CBSE Class 8 - Maths - CH 6 - Square and Square Roots (Ex 6.4)

Square and Square roots

Exercise 6.4

Q1: 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

Answer:

(i)The square root of 2304 is calculated as follows.

	48
4	23 04 -16
88	704 704
	0

∴√ 2304  = 48

(ii)The square root of 4489 is calculated as follows.

	67
6	44 89 -36
127	889 889
	0

∴√ 4489  = 67

(iii)The square root of 3481 is calculated as follows.

	59
5	34 81 -25
109	981 981
	0

∴√ 3481  = 59

(iv)The square root of 529 is calculated as follows.

	23
5	5 29 -4
43	129 129
	0

∴√ 529  = 23

(v)The square root of 3249 is calculated as follows.

	57
5	32 49 -25
107	749 749
	0

∴√ 3249  = 57

(vi)The square root of 1369 is calculated as follows.

	37
3	13 69 -9
67	469 469
	0

∴√ 1369  = 37

(vii)The square root of 5776 is calculated as follows.

	76
7	57 76 -49
146	876 876
	0

∴√ 5776  = 76



(viii)The square root of 7921 is calculated as follows.

	89
8	79 21 -64
169	1521 1521
	0

∴√ 7921  = 89

(ix)The square root of 576 is calculated as follows.

	24
2	5 76 -4
44	176 176
	0

∴√ 576  = 24

(x)The square root of 1024 is calculated as follows.

	32
3	10 24 -9
62	124 124
	0

∴√ 1024  = 32

(xi)The square root of 3136 is calculated as follows.

	56
5	31 36 -25
106	636 636
	0

∴√ 3136  = 56

(xii)The square root of 900 is calculated as follows.

	30
3	9 00 -9
60	00 00
	0

∴√ 900  = 30

Q2: 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

Answer:

(i) Let us place bars and pair the digits from right to left, we get

64 =  64 

Since there is only one bar, the square root of 64 will have only one digit in it.

(ii) Let us place bars, we get

144 =  1   44 

Since there are two bars, the square root of 144 will have two digits in it.

(iii)By pairing digits from right to left, we have

4489 =  44   89 

Since there are two bars, the square root of 4489 will be of two digits.

(iv) By placing bars, we obtain

27225 =  2   72   25 

Since there are three bars, the square root of 27225 will be of three digits.

(v) By placing bars, we get

390625 =  39   06   25 

Since there are three bars, the square root of 27225 will be of three digits.


Q3: Find the square root of the following decimal numbers.

(i) 2.56  (ii) 7.29  (iii) 51.84  (iv) 42.25  (v) 31.36

Answer:

(i) The square root of 2.56 is calculated as follows.

	1.6
1	2 .56 -1
26	156 156
	0

⇒√ 2.56  = 1.6

(ii)The square root of 7.29 is calculated as follows.

	2.7
2	7 .29 -4
47	329 329
	0

⇒√ 7.29  = 2.7

(iii) The square root of 51.84 is calculated as follows.

	7.2
7	51 .84 -49
142	284 284
	0

⇒√ 51.84  = 7.2

(iv) The square root of 42.25 is calculated as follows.

	6.5
6	42 .25 -36
125	625 625
	0

⇒√ 42.25  = 6.5

(v) The square root of 31.36 is calculated as follows.

	5.6
5	31 .36 -25
106	636 636
	0

⇒√ 31.36  = 5.6


Q4: 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

Answer:

(i) The square root of 402 is calculated as follows:

	20
2	4 02 -4
40	02 00
	2

The remainder is 2. 
⇒ That the square of 20 is less than 402 by 2. 
∴  the required perfect square = 402 − 2 = 400
i.e. √ 400  = 20
(ii) The square root of 1989 is calculated as:

	44
4	19 89 -16
84	389 336
	53

The remainder is 53. 
⇒ That the square of 44 is less than 1989 by 53. 
∴ required perfect square = 1989 − 53 = 1936

i.e. √ 1936  = 44

(iii) The square root of 3250 is calculated as follows:

	57
5	32 50 -25
107	750 749
	1

The remainder is 1. 
⇒ that the square of 57 is less than 3250 by 1.

