Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

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

Question (i).

2304

Solution:

Question (ii).

4489

Solution:

Question (iii).

3481

Solution:

Question (iv).

529

Solution:

Question (v).

3249

Solution:

Question (vi).

1369

Solution:

Question (vii).

5776

Solution:

Question (viii).

7921

Solution:

Question (ix).

576

Solution:

Question (x).

1024

Solution:

Question (xi).

3136

Solution:

Question (xii).

900

Solution:

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

Question (i).

64

Solution:

Here, number of digits, n = 2

(Which is an even number.)

∴ Number of digits in the square root of 64 = \(\frac{n}{2}=\frac{2}{2}\) = 1

Question (ii).

144

Solution:

Here, number of digits, n = 3

(Which is an odd number.)

∴ Number of digits in the square root of 144 = \(\frac{n+1}{2}=\frac{3+1}{2}\)

= \(\frac {4}{2}\)

= 2

Question (iii).

4489

Solution:

Here, number of digits, n = 4

(Which is an even number.)

∴ Number of digits in the square root of 4489 = \(\frac{n}{2}=\frac{4}{2}\)

= 2

Question (iv).

27225

Solution:

Here, number of digits, n = 5

(Which is an odd number.)

∴ Number of digits in the square root of 27225 = \(\frac{n+1}{2}=\frac{5+1}{2}\)

= \(\frac {6}{2}\)

= 3

Question (v).

390625

Solution:

Here, number of digits, n = 6

(Which is an even number.)

∴ Number of digits in the square root of 390625 = \(\frac{n}{2}=\frac{6}{2}\)

= 3

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

Question (i).

2.56

Solution:

Here, number of decimal places are two.

∴ The number of decimal places in square root should be one.

Question (ii).

7.29

Solution:

Here, number of decimal places are two.

∴ The number of decimal places in square root should be one.

Question (iii).

51.84

Solution:

Here, number of 51.84 decimal places are two.

∴ The number of decimal places in square root should be one.

Question (iv).

42.25

Solution:

Question (v).

31.36

Solution:

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.

Question (i).

402

Solution:

Here, the remainder is 2. It shows that 20^{2} is less than 402 by 2.

So, to get a perfect square, 2 must be subtracted from given number.

∴ Required perfect square number = 402 – 2 = 400

\(\sqrt{400}\) = 20

Question (ii).

1989

Solution:

Here, the remainder is 53. It shows that 44^{2} is less than 1989 by 53.

So, to get a perfect square, 53 must be subtracted from the given number.

∴ Required perfect square number = 1989 – 53 = 1936

\(\sqrt{1936}\) = 44

Question (iii).

3250

Solution:

Here, the remainder is 1. It shows that 57^{2} is less them 3250 by 1.

So, to get a perfect square, 1 must be subtracted from the given number.

∴ Required perfect square number = 3250 – 1 = 3249

\(\sqrt{3249}\) = 57

Question (iv).

825

Solution:

Here, the remainder is 41. It shows that 28^{2} is less than 825 by 41.

So, to get a perfect square, 41 must be subtracted from the given number.

∴ Required perfect square number = 825 – 41 = 784

\(\sqrt{784}\) = 28

Question (v).

4000

Solution:

Here, the remainder is 31. It shows that 63^{2} is less than 4000 by 31.

So, to get a perfect square, 31 must be subtracted from the given number.

∴ Required perfect square number = 4000 – 31 = 3969

\(\sqrt{3969}\) = 63

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.

Question (i).

525

Solution:

Here, the remainder is 41.

525 > 22^{2},

and the number next to 22 is 23.

23^{2} = 529.

∴ The required number to be added

= 23^{2} – 525

= 529 – 525 = 4

Now, 525 + 4 = 529

\(\sqrt{529}\) = 23

Thus, 4 is the least number which must be added to 525 to get a perfect square.

Question (ii).

1750

Solution:

Here, the remainder is 69.

1750 > 41^{2},

and the number next to 41 is 42.

42^{2} = 1764.

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

= 1764 – 1750

= 14

Now, 1750 + 14 = 1764

\(\sqrt{1764}\) = 42

Thus, 14 is the least number which must be added to 1750 to get a perfect square.

Question (iii).

252

Solution:

Here, the remainder is 27.

252 > 15^{2},

and the number next to 15 is 16.

16^{2} = 256.

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

= 256 – 252

= 4

Now, 252 + 4 = 256

\(\sqrt{256}\) = 16

Thus, 4 is the least number which must be added to 252 to get a perfect square.

Question (iv).

1825

Solution:

Here, the remainder is 61.

1825 > 42^{2},

and the number next to 42 is 43.

43^{2} = 1849.

∴ The required number to be added 43^{2} – 1825

= 1849 – 1825

= 24

Now, 1825 + 24 = 1849

\(\sqrt{1849}\) = 43

Thus, 24 is the least number which must be added to 1825 to get a perfect square.

Question (v).

6412

Solution:

Here, the remainder is 12.

6412 > 80^{2},

and the number next to 80 is 81.

81^{2} = 6561.

∴ The required number to be added = 81^{2} – 6412

= 6561 – 6412

= 149

Now, 6412 + 149 = 6561

\(\sqrt{6561}\) = 81

Thus, 149 is the least number which must be added to 6412 to get a perfect square.

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

Solution:

Let the side of a square be x m.

Area of a square = x × x = x^{2}

Area of a square = 441 (given)

∴x^{2} = 441

∴ x = \(\sqrt{441}\)

∴ x = 21

Thus, the length of the side of the square is 21 m.

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:

[Note: In a right triangle, the longest side is called the hypotenuse. The square of the hypotenuse is equal to the sum of the squares of the remaining two sides.]

Let us use this theorem here.

(a) Here, ∠B = 90°, AB = 6 cm, BC = 8 cm

In ΔABC, AC is hypotenuse.

AC^{2} = AB^{2} + BC^{2}

= (6)^{2} + (8)^{2}

= 36 + 64

=100

∴ AC = \(\sqrt{100}\)

= 10cm

(b) Here, ∠B= 90°, AC = 13cm, BC = 5cm

In ΔABC, AC Is hypotenuse.

AC^{2} = AB^{2} + BC^{2}

∴ (13)^{2} = AB^{2} + (5)^{2}

∴ 169 = AB^{2} + 25

∴ AB^{2} = 169 – 25

= 144

∴ AB = \(\sqrt{144}\)

= 12 cm

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:

Total number of plants = 1000

The number of plants in a row = The number of plants in a column

Let the number of plants planted in a row be x.

So, the number of plants planted in a column is x.

∴ Total plants = x × x = x^{2}

∴ x^{2} > 1000

∴ x > \(\sqrt{1000}\)

Here, the remainder is 39.

(31)^{2} < 1000

The next square number would be 32.

32^{2} = 1024

∴ The number of plants required to be added = 1024 – 1000

= 24

Thus, 24 more plants needed.

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:

Total number of children in a school = 500

The number of rows = The number of columns

Let the number of children in a row be x.

So, the number of children in a column is x.

Total number of children = x × x = x^{2}.

∴ x^{2} < 500

∴ x < \(\sqrt{500}\)

Here, the remainder is 16.

500 > 22^{2}

∴ 500 > 484

500 – 484 = 16

Thus, 16 children would be left out in this arrangement.