Sunday, January 12, 2014

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
423 04
-16
88704
704

0

∴√ 2304  = 48

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



67
644 89
-36
127889
889

0

∴√ 4489  = 67

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



59
534 81
-25
109981
981

0

∴√ 3481  = 59

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



23
55 29
-4
43129
129

0

∴√ 529  = 23

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



57
532 49
-25
107749
749

0

∴√ 3249  = 57

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



37
313 69
-9
67469
469

0

∴√ 1369  = 37

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



76
757 76
-49
146876
876

0

∴√ 5776  = 76



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



89
879 21
-64
1691521
1521

0

∴√ 7921  = 89

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



24
25 76
-4
44176
176

0

∴√ 576  = 24

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



32
310 24
-9
62124
124

0

∴√ 1024  = 32

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



56
531 36
-25
106636
636

0

∴√ 3136  = 56

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



30
39 00
-9
6000
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
12 .56
-1
26156
156

0

⇒√ 2.56  = 1.6

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



2.7
27 .29
-4
47329
329

0

⇒√ 7.29  = 2.7

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



7.2
751 .84
-49
142284
284

0

⇒√ 51.84  = 7.2

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



6.5
642 .25
-36
125625
625

0

⇒√ 42.25  = 6.5

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



5.6
531 .36
-25
106636
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
24 02
-4
4002
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
419 89
-16
84389
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
532 50
-25
107750
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
28 25
-4
48425
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
640 00
-36
123400
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
25 25
-4
42125
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
417 50
-16
81150
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
12 52
-1
25152
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
418 25
-16
82225
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
864 12
-64
160012
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
41041
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
310 00
-9
61100
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
25 00
-4
42100
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.

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