RD Sharma Solutions Class 8 Squares And Square Roots Exercise 3.1

RD Sharma Class 8 Solutions Chapter 3 Ex 3.1 PDF Free Download

RD Sharma Solutions Class 8 Chapter 3 Exercise 3.1

Exercise 3.1

1.) Which of the following numbers are perfect squares?

(i) 484

Answer: 484 = 222

484 is a perfect square

(ii) 625

Answer: 625 = 252

625 is a perfect square

(iii) 576

Answer: 576 = 242

576 is a perfect square

(iv) 941

Answer: 941 cannot be written as a square. Hence, 941 is not a perfect square.

Also, Perfect squares closest to 941 is 900 (302) and 961 (312). Since 30 and 31 are consecutive numbers, there are no perfect squares between 900 and 961.

(v) 961

Answer: 961 = 312

961 is a perfect square.

(vi)2500

Answer: 2500 = 502

2500 is a perfect square.

2.) Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number in each case:

Answer:

First factorize each number into its prime factors.

(i) 1156 = 2 x 2 x 17 x 17

Grouping the factors into pairs of equal factors,

1156 = (2 x 2) x (17 x 17)

Or 1156 = (2 x 17) x (2 x 17) = (2 x 17)2

Hence, 1156 is the square of 34, which is equal to 2 x 17.

(ii) 2025 = 3x3x3x3x5x5

Grouping the factors into pairs of equal factors,

2025 = (3 x 3) x (3 x 3) x (5 x 5)

2025 = (3 x 3 x 5) x (3 x 3 x 5) = (3 x 3 x 5)2

Hence, 2025 is a perfect square.

And 2025 is the square of 45, which is equal to 3 x 3 x 5.

(iii) 14641 = 11 x 11 x 11 x 11

Grouping the factors into pairs of equal factors, we obtain:

14641 = (11 x 11) x (11 x 11)

Hence, 14641 is a perfect square.

14641 = (11 x 11) x (11 x 11) = (11 x 11)2

14641 is the square of 121, which is equal to 11 x 11.

(iv) 4761 = 3 x 3 x 23 x 23

4761 = (3 x 3) x (23 x 23)

4761 is a perfect square.

The above expression is already grouped into equal factors:

4761 = (3 x 23) x (3 x 23) = (3 x 23)2

Hence, 4761 is the square of 69, which is equal to 3 x 23.

3.) Find the smallest number by which of the following number must be multiplied so that the product is a perfect square:

Answer:

Find factor of each number

(i) 23805 = 3 x 3 x 5 x 23 x 23

Grouping 23805 into pairs of equal factors:

23805 = (3 x 3) x (23 x 23) x 5

Here, the factor 5 does not occur in pairs.

Hence, the smallest number by which 23805 must be multiplied is 5.

(ii) 12150 = 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5

12150 = (3 x 3 x 3 x 3) x (5 x 5) x 2 x 3

Here, 2 and 3 do not occur in pairs.

To be a perfect square, every prime factor has to be in pairs.

Hence, the smallest number by which 12150 must be multiplied is 2 x 3, i.e. by 6.

(iii) 7688 = 2 x 2 x 2 x 31 x 31

7688 = (2 x 2 ) x (31 x 31 ) x 2

Here, 2 do not occur in pairs.

Hence the smallest number by which 7688 must be multiplied is 2.

4.) Find the smallest number by which the given number must be divided so that the resulting number is a perfect square:

Answer:

For each question, factorize the number into its prime factors.

(i) 14283 = 3 x 3 x 3 x 23 x23

14283 = (3 x 3) x (23 x 23) x 3

The factor 3 does not occur in pairs.

Hence, the smallest number by which 14283 must be divided for it to be a perfect square is 3.

(ii) 1800 = 2 x 2 x 2 x 3 x 3 x 5 x 5

Grouping the factors into pairs:

1800 = (2 x2 ) x (3 x 3) x (5 x 5) x2

Here the factor 2 does not occur in pairs.

Hence, the smallest number by which 1800 must be divided for it to be a perfect square is 2.

(iii) 2904 = 2 x 2 x 2 x 3 x 11 x 11

2904= (2 x 2) x (11 x 11) x 2 x 3

Here the factor 2 and 3 does not occur in pairs.

Hence, the smallest number by which 2304 must be divided for it to be a perfect square is 2 x 3, i.e. 6.

5.) Which of the following numbers are perfect squares?

Answer:

11, 12, 16, 32, 36, 50, 79, 81, 111, 121.

For number 11: The perfect squares closest to 11: 9 (9 = 32) and 16 (16 = 42).

Since 3 and 4 are consecutive numbers, there are no perfect squares between 9 and 16

Which shows 11 is not a perfect square.

For number 12: The perfect squares closest to 12 are 9 (9 =32) and 16 (16 = 42). Since 3 and 4 are consecutive numbers, there are no perfect squares between 9 and 16.

12 is not a perfect square.

For number 16: 16 = 42, perfect square

For number 32: The perfect squares closest to 32 are 25 (25 = 52) and 36 (36 = 62). Since 5 and 6 are consecutive numbers, there are no perfect squares between 25 and 3.

