The document RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 1, Class 8 Math Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

All you need of Class 8 at this link: Class 8

**Question 1:Which of the following numbers are perfect squares?(i) 484(ii) 625(iii) 576(iv) 941(v) 961(vi) 2500**

(i) 484 = 22

(ii) 625 = 25

(iii) 576 = 24

(iv) Perfect squares closest to 941 are 900 (30

(v) 961 = 31

(vi) 2500 = 50

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

(i) 1156

(ii) 2025

(iii) 14641

(iv) 4761

(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)

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)

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)

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)

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

(i) 23805

(ii) 12150

(iii) 7688

Factorise each number into its prime factors.

(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. 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

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

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.

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.

(i) 14283

(ii) 1800

(iii) 2904

**Answer 4:**

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

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

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

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

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.

**Question 5:****Which of the following numbers are perfect squares?11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121**

**Answer 5:**

11: The perfect squares closest to 11 are 9 (9 = 3^{2}) and 16 (16 = 4^{2}). 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 =3^{2}) and 16 (16 = 4^{2}). 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 = 4^{2}

32: The perfect squares closest to 32 are 25 (25 = 5^{2}) and 36 (36 = 6^{2}). 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 = 6^{2}

50: The perfect squares closest to 50 are 49 (49 = 7^{2}) and 64 (64 = 8^{2}). 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 = 8^{2}) and 81 (81 = 9^{2}). 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 = 9^{2}

111: The perfect squares closest to 111 are 100 (100 = 10^{2}) and 121 (121 = 11^{2}). 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.

**Question 6:****Using prime factorization method, find which of the following numbers are perfect squares?****189, 225, 2048, 343, 441, 2916, 11025, 3549**

**Answer 6:**

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

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

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

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

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

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

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

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

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.

### RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math

- Doc | 9 pages
### RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.2) Part - 1, Class 8 Math

- Doc | 4 pages