### NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 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 = n/2 if n is even and n+1/2 if n is odd.

(i) 64

Here n = 2 (even)

Number of digits in √64 = 2/2 = 1

(ii) 144

Here n = 3 (odd)

Number of digits in square root = 3+1/2 = 2

(iii) 4489

Here n = 4 (even)

Number of digits in square root = 4/2 = 2

(iv) 27225

Here n = 5 (odd)

Number of digits in square root = 5+1/2 = 3

(iv) 390625

Here n = 6 (even)

Number of digits in square root = 6/2 = 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 23² = 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 42² = 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 16² = 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 43² = 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 81² = 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 m²**

**Solution:**

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

Area of the square = (side)² = x² m²

x² = 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

AC2 = AB² + BC² [By Pythagoras Theorem]

⇒ AC² = (6)² + (8)² = 36 + 64 = 100

⇒ AC = √100 = 10

Thus, AC = 10 cm.

(b) In right triangle ABC

AC² = AB² + BC² [By Pythagoras Theorem]

⇒ (13)² = AB² + (5)²

⇒ 169 = AB² + 25

⇒ 169 – 25 = AB²

⇒ 144 = AB²

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 = x²

x² = 1000 ⇒ x = √1000

Here the remainder is 39

So the square of 31 is less than 1000.

Next number is 32 and 32² = 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 = 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.