32 is not a perfect square.

For number 36: 36 = 62, perfect square

For number 50: The perfect squares closest to 50 are 49 (49 = 72) and 64 (64 = 82). Here 7 and 8 are consecutive numbers, there are no perfect squares between 49 and 64.

50 is not a perfect square.

For number 79: The perfect squares closest to 79 are 64 (64 = 82) and 81 (81 = 92). Since 8 and 9 are consecutive numbers, there are no perfect squares between 64 and 81.

79 is not a perfect square.

For number 81: 81= 92, perfect square

For number 111: The perfect squares closest to 111 are 100 (100 = 102) and 121 (121 = 112).

10 and 11 are consecutive numbers,

Not a perfect squares between 100 and 121.

111 not a perfect square.

For number 121: 121 = 112 , perfect square.

Hence, the perfect squares are 16, 36, 64, 81 and 121.

6.) Using prime factorization method, find which of the following numbers are perfect squares?

(i) 189 = 3 x 3 x 3 x 7

Prime Factorization Method

189 = (3 x 3) x 3 x 7

The factors 3 and 7 cannot be paired. Hence, 189 is not a perfect square.

(ii) 225 = 3 x 3 x 5 x 5

Prime factorization

225 = (3 x 3) x (5 x 5)

All factors are paired. Hence, 225 is a perfect square.

(iii) 2048 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Perfect Square

2048 = (2 x 2) x (2 x 2) x (2 x 2 ) x (2 x 2) x 2

The last factor, 2 cannot be paired.

All factors are paired. 2048 is a perfect square.

(iv) 343 = 7 x 7 x 7

Perfect Square Problem

Grouping them into pairs of equal factors:

343 = (7 x 7) x 7

The last factor, 7 cannot be paired.

343 is not a perfect square.

(v) 441 = 3 x 3 x 7 x 7

Perfect Square Problems

441 = ( 3 x 3) x ( 7 x 7)

All factors are paired. A perfect square.

(vi) 2916 = 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 x 3

Problem on Perfect Square

2916 = ( 2 x 2) x ( 3 x 3) x ( 3 x 3 ) x ( 3 x 3)

All factors are paired. 2916 is a perfect square.

(vii) 11025= 3 x 3 x 5 x 7 x 7

Factorization Problems

11025 = (3 x 3) x (5 x 5) x ( 7 x 7)

All factors are paired. 11025 is a perfect square.

(viii) 3549 = 3 x 7 x 13 x 13

Number not a Perfect Square

3549 = (13 x 13) x 3 x 7

All factors cannot be paired. Hence, 3549 is not a perfect square.

Hence, the perfect squares are 225, 441, 2916 and 11025.

7.) By what number should each of the following numbers be multiplied to get a perfect square in each case? Also, find the number whose square is the new number.

Sol:

Factorizing each number

(i) 8820 = 2 x 2 x 2 x 3 x 5 x 7 x 7

Factorization of a Number

To Find: A number whose square is a perfect square.

Form figure, 8820= (2 x 2) x (3 x 3) x ( 7 x 7) x 5

All factors cannot be paired.

Hence, 8820 must be multiplied by 5 to be a perfect square.

The new number would be (2×2) x (3 x 3) x (7 x7) x (5 x 5).

Furthermore, we have:

(2 x 2) x (3 x 3) x ( 7 x 7) x (5 x 5) = ( 2 x 3 x 5 x 7 ) x ( 2 x 3 x 5 x 7)

So, the new number is:

2 x 3 x 5 x 7 = 210

(ii) 3675 = 3 x 5 x 5 x 7 x 7

Prime Factors of a Number

3675 = (5 x 5) x ( 7 x 7 ) x 3

The factor 3 is not paired.

3675 must be multiplied by 3 for it to be a perfect square.

The new number would be (5 x 5) x (7 x 7) x (3 x 3).

or

(5 x 5) x (7 x 7) x (3 x 3) = ( 3 x 5 x 7 ) x (3 x 5 x 7)

New number is:

3 x 5 x 7 = 105

(iii) 605 = 5 x 11 x 11

Perfect Square Example

605 = 5 x (11 x 11)

The factor 5 is not paired.

So, 605 must be multiplied by 5 for it to be a perfect square.

The new number would be (5 x 5) x (11 x 11)

or

(5 x 5) x (11 x 11) = (5 x 11) x (5x 11)

The new number whose square is a perfect square:

5 x 11 = 55

(iv) 2880 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5

Find Missing Number for  Perfect Square

2880 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 5

factor 5 is not paired.

Which implies, 2880 must be multiplied by 5 to be a perfect square.

The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5).

Furthermore,

(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5) = (2 x2x2x3x 5) x (2 x2x2x3x 5)

2x2x2x3x5= 120

The number whose square is a perfect square 120.

(v) 4056 = 2 x 2 x 2 x 3 x 13 x 13

How to make a Perfect Square

4056 = ( 2 x 2) x( 13 x 13) x 2 x 3

The factors at the end, 2 and 3 are not paired.

