RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots are available here. The subject experts at BYJUâ€™S outline the concepts in a clear and precise manner based on the IQ level of students. Our solution module utilizes numerous shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language. If you wish to obtain an excellent score, solving RD Sharma Class 8 Solutions is a must.

Here the students will learn various techniques to determine whether a given natural number is a perfect square or not. The solutions to all questions in RD Sharma books are given here in a detailed and step by step way to help the students understand more effectively. Students can download the solutions of these exercises from the provided links.

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### EXERCISE 3.1 PAGE NO: 3.4

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

**(i) 484**

**(ii) 625**

**(iii) 576**

**(iv) 941**

**(v) 961**

**(vi) 2500**

**Solution:**

**(i) **484

First find the prime factors for 484

484 = 2Ã—2Ã—11Ã—11

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (11Ã—11)

By observation, none of the prime factors are left out.

âˆ´ 484 is a perfect square.

**(ii)** 625

First find the prime factors for 625

625 = 5Ã—5Ã—5Ã—5

By grouping the prime factors in equal pairs we get,

= (5Ã—5) Ã— (5Ã—5)

By observation, none of the prime factors are left out.

âˆ´ 625 is a perfect square.

**(iii)** 576

First find the prime factors for 576

576 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3)

By observation, none of the prime factors are left out.

âˆ´ 576 is a perfect square.

**(iv)** 941

First find the prime factors for 941

941 = 941 Ã— 1

We know that 941 itself is a prime factor.

âˆ´ 941 is not a perfect square.

**(v)** 961

First find the prime factors for 961

961 = 31Ã—31

By grouping the prime factors in equal pairs we get,

= (31Ã—31)

By observation, none of the prime factors are left out.

âˆ´ 961 is a perfect square.

**(vi)** 2500

First find the prime factors for 2500

2500 = 2Ã—2Ã—5Ã—5Ã—5Ã—5

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (5Ã—5) Ã— (5Ã—5)

By observation, none of the prime factors are left out.

âˆ´ 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:
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761**

**Solution:**

**(i)** 1156

First find the prime factors for 1156

1156 = 2Ã—2Ã—17Ã—17

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (17Ã—17)

By observation, none of the prime factors are left out.

âˆ´ 1156 is a perfect square.

To find the square of the given number

1156 = (2Ã—17) Ã— (2Ã—17)

= 34 Ã— 34

= (34)^{2}

âˆ´ 1156 is a square of 34.

**(ii)** 2025

First find the prime factors for 2025

2025 = 3Ã—3Ã—3Ã—3Ã—5Ã—5

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (3Ã—3) Ã— (5Ã—5)

By observation, none of the prime factors are left out.

âˆ´ 2025 is a perfect square.

To find the square of the given number

2025 = (3Ã—3Ã—5) Ã— (3Ã—3Ã—5)

= 45 Ã— 45

= (45)^{2}

âˆ´ 2025 is a square of 45.

**(iii)** 14641

First find the prime factors for 14641

14641 = 11Ã—11Ã—11Ã—11

By grouping the prime factors in equal pairs we get,

= (11Ã—11) Ã— (11Ã—11)

By observation, none of the prime factors are left out.

âˆ´ 14641 is a perfect square.

To find the square of the given number

14641 = (11Ã—11) Ã— (11Ã—11)

= 121 Ã— 121

= (121)^{2}

âˆ´ 14641 is a square of 121.

**(iv)** 4761

First find the prime factors for 4761

4761 = 3Ã—3Ã—23Ã—23

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (23Ã—23)

By observation, none of the prime factors are left out.

âˆ´ 4761 is a perfect square.

To find the square of the given number

4761 = (3Ã—23) Ã— (3Ã—23)

= 69 Ã— 69

= (69)^{2}

âˆ´ 4761 is a square of 69.

**3. Find the smallest number by which the given number must be multiplied so that the product is a perfect square:
(i) 23805
(ii) 12150
(iii) 7688**

**Solution:**

**(i)** 23805

First find the prime factors for 23805

23805 = 3Ã—3Ã—23Ã—23Ã—5

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (23Ã—23) Ã— 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

23805 Ã— 5 = (3Ã—3) Ã— (23Ã—23) Ã— (5Ã—5)

= (3Ã—5Ã—23) Ã— (3Ã—5Ã—23)

= 345 Ã— 345

= (345)^{2}

âˆ´ Product is the square of 345.

**(ii)** 12150

First find the prime factors for 12150

12150 = 2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—5Ã—5

By grouping the prime factors in equal pairs we get,

= 2Ã—3 Ã— (3Ã—3) Ã— (3Ã—3) Ã— (5Ã—5)

By observation, prime factor 2 and 3 are left out.

So, multiply by 2Ã—3 = 6 we get,

12150 Ã— 6 = 2Ã—3 Ã— (3Ã—3) Ã— (3Ã—3) Ã— (5Ã—5) Ã— 2 Ã— 3

= (2Ã—3Ã—3Ã—3Ã—5) Ã— (2Ã—3Ã—3Ã—3Ã—5)

= 270 Ã— 270

= (270)^{2}

âˆ´ Product is the square of 270.

**(iii)** 7688

First find the prime factors for 7688

7688 = 2Ã—2Ã—31Ã—31Ã—2

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (31Ã—31) Ã— 2

By observation, prime factor 2 is left out.

So, multiply by 2 we get,

7688 Ã— 2 = (2Ã—2) Ã— (31Ã—31)Ã— (2Ã—2)

= (2Ã—31Ã—2) Ã— (2Ã—31Ã—2)

= 124 Ã— 124

= (124)^{2}

âˆ´ Product is the square of 124.

**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**

**Solution:**

**(i)** 14283

First find the prime factors for 14283

14283 = 3Ã—3Ã—3Ã—23Ã—23

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (23Ã—23) Ã— 3

By observation, prime factor 3 is left out.

So, divide by 3 to eliminate 3 we get,

14283/3 = (3Ã—3) Ã— (23Ã—23)

= (3Ã—23) Ã— (3Ã—23)

= 69 Ã— 69

= (69)^{2}

âˆ´ Resultant is the square of 69.

**(ii)** 1800

First find the prime factors for 1800

1800 = 2Ã—2Ã—5Ã—5Ã—3Ã—3Ã—2

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (5Ã—5) Ã— (3Ã—3) Ã— 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

1800/2 = (2Ã—2) Ã— (5Ã—5) Ã— (3Ã—3)

= (2Ã—5Ã—3) Ã— (2Ã—5Ã—3)

= 30 Ã— 30

= (30)^{2}

âˆ´ Resultant is the square of 30.

**(iii)** 2904

First find the prime factors for 2904

2904 = 2Ã—2Ã—11Ã—11Ã—2Ã—3

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (11Ã—11) Ã— 2 Ã— 3

By observation, prime factor 2 and 3 are left out.

So, divide by 6 to eliminate 2 and 3 we get,

2904/6 = (2Ã—2) Ã— (11Ã—11)

= (2Ã—11) Ã— (2Ã—11)

= 22 Ã— 22

= (22)^{2}

âˆ´ Resultant is the square of 22.

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

**Solution:**

11 it is a prime number by itself.

So it is not a perfect square.

12 is not a perfect square.

16= (4)^{2}

16 is a perfect square.

32 is not a perfect square.

36= (6)^{2}

36 is a perfect square.

50 is not a perfect square.

64= (8)^{2}

64 is a perfect square.

79 it is a prime number.

So it is not a perfect square.

81= (9)^{2}

81 is a perfect square.

111 it is a prime number.

So it is not a perfect square.

121= (11)^{2}

121 is a perfect square.

**6. Using prime factorization method, find which of the following numbers are perfect squares?
189, 225, 2048, 343, 441, 2961, 11025, 3549**

**Solution:**

189 prime factors are

189 = 3^{2}Ã—3Ã—7

Since it does not have equal pair of factors 189 is not a perfect square.

225 prime factors are

225 = (5Ã—5) Ã— (3Ã—3)

Since 225 has equal pair of factors. âˆ´ It is a perfect square.

2048 prime factors are

2048 = (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— 2

Since it does not have equal pair of factors 2048 is not a perfect square.

343 prime factors are

343 = (7Ã—7) Ã— 7

Since it does not have equal pair of factors 2048 is not a perfect square.

441 prime factors are

441 = (7Ã—7) Ã— (3Ã—3)

Since 441 has equal pair of factors. âˆ´ It is a perfect square.

2961 prime factors are

2961 = (3Ã—3) Ã— (3Ã—3) Ã— (3Ã—3) Ã— (2Ã—2)

Since 2961 has equal pair of factors. âˆ´ It is a perfect square.

11025 prime factors are

11025 = (3Ã—3) Ã— (5Ã—5) Ã— (7Ã—7)

Since 11025 has equal pair of factors. âˆ´ It is a perfect square.

3549 prime factors are

3549 = (13Ã—13) Ã— 3 Ã— 7

Since it does not have equal pair of factors 3549 is not a perfect square.

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

**(i)Â 8820**

**(ii) 3675
(iii) 605 **

**(iv) 2880
(v) 4056 **

**(vi) 3468
(vii) 7776**

**Solution:**

**(i)** 8820

First find the prime factors for 8820

8820 = 2Ã—2Ã—3Ã—3Ã—7Ã—7Ã—5

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (3Ã—3) Ã— (7Ã—7) Ã— 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

8820 Ã— 5 = (2Ã—2) Ã— (3Ã—3) Ã— (7Ã—7) Ã— (5Ã—5)

= (2Ã—3Ã—7Ã—5) Ã— (2Ã—3Ã—7Ã—5)

= 210 Ã— 210

= (210)^{2}

âˆ´ Product is the square of 210.

