**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 Ex 6.4 are the part of NCERT Solutions for Class 8 Maths. Here you can
find the 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 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:**

**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)

Thus, 2 is the least number to be subtracted from 402 to get a perfect square.

New square number = 402 – 2 = 400

Thus, √400 = 20

(ii)

Thus, 53 is the least number to be subtracted from 1989 to get a perfect square.

New square number = 1989 – 53 = 1936

Thus, √1936 = 44

(iii)

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

New square number = 3250 – 1 = 3249

Thus, √3249 = 57

(iv)

Thus, 41 is the least number to be subtracted from 825 to get a perfect square.

New square number = 825 – 41 = 784

Thus, √784 = 28

(v)

Thus, 31 is the least number to be subtracted from 4000 to get a perfect square.

New square 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)

It represents that the square of 22 is less than 525.

The next square number is 23 and 23

^{2}= 529

Hence, the number to be added to 525 to get a perfect square = 529 – 525 = 4

New perfect square number = 529

Thus, √529 = 23

(ii)

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

The next square number is 42 and 42

^{2}= 1764

Hence, the number to be added to 1750 to get a perfect square = 1764 – 1750 = 14

New perfect square number = 1764

Thus, √1764 = 42

(iii)

Here, the remainder is 27.

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

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

Hence, the number to be added to 252 to get a perfect square = 256 – 252 = 4

New perfect square number = 252 + 4 = 256

Thus, √256 = 16

(iv)

Here, the remainder is 61.

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

The next square number is 43 and 43^{2} = 1849

Hence, the number to be added to 1825 to get a perfect square = 1849 – 1825 =
24

New perfect square number = 1849

Thus, √1849 = 43

(v)

Here, the remainder is 12.

It represents that the square of 80 is less than 6412.

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

Hence, the number to be added to 6412 to get a perfect square = 6561 – 6412 =
149

New perfect square number = 6561

Thus, √6561 = 81

**Ex 6.4 Class 8
Maths Question 6.**

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

**Solution:
**Let the length of each side
of the square be x m.

Area of the square = (side)

^{2}= x

^{2}m

^{2}

x

^{2}= 441 ⇒ x = √441 = 21

Hence, the length of the side of the square is 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 a right-angled triangle ABC,

AC

^{2}= AB

^{2}+ BC

^{2}[By Pythagoras Theorem]

^{2}= (6)

^{2}+ (8)

^{2}= 36 + 64 = 100

⇒ AC = √100 = 10

Thus, AC = 10 cm

(b) In a right-angled triangle ABC,

^{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}

or, 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.
Then the number of columns also be x.

Total number of plants = x × x = x

^{2}

x

^{2}= 1000 ⇒ x = √1000

So, the square of 31 is less than 1000.

Next square number is 32 and 32

^{2}= 1024

Hence, the number to be added to 1000 to get a perfect square = 1024 – 1000 = 24

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

**Alternative method:**

**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. Then the number of children in a column also be x.

Total number of children = x × x = x

^{2}

x

^{2}= 500 ⇒ x = √500

New number = 500 – 16 = 484

And, √484 = 22

Thus, 16 children would be left out in this arrangement.

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