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Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8 PDF Download

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

(i) 484

(ii) 625

(iii) 576

(iv) 941

(v) 961

(vi) 2500

Ans:

(i) 484 = 222

(ii) 625 = 252

(iii) 576 = 242

(iv) Perfect squares closest to 941 are 900 (302) and 961 (312). Since 30 and 31 are consecutive numbers, there are no perfect squares between 900 and 961. Hence, 941 is not a perfect square.

(v) 961 = 312

(vi) 2500 = 502

Hence, all numbers except that in (iv), i.e. 941, are perfect squares.

Q.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:

(i) 1156

(ii) 2025

(iii) 14641

(iv) 4761

Ans: In each problem, factorise the number into its prime factors.

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

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

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

No factors are left over. Hence, 1156 is a perfect square. Moreover, by grouping 1156 into equal factors:

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 = 3 x 3 x 3 x 3 x 5 x 5

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

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

No factors are left over. Hence, 2025 is a perfect square. Moreover, by grouping 2025 into equal factors:

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

= (3 x 3 x 5)²

Hence, 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)

No factors are left over. Hence, 14641 is a perfect square. The above expression is already grouped into equal factors:

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

= (11 x 11)²

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

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

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

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

No factors are left over. Hence, 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.

Q.3. Find the smallest number by which the given number must bew multiplied so that the product is a perfect square:

(i) 23805

(ii) 12150

(iii) 7688

Ans: Factorise each number into its prime factors.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping 12150 into pairs of equal factors:

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping 7688 into pairs of equal factors:

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

Here, 2 does not occur in pairs. To be a perfect square, every prime factor has to be in pairs. Hence, the smallest number by which 7688 must be multiplied is 2.

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

(i) 14283

(ii) 1800

(iii) 2904

Ans: For each question, factorise the number into its prime factors.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping the factors into pairs:

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

Here, the factor 3 does not occur in pairs. To be a perfect square, all the factors have to be 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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping the factors into pairs:

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

Here, the factor 2 does not occur in pairs. To be a perfect square, all the factors have to be 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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping the factors into pairs:

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

Here, the factors 2 and 3 do not occur in pairs. To be a perfect square, all the factors have to be in pairs. Hence, the smallest number by which 2904 must be divided for it to be a perfect square is 2 x 3, i.e. 6.

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

11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121

Ans:

11: The perfect squares closest to 11 are 9 (9 = 3²) and 16 (16 = 4²). Since 3 and 4 are consecutive numbers, there are no perfect squares between 9 and 16, which means that 11 is not a perfect square.

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, which means that 12 is not a perfect square.

16 = 4²

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 36, which means that 32 is not a perfect square.

36 = 6²

50: The perfect squares closest to 50 are 49 (49 = 7²) and 64 (64 = 8²). Since 7 and 8 are consecutive numbers, there are no perfect squares between 49 and 64, which means that 50 is not a perfect square.

64 = 8²

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, which means that 79 is not a perfect square.

81 = 9²

111: The perfect squares closest to 111 are 100 (100 = 10²) and 121 (121 = 11²). Since 10 and 11 are consecutive numbers, there are no perfect squares between 100 and 121, which means that 111 is not a perfect square.

121 = 11²

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

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

189, 225, 2048, 343, 441, 2916, 11025, 3549

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

There are no left out of pairs. 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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The last factor, 2 cannot be paired. Hence, 2048 is not a perfect square.

(iv) 343 = 7 x 7 x 7

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

343 = (7 x 7) x 7

The last factor, 7 cannot be paired. Hence, 343 is not a perfect square.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

There are no left out of pairs. Hence, 441 is a perfect square.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

There are no left out of pairs. Hence, 2916 is a perfect square.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

There are no left out of pairs. Hence, 11025 is a perfect square.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

3549 = (13 x 13) x 3 x 7

The last factors, 3 and 7 cannot be paired. Hence, 3549 is not a perfect square.

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

Q.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.

(i) 8820

(ii) 3675

(iii) 605

(iv) 2880

(v) 4056

(vi) 3468

(vii) 7776

Ans: Factorising each number.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 8820 must be multiplied by 5 for it to be a perfect square.

The new number would be (2 x 2) x (3 x 3) x (7 x 7) 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)

Hence, the number whose square is the new number is:

2 x 3 x 5 x 7 = 210

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor, 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 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).

Furthermore, we have:

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

Hence, the number whose square is the new number is:

3 x 5 x 7 = 105

(iii) 605 = 5 x 11 x 11

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

605 = 5 x (11 x 11)

The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 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).