**(ii)** 3675

First find the prime factors for 3675

3675 = 5Ã—5Ã—7Ã—7Ã—3

By grouping the prime factors in equal pairs we get,

= (5Ã—5) Ã— (7Ã—7) Ã— 3

By observation, prime factor 3 is left out.

So, multiply by 3 we get,

3675 Ã— 3 = (5Ã—5) Ã— (7Ã—7) Ã— (3Ã—3)

= (5Ã—7Ã—3) Ã— (5Ã—7Ã—3)

= 105 Ã— 105

= (105)^{2}

âˆ´ Product is the square of 105.

**(iii)** 605

First find the prime factors for 605

605 = 5Ã—11Ã—11

By grouping the prime factors in equal pairs we get,

= (11Ã—11) Ã— 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

605 Ã— 5 = (11Ã—11) Ã— (5Ã—5)

= (11Ã—5) Ã— (11Ã—5)

= 55 Ã— 55

= (55)^{2}

âˆ´ Product is the square of 55.

**(iv)** 2880

First find the prime factors for 2880

2880 = 5Ã—3Ã—3Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

2880 Ã— 5 = (3Ã—3) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (5Ã—5)

= (3Ã—2Ã—2Ã—2Ã—5) Ã— (3Ã—2Ã—2Ã—2Ã—5)

= 120 Ã— 120

= (120)^{2}

âˆ´ Product is the square of 120.

**(v)** 4056

First find the prime factors for 4056

4056 = 2Ã—2Ã—13Ã—13Ã—2Ã—3

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (13Ã—13) Ã— 2 Ã— 3

By observation, prime factors 2 and 3 are left out.

So, multiply by 6 we get,

4056 Ã— 6 = (2Ã—2) Ã— (13Ã—13) Ã— (2Ã—2) Ã— (3Ã—3)

= (2Ã—13Ã—2Ã—3) Ã— (2Ã—13Ã—2Ã—3)

= 156 Ã— 156

= (156)^{2}

âˆ´ Product is the square of 156.

**(vi)** 3468

First find the prime factors for 3468

3468 = 2Ã—2Ã—17Ã—17Ã—3

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (17Ã—17) Ã— 3

By observation, prime factor 3 is left out.

So, multiply by 3 we get,

3468 Ã— 3 = (2Ã—2) Ã— (17Ã—17) Ã— (3Ã—3)

= (2Ã—17Ã—3) Ã— (2Ã—17Ã—3)

= 102 Ã— 102

= (102)^{2}

âˆ´ Product is the square of 102.

**(vii)** 7776

First find the prime factors for 7776

7776 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—2Ã—3

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3) Ã— (3Ã—3) Ã— 2 Ã— 3

By observation, prime factors 2 and 3 are left out.

So, multiply by 6 we get,

7776 Ã— 6 = (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3) Ã— (3Ã—3) Ã— (2Ã—2) Ã— (3Ã—3)

= (2Ã—2Ã—3Ã—3Ã—2Ã—3) Ã— (2Ã—2Ã—3Ã—3Ã—2Ã—3)

= 216 Ã— 216

= (216)^{2}

âˆ´ Product is the square of 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.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575**

**Solution:**

**(i)** 16562

First find the prime factors for 16562

16562 = 7Ã—7Ã—13Ã—13Ã—2

By grouping the prime factors in equal pairs we get,

= (7Ã—7) Ã— (13Ã—13) Ã— 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

16562/2 = (7Ã—7) Ã— (13Ã—13)

= (7Ã—13) Ã— (7Ã—13)

= 91 Ã— 91

= (91)^{2}

âˆ´ Resultant is the square of 91.

**(ii)** 3698

First find the prime factors for 3698

3698 = 2Ã—43Ã—43

By grouping the prime factors in equal pairs we get,

= (43Ã—43) Ã— 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

3698/2 = (43Ã—43)

= (43)^{2}

âˆ´ Resultant is the square of 43.

**(iii)** 5103

First find the prime factors for 5103

5103 = 3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—7

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (3Ã—3) Ã— (3Ã—3) Ã— 7

By observation, prime factor 7 is left out.

So, divide by 7 to eliminate 7 we get,

5103/7 = (3Ã—3) Ã— (3Ã—3) Ã— (3Ã—3)

= (3Ã—3Ã—3) Ã— (3Ã—3Ã—3)

= 27 Ã— 27

= (27)^{2}

âˆ´ Resultant is the square of 27.

**(iv)** 3174

First find the prime factors for 3174

3174 = 2Ã—3Ã—23Ã—23

By grouping the prime factors in equal pairs we get,

= (23Ã—23) Ã— 2 Ã— 3

By observation, prime factor 2 and 3 are left out.

So, divide by 6 to eliminate 2 and 3 we get,

3174/6 = (23Ã—23)

= (23)^{2}

âˆ´ Resultant is the square of 23.

**(v)** 1575

First find the prime factors for 1575

1575 = 3Ã—3Ã—5Ã—5Ã—7

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (5Ã—5) Ã— 7

By observation, prime factor 7 is left out.

So, divide by 7 to eliminate 7 we get,

1575/7 = (3Ã—3) Ã— (5Ã—5)

= (3Ã—5) Ã— (3Ã—5)

= 15 Ã— 15

= (15)^{2}

âˆ´ Resultant is the square of 15.

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

**Solution:**

We know that the two digit greatest number is 99

âˆ´ Greatest two digit perfect square number is 99-18 = 81

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

**Solution: **

We know that the three digit greatest number is 100

To find the square root of 100

âˆ´ the least number of three digits which is a perfect square is 100 itself.

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

**Solution: **

First find the prime factors for 4851

4851 = 3Ã—3Ã—7Ã—7Ã—11

By grouping the prime factors in equal pairs we get,

= (3Ã—3) Ã— (7Ã—7) Ã— 11

âˆ´ The smallest number by which 4851 must be multiplied so that the product becomes a perfect square is 11.

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

**Solution:**

First find the prime factors for 28812

28812 = 2Ã—2Ã—3Ã—7Ã—7Ã—7Ã—7

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— 3 Ã— (7Ã—7) Ã— (7Ã—7)

âˆ´ The smallest number by which 28812 must be divided so that the quotient becomes a perfect square is 3.

**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.**

**Solution:**

First find the prime factors for 1152

1152 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3) Ã— 2

âˆ´ The smallest number by which 1152 must be divided so that the quotient becomes a perfect square is 2.

The number after division, 1152/2 = 576

prime factors for 576 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3

By grouping the prime factors in equal pairs we get,

= (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3)

= 2^{6} Ã— 3^{2}

= 24^{2}

âˆ´ The resulting number is the square of 24.

### EXERCISE 3.2 PAGE NO: 3.18

**1. The following numbers are not perfect squares. Give reason.
(i) 1547 **

**(ii) 45743
(iii)8948 **

**(iv) 333333**

**Solution:**

The numbers ending with 2, 3, 7 or 8 is not a perfect square.

So, **(i) **1547

** (ii) **45743**
(iii) **8948

** (iv) **333333

Are not perfect squares.

**2. Show that the following numbers are not, perfect squares:
(i) 9327 **

**(ii) 4058
(iii)22453 **

**(iv) 743522**

**Solution:**

The numbers ending with 2, 3, 7 or 8 is not a perfect square.

So, **(i)** 9327

**(ii)** 4058

**(iii)** 22453

**(iv)** 743522

Are not perfect squares.

**3. The square of which of the following numbers would be an old number?
(i) 731 **

**(ii) 3456
(iii)5559 **

**(iv) 42008**

**Solution:**

We know that square of an even number is even number.

Square of an odd number is odd number.

**(i)** 731

Since 731 is an odd number, the square of the given number is also odd.

**(ii)** 3456

Since 3456 is an even number, the square of the given number is also even.

**(iii)** 5559

Since 5559 is an odd number, the square of the given number is also odd.

**(iv)** 42008

Since 42008 is an even number, the square of the given number is also even.

**4. What will be the unitâ€™s digit of the squares of the following numbers?
(i) 52 **

**(ii) 977
(iii) 4583 **

**(iv) 78367
(v) 52698 **

**(vi) 99880
(vii) 12796 **

**(viii) 55555
(ix) 53924**

**Solution: **

**(i)** 52

Unit digit of (52)^{2} = (2^{2}) = 4

**(ii)** 977

Unit digit of (977)^{2} = (7^{2}) = 49 = 9

**(iii)** 4583

Unit digit of (4583)^{2} = (3^{2}) = 9

**(iv)** 78367

Unit digit of (78367)^{2} = (7^{2}) = 49 = 9

**(v)** 52698

Unit digit of (52698)^{2} = (8^{2}) = 64 = 4

**(vi)** 99880

Unit digit of (99880)^{2} = (0^{2}) = 0

**(vii)** 12796

Unit digit of (12796)^{2} = (6^{2}) = 36 = 6

**(viii)** 55555

Unit digit of (55555)^{2} = (5^{2}) = 25 = 5

**(ix)** 53924

Unit digit of (53924)^{2} = (4^{2}) = 16 = 6

**5. Observe the following pattern
1+3 = 2 ^{2}**

**1+3+5 = 3 ^{2}**

**1+3+5+7 = 4 ^{2}
And write the value of 1+3+5+7+9+â€¦â€¦â€¦ up to n terms.**

**Solution:**

We know that the pattern given is the square of the given number on the right hand side is equal to the sum of the given numbers on the left hand side.

âˆ´ The value of 1+3+5+7+9+â€¦â€¦â€¦ up to n terms = n^{2} (as there are only n terms).

