NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Roots Exercise 5.3

Squares And Square Roots 

Exercise 5.3

 

1). What could be the possible ‘one’s’ digits of the square root of each of the following

numbers?

(i) 9801                                          (ii) 99856

(iii) 998001                                     (iv) 657666025

Solution:

(i)  9801       

The possible one’s digit of the square root of 9801 may be 1 or 9.                                  (ii) 99856

The possible one’s digit of the square root of 99856 may be 4 or 6.

(iii) 998001

The possible one’s digit of the square root of 998001 may be 1 or 9.

(iv) 657666025

The possible one’s digit of the square root of 657666025 is 5.

2). Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153                                            (ii) 257

(iii) 408                                          (iv) 441

Solution:

(i) 153

We know that numbers ending with 2, 3, 7 and 8 are not perfect squares.

Therefore 153 is not a perfect square.                                           

(ii) 257

We know that numbers ending with 2, 3, 7 and 8 are not perfect squares.

Therefore 257 is not a perfect square.                                    

(iii) 408

We know that numbers ending with 2, 3, 7 and 8 are not perfect squares.

Therefore 408 is not a perfect square.                                                                         

(iv) 441

 441 is a perfect square.                                    

3). Find the square roots of 100 and 169 by the method of repeated subtraction.

(i) 100

100 – 1 = 99

99 – 3 = 96

96 – 5 = 91

91 – 7 = 84

84 – 9 = 75

75 – 11 = 64

64 – 13 = 51

51 – 15 = 36

36 – 17 = 19

19 – 19 = 0

In the tenth step we got 0 therefore square root of 100 is 10.

(ii) 169

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 98

98 – 19 = 79

79 – 21 = 58

58 – 23 = 25

25 – 25 = 0

In the thirteenth step we got 0 therefore square root of 169 is 13.

4). Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729                                            (ii) 400

(iii) 1764                                        (iv) 4096

(v) 7744                                         (vi) 9604

(vii) 5929                                        (viii) 9216

(ix) 529                                          (x) 8100

Solution:

(i) 729    

                                       

 

729 = 3 X 3 X 3 X 3 X 3 X 3

       = 3 X 3 X 3 X 3 X 3 X 3

       = 32 X 32 X 32

√729 = 3 X 3 X 3 = 27      

(ii) 400

400 = 2 X 2 X 2 X 2 X 5 X 5

       = 2 X 2 X 2 X 2 X 5 X 5

       = 22 X 22 X 52

√400 = 2 X 2 X 5 = 20

(iii) 1764

1764   = 2 X 2 X 3 X 3 X 7 X 7

           = 2 X 2 X 3 X 3 X 7 X 7

           = 22 X 32 X 72

√1764  = 2 X 3 X 7 = 42                                       

(iv) 4096

4096   = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2

           = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2

           = 22 X 22 X 22 X 22 X 22 X 22

√4096 = 2 X 2 X 2 X 2 X 2 X 2  = 64

(v) 7744

7744      = 2 X 2 X 2 X 2 X 2 X 2 X 11 X 11

              = 2 X 2 X 2 X 2 X 2 X 2 X 11 X 11

              = 22 X 22 X 22 X 112

√7744    = 2 X 2 X 2 X 11 = 88                                       

(vi) 9604

9604 = 2 X 2 X 7 X 7 X 7 X 7

         =  2 X 2 X 7 X 7 X 7 X 7

         = 22 X 72 X 72

√9604 = 2 X 7 X 7 = 98

(vii) 5929

5929   = 7 X 7 X 11 X 11

          = 7 X 7 X 11 X 11

          = 72 X 112

√5929 = 7 X 11 = 77                                       

(viii) 9216

9216      = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3

              = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3

              = 22 X 22 X 22 X 22 X 22 X 32

√4096 = 2 X 2 X 2 X 2 X 2 X 3  = 96

(ix) 529

 

529 = 23 X 23

       = 23 X 23

       = 232

√529 = 23                                              

(x) 8100

8100      = 2 X 2 X 3 X 3 X 3 X 3 X 5 X 5

              = 2 X 2 X 3 X 3 X 3 X 3 X 5 X 5

              = 22 X 32 X 32 X 52

√8100 = 2 X 3 X 3 X 5 = 90

5). For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252                                            (ii) 180

(iii) 1008                                        (iv) 2028

(v) 1458                                         (vi) 768

Solution:

(i) 252   

Prime factors of 252 = 2 X 2 X 3 X 3 X 7

Here 7 has no pair

7 is the smallest whole number by which 252 is multiplied to get a perfect square number.

