GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

   

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

Question 1.
Find the square root of each of the following numbers by Division method?

  1. 2304
  2. 4489
  3. 3481
  4. 529
  5. 3249
  6. 1369
  7. 5776
  8. 7921
  9. 576
  10. 1024
  11. 3136
  12. 900

Solution:
1. We have:
∴ \(\sqrt{2304}\) = 48
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 1

2. We have:
∴ \(\sqrt{4489}\) = 67
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 2

3. We have:
∴ \(\sqrt{3481}\) = 59
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 3

4. We have:
∴ \(\sqrt{529}\) = 23
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 4

5. We have:
∴ \(\sqrt{3249}\) = 57
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 5

6. We have:
∴ \(\sqrt{1369}\) = 37
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 6

7. We have:
∴ \(\sqrt{5776}\) = 76
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 7

8. We have:
∴ \(\sqrt{7921}\) = 89
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 8

9. We have:
∴ \(\sqrt{576}\) = 24
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 9

10. We have:
∴ \(\sqrt{1024}\) = 32
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 10

11. We have:
∴ \(\sqrt{3136}\) = 56
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 11

12. We have:
∴ \(\sqrt{900}\) = 30
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 12

GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation).

  1. 64
  2. 144
  3. 4489
  4. 27225
  5. 390625

Solution:
If ‘V’ stands for number of digits in the given number, then

1. For 64, n = 2 [even number]
∴ Number of digit is its square root
= \(\frac{n}{2}\) = \(\frac{2}{2}\) = 1

2. For 144, n = 3 [odd number]
∴ Number of digits in its square root
= \(\frac{n+1}{2}\) = \(\frac{3+1}{2}\) = \(\frac{4}{2}\) = 2

3. For 4489, n = 4 [even number]
Number of digits in its square root
= \(\frac{n}{2}\) = \(\frac{4}{2}\) = 2

4. For 27225, n = 5 [odd number]
∴ Number of digits in its square root
= \(\frac{n+1}{2}\) = \(\frac{5+1}{2}\)
= \(\frac{6}{2}\) = 3

5. For 390625, n = 6 [even number]
∴ Number of digits in its square root
= \(\frac{n}{2}\) = \(\frac{6}{2}\) = 3

GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

Question 3.
Find the square root of the following decimal numbers?

  1. 2.56
  2. 7.29
  3. 51.84
  4. 42.25
  5. 31.36

Solution:
1. \(\sqrt{2.56}\)
Here, number of decimal places, are already even.
∴ We mark off’the periods and find the square root.
∴ \(\sqrt{2.56}\) = 1.6
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 13

2. \(\sqrt{7.29}\)
Here, number of decimal places are already even. Therefore, we mark off the periods and find the square root.
∴ \(\sqrt{7.29}\) = 2.7
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 14

3. \(\sqrt{51.84}\)
Here, the decimal places are already even.
∴ We mark off the periods and find the square root:
∴ \(\sqrt{51.84}\) = 7.2
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 15

4. \(\sqrt{42.25}\)
Here, the decimal places are already even.
∴ We mark off periods and find the square root:
∴ \(\sqrt{42.25}\) = 6.5
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 16

5. \(\sqrt{31.36}\)
Here, the decimal places are already even.
∴ We mark off the periods and find the square root:
∴ \(\sqrt{31.36}\) = 5.6
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 17

GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

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?

  1. 402
  2. 1989
  3. 3250
  4. 825
  5. 4000

Solution:
1. Since, we get a remainder 2
∴ The required least number to be subtracted from 402 is 2.
∴ 402 – 2 = 400, and \(\sqrt{400}\) = 20
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 18

2. Since, we get a remainder of 53
∴ The least number to be subtracted from the given number = 53
1989 – 53 = 1936,
and \(\sqrt{1936}\) = 44
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 19

3. Since, we get a remainder 1
∴ The smallest number to 53250 be subtracted from the given number = 1
Now, 3250 – 1 = 3249,
and \(\sqrt{3249}\) = 57
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 20

4. Since, we get a remainder 41
∴ The required smallest number to be subtracted from the given number = 41
Now, 825 – 41 = 784,
and \(\sqrt{784}\) = 28
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 21

5. Since, we get a remainder 31
∴ The required smallest number to be subtracted from the given number = 31
Now, 4000 – 31 = 3969,
and \(\sqrt{3969}\) = 63
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 22

GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

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?

  1. 525
  2. 1750
  3. 252
  4. 1825
  5. 6412

Solution:
1. Since, we get a remainder 41.
i.e; 525 > 222
and next square number is 23.
The required number to be added
= 232 – 525
= 529 – 525 = 4
Now, 524 + 4 = 529, and \(\sqrt{529}\) = 23.
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 23

2. Since, we get a remainder 69.
i.e., 1750 > (41)2
and next square number is 422.
∴ The required number to be added = 422 – 1750
= 1764 – 1750 = 14
Now, 1750 + 14 = 1764, and \(\sqrt{1764}\)
and \(\sqrt{1764}\) = 42
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 24

3. Since, we get a remainder 27.
Since, 252 > (15)2 and next
square number = 16
∴ The required number to be added = 162 – 252
= 256 – 252 = 4
Now, 252 + 4 = 256, and \(\sqrt{256}\) = 16
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 25

4. Since, we get a remainder, 61.
∴ 1825 > (42)2
∵ Next square number = 43
∴ The required number to be added = (43)2 – 1825
= 1849 – 1825 = 24
Now, 1825 + 24 = 1849
and \(\sqrt{1849}\) = 43
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 26

5. Since, we get a remainder 12.
∴ 6412 > (80)2
∵ Next square number = 81
∴ Required number to be added = (81)2 – 6412
= 6561 – 6412 = 149
Now, 6412 + 149 = 6561 and \(\sqrt{6561}\) = 81

GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

Question 6.
Find the length of the side of a square whose area is 441 m2?
Solution:
Let the side of the square = x metre
∴ Area = side × side
= x × x = x2 metre2
∴ x2 = 441 ⇒ \(\sqrt{x^{2}}=\sqrt{441}\)
x = \(\sqrt{441}\) = 21
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 28
Thus, the required side is 21m.

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:
I. In a right triangle, the the hypotenuse.
II. (Hypotenuse)2 = [Sum of the squares of other two sides]
(a) ∵ ∠B = 90°
∴ Hypotenuse = AC
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 29
∴ AC2 = AB2 + BC2 = 82 + 62 = 64 = 100
\(\sqrt{A C^{2}}\) = \(\sqrt{100}\)
AC = 10
Thus, AC = 10 cm

(b) Here ∠B = 90°
∴ Hypotenuse = AC
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 30
∵ AC2 = AB2 + BC2 or 132 = AB2 + 52
or AB2 = 132 – 52 = 169 – 25 = 144
Now \(\sqrt{A B^{2}}\) = \(\sqrt{144}\) or AB = 12
Thus, 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?
Solution:
Since, the number of plants in a row and the number of columns are the same.
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 31
∴ Their product must be a square number.
∵ The gardener has 1000 plants.
∴ 1000 is not a perfect square, and (31)2 < 1000
(∵ There is a remainder of 39).
Obviously the next square number = 32
∴ Number of plants required to be added
= (32)2 – 1000
= 1024 – 1000 = 24

GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4

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 number of columns. How many children would be left out in this arrangement?
Solution:
Since, the number of rows and the number of columns are same.
GSEB Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 img 32
∴ Total number (i.e. their product) must be a square number, we have
Since, we get a remainder of 16
∴ 500 > (22)2 or 500 – 16 = (22)2
Thus, the required number of children to be left out = 16.

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