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?

- 2304
- 4489
- 3481
- 529
- 3249
- 1369
- 5776
- 7921
- 576
- 1024
- 3136
- 900

Solution:

1. We have:

∴ \(\sqrt{2304}\) = 48

2. We have:

∴ \(\sqrt{4489}\) = 67

3. We have:

∴ \(\sqrt{3481}\) = 59

4. We have:

∴ \(\sqrt{529}\) = 23

5. We have:

∴ \(\sqrt{3249}\) = 57

6. We have:

∴ \(\sqrt{1369}\) = 37

7. We have:

∴ \(\sqrt{5776}\) = 76

8. We have:

∴ \(\sqrt{7921}\) = 89

9. We have:

∴ \(\sqrt{576}\) = 24

10. We have:

∴ \(\sqrt{1024}\) = 32

11. We have:

∴ \(\sqrt{3136}\) = 56

12. We have:

∴ \(\sqrt{900}\) = 30

Question 2.

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

- 64
- 144
- 4489
- 27225
- 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

Question 3.

Find the square root of the following decimal numbers?

- 2.56
- 7.29
- 51.84
- 42.25
- 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

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

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

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

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

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?

- 402
- 1989
- 3250
- 825
- 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

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

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

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

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

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?

- 525
- 1750
- 252
- 1825
- 6412

Solution:

1. Since, we get a remainder 41.

i.e; 525 > 22^{2}

and next square number is 23.

The required number to be added

= 23^{2} – 525

= 529 – 525 = 4

Now, 524 + 4 = 529, and \(\sqrt{529}\) = 23.

2. Since, we get a remainder 69.

i.e., 1750 > (41)^{2}

and next square number is 42^{2}.

∴ The required number to be added = 42^{2} – 1750

= 1764 – 1750 = 14

Now, 1750 + 14 = 1764, and \(\sqrt{1764}\)

and \(\sqrt{1764}\) = 42

3. Since, we get a remainder 27.

Since, 252 > (15)^{2} and next

square number = 16

∴ The required number to be added = 16^{2} – 252

= 256 – 252 = 4

Now, 252 + 4 = 256, and \(\sqrt{256}\) = 16

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

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

Question 6.

Find the length of the side of a square whose area is 441 m^{2}?

Solution:

Let the side of the square = x metre

∴ Area = side × side

= x × x = x^{2} metre^{2}

∴ x^{2} = 441 ⇒ \(\sqrt{x^{2}}=\sqrt{441}\)

x = \(\sqrt{441}\) = 21

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

∴ AC^{2} = AB^{2} + BC^{2} = 8^{2} + 6^{2} = 64 = 100

\(\sqrt{A C^{2}}\) = \(\sqrt{100}\)

AC = 10

Thus, AC = 10 cm

(b) Here ∠B = 90°

∴ Hypotenuse = AC

∵ AC^{2} = AB^{2} + BC^{2} or 13^{2} = AB^{2} + 5^{2}

or AB^{2} = 13^{2} – 5^{2} = 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.

∴ 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

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.

∴ 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.