2}

which is a perfect square

∴ 500 = x^{2}

The remainder is 16.

Number of children left out in this arrangement = 16.