**6. Observe the following pattern**

**2 ^{2} -1^{2} = 2 + 1**

**3 ^{2} â€“ 2^{2} = 3 + 2**

**4 ^{2} â€“ 3^{2} = 4 + 3**

**5 ^{2} â€“ 4^{2} = 5 + 4**

**And find the value of**

**(i) 100 ^{2} -99^{2}**

**(ii)111 ^{2} – 109^{2}**

**(iii) 99 ^{2} â€“ 96^{2}**

**Solution:**

**(i)** 100^{2} -99^{2}

100 + 99 = 199

**(ii)** 111^{2} â€“ 109^{2}

(111^{2} â€“ 110^{2}) + (110^{2} â€“ 109^{2})

(111 + 110) + (100 + 109)

440

**(iii)** 99^{2} â€“ 96^{2}

(99^{2} â€“ 98^{2}) + (98^{2} â€“ 97^{2}) + (97^{2} â€“ 96^{2})

(99 + 98) + (98 + 97) + (97 + 96)

585

**7. Which of the following triplets are Pythagorean?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(v) (16, 63, 65)
(vi) (12, 35, 38)**

**Solution:**

**(i)** (8, 15, 17)

LHS = 8^{2} + 15^{2}

= 289

RHS = 17^{2}

= 289

LHS = RHS

âˆ´ The given triplet is a Pythagorean.

**(ii)** (18, 80, 82)

LHS = 18^{2} + 80^{2}

= 6724

RHS = 82^{2}

= 6724

LHS = RHS

âˆ´ The given triplet is a Pythagorean.

**(iii)** (14, 48, 51)

LHS = 14^{2} + 48^{2}

= 2500

RHS = 51^{2}

= 2601

LHS â‰ RHS

âˆ´ The given triplet is not a Pythagorean.

**(iv)** (10, 24, 26)

LHS = 10^{2} + 24^{2}

= 676

RHS = 26^{2}

= 676

LHS = RHS

âˆ´ The given triplet is a Pythagorean.

**(v)** (16, 63, 65)

LHS = 16^{2} + 63^{2}

= 4225

RHS = 65^{2}

= 4225

LHS = RHS

âˆ´ The given triplet is a Pythagorean.

**(vi)** (12, 35, 38)

LHS = 12^{2} + 35^{2}

= 1369

RHS = 38^{2}

= 1444

LHS â‰ RHS

âˆ´ The given triplet is not a Pythagorean.

**8. Observe the following pattern**

**(1Ã—2) + (2Ã—3) = (2Ã—3Ã—4)/3**

**(1Ã—2) + (2Ã—3) + (3Ã—4) = (3Ã—4Ã—5)/3**

**(1Ã—2) + (2Ã—3) + (3Ã—4) + (4Ã—5) = (4Ã—5Ã—6)/3**

**And find the value of**

**(1Ã—2) + (2Ã—3) + (3Ã—4) + (4Ã—5) + (5Ã—6)**

**Solution:**

(1Ã—2) + (2Ã—3) + (3Ã—4) + (4Ã—5) + (5Ã—6) = (5Ã—6Ã—7)/3 = 70

**9. Observe the following pattern**

**1 = 1/2 (1Ã—(1+1))**

**1+2 = 1/2 (2Ã—(2+1))**

**1+2+3 = 1/2 (3Ã—(3+1))**

**1+2+3+4 = 1/2 (4Ã—(4+1))**

**And find the values of each of the following:**

**(i) 1+2+3+4+5+â€¦+50**

**(ii) 31+32+â€¦.+50**

**Solution:**

We know that R.H.S =Â 1/2Â [No. of terms in L.H.S Ã— (No. of terms + 1)] (if only when L.H.S starts with 1)

**(i)** 1+2+3+4+5+â€¦+50 = 1/2 (5Ã—(5+1))

25 Ã— 51 = 1275

**(ii)** 31+32+â€¦.+50 = (1+2+3+4+5+â€¦+50) â€“ (1+2+3+â€¦+30)

1275 â€“ 1/2 (30Ã—(30+1))

1275 â€“ 465

810

**10. Observe the following pattern**

**1 ^{2} = 1/6 (1Ã—(1+1)Ã—(2Ã—1+1))**

**1 ^{2}+2^{2} = 1/6 (2Ã—(2+1)Ã—(2Ã—2+1)))**

**1 ^{2}+2^{2}+3^{2} = 1/6 (3Ã—(3+1)Ã—(2Ã—3+1)))**

**1 ^{2}+2^{2}+3^{2}+4^{2} = 1/6 (4Ã—(4+1)Ã—(2Ã—4+1)))**

**And find the values of each of the following:**

**(i) 1 ^{2}+2^{2}+3^{2}+4^{2}+â€¦+10^{2}**

**(ii) 5 ^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2}**

**Solution:**

RHS = 1/6Â [(No. of terms in L.H.S) Ã— (No. of terms + 1) Ã— (2 Ã— No. of terms + 1)]

**(i)** 1^{2}+2^{2}+3^{2}+4^{2}+â€¦+10^{2} = 1/6 (10Ã—(10+1)Ã—(2Ã—10+1))

= 1/6 (2310)

= 385

**(ii)** 5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2} = 1^{2}+2^{2}+3^{2}+â€¦+12^{2} â€“ (1^{2}+2^{2}+3^{2}+4^{2})

1/6 (12Ã—(12+1)Ã—(2Ã—12+1)) – 1/6 (4Ã—(4+1)Ã—(2Ã—4+1))

650-30

620

**11. Which of the following numbers are squares of even numbers?
121, 225, 256, 324, 1296, 6561, 5476, 4489, 373758**

**Solution:**

We know that only even numbers be the squares of even numbers.

So, 256, 324, 1296, 5476, 373758 are even numbers, since 373758 is not a perfect square

âˆ´ 256, 324, 1296, 5476 are squares of even numbers.

**12. By just examining the units digits, can you tell which of the following cannot be whole squares?**

(i) 1026

**(ii) 1028
(iii)1024 **

**(iv) 1022
(v) 1023 **

**(vi) 1027**

**Solution:**

We know that numbers ending with 2, 3, 7, 8 cannot be a perfect square.

âˆ´ 1028, 1022, 1023, and 1027 cannot be whole squares.

**13. Which of the numbers for which you cannot decide whether they are squares.**

**Solution:**

We know that the natural numbers such as 0, 1, 4, 5, 6 or 9 cannot be decided surely whether they are squares or not.

**14. Write five numbers which you cannot decide whether they are square just by looking at the unitâ€™s digit.**

**Solution:**

We know that any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number.

Here are the five examples which you cannot decide whether they are square or not just by looking at the units place:

**(i)** 2061

The unit digit is 1. So, it may or may not be a square number

**(ii)** 1069

The unit digit is 9. So, it may or may not be a square number

**(iii)** 1234

The unit digit is 4. So, it may or may not be a square number

**(iv)** 56790

The unit digit is 0. So, it may or may not be a square number

**(v)** 76555

The unit digit is 5. So, it may or may not be a square number

**15.** **Write true (T) or false (F) for the following statements.
(i) The number of digits in a square number is even.
(ii) The square of a prime number is prime.
(iii) The sum of two square numbers is a square number.
(iv) The difference of two square numbers is a square number.
(v) The product of two square numbers is a square number.
(vi) No square number is negative.
(vii) There is no square number between 50 and 60.
(viii) There are fourteen square number up to 200.**

**Solution:**

**(i)** False, because 169 is a square number with odd digit.

**(ii)** False, because square of 3(which is prime) is 9(which is not prime).

**(iii)** False, because sum of 2^{2}Â and 3^{2}Â is 13 which is not square number.

**(iv)** False, because difference of 3^{2}Â and 2^{2}Â is 5, which is not square number.

**(v)** True, because the square of 2^{2}Â and 3^{2}Â is 36 which is square of 6

**(vi)** True, because (-2)^{2} is 4, which is not negative.

**(vii)** True, because as there is no square number between them.

**(viii)** True, because the fourteen numbers up to 200 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196.

### EXERCISE 3.3 PAGE NO: 3.32

**1. Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication:
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96**

**Solution:**

**(i) **25

So here, a = 2 and b = 5

Column I | Column II | Column III |

a^{2}
4 +2 6 |
2ab
20 +2 22 |
b^{2}
25 |

6 | 2 | 5 |

âˆ´ 25^{2}Â = 625

Where, it can be expressed as

25^{2}Â = 25Ã— 25 = 625

**(ii)** 37

So here, a = 3 and b = 7

Column I | Column II | Column III |

a^{2}
9 +4 13 |
2ab
42 +4 46 |
b^{2}
49 |

13 | 6 | 9 |

âˆ´ 37^{2}Â = 1369

Where, it can be expressed as

25^{2}Â = 37Ã— 37 = 1369

**(iii) 54**

So here, a = 5 and b = 4

Column I | Column II | Column III |

a^{2}
25 +4 29 |
2ab
40 +1 41 |
b^{2}
16 |

29 | 1 | 6 |

âˆ´ 54^{2}Â = 2916

Where, it can be expressed as

54^{2}Â = 54 Ã— 54 = 2916

**(iv) **71

So here, a = 7 and b = 1

Column I | Column II | Column III |

a^{2}
49 +1 50 |
2ab
14 +0 14 |
b^{2}
01 |

50 | 4 | 1 |

âˆ´ 71^{2}Â = 5041

Where, it can be expressed as

71^{2}Â = 71 Ã— 71 = 5041

**(v)** 96

So here, a = 9 and b = 6

Column I | Column II | Column III |

a^{2}
81 +11 92 |
2ab
108 +3 111 |
b^{2}
36 |

92 | 1 | 6 |

âˆ´ 96^{2}Â = 9216

Where, it can be expressed as

96^{2}Â = 96 Ã— 96 = 9216

**2. Find the squares of the following numbers using diagonal method:
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171**

**Solution:**

**(i)** 98

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.