∴ required perfect square = 3250 − 1 = 3249

i.e. √ 3249  = 57


(iv) The square root of 825 is calculated as follows:

	28
2	8 25 -4
48	425 384
	41

The remainder is 41. 
⇒ that the square of 28 is less than 825 by 41. 

∴ required perfect square = 825 − 41 = 784

⇒ √ 784  = 28

(v) The square root of 4000 is calculated as follows:

	63
6	40 00 -36
123	400 369
	31

The remainder is 31.

⇒ the required perfect square = 4000 − 31 = 3969

⇒ √ 3969  = 63


Q5: 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

Answer:

(i)The square root of 525 is calculated as follows:

	22
2	5 25 -4
42	125 84
	41

The remainder is 41.
And the next number is 23 ⇒ 23² = 529

∴ number to be added to 525 = 23² − 525 = 529 − 525 = 4

∴ The required perfect square is 529  i.e. √ 529  = 23

(ii) The square root of 1750 is calculated as follows:

	41
4	17 50 -16
81	150 81
	69

The remainder is 69 and the next number is 42 
⇒ 42² = 1764

∴ number to be added to 1750 = 42² − 1750 = 1764 − 1750 = 14

∴ The required perfect square is 1764 and √ 1764  = 42


(iii) The square root of 252 is calculated as:

	15
1	2 52 -1
25	152 125
	27

The remainder is 27 and the next number is 16  i.e. 16² = 256

∴ the number to be added to 252 = 16² − 252 = 256 − 252 = 4

∴ The required perfect square is = 256 i.e. √ 256  = 16

(iv) The square root of 1825 is calculated as follows:

	42
4	18 25 -16
82	225 164
	61

The remainder = 61. 
The next number = 43 i.e. 43² = 1849

∴ the number to be added = 1825 = 43² − 1825 = 1849 − 1825 = 24

∴ The required perfect square = 1849 i.e. √ 1849  = 43

(v) The square root of 6412 is calculated as:

	80
8	64 12 -64
160	012 0
	12

The remainder is 12 and the next number is 81 ⇒ 81² = 6561

∴ the number to be added to 6412 = 81² − 6412 = 6561 − 6412 = 149

∴ The required perfect square = 6561 i.e. √ 6561  = 81

Q6: Find the length of the side of a square whose area is 441 m².

Answer:

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

∴ Area of square = (x)² = 441 m²

x = √ 441 

The square root of 441 is:

	21
2	4  41 -4
41	041 41
	0

∴ x = 21m

Thus the length of the side of the square is 21 m.

Q7: 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

Answer:

(a) Since ∆ABC is right-angled at B, ∴ applying Pythagoras theorem:

AC² = AB² + BC²

AC² = (6 cm)² + (8 cm)²

AC² = (36 + 64) cm² =100 cm²

AC = (√ 100 )cm = (√ 10 x 10 )cm

AC = 10 cm

(b) ∴ ABC is right-angled at B. By applying Pythagoras theorem:

AC² = AB² + BC²

(13 cm)² = (AB)² + (5 cm)²

AB² = (13 cm)² − (5 cm)² = (169 − 25) cm² = 144 cm²

AB = (√ 144 )cm = (√ 12 x 12 )cm

AB = 12 cm

Q8: 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.

Answer:

It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same.

Let the number of rows = number of cols = x

⇒ x² = perfect square number  
⇒ The number should be added to 1000 to make it a perfect square.

Let x = 
The Let x = √1000 i.e.

	31
3	10 00 -9
61	100 61
	39

The remainder is 39 i.e. the square of 31 is less than 1000.

The next number after 31 is 32 ⇒ 32² = 1024

∴ the number to be added to 1000 to make it a perfect square

= 32² − 1000 = 1024 − 1000 = 24

Thus, the required number of plants is 24.

Q9: These 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?

Answer:

Given total children in a school = 500 
For P.T. drill, the number of rows = number of columns = x.

The number of children who will be left out in PT drill = 500 - x² 

The square root of 500 is as follows:

	22
2	5 00 -4
42	100 84
	16

The remainder is 16. i.e. the square of 22 is less than 500 by 16. 

To make a perfect square = 500 − 16 = 484

∴ the number of children who will be left out is 16.