Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
Exercise 6.4
Question 1
Find the square root of each of the following numbers by division method.
(i) 2304
Sol :
(i) The square root of 2304 can be calculated as follows.
48 

4 

88 
704 704 
0 
∴
(ii) 4489
Sol :
(ii) The square root of 4489 can be calculated as follows.
67 

6 

127 
889 889 
0 
∴
(iii) 3481
Sol :
(iii) The square root of 3481 can be calculated as follows.
59 

5 

109 
981 981 
0 
Therefore,
(iv) 529
Sol :
(iv) The square root of 529 can be calculated as follows.
23 

2 

43 
129 129 
0 
∴
(v) 3249
Sol :
(v) The square root of 3249 can be calculated as follows.
57 

5 

107 
749 749 
0 
∴
(vi) 1369
Sol :
(vi) The square root of 1369 can be calculated as follows.
37 

3 

67 
469 469 
0 
∴
(vii) 5776
Sol :
(vii) The square root of 5776 can be calculated as follows.
76 

7 

146 
876 876 
0 
∴
(viii) 7921
Sol :
(viii) The square root of 7921 can be calculated as follows.
89 

8 

169 
1521 1521 
0 
∴
(ix) 576
Sol :
(ix) The square root of 576 can be calculated as follows.
24 

2 

44 
176 176 
0 
∴
(x) 1024
Sol :
(x) The square root of 1024 can be calculated as follows.
32 

3 

62 
124 124 
0 
∴
(xi) 3136
Sol :
(xi) The square root of 3136 can be calculated as follows.
56 

5 

106 
636 636 
0 
∴
(xii) 900
Sol :
(xii) The square root of 900 can be calculated as follows.
30 

3 

60 
00 00 
0 
∴
Question 2
Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64
Sol :
(i) By placing bars, we obtain
Since there is only one bar, the square root of 64 will have only one digit in it.
(ii) 144
Sol :
(ii) By placing bars, we obtain
Since there are two bars, the square root of 144 will have 2 digits in it.
(iii) 4489
Sol :
(iii) By placing bars, we obtain
Since there are two bars, the square root of 4489 will have 2 digits in it.
(iv) 27225
Sol :
(iv) By placing bars, we obtain
Since there are three bars, the square root of 27225 will have three digits in it.
(v) 390625
Sol :
(v) By placing the bars, we obtain
Since there are three bars, the square root of 390625 will have 3 digits in it.
Question 3
Find the square root of the following decimal numbers.
(i) 2.56
Sol :
(i) The square root of 2.56 can be calculated as follows.
1. 6 

1 

26 
156 156 
0 
∴
(ii) 7.29
Sol :
(ii) The square root of 7.29 can be calculated as follows.
2. 7 

2 

47 
329 329 
0 
∴
(iii) 51.84
Sol :
(iii) The square root of 51.84 can be calculated as follows.
7.2 

7 

142 
284 284 
0 
∴
(iv) 42.25
Sol :
(iv) The square root of 42.25 can be calculated as follows.
6.5 

6 

125 
625 625 
0 
∴
(v) 31.36
Sol :
(v) The square root of 31.36 can be calculated as follows.
5.6 

5 

106 
636 636 
0 
∴
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
Sol :
(i) The square root of 402 can be calculated by long division method as follows.
20 

2 

40 
02 00 
2 
The remainder is 2. It represents that the square of 20 is less than 402 by 2. Therefore, a perfect square will be obtained by subtracting 2 from the given number 402.
Therefore, required perfect square = 402 − 2 = 400
And,
(ii) 1989
Sol :
(ii) The square root of 1989 can be calculated by long division method as follows.
44 

4 

84 
389 336 
53 
The remainder is 53. It represents that the square of 44 is less than 1989 by 53. Therefore, a perfect square will be obtained by subtracting 53 from the given number 1989.
Therefore, required perfect square = 1989 − 53 = 1936
And,
(iii) 3250
Sol :
(iii) The square root of 3250 can be calculated by long division method as follows.
57 

5 

107 
750 749 
1 
The remainder is 1. It represents that the square of 57 is less than 3250 by 1. Therefore, a perfect square can be obtained by subtracting 1 from the given number 3250.
Therefore, required perfect square = 3250 − 1 = 3249
And,
(iv) 825
Sol :
(iv) The square root of 825 can be calculated by long division method as follows.
28 

2 

48 
425 384 
41 
The remainder is 41. It represents that the square of 28 is less than 825 by 41. Therefore, a perfect square can be calculated by subtracting 41 from the given number 825.
Therefore, required perfect square = 825 − 41 = 784
And,
(v) 4000
Sol :
(v) The square root of 4000 can be calculated by long division method as follows.
63 