Step 3: Draw the diagonals of each sub-square.

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step 7: Obtain the required square by writing the digits from the left-most side.

âˆ´ 98^{2}Â = 9604

**(ii)** 273

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.

Step 3: Draw the diagonals of each sub-square.

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step 7: Obtain the required square by writing the digits from the left-most side.

âˆ´ 273^{2}Â = 74529

**(iii)** 348

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.

Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.

Step 3: Draw the diagonals of each sub-square.

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.

Step 7: Obtain the required square by writing the digits from the left-most side.

âˆ´ 348^{2}Â = 121104

**(iv)** 295

^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.

Step 3: Draw the diagonals of each sub-square.

Step 7: Obtain the required square by writing the digits from the left-most side.

âˆ´ 295^{2}Â = 87025

**(v)** 171

^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.

Step 3: Draw the diagonals of each sub-square.

Step 7: Obtain the required square by writing the digits from the left-most side.

âˆ´ 171^{2}Â = 29241

**3. Find the squares of the following numbers:
(i) 127 **

**(ii) 503
(iii) 450 **

**(iv) 862
(v) 265**

**Solution:**

**(i) **127

127^{2} = 127 Ã— 127 = 16129

**(ii) **503

503^{2} = 503 Ã— 503 = 253009

**
(iii) **450

450^{2} = 450 Ã— 450 = 203401

**(iv) **862

862^{2} = 862 Ã— 862 = 743044

**
(v) **265

265^{2} = 265 Ã— 265 = 70225

**4. Find the squares of the following numbers:
(i) 425 **

**(ii) 575
(iii)405 **

**(iv) 205
(v) 95 **

**(vi) 745
(vii) 512 **

**(viii) 995**

**Solution:**

**(i)**425

425^{2} = 425 Ã— 425 = 180625

**(ii) 575**

575^{2} = 575 Ã— 575 = 330625

(iii)405

405^{2} = 405 Ã— 405 = 164025

**(iv) 205**

205^{2} = 205 Ã— 205 = 42025

(v) 95

95^{2} = 95 Ã— 95 = 9025

**(vi) 745**

745^{2} = 745 Ã— 745 = 555025

(vii) 512

512^{2} = 512 Ã— 512 = 262144

**(viii) 995**

995^{2} = 995 Ã— 995 = 990025

**5. Find the squares of the following numbers using the identityÂ (a+b) ^{ 2}= a^{2}+2ab+b^{2}:**

(i) 405

(ii) 510

(iii) 1001

(iv) 209

(v) 605

**Solution:**

**(i)** 405

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

405 = (400+5)^{2}

= (400)^{2}Â + 5^{2}Â + 2 (400) (5)

= 160000 + 25 + 4000

= 164025

**(ii) **510

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

510 = (500+10)^{2}

= (500)^{2}Â + 10^{2}Â + 2 (500) (10)

= 250000 + 100 + 10000

= 260100

**
(iii) **1001

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

1001 = (1000+1)^{2}

= (1000)^{2}Â + 1^{2}Â + 2 (1000) (1)

= 1000000 + 1 + 2000

= 1002001

**
(iv) **209

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

209 = (200+9)^{2}

= (200)^{2}Â + 9^{2}Â + 2 (200) (9)

= 40000 + 81 + 3600

= 43681

**
(v)** 605

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

605 = (600+5)^{2}

= (600)^{2}Â + 5^{2}Â + 2 (600) (5)

= 360000 + 25 + 6000

= 366025

**6. Find the squares of the following numbers using the identityÂ (a-b) ^{ 2}= a^{2}-2ab+b^{2}**

(i) 395

**(ii) 995
(iii)495 **

**(iv) 498
(v) 99 **

**(vi) 999
(vii)599**

**Solution:**

**(i) **395

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

395 = (400-5)^{2}

= (400)^{2}Â + 5^{2}Â – 2 (400) (5)

= 160000 + 25 â€“ 4000

= 156025

**(ii) **995

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

995 = (1000-5)^{2}

= (1000)^{2}Â + 5^{2}Â – 2 (1000) (5)

= 1000000 + 25 â€“ 10000

= 990025

**
(iii) **495

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

495 = (500-5)^{2}

= (500)^{2}Â + 5^{2}Â – 2 (500) (5)

= 250000 + 25 â€“ 5000

= 245025

**(iv) **498

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

498 = (500-2)^{2}

= (500)^{2}Â + 2^{2}Â – 2 (500) (2)

= 250000 + 4 â€“ 2000

= 248004

**
(v) **99

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

99 = (100-1)^{2}

= (100)^{2}Â + 1^{2}Â – 2 (100) (1)

= 10000 + 1 â€“ 200

= 9801

**(vi) **999

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

999 = (1000-1)^{2}

= (1000)^{2}Â + 1^{2}Â – 2 (1000) (1)

= 1000000 + 1 â€“ 2000

= 998001

**
(vii) **599

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

599 = (600-1)^{2}

= (600)^{2}Â + 1^{2}Â – 2 (600) (1)

= 360000 + 1 â€“ 1200

= 358801

**7. Find the squares of the following numbers by visual method:
(i) 52 **

**(ii) 95
(iii) 505 **

**(iv) 702
(v) 99**

**Solution:**

**(i) **52

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

52 = (50+2)^{2}

= (50)^{2}Â + 2^{2}Â + 2 (50) (2)

= 2500 + 4 + 200

= 2704

**(ii) **95

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

95 = (100-5)^{2}

= (100)^{2}Â + 5^{2}Â – 2 (100) (5)

= 10000 + 25 – 1000

= 9025

**
(iii) **505

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

505 = (500+5)^{2}

= (500)^{2}Â + 5^{2}Â + 2 (500) (5)

= 250000 + 25 + 5000

= 255025

**(iv) **702

We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}

702 = (700+2)^{2}

= (700)^{2}Â + 2^{2}Â + 2 (700) (2)

= 490000 + 4 + 2800

= 492804

**
(v) **99

We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}

99 = (100-1)^{2}

= (100)^{2}Â + 1^{2}Â – 2 (100) (1)

= 10000 + 1 – 200

= 9801

### EXERCISE 3.4 PAGE NO: 3.38

**1.Write the possible unitâ€™s digits of the square root of the following numbers. Which of these numbers are odd square roots?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025**

**Solution:**

**(i)**Â 9801

We know that unit digit of 9801 is 1

Unit digit of square root = 1 or 9

Since the number is odd, square root is also odd

**(ii)**Â 99856

We know that unit digit of 99856 = 6

Unit digit of square root = 4 or 6

Since the number is even, square root is also even

**(iii)**Â 998001

We know that unit digit of 998001 = 1

Unit digit of square root = 1 or 9

Since the number is odd, square root is also odd

**(iv)**Â 657666025

We know that unit digit of 657666025 = 5

Unit digit of square root = 5

Since the number is odd, square root is also odd

**2. Find the square root of each of the following by prime factorization.
(i) 441 (ii) 196
(iii) 529 (iv) 1764
(v) 1156 (vi) 4096
(vii) 7056 (viii) 8281
(ix) 11664 (x) 47089
(xi) 24336 (xii) 190969
(xiii) 586756 (xiv) 27225
(xv) 3013696**

