# Squares And Square Roots

## Exercise 5.4

1). 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

Solution:

(i) 2304

√2304 = 48

(ii) 4489

√4489 = 67

(iii) 3481

√3481 = 59

(iv) 529

√529 = 23

(v) 3249

√3249 = 57

(vi) 1369

√1369 = 37

(vii) 5776

√5776 = 76

(viii) 7921

√7921 = 89

(ix) 576

√576 = 24

(x) 1024

√1024 = 32

(xi) 3136

√3136 = 56

(xii) 900

√300 = 30

2). 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

Solution:

(i) 64

Number of digits in 64 is 2

Here, *n* = 2 (even)

Number of digits in square root = * ^{n}*/

_{2}=

^{2}/

_{2}= 1

(ii) 144

Number of digits in 144 is 3

Here, *n* = 3 (even)

Number of digits in square root = ^{n}^{+1}/_{2} = ^{3+1}/_{2} = ^{4}/_{2} = 2

(iii) 4489

Number of digits in 4489 is 4

Here, *n* = 4 (even)

Number of digits in square root = * ^{n}*/

_{2}=

^{4}/

_{2}= 2

(iv) 27225

Number of digits in 27225 is 5

Here, *n* = 5 (odd)

Number of digits in square root = ^{n}^{+1}/_{2} = ^{5+1}/_{2} = ^{6}/_{2} = 3

(v) 390625

Number of digits in 390625 is 6

Here, *n* = 6 (even)

Number of digits in square root = * ^{n}*/

_{2}=

^{6}/

_{2}= 3

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

(i) 2.56 (ii) 7.29

(iii) 51.84 (iv) 42.25

(v) 31.36

Solution:

(i) 2.56

√2.56 = 1.6

(ii) 7.29

√7.29 = 2.7

(iii) 51.84

√51.84 = 7.2

(iv) 42.25

√42.25 = 6.5

(v) 31.36

√31.36 = 5.6

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 (ii) 1989

(iii) 3250 (iv) 825

(v) 4000

Solution:

(i) 402

Here remainder is 2

Therefore, 2 is the least number to be subtracted

402 – 2 = 400

√400 = 20

(ii) 1989

Here remainder is 53

Therefore, 53 is the least number to be subtracted

1989 – 53 = 1936

√1936 = 44

(iii) 3250

Here remainder is 1

Therefore, 1 is the least number to be subtracted

3250 – 1 = 3249

√3249 = 57

(iv) 825

Here remainder is 41

Therefore, 41 is the least number to be subtracted

825 – 41 = 784

√784 = 28

(v) 4000

Here remainder is 31

Therefore, 31 is the least number to be subtracted

4000 – 31 = 3969

√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.

(i) 525 (ii) 1750

(iii) 252 (iv) 1825

(v) 6412

Solution:

(i) 525

Here remainder is 41 and quotient is 22

Square of 22 is less than 525

Next number is 23

Square of 23 = 529

The least number to be added = 529 – 525 = 4

525 + 4 = 529

√529 = 23

(ii) 1750

Here remainder is 69 and quotient is 41

Square of 41 is less than 1750

Next number is 42

Square of 42 = 1764

The least number to be added = 1764 – 1750 = 14

√1764 = 42

(iii) 252

Here remainder is 27 and quotient is 15

Square of 15 is less than 252

Next number is 16

Square of 16 = 256

The least number to be added = 256 – 252 = 4

√256 = 16

(iv) 1825

Here remainder is 61 and quotient is 42

Square of 42 is less than 1825

Next number is 43

Square of 43 = 1849

The least number to be added = 1849 – 1825 = 24

√1849 = 43

(v) 6412

Here remainder is 12 and quotient is 80

Square of 80 is less than 6412

Next number is 81

Square of 81 =6561

The least number to be added = 6561 – 6412 = 149

√6561 = 81

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

Given: Area of square = 441 m^{2}

Area of square = side^{2}

Side^{2} = 441

taking square root on both the sides

√side^{2} = √441

side = 21

Ans: Side of the square = 21m

7). In a right triangle ABC, ÐB = 90°.

(a) If AB = 6 cm, BC = 8 cm, find AC

(b) If AC = 13 cm, BC = 5 cm, find AB

Solution:

(a) If AB = 6 cm, BC = 8 cm, find AC

In ∆ABC by Pythagoras theorem

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

AC^{2} = 6^{2} + 8^{2}

= 36 + 64

= 100

taking square root on both the sides

√AC^{2} = √100

AC = 10

(b) If AC = 13 cm, BC = 5 cm, find AB

In ∆ABC by Pythagoras theorem

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

AB^{2} + 5^{2 }= 13^{2}

AB^{2} + 25 = 169

AB^{2} = 169 – 25

AB^{2} ^{ }= 144

taking square root on both the sides

√AB^{2} = √144

AB = 12

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.

Total plants = 1000

Number of columns = number of rows

1000 is not a perfect square.

Here remainder is 39 and quotient is 31

Square of 31 is less than 1000

Next number is 32

Square of 32 = 1024

The least number to be added = 1024 – 1000 = 24

Ans: Gardener requires 24 more plants

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.

Total number of students = 500

Number of rows = number of columns

500 is not a perfect square

Here remainder is 16

500 – 16= 484

√484 = 22

Ans: students left out in this arrangement are 16

Click here for the solutions of Std 8 Maths

1). Rational Numbers

2). Linear Equations in One Variable

3). Understanding Quadrilaterals

4). Data Handling

5). Squares and Square Roots