NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

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

Ex 6.4 Class 8 Maths Question 1.

Find the square root of each of the following numbers by Long 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:

Ex 6.4 Class 8 Maths 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:

We know that if n is number of digits in a square number then

Number of digits in the square root = n2 if n is even and n+12 if n is odd.

(i) 64

Here n = 2 (even)

Number of digits in √64 = 22 = 1

(ii) 144

Here n = 3 (odd)

Number of digits in square root = 3+12 = 2

(iii) 4489

Here n = 4 (even)

Number of digits in square root = 42 = 2

(iv) 27225

Here n = 5 (odd)

Number of digits in square root = 5+12 = 3

(iv) 390625

Here n = 6 (even)

Number of digits in square root = 62 = 3

Ex 6.4 Class 8 Maths 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:

Ex 6.4 Class 8 Maths 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)

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)

Here remainder is 53

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

New number = 1989 – 53 = 1936

Thus, √1936 = 44

(iii)

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)

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)

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

Ex 6.4 Class 8 Maths 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)

Here remainder is 41

It represents that square of 22 is less than 525.

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

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

New number = 529

Thus, √529 = 23

(ii)

Here the remainder is 69

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

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

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

Require perfect square = 1764

√1764 = 42

(iii)

Here the remainder is 27.

It represents that a square of 15 is less than 252.

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

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

New number = 252 + 4 = 256

Required perfect square = 256

and √256 = 16

(iv)

The remainder is 61.

It represents that square of 42 is less than in 1825.

Next number is 43 and 43^{2} = 1849

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

The required perfect square is 1848 and √1849 =43

(v)

Here, the remainder is 12.

It represents that a square of 80 is less than in 6412.

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

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

The require perfect square is 6561 and √6561 = 81

Ex 6.4 Class 8 Maths Question 6.

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

Solution:

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

Area of the square = (side)^{2} = x^{2} m^{2}

x^{2} = 441 ⇒ x = √441 = 21

Thus, x = 21 m.

Hence the length of the side of square = 21 m.

Ex 6.4 Class 8 Maths 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 right triangle ABC

AC^{2} = AB^{2} + BC^{2} [By Pythagoras Theorem]

⇒ AC^{2} = (6)^{2} + (8)^{2} = 36 + 64 = 100

⇒ AC = √100 = 10

Thus, AC = 10 cm.

(b) In right triangle ABC

AC^{2} = AB^{2} + BC^{2} [By Pythagoras Theorem]

⇒ (13)^{2} = AB^{2} + (5)^{2}

⇒ 169 = AB^{2} + 25

⇒ 169 – 25 = AB^{2}

⇒ 144 = AB^{2}

AB = √144 = 12 cm

Thus, AB = 12 cm.

Ex 6.4 Class 8 Maths 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 the same. Find the minimum number of plants he needs more for this.

Solution:

Let the number of rows be x.

And the number of columns also be x.

Total number of plants = x × x = x^{2}

x^{2} = 1000 ⇒ x = √1000

Here the remainder is 39

So the square of 31 is less than 1000.

Next number is 32 and 32^{2} = 1024

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

Thus the minimum number of plants required by him = 24.

Alternative method:

The minimum number of plants required by him = 24.

Ex 6.4 Class 8 Maths 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 the number of columns. How many children would be left out in this arrangement?

Solution:

Let the number of children in a row be x. And also that of in a column be x.

Total number of students = x × x = x^{2}

x^{2} = 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.