**Solution:**

**(i) **441

Firstly letâ€™s find the prime factors for

441 = 3Ã—3Ã—7Ã—7

= 3^{2}Ã—7^{2}

âˆš441 = 3Ã—7

= 21

**(ii)** 196

Firstly letâ€™s find the prime factors for

196 = 2Ã—2Ã—7Ã—7

= 2^{2}Ã—7^{2}

âˆš196 = 2Ã—7

= 14

**
(iii) **529

Firstly letâ€™s find the prime factors for

529 = 23Ã—23

= 23^{2}

âˆš529 = 23

**(iv) **1764

Firstly letâ€™s find the prime factors for

1764 = 2Ã—2Ã—3Ã—3Ã—7Ã—7

= 2^{2}Ã—3^{2}Ã—7^{2}

âˆš1764 = 2Ã—3Ã—7

= 42

**
(v) **1156

Firstly letâ€™s find the prime factors for

1156 = 2Ã—2Ã—17Ã—17

= 2^{2}Ã—17^{2}

âˆš1156 = 2Ã—17

= 34

**(vi) **4096

Firstly letâ€™s find the prime factors for

4096 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2

= 2^{12}

âˆš4096 = 2^{6}

= 64

**(vii) **7056

Firstly letâ€™s find the prime factors for

7056 = 2Ã—2Ã—2Ã—2Ã—21Ã—21

= 2^{2}Ã—2^{2}Ã—21^{2}

âˆš7056 = 2Ã—2Ã—21

= 84

**(viii) **8281

Firstly letâ€™s find the prime factors for

8281 = 91Ã—91

= 91^{2}

âˆš8281 = 91

**
(ix) **11664

Firstly letâ€™s find the prime factors for

11664 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3

= 2^{2}Ã—2^{2}Ã—3^{2}Ã—3^{2}Ã—3^{2}

âˆš11664 = 2Ã—2Ã—3Ã—3Ã—3

= 108

**(x) **47089

Firstly letâ€™s find the prime factors for

47089 = 217Ã—217

= 217^{2}

âˆš47089 = 217

**
(xi)** 24336

Firstly letâ€™s find the prime factors for

24336 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—13Ã—13

= 2^{2}Ã—2^{2}Ã—3^{2}Ã—13^{2}

âˆš24336 = 2Ã—2Ã—3Ã—13

= 156

**(xii) **190969

Firstly letâ€™s find the prime factors for

190969 = 23Ã—23Ã—19Ã—19

= 23^{2}Ã—19^{2}

âˆš190969 = 23Ã—19

= 437

**
(xiii) **586756

Firstly letâ€™s find the prime factors for

586756 = 2Ã—2Ã—383Ã—383

= 2^{2}Ã—383^{2}

âˆš586756 = 2Ã—383

= 766

**(xiv) **27225

Firstly letâ€™s find the prime factors for

27225 = 5Ã—5Ã—3Ã—3Ã—11Ã—11

= 5^{2}Ã—3^{2}Ã—11^{2}

âˆš27225 = 5Ã—3Ã—11

= 165

**
(xv) **3013696

Firstly letâ€™s find the prime factors for

3013696 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—217Ã—217

= 2^{6}Ã—217^{2}

âˆš3013696 = 2^{3}Ã—217

= 1736

**3.Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.**

**Solution:**

Firstly letâ€™s find the prime factors for

180 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— 5

=2^{2}Â Ã— 3^{2}Â Ã— 5

To make the unpaired 5 into paired, multiply the number with 5

180 Ã— 5 = 2^{2}Â Ã— 3^{2}Â Ã— 5^{2}

âˆ´ Square root of âˆš (180 Ã— 5) = 2Â Ã— 3Â Ã— 5

= 30

**4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.**

**Solution:**

Firstly letâ€™s find the prime factors for

147 = (7 Ã— 7) Ã— 3

=7^{2}Â Ã— 3

To make the unpaired 3 into paired, multiply the number with 3

147 Ã— 3 = 7^{2}Â Ã— 3^{2}

âˆ´ Square root of âˆš (147 Ã— 3) = 7Â Ã— 3

= 21

**5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.**

**Solution:**

Firstly letâ€™s find the prime factors for

3645 = (3 Ã— 3) Ã— (3 Ã— 3) Ã— (3 Ã— 3) Ã— 5

=3^{2}Â Ã— 3^{2} Ã— 3^{2}Â Ã— 5

To make the unpaired 5 into paired, the number 3645 has to be divided by 5

3645 Ã· 5 = 3^{2} Ã— 3^{2} Ã— 3^{2}

âˆ´ Square root of âˆš (3645 Ã· 5) = 3Â Ã— 3 Ã— 3

= 27

**6. Find the smallest number by which 1152 must be divided so that it becomes a square. Also, find the square root of the number so obtained.**

**Solution:**

Firstly letâ€™s find the prime factors for

1152 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— 2 Ã— (3 Ã— 3)

=2^{2}Â Ã— 2^{2} Ã— 2^{2} Ã— 3^{2} Ã— 2

To make the unpaired 2 into paired, the number 1152 has to be divided by 2

1152 Ã· 2 = 2^{2}Â Ã— 2^{2 }Ã— 2^{2} Ã— 3^{2}

âˆ´ Square root of âˆš (1152 Ã· 2) = 2Â Ã— 2 Ã— 2 Ã— 3

= 24

**7. The product of two numbers is 1296. If one number is 16 times the other, find the numbers.**

**Solution:**

Let us consider two numbers a and b

So we know that one of the number, a =16b

a Ã— b = 1296

16b Ã— b = 1296

16b^{2} = 1296

b^{2} = 1296/16 = 81

b = 9

a = 16b

= 16(9)

= 144

âˆ´ a =144 and b =9

**8. A welfare association collected Rs 202500 as donation from the residents. If each paid as many rupees as there were residents, find the number of residents.**

**Solution:**

Let us consider total residents as a

So, each paid Rs. a

Total collection = a (a) = a^{2}

We know that the total Collection = 202500

a =Â âˆš 202500

a = âˆša(2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— 5 Ã— 5)

= 2 Ã— 3 Ã— 3 Ã— 5 Ã— 5a

= 450

âˆ´ Total residents = 450

**9. A society collected Rs 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?**

**Solution:**

Let us consider there were few members, each attributed a paise

a (a), i.e. total cost collected = 9216 paise

a^{2}Â = 9216

a =Â âˆš9216

= 2 Ã— 2 Ã— 2 Ã— 12

= 96

âˆ´ There were 96 members in the society and each contributed 96 paise

**10. A society collected Rs 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school?**

**Solution:**

Let us consider number of school students as a

each student contributed a paise

Total money obtained = a^{2}paise

= 2304 paise

a =Â âˆš2304

a = âˆš2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

a = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

a = 48

âˆ´ There were 48 students in the school

**11. The area of a square field is 5184 m ^{2}. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.**

**Solution:**

Let us consider the side of square field as a

a^{2}Â = 5184 m^{2}

a =Â âˆš5184m

a = 2 Ã— 2 Ã—2 Ã— 9

= 72 m

Perimeter of square = 4a

= 4(72)

= 288 m

Perimeter of rectangle = 2 (l + b) = perimeter of the square field

= 288 m

2 (2b + b) = 288

b = 48 and l = 96

Area of rectangle = 96 Ã— 48 m^{2}

= 4608 m^{2}

**12. Find the least square number, exactly divisible by each one of the numbers: (i) 6, 9, 15 and 20 (ii) 8, 12, 15 and 20**

**Solution:**

**(i)** 6, 9, 15 and 20

Firstly take L.C.M for 6, 9, 15, 20 which is 180

So the prime factors of 180 = 2^{2}Â Ã— 3^{2}Â Ã— 5

To make it a perfect square, we have to multiply the number with 5

180 Ã— 5 = 2^{2}Â Ã— 3^{2}Â Ã— 5^{2}

âˆ´ 900 is the least square number divisible by 6, 9, 15 and 20

**(ii)** 8, 2, 15 and 20

Firstly take L.C.M for 8, 2, 15, 20 which is 360

So the prime factors of 360 = 2^{2}Â Ã— 3^{2}Â Ã—2 Ã— 5

To make it a perfect square, we have to multiply the number with 2 Ã— 5 = 10

360 Ã— 10 = 2^{2}Â Ã— 3^{2}Â Ã— 5^{2}Â Ã— 2^{2}

âˆ´ 3600 is the least square number divisible by 8, 12, 15 and 20

**13. Find the square roots of 121 and 169 by the method of repeated subtraction.**

**Solution:**

Let us find the square roots of 121 and 169 by the method of repeated subtraction

121 â€“ 1 = 120

120 â€“ 3 = 117

117 â€“ 5 = 112

112 â€“ 7 = 105

105 â€“ 9 = 96

96 â€“ 11 = 85

85 â€“ 13 = 72

72 â€“ 15 = 57

57 â€“ 17 = 40

40 â€“ 19 = 21

21 â€“ 21 = 0

Clearly, we have performed operation 11 times

âˆ´ âˆš121Â = 11

169 â€“ 1 = 168

168 â€“ 3 = 165

165 â€“ 5 = 160

160 â€“ 7 = 153

153 â€“ 9 = 144

144 â€“ 11 = 133

133 â€“ 13 = 120

120 â€“ 15 = 105

105 â€“ 17 = 88

88 â€“ 19 = 69

69 â€“ 21 = 48

48 â€“ 23 = 25

25 â€“ 25 = 0

Clearly, we have performed subtraction 13 times

âˆ´ âˆš169Â = 13

**14. Write the prime factorization of the following numbers and hence find their square roots.
(i) 7744**

**(ii) 9604
(iii) 5929
(iv) 7056**

**Solution:**

**(i)**Â 7744

Prime factors of 7744 is

7744 = 2^{2}Â Ã— 2^{2}Â Ã— 2^{2}Â Ã— 11^{2}

âˆ´ The square root of 7744 is

âˆš7744Â = 2 Ã— 2 Ã— 2 Ã— 11

= 88

**(ii)**Â 9604

Prime factors of 9604 is

9604 = 2^{2}Â Ã— 7^{2}Â Ã— 7^{2}

âˆ´ The square root of 9604 is

âˆš9604Â = 2 Ã— 7 Ã— 7

= 98

**(iii)**Â 5929

Prime factors of 5929 is

5929 = 11^{2}Â Ã— 7^{2}

âˆ´ The square root of 5929 is

âˆš5929Â = 11 Ã— 7

= 77

**(iv)**Â 7056

Prime factors of 7056 is

7056 = 2^{2}Â Ã— 2^{2}Â Ã— 7^{2}Â Ã— 3^{2}

âˆ´ The square root of 7056 is

âˆš7056Â = 2 Ã— 2 Ã— 7 Ã— 3

= 84

**15. The students of class VIII of a school donated Rs 2401 for PMâ€™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.**

**Solution:**

Let us consider number of students as a

Each student denoted a rupee

So, total amount collected is a Ã— a rupees = 2401

a^{2}Â = 2401

a =Â âˆš2401

a = 49

âˆ´ There are 49 students in the class.

**16. A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.**

**Solution:**

Let us consider number of rows as a

No. of columns = a

Total number of students who sat in the field = a^{2}

Total students a^{2}Â + 71 = 6000

a^{2}Â = 5929

a =Â âˆš5929

a = 77

âˆ´ total number of rows are 77.