New square number = 252 X 7 = 1764

√1764 = 2 X 3 X 7 = 42                                   

(ii) 180

Prime factors of 180 = 2 X 2 X 3 X 3 X 5

Here 5 has no pair

5 is the smallest whole number by which 180 is multiplied to get a perfect square number.

New square number = 180 X 5 = 900

√900 = 2 X 3 X 5 = 30

(iii) 1008

Prime factors of 1008 = 2 X 2 X 2 X 2 X 3 X 3 X 7

Here 7 has no pair

7 is the smallest whole number by which 1008 is multiplied to get a perfect square number.

New square number = 1008 X 7 = 7056

√7056 = 2 X 2 X 3 X 7 = 84

(iv) 2028

Prime factors of 2028 = 2 X 2 X 3 X 13 X 13

Here 3 has no pair

3 is the smallest whole number by which 2028 is multiplied to get a perfect square number.

New square number = 2028 X 3 = 6084

√6084 = 2 X 3 X 13 = 78   

(v) 1458

Prime factors of 1458 = 2 X 3 X 3 X 3 X 3 X 3 X 3

Here 2 has no pair

2 is the smallest whole number by which 1458 is multiplied to get a perfect square number.

New square number = 1458 X 2 = 2916

√2916 = 2 X 3 X 3 X 3 = 54

(vi) 768

Prime factors of 768 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3

Here 3 has no pair

3 is the smallest whole number by which 768 is multiplied to get a perfect square number.

New square number = 768 X 3 = 2304

√2304 = 2 X 2 X 2 X 2 X 3 = 48

6). For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252                                            (ii) 2925

(iii) 396                                          (iv) 2645

(v) 2800                                         (vi) 1620

 

Solution:

(i) 252

Prime factors of 252 = 2 X 2 X 3 X 3 X 7

Here 7 has no pair

7 is the smallest whole number by which 252 is divided to get a perfect square number.

New square number = 252 ÷ 7 = 36

√36 = 6 

(ii) 2925

Prime factors of 2925 = 3 X 3 X 5 X 5 X 13

Here 13 has no pair

13 is the smallest whole number by which 2925 is divided to get a perfect square number.

New square number = 2925 ÷ 13 = 225

√225 = 15

(iii) 396

Prime factors of 396 = 2 X 2 X 3 X 3 X 11

Here 11 has no pair

11 is the smallest whole number by which 396 is divided to get a perfect square number.

New square number = 396 ÷ 11 = 36

√36 = 6 

(iv) 2645

Prime factors of 2645 = 5 X 23 X 23

Here 5 has no pair

5 is the smallest whole number by which 2645 is divided to get a perfect square number.

New square number = 2645 ÷ 5 = 529

√529 = 23    

(v) 2800

Prime factors of 2800 = 2 X 2 X 2 X 2 X 5 X 5 X 7

Here 7 has no pair

7 is the smallest whole number by which 2800 is divided to get a perfect square number.

New square number = 2800 ÷ 7 = 400

√400 = 20    

(vi) 1620

Prime factors of 1620 = 2 X 2 X 3 X 3 X 3 X 3 X 5

Here 5 has no pair

5 is the smallest whole number by which 1620 is divided to get a perfect square number.

New square number = 1620 ÷ 5 = 324

√324 = 18    

7). The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Amount donated = 2401

Number of students = amount donated by each student

 

Number of students = √2401

                              = √(7 X 7 X 7 X 7)

                              = √(72 X 72)

                            = 7 X 7

                            = 49

Ans: number of students = 49

8). 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Number of plants = 2025

Number of plants in a row  = number of rows

Number of plants in each row  = √2025

                              = √(3 X 3 X 3 X 3 X 5 X 5)

                              = √(32 X 32 X 52)

                            = 3 X 3 X 5

                            = 45

Ans: number of plants in each row = 45

9). Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Smallest number divisible by each of the numbers 4, 9 and 10 is

LCM of 4 , 9 , 10 =  180

Prime factors of  180 = 2 X 2 X 3 X 3 X 5

Here, 5 has no pair.

The required smallest square number = 180 X 5 = 900

10). Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Smallest number divisible by each of the numbers 8, 15 and 20 is

LCM of 8, 15 and 20 =  120

Prime factors of  120 = 2 X 2 X 2 X 3 X 5

Here, 2 X 3 X 5 has no pair.

The required smallest square number = 120 X 30 = 3600

Click here for the solutions of Std 8 Maths

1). Rational Numbers

 Exercise 1.1

2). Linear Equations in One Variable

Exercise 2.1

Exercise 2.2

3). Understanding Quadrilaterals

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

4). Data Handling

Exercise 4.1

Exercise 4.2

5). Squares and Square Roots

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

 

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