## 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

(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

(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

(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

(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

(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.

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

(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.

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?

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.  #### 1 comment:

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