### EXERCISE 3.5 PAGE NO: 3.43

**1.Find the square root of each of the following by long division method:
(i) 12544 (ii) 97344
(iii) 286225 (iv) 390625
(v) 363609 (vi) 974169
(vii) 120409 (viii) 1471369
(ix) 291600 (x) 9653449
(xi) 1745041 (xii) 4008004
(xiii) 20657025 (xiv) 152547201
(xv) 20421361 (xvi)62504836
(xvii) 82264900 (xviii) 3226694416
(xix) 6407522209 (xx) 3915380329**

**Solution:**

**(i) **12544

By using long division method

âˆ´ the square root of 12544

âˆš12544 = 112

**(ii) **97344

By using long division method

âˆ´ the square root of 97344

âˆš97344 = 312

**
(iii) **286225

By using long division method

âˆ´ the square root of 286225

âˆš286225 = 535

**(iv) **390625

By using long division method

âˆ´ the square root of 390625

âˆš390625 = 625

**
(v) **363609

By using long division method

âˆ´ the square root of 363609

âˆš36369 = 603

**(vi)** 974169

By using long division method

âˆ´ the square root of 974169

âˆš974169 = 987

**(vii) **120409

By using long division method

âˆ´ the square root of 120409

âˆš120409 = 347

**(viii) **1471369

By using long division method

âˆ´ the square root of 1471369

âˆš1471369 = 1213

**
(ix) **291600

By using long division method

âˆ´ the square root of 291600

âˆš291600 = 540

**(x) **9653449

By using long division method

âˆ´ the square root of 9653449

âˆš9653449 = 3107

**
(xi) **1745041

By using long division method

âˆ´ the square root of 1745041

âˆš1745041 = 1321

** (xii) **4008004

By using long division method

âˆ´ the square root of 4008004

âˆš4008004 = 2002

**
(xiii) **20657025

By using long division method

âˆ´ the square root of 20657025

âˆš20657025 = 4545

**(xiv) **152547201

By using long division method

âˆ´ the square root of 152547201

âˆš152547201 = 12351

**
(xv) **20421361

By using long division method

âˆ´ the square root of 20421361

âˆš20421361 = 4519

**(xvi) **62504836

By using long division method

âˆ´ the square root of 62504836

âˆš62504836 = 7906

**
(xvii) **82264900

By using long division method

âˆ´ the square root of 82264900

âˆš82264900 = 9070

**(xviii) **3226694416

By using long division method

âˆ´ the square root of 3226694416

âˆš3226694416 = 56804

**
(xix) **6407522209

By using long division method

âˆ´ the square root of 6407522209

âˆš6407522209 = 80047

**(xx) **3915380329

By using long division method

âˆ´ the square root of 3915380329

âˆš3915380329 = 62573

**2. Find the least number which must be subtracted from the following numbers to make them a perfect square:
(i) 2361
(ii) 194491
(iii) 26535
(iv) 161605
(v) 4401624**

**Solution:**

**(i) **2361

By using long division method

âˆ´ 57 has to be subtracted from 2361 to get a perfect square.

**
(ii) **194491

By using long division method

âˆ´ 10 has to be subtracted from 194491 to get a perfect square.

**
(iii) **26535

By using long division method

âˆ´ 291 has to be subtracted from 26535 to get a perfect square.

**
(iv) **161605

By using long division method

âˆ´ 1 has to be subtracted from 161605 to get a perfect square.

**
(v)** 4401624

By using long division method

âˆ´ 20 has to be subtracted from 4401624 to get a perfect square.

**3. Find the least number which must be added to the following numbers to make them a perfect square:
(i) 5607
(ii)4931
(iii) 4515600
(iv) 37460
(v) 506900**

**Solution:**

**(i) **5607

By using long division method

The remainder is 131

Since, (74)^{2}Â < 5607

We take, the next perfect square number i.e., (75)^{2}

(75)^{2}Â = 5625 > 5607

So, the number to be added = 5625 â€“ 5607 = 18

**
(ii) **4931

By using long division method

The remainder is 31

Since, (70)^{2}Â < 4931

We take, the next perfect square number i.e., (71)^{2}

(71)^{2}Â = 5041 > 4931

So, the number to be added = 5041 â€“ 4931 = 110

**
(iii) **4515600

By using long division method

The remainder is 4224

Since, (2124)^{2}Â < 4515600

We take, the next perfect square number i.e., (2125)^{2}

(2125)^{2}Â = 4515625 > 4515600

So, the number to be added = 4515625 â€“ 4515600 = 25

**
(iv)** 37460

By using long division method

The remainder is 211

Since, (193)^{2}Â < 37460

We take, the next perfect square number i.e., (194)^{2}

(194)^{2}Â = 37636 > 37460

So, the number to be added = 37636 â€“ 37460 = 176

**
(v)** 506900

By using long division method

The remainder is 1379

Since, (711)^{2}Â < 506900

We take, the next perfect square number i.e., (712)^{2}

(712)^{2}Â = 506944 > 506900

So, the number to be added = 506944 â€“ 506900 = 44

**4. Find the greatest number of 5 digits which is a perfect square.**

**Solution:**

We know that the greatest 5 digit number is 99999

By using long division method

The remainder is 143

So, the greatest 5 digit perfect square number is:

99999 â€“ 143 = 99856

âˆ´ 99856 is the required greatest 5 digit perfect square number.

**5. Find the least number of 4 digits which is a perfect square.**

**Solution:**

We know that the least 4 digit number is 1000

By using long division method

The remainder is 39

Since, (31)^{2}Â < 1000

We take, the next perfect square number i.e., (32)^{2}

(32)^{2}Â = 1024 > 1000

âˆ´ 1024 is the required least number 4 digit number which is a perfect square.

**6. Find the least number of six digits which is a perfect square.**

**Solution:**

We know that the least 6 digit number is 100000

By using long division method

The remainder is 144

Since, (316)^{2}Â < 100000

We take, the next perfect square number i.e., (317)^{2}

(317)^{2}Â = 100489 > 100000

âˆ´ 100489 is the required least number 6 digit number which is a perfect square.

**7. Find the greatest number of 4 digits which is a perfect square.**

**Solution:**

We know that the greatest 4 digit number is 9999

By using long division method

The remainder is 198

So, the greatest 4 digit perfect square number is:

9999 â€“ 198 = 9801

âˆ´ 9801 is the required greatest 4 digit perfect square number.

**8. A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row**

**Solution:**

We know that the total number of soldiers = 8160

Number of soldiers left out = 60

Number of soldiers arranged in rows to form a perfect square = 8160 â€“ 60 = 8100

âˆ´ number of soldiers in each row =Â âˆš8100

=Â âˆš (9Ã—9Ã—10Ã—10)

= 9Ã—10

= 90

**9. The area of a square field is 60025m ^{2}. A man cycles along its boundary at 18 Km/hr. In how much time will he return at the starting point?**

**Solution:**

We know that the area of square field = 60025 m^{2}

Speed of cyclist = 18 km/h

= 18 Ã—Â (1000/60Ã—60)

= 5 m/s^{2}

Area = 60025 m^{2}

Side^{2}Â = 60025

Side =Â âˆš60025

= 245

We know, Total length of boundary = 4 Ã— Side

= 4 Ã— 245

= 980 m

âˆ´ Time taken to return to the starting point =Â 980/5

= 196 seconds

= 3 minutes 16 seconds

**10. The cost of levelling and turning a square lawn at Rs 2.50 per m ^{2}Â is Rs13322.50 Find the cost of fencing it at Rs 5 per metre.**

**Solution:**

We know that the cost of levelling and turning a square lawn = 2.50 per m^{2}

Total cost of levelling and turning = Rs. 13322.50

Total area of square lawn =Â 13322.50/2.50

= 5329 m^{2}

Side^{2}Â = 5329

Side of square lawn =Â âˆš5329

= 73 m

So, total length of lawn = 4 Ã— 73

= 292 m

âˆ´ Cost of fencing the lawn at Rs 5 per metre = 292 Ã— 5

= Rs. 1460

**11. Find the greatest number of three digits which is a perfect square.**

**Solution:**

We know that the greatest 3 digit number is 999

By using long division method

The remainder is 38

So, the greatest 3 digit perfect square number is:

999 â€“ 38 = 961

âˆ´ 961 is the required greatest 3 digit perfect square number.