Furthermore, we have:

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

Hence, the number whose square is the new number is:

5 x 11 = 55

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

There is a 5 as the leftover. For a number to be a perfect square, each prime factor has to be paired. Hence, 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, we have:

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

Hence, the number whose square is the new number is:

2 x 2 x 2 x 3 x 5 = 120

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factors at the end, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 4056 must be multiplied by 6 (2 x 3) for it to be a perfect square.

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

Furthermore, we have:

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

Hence, the number whose square is the new number is:

2 x 2 x 3 x 13 = 156

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3468 must be 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).

Furthermore, we have:

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

Hence, the number whose square is the new number is:

2 x 3 x 17 = 102

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factors, 2 and 3 at the end are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 7776 must be multiplied by 6 (2 x 3) for it 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 (3 x 3) x (3 x 3).

Furthermore, we have:

(2 x 2) x (2 x 2) x (2 x 2) x (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)

Hence, the number whose square is the new number is:

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

Q.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.

(i) 16562

(ii) 3698

(iii) 5103

(iv) 3174

(v) 1575

Ans: Factorising each number.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 16562 must be divided by 2 for it to be a perfect square.

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

Furthermore, we have:

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

Hence, the number whose square is the new number is:

7 x 13 = 91

(ii) 3698 = 2 x 43 x 43

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

3698 = 2 x (43 x 43)

The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3698 must be divided by 2 for it to be a perfect square.

The new number would be (43 x 43).

Hence, the number whose square is the new number is 43.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 5103 must be divided by 7 for it to be a perfect square.

The new 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)

Hence, the number whose square is the new number is:

3 x 3 x 3 = 27

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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. Hence, 3174 must be divided by 6 (2 x 3) for it to be a perfect square.

The new number would be (23 x 23).

Hence, the number whose square is the new number is 23.

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 1575 must be divided by 7 for it to be a perfect square.

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

Furthermore, we have:

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

Hence, the number whose square is the new number is:

3 x 5 = 15

Q.9. Fnd the greatest number of two digits which is a perfect square.

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

Since 10 and 9 are consecutive numbers, there is no perfect square between 100 and 81. Since 100 is the first perfect square that has more than two digits, 81 is the greatest two-digit perfect square.

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

Ans: Let us make a 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. Hence, the least three-digit perfect square is 100.

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

Ans: Prime factorisation of 4851:

4851 = 3 x 3 x 7 x 7 x 11

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor, 11 is not paired. The smallest number by which 4851 must be multiplied such that the resulting number is a perfect square is 11.

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

Ans: Prime factorisation of 28812:

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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

The factor, 3 is not paired. Hence, the smallest number by which 28812 must be divided such that the resulting number is a perfect square is 3.

Q.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.

Ans: Prime factorisation of 1152:

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

Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Grouping them into pairs of equal factors:

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. For a number to be a perfect square, each prime factor has to be paired. Hence, 1152 must be divided by 2 for it to be a perfect square.

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

Furthermore, we have:

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

Hence, the number whose square is the resulting number is:

2 x 2 x 2 x 3 = 24

The document Exercise 3.1 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8 is a part of the Class 8 Course Mathematics (Maths) Class 8.

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FAQs on Exercise 3.1 - Squares And Square Roots RD Sharma Solutions - Mathematics (Maths) Class 8

1. What is the formula to find the square of a number?

Ans. The formula to find the square of a number is to multiply the number by itself. For example, the square of 5 is 5 multiplied by 5, which equals 25.

2. How can I find the square root of a number?

Ans. To find the square root of a number, you can use the prime factorization method or the long division method. By prime factorization, you need to write the given number as a product of its prime factors, and then take one factor from each pair. By long division, you repeatedly divide the number by a suitable divisor until you obtain a quotient with a remainder of zero.

3. What is the difference between a perfect square and a square root?

Ans. A perfect square is the result of multiplying a number by itself, while a square root is the number that, when multiplied by itself, gives the original number. For example, 9 is a perfect square because 3 multiplied by 3 equals 9. The square root of 9 is 3 because 3 multiplied by itself equals 9.

4. How do I simplify the square root of a number?

Ans. To simplify the square root of a number, you need to find the prime factors of the number and pair them up. Take one factor from each pair and multiply them together. The product of the factors outside the square root sign becomes the simplified form of the square root.

5. What are some real-life applications of squares and square roots?

Ans. Squares and square roots have various applications in real life, such as in construction to determine the area of a square or rectangular space, in finance to calculate compound interest or mortgage payments, in physics to solve problems involving velocity or acceleration, and in computer science for algorithms and encryption techniques.

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RD Sharma Solutions: Exercise 3.1 - Squares And Square Roots Notes

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RD Sharma Solutions: Exercise 3.1 - Squares And Square Roots Class 8 Questions

The "RD Sharma Solutions: Exercise 3.1 - Squares And Square Roots Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

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