4056 must me multiplied by 6 for it to be a perfect square.

The new number would be (2 x 2) x (2 x2) x (3 x 3 ) x (13 x 13).

Furthermore, we have

(2 x 2) x (2 x2) x (3 x 3 ) x (13 x 13) = (2 x 2 x 3 x 13) x ( 2 x 2 x 3 x13)

2 x 2 x 3 x13 = 156

The number whose square is a perfect square 156

(vi) 3468 = 2 x 2 x 3 x 17 x 17

3468 = (2 x 2) x (17 x 17) x 3

The factor at the end, 3 is not paired.

3468 must me multiplied by 3 for it to be a perfect square.

The new number would be (2 x 2) x (17 x 17) x (3 x 3).

(2 x 2) x (17 x17) x (3 x3) = (2 x 3 x 17) x ( 2 x 3 x17)

2 x 3 x 17 = 102

The number whose square is a perfect square 102.

(viii) 7776 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3

7776 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x 2 x 3

The factor at the end, 2 and 3 are not paired.

7776 must me multiplied by 6 to be a perfect square.

The new number would be (2 x 2) x (2 x 2) x (2 x 2 )(3 x 3) x ( 3 x 3) x (3 x 3) .

Furthermore, we have

(2 x 2) x (2 x 2) x (2 x 2) (3 x 3) x ( 3 x 3) x (3 x 3)= (2 x 2 x 2 x 3 x 3 x 3 ) x (2 x 2 x 2 x 3 x 3 x 3)

The new number is:

2 x 2 x 2 x 3 x 3 x 3 = 216

8.) By what numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.

Answer:

Factoring each number

(i) 16562 = 2 x 7 x 7 x 13 x13

Group all the factors into pairs :

16562= 2 x (7 x 7 ) x (13 x 13)

The factor at the end, 2 is not paired.

16652 must be divided by 2 for it to be a perfect square.

The new number would be (7x 7) x (13 x13).

Furthermore, we have

(7x 7) x (13 x13) = (7 x 13) x (7 x 13)

The new number is:

7 x 13= 91

(ii) 3698 = 2 x 43 x 43

3698 = 2 x (43 x 43)

The factor at the end, 2 is not paired.

3698 must be divided by 2 for it to be a perfect square.

The new number is 43.

(iii) 5103 = 3 x 3 x 3 x 3 x 3 x 3 x 7

5103 = (3 x 3) x (3 x 3) x (3 x 3) x 7

The factor, 7 is not paired. As per perfect square definition, each prime factor has to be paired.

5103 must be divided by 7 for it to be a perfect square.

The required number would be (3 x 3) x (3 x 3) x (3 x 3).

Furthermore, we have: (3 x 3) x (3 x 3) x (3 x 3) = (3 x 3 x 3) x (3 x 3 x 3) The new number is:

3 x 3 x 3 = 27

(iv) 3174 = 2 x 3 x 23 x 23

Or 3174 = 2 x 3 x (23 x 23)

The factors, 2 and 3 are not paired.

For a number to be a perfect square, each prime factor has to be paired.

3174 must be divided by 6 to be a perfect square.

The new number is 23.

(v) 1575= 3 x 3 x 5 x 7

1575 = (3 x 3) x (5 x 5) x 7

The factor, 7 is not paired.

1575 must be divided by 7 for it to be a perfect square.

The new number would be 15.

9.) Find the greatest number of two digits which is a perfect square.

Answer:

We know that 102 is equal to 100 and 92 is equal to 81.

9 and 10 are consecutive numbers, there is no perfect square between 81 and 100.

Here we can see that, 100 is the first perfect square that has more than two digits and 81 is the greatest two-digit perfect square.

10.) Find the least number of three digits which is a perfect square.

Sol:

Below is the list of the squares starting from 1.

12=1

22 = 4

32 = 9

42 = 16

52=25

62 = 36

72= 49

82 = 64

92 = 81

102=100

The square of 10 has three digits. The least 3 digit perfect square is 100.

11.) Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.

Sol:

4581 = 3 x 3 x 7 x 7 x11

Find a Perfect Square

4851 = (3 x 3) x (7 x 7) x 11

The factor, 11 is not paired.

So, 11 is the smallest number by which 4851 must be multiplied to get a perfect square.

12.) Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.

Sol:

Prime factorization of 28812:

28812 = 2 x 2 x 3 x 7 x 7 x 7 x7

28812 = (2 x2) x (7 x 7) x (7x 7) x 3

The factor, 3 is not paired.

So, 3 is the smallest number by which 28812 must be divided to get a perfect square.

13.) Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the number whose square is the resulting number.

Sol:

Prime factorization of 1152:

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

1152=(2 x 2)x(2 x 2)x(2 x 2)x(3 x 3)x 2

The factor, 2 at the end is not paired.

Therefore, 1152 must be divided by 2 for it to be a perfect square.

And 24 is the required number.

1 Comment

  1. Thanks for helping me.

Leave a Comment

Your email address will not be published. Required fields are marked *