**12. Find the smallest number which must be added to 2300 so that it becomes a perfect square.**

**Solution:**

By using long division method letâ€™s find the square root of 2300

The remainder is 91

Since, (47)^{2}Â < 2300

We take, the next perfect square number i.e., (48)^{2}

(48)^{2}Â = 2304 > 2300

âˆ´ The smallest number required to be added to 2300 to get a perfect square is

2304 â€“ 2300 = 4

### EXERCISE 3.6 PAGE NO: 3.48

**1. Find the square root of:**

**(i) 441/961**

**(ii) 324/841**

**(iii) 4 29/29**

**(iv) 2 14/25**

**(v) 2 137/196**

**(vi) 23 26/121**

**(vii) 25 544/729**

**(viii) 75 46/49**

**(ix) 3 942/2209**

**(x) 3 334/3025**

**(xi) 21 2797/3364**

**(xii) 38 11/25**

**(xiii) 23 394/729**

**(xiv) 21 51/169**

**(xv) 10 151/225**

**Solution:**

**(i) **441/961

The square root of

âˆš441/961 = 21/31

**(ii) **324/841

The square root of

âˆš324/841= 18/29

**(iii) **4 29/29

The square root of

âˆš(4 29/29) = âˆš(225/49) = 15/7

**(iv) **2 14/25

The square root of

âˆš(2 14/25) = âˆš(64/25) = 8/5

**(v) **2 137/196

The square root of

âˆš2 137/196 = âˆš (529/196) = 23/14

**(vi) **23 26/121

The square root of

âˆš(23 26/121) = âˆš(2809/121) = 53/11

**(vii) **25 544/729

The square root of

âˆš(25 544/729) = âˆš(18769/729) = 137/27

**(viii) **75 46/49

The square root of

âˆš(75 46/49) = âˆš(3721/49) = 61/7

**(ix) **3 942/2209

The square root of

âˆš(3 942/2209) = âˆš(7569/2209) = 87/47

**(x) **3 334/3025

The square root of

âˆš(3 334/3025) = âˆš(9409/3025) = 97/55

**(xi) **21 2797/3364

The square root of

âˆš(21 2797/3364) = âˆš(73441/3364) = 271/58

**(xii) **38 11/25

The square root of

âˆš(38 11/25) = âˆš(961/25) = 31/5

**(xiii) **23 394/729

The square root of

âˆš(23 394/729) = âˆš(17161/729) = 131/27 = 4 23/27

**(xiv) **21 51/169

The square root of

âˆš(21 51/169) = âˆš(3600/169) = 60/13 = 4 8/13

**(xv) **10 151/225

The square root of

âˆš(10 151/225) = âˆš(2401/225) = 49/15 = 3 4/15

**2. Find the value of:**

**(i) âˆš80/âˆš405**

**(ii) âˆš441/âˆš625**

**(iii) âˆš1587/âˆš1728**

**(iv) âˆš72 Ã—âˆš338**

**(v) âˆš45 Ã— âˆš20**

**Solution:**

**(i) **âˆš80/âˆš405

âˆš80/âˆš405 = âˆš16/âˆš81 = 4/9

**(ii) **âˆš441/âˆš625

âˆš441/âˆš625 = 21/25

**(iii) **âˆš1587/âˆš1728

âˆš1587/âˆš1728 = âˆš529/âˆš576 = 23/24

** (iv) **âˆš72 Ã—âˆš338

âˆš72 Ã—âˆš338 = âˆš(2Ã—2Ã—2Ã—3Ã—3) Ã—âˆš(2Ã—13Ã—13)

By using the formula âˆša Ã— âˆšb = âˆš(aÃ—b)

= âˆš(2Ã—2Ã—2Ã—3Ã—3Ã—2Ã—13Ã—13)

= 2^{2} Ã— 3 Ã— 13

= 156

**(v) **âˆš45 Ã— âˆš20

âˆš45 Ã— âˆš20 = âˆš(5Ã—3Ã—3) Ã— âˆš(5Ã—2Ã—2)

By using the formula âˆša Ã— âˆšb = âˆš(aÃ—b)

= âˆš(5Ã—3Ã—3Ã—5Ã—2Ã—2)

= 5 Ã— 3 Ã— 2

= 30

**3. The area of a square field is 80 244/729Â square metres. Find the length of each side of the field.**

**Solution:**

We know that the given area = 80 244/729Â m^{2}

=Â 58564/729Â m^{2}

If L is length of each side

L^{2}Â =Â 58564/729

L =Â âˆš (58564/729) = âˆš58564/âˆš729

=Â 242/27

= 8 26/27

âˆ´ Length isÂ 8 26/27

**4. The area of a square field isÂ 30 1/4m ^{2}. Calculate the length of the side of the square.**

**Solution:**

We know that the given area = 30 1/4Â m^{2}

=Â 121/4Â m^{2}

If L is length of each side then,

L^{2}Â =Â 121/4

L = âˆš(121/4)Â Â =Â âˆš121/âˆš4

=Â 11/2

âˆ´ Length isÂ 11/2

**5. Find the length of a side of a square playground whose area is equal to the area of a rectangular field of dimensions 72m and 338 m.**

**Solution:**

By using the formula

Area of rectangular field = l Ã— b

= 72 Ã— 338 m^{2}

= 24336 m^{2}

Area of square, L^{2}Â = 24336 m^{2}

L =Â âˆš24336

= 156 m

âˆ´ Length of side of square playground is 156 m.

### EXERCISE 3.7 PAGE NO: 3.52

**Find the square root of the following numbers in decimal form:
1. 84.8241**

**Solution:**

By using long division method

âˆ´ the square root of 84.8241

âˆš84.8241 = 9.21

**2. 0.7225**

**Solution:**

By using long division method

âˆ´ the square root of 0.7225

âˆš0.7225 = 0.85

**3. 0.813604**

**Solution:**

By using long division method

âˆ´ the square root of 0.813604

âˆš0.813604 = 0.902

**4. 0.00002025**

**Solution:**

By using long division method

âˆ´ the square root of 0.00002025

âˆš0.00002025 = 0.0045

**5. 150.0625**

**Solution:**

By using long division method

âˆ´ the square root of 150.0625

âˆš150.0625 = 12.25

**6. 225.6004**

**Solution:**

By using long division method

âˆ´ the square root of 225.6004

âˆš225.6004 = 15.02

**7. 3600.720036**

**Solution:**

By using long division method

âˆ´ the square root of 3600.720036

âˆš3600.720036 = 60.006

**8. 236.144689**

**Solution:**

By using long division method

âˆ´ the square root of 236.144689

âˆš236.144689 = 15.367

**9. 0.00059049**

**Solution:**

By using long division method

âˆ´ the square root of 0.00059049

âˆš0.00059049 = 0.0243

**10. 176.252176**

**Solution:**

By using long division method

âˆ´ the square root of 176.252176

âˆš176.252176 = 13.276

**11. 9998.0001**

**Solution:**

By using long division method

âˆ´ the square root of 9998.0001

âˆš9998.0001 = 99.99

**12. 0.00038809**

**Solution:**

By using long division method

âˆ´ the square root of 0.00038809

âˆš0.00038809 = 0.0197

**13. What is that fraction which when multiplied by itself gives 227.798649?**

**Solution:**

Let us consider a number a

Where, a = âˆš227.798649

= 15.093

By using long division method let us verify

âˆ´ 15.093 is the fraction which when multiplied by itself gives 227.798649.

**14. The area of a square playground is 256.6404 square meter. Find the length of one side of the playground.**

**Solution:**

We know that the given area of a square playground = 256.6404

i.e., L^{2} = 256.6404 m^{2}

L = âˆš256.6404

= 16.02m

By using long division method let us verify

âˆ´ length of one side of the playground is 16.02m.

**15. What is the fraction which when multiplied by itself gives 0.00053361?**

**Solution:**

Let us consider a number a

Where, a = âˆš0.00053361

= 0.0231

By using long division method let us verify

âˆ´ 0.0231 is the fraction which when multiplied by itself gives 0.00053361.

**16. Simplify:**

**(i) (âˆš59.29 – âˆš5.29)/ (âˆš59.29 + âˆš5.29)**

**(ii) (âˆš0.2304 + âˆš0.1764)/ (âˆš0.2304 – âˆš0.1764)**

**Solution:**

**(i) **(âˆš59.29 – âˆš5.29)/ (âˆš59.29 + âˆš5.29)

Firstly let us find the square root âˆš59.29 and âˆš5.29

âˆš59.29 = âˆš5929/ âˆš100

= 77/10

= 7.7

âˆš5.29 = âˆš5.29/ âˆš100

= 23/10

= 2.3

So, (7.7 â€“ 2.3)/ (7.7 + 2.3)

= 54/10

= 0.54

**(ii)** (âˆš0.2304 + âˆš0.1764)/ (âˆš0.2304 – âˆš0.1764)

Firstly let us find the square root âˆš0.2304 and âˆš0.1764

âˆš0.2304 = âˆš2304/ âˆš10000

= 48/100

= 0.48

âˆš0.1764 = âˆš1764/ âˆš10000

= 42/100

= 0.42

So, (0.48 + 0.42)/ (0.48 â€“ 0.42)

= 0.9/0.06

= 15

**17. Evaluate âˆš50625Â and hence find the value of âˆš506.25 + âˆš5.0625**

**Solution:**

By using long division method let us find the âˆš50625**Â **

So now, **âˆš**506.25** = **âˆš50625/ âˆš100

= 225/10

= 22.5

âˆš5.0625 = âˆš50625/ âˆš10000

= 225/100

= 2.25

So equating in the above equation we get,

âˆš506.25 + âˆš5.0625 = 22.5 + 2.25

= 24.75

**18. Find the value of âˆš103.0225Â and hence find the value of
(i) âˆš10302.25
(ii)Â âˆš1.030225**

**Solution:**

By using long division method let us find the

âˆš103.0225 = âˆš(1030225/10000) = âˆš1030225/âˆš10000

So now, (i)**âˆš**10302.25** = **âˆš(1030225/ 100)

= 1015/ 10

= 101.5

(ii)âˆš1.030225 = âˆš1030225/ âˆš1000000

= 1015/1000

= 1.015

### EXERCISE 3.8 PAGE NO: 3.56

**1. Find the square root of each of the following correct to three places of decimal.
(i) 5 (ii) 7
(iii) 17 (iv) 20
(v) 66 (vi) 427
(vii) 1.7 (viii) 23.1
(ix) 2.5 (x) 237.615
(xi) 15.3215 (xii) 0.9
(xiii) 0.1 (xiv) 0.016
(xv) 0.00064 (xvi) 0.019
(xvii)Â 7/8Â (xviii)Â 5/12
(xix)Â 2 1/2Â (xx)Â 287 5/8**