6 

123 
400 369 
31 
The remainder is 31. It represents that the square of 63 is less than 4000 by 31. Therefore, a perfect square can be obtained by subtracting 31 from the given number 4000.
Therefore, required perfect square = 4000 − 31 = 3969
And,
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
Sol :
(i) The square root of 525 can be calculated by long division method as follows.
22 

2 

42 
125 84 
41 
The remainder is 41.
It represents that the square of 22 is less than 525.
Next number is 23 and 23^{2} = 529
Hence, number to be added to 525 = 23^{2} − 525 = 529 − 525 = 4
The required perfect square is 529 and
(ii) 1750
Sol :
(ii) The square root of 1750 can be calculated by long division method as follows.</p
41 

4 

81 
150 81 
69 
The remainder is 69.
It represents that the square of 41 is less than 1750.
The next number is 42 and 42^{2} = 1764
Hence, number to be added to 1750 = 42^{2} − 1750 = 1764 − 1750 = 14
The required perfect square is 1764 and
(iii) 252
Sol :
(iii) The square root of 252 can be calculated by long division method as follows.
15 

1 

25 
152 125 
27 
The remainder is 27. It represents that the square of 15 is less than 252.
The next number is 16 and 16^{2} = 256
Hence, number to be added to 252 = 16^{2} − 252 = 256 − 252 = 4
The required perfect square is 256 and
(iv) 1825
Sol :
(iv) The square root of 1825 can be calculated by long division method as follows.
42 

4 

82 
225 164 
61 
The remainder is 61. It represents that the square of 42 is less than 1825.
The next number is 43 and 43^{2} = 1849
Hence, number to be added to 1825 = 43^{2} − 1825 = 1849 − 1825 = 24
The required perfect square is 1849 and
(v) 6412
Sol :
(v) The square root of 6412 can be calculated by long division method as follows.
80 

8 

160 
012 0 
12 
The remainder is 12.
It represents that the square of 80 is less than 6412.
The next number is 81 and 81^{2} = 6561
Hence, number to be added to 6412 = 81^{2} − 6412 = 6561 − 6412 = 149
The required perfect square is 6561 and
Question 6
Find the length of the side of a square whose area is 441 m^{2}.
Sol :
Let the length of the side of the square be x m.
Area of square = (x)^{2} = 441 m^{2}
The square root of 441 can be calculated as follows.
21 

2 

41 
041 41 
0 
∴
Hence, the length of the side of the square is 21 m.
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
Sol :
(a) ΔABC is rightangled at B.
Therefore, by applying Pythagoras theorem, we obtain
AC^{2} = AB^{2} + BC^{2}
AC^{2} = (6 cm)^{2} + (8 cm)^{2}
AC^{2} = (36 + 64) cm^{2} =100 cm^{2}
AC =
AC = 10 cm
(b) ΔABC is rightangled at B.
Therefore, by applying Pythagoras theorem, we obtain
AC^{2} = AB^{2} + BC^{2}
(13 cm)^{2} = (AB)^{2} + (5 cm)^{2}
AB^{2} = (13 cm)^{2} − (5 cm)^{2} = (169 − 25) cm^{2} = 144 cm^{2}
AB =
AB = 12 cm
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 same. Find the minimum number of plants he needs more for this.
Sol :
It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same.
We have to find the number of more plants that should be there, so that when the gardener plants them, the number of rows and columns are same.
That is, the number which should be added to 1000 to make it a perfect square has to be calculated.
The square root of 1000 can be calculated by long division method as follows.
31 

3 

61 
100 61 
39 
The remainder is 39. It represents that the square of 31 is less than 1000.
The next number is 32 and 32^{2} = 1024
Hence, number to be added to 1000 to make it a perfect square
= 32^{2} − 1000 = 1024 − 1000 = 24
Thus, the required number of plants is 24.
Question 9
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?
Sol :
It is given that there are 500 children in the school. They have to stand for a P.T. drill such that the number of rows is equal to the number of columns.
The number of children who will be left out in this arrangement has to be calculated. That is, the number which should be subtracted from 500 to make it a perfect square has to be calculated.
The square root of 500 can be calculated by long division method as follows.
22 

2 

42 
100 84 
16 
The remainder is 16.
It shows that the square of 22 is less than 500 by 16. Therefore, if we subtract 16 from 500, we will obtain a perfect square.
Required perfect square = 500 − 16 = 484
Thus, the number of children who will be left out is 16.