**Solution:**

**(i) **5

By using long division method

âˆ´ the square root of 5 is 2.236

**(ii) **7

By using long division method

âˆ´ the square root of 7 is 2.646

**
(iii) **17

By using long division method

âˆ´ the square root of 17 is 4.123

**(iv) **20

By using long division method

âˆ´ the square root of 20 is 4.472

**
(v) **66

By using long division method

âˆ´ the square root of 66 is 8.124

**(vi) **427

By using long division method

âˆ´ the square root of 427 is 20.664

**
(vii) **1.7

By using long division method

âˆ´ the square root of 1.7 is 1.304

**(viii) **23.1

By using long division method

âˆ´ the square root of 23.1 is 4.806

**
(ix) **2.5

By using long division method

âˆ´ the square root of 2.5 is 1.581

**(x) **237.615

By using long division method

âˆ´ the square root of 237.615 is 15.415

**
(xi) **15.3215

By using long division method

âˆ´ the square root of 15.3215 is 3.914

**(xii) **0.9

By using long division method

âˆ´ the square root of 0.9 is 0.949

**
(xiii) **0.1

By using long division method

âˆ´ the square root of 0.1 is 0.316

**(xiv) **0.016

By using long division method

âˆ´ the square root of 0.016 is 0.126

**
(xv) **0.00064

By using long division method

âˆ´ the square root of 0.00064 is 0.025

**(xvi) **0.019

By using long division method

âˆ´ the square root of 0.019 is 0.138

**
(xvii)Â **7/8

By using long division method

âˆ´ the square root of 7/8 is 0.935

**(xviii)Â **5/12

By using long division method

âˆ´ the square root of 5/12 is 0.645

**
(xix)Â **2 1/2

**Â**

By using long division method

âˆ´ the square root of 5/2 is 1.581

**(xx)Â **287 5/8

By using long division method

âˆ´ the square root of 2301/8 is 16.960

**2. Find the square root of 12.0068 correct to four decimal places.**

**Solution:**

By using long division method

âˆ´ the square root of 12.0068 is 3.4651

**3. Find the square root of 11 correct to five decimal places.**

**Solution:**

By using long division method

âˆ´ the square root of 11 is 3.31662

**4. Give that: âˆš2 = 1.414, âˆš3 = 1.732, âˆš5 = 2.236 and âˆš7 = 2.646, evaluate each of the following:
(i)Â âˆš (144/7)
(ii)Â âˆš (2500/3)**

**Solution:**

**(i)Â **âˆš (144/7)

Now let us simplify the given equation

âˆš (144/7) = âˆš (12Ã—12)/ âˆš7

= 12/ 2.646

= 4.535

**(ii)Â **âˆš (2500/3)

Now let us simplify the given equation

âˆš (2500/3) = âˆš (5Ã—5Ã—10Ã—10) /âˆš3

= 5Ã—10/ 1.732

= 50/1.732

= 28.867

**5. Given that âˆš2 = 1.414, âˆš3 = 1.732, âˆš5 = 2.236 and âˆš7 = 2.646Â find the square roots of the following:
(i)Â 196/75
(ii)Â 400/63
(iii)Â 150/7
(iv)Â 256/5
(v)Â 27/50**

**Solution:**

**(i)Â **196/75

Let us find the square root for196/75

**âˆš**(196/75) = **âˆš**(196)/** âˆš** (75)

= **âˆš**(14Ã—14)/** âˆš** (5Ã—5Ã—3)

= 14/ (5**âˆš**3)

= 14/ (5Ã—1.732)

= 14/8.66

= 1.617

**
(ii)Â **400/63

Let us find the square root for400/63

**âˆš**(400/63) = **âˆš**(400)/** âˆš** (63)

= **âˆš**(20Ã—20)/** âˆš** (3Ã—3Ã—7)

= 20/ (3**âˆš**7)

= 20/ (3Ã—2.646)

= 20/7.938

= 2.520

**(iii)Â **150/7

Let us find the square root for150/7

**âˆš**(150/7) = **âˆš**(150)/** âˆš** (7)

= **âˆš**(3Ã—5Ã—5Ã—2)/** âˆš** (7)

= (5**âˆš**3Ã—**âˆš**2)/ (**âˆš**7)

= 5Ã—1.732Ã—1.414/ (2.646)

= 12.245/2.646

= 4.628

**(iv)Â **256/5

Let us find the square root for256/5

**âˆš**(256/5) = **âˆš**(256)/** âˆš** (5)

= **âˆš**(16Ã—16)/** âˆš** (5)

= 16/ (**âˆš**5)

= 16/2.236

= 7.155

**
(v)Â **27/50

Let us find the square root for27/50

**âˆš**(27/50) = **âˆš**(27)/** âˆš** (50)

= **âˆš**(3Ã—3Ã—3)/** âˆš** (5Ã—5Ã—2)

= (3**âˆš**3)/ (5**âˆš**2)

= (3Ã—1.732)/ (5Ã—1.414)

= 5.196/7.07

= 0.735

### EXERCISE 3.9 PAGE NO: 3.61

**Using square root table, find the square roots of the following:
1. 7**

**Solution:**

From square root table we know,

Square root of 7 is:

**âˆš7**Â = 2.645

âˆ´ The square root of 7 is 2.645

**2. 15**

**Solution:**

From square root table we know,

Square root of 7 is:

**âˆš**15Â = 3.8729

âˆ´ The square root of 15 is 3.873

**3. 74**

**Solution:**

From square root table we know,

Square root of 74 is:

**âˆš**74Â = 8.6023

âˆ´ The square root of 74 is 8.602

**4. 82**

**Solution:**

From square root table we know,

Square root of 82 is:

**âˆš**82Â = 9.0553

âˆ´ The square root of 82 is 9.055

**5. 198**

**Solution:**

From square root table we know,

Square root of 198 is:

**âˆš**198Â = 14.0712

âˆ´ The square root of 198 is 14.071

**6. 540**

**Solution:**

From square root table we know,

Square root of 540 is:

**âˆš**540Â = 23.2379

âˆ´ The square root of 540 is 23.24

**7. 8700**

**Solution:**

From square root table we know,

Square root of 8700 is:

**âˆš**8700Â = 93.2737

âˆ´ The square root of 8700 is 93.27

**8. 3509**

**Solution:**

From square root table we know,

Square root of 3509 is:

**âˆš**3509Â = 59.2368

âˆ´ The square root of 3509 is 59.235

**9. 6929**

**Solution:**

From square root table we know,

Square root of 6929 is:

**âˆš**6929Â = 83.2406

âˆ´ The square root of 6929 is 83.239

**10. 25725**

**Solution:**

From square root table we know,

Square root of 25725 is:

**âˆš**25725Â = 160.3901

âˆ´ The square root of 25725 is 160.41

**11. 1312.**

**Solution:**

From square root table we know,

Square root of 1312 is:

**âˆš**1312Â = 36.2215

âˆ´ The square root of 1312 is 36.22

**12. 4192**

**Solution:**

From square root table we know,

Square root of 4192 is:

**âˆš**4192Â = 64.7456

âˆ´ The square root of 4192 is 64.75

**13. 4955**

**Solution:**

From square root table we know,

Square root of 4955 is:

**âˆš**4955Â = 70.3917

âˆ´ The square root of 4955 is 70.39

**14. 99/144**

**Solution:**

From square root table we know,

Square root of 99/144 is:

**âˆš**(99/144)Â = 0.82915

âˆ´ The square root of 99/144 is 0.829

**15. 57/169**

**Solution:**

From square root table we know,

Square root of 57/169 is:

**âˆš(**57/169)Â = 0.58207

âˆ´ The square root of 57/169 is 0.581

**16. 101/169**

**Solution:**

From square root table we know,

Square root of 101/169 is:

**âˆš(**101/169)Â = 0.77306

âˆ´ The square root of 57/169 is 0.773

**17. 13.21**

**Solution:**

From square root table we know,

Square root of 13.21 is:

âˆš13.21Â = 3.6345

âˆ´ The square root of 13.21 is 3.635

**18. 21.97**

**Solution:**

From square root table we know,

Square root of 21.97 is:

âˆš21.97Â = 4.6872

âˆ´ The square root of 21.97 is 4.6872

**19. 110**

**Solution:**

From square root table we know,

Square root of 110 is:

âˆš110Â = 10.4880

âˆ´ The square root of 110 is 10.488

**20. 1110**

**Solution:**

From square root table we know,

Square root of 1110 is:

âˆš1110Â = 33.3166

âˆ´ The square root of 1110 is 33.317

**21. 11.11**

**Solution:**

From square root table we know,

Square root of 11.11 is:

âˆš11.11Â = 3.33316

âˆ´ The square root of 11.11 is 3.333

**22. The area of a square field is 325m ^{2}. Find the approximate length of one side of the field.**

**Solution:**

We know that the given area of the field = 325 m^{2}

To find the approximate length of the side of the field we will have to calculate the square root of 325

âˆš325Â = 18.027 m

âˆ´ The approximate length of one side of the field is 18.027 m

**23. Find the length of a side of a square, whose area is equal to the area of a rectangle with sides 240 m and 70 m.**

**Solution:**

We know that from the question,

Area of square = Area of rectangle

Side^{2}Â = 240 Ã— 70

Side =Â âˆš(240 Ã— 70)

=Â âˆš(10Ã—10Ã—2Ã—2Ã—2Ã—3Ã—7)

= 20âˆš(42)

= 20 Ã— 6.48

= 129.60 m

âˆ´ The length of side of the square is 129.60 m

## RD Sharma solutions For Class 8 Maths Chapter 3 – Squares and Square Roots

Chapter 3, Squares and Square Roots contains nine exercises. RD Sharma Solutions are given here which include answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

- Perfect Square or Square numbers.
- Properties and patterns of some Square numbers.
- Product of two consecutive odd or consecutive even numbers.
- Column method for squaring two-digit numbers.
- Diagonal method for squaring a number.
- Definition of square roots.
- The square root of a perfect square by the prime factorisation method.
- The square root of a perfect square by the long division method.
- Square roots of a number in decimal form.
- The square root of fractions.

### Chapter Brief of RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square roots

The RD Sharma Solutions for Class 8 Maths Chapter 3 â€“ Squares and Square Roots deal with the properties of perfect squares, which in turn help to solve problems easily and quickly. By learning these concepts thoroughly, we can find squares and square roots of the given number without using a calculator.