RD Sharma Solutions for Class 8 Chapter 3 - Squares and Square Roots

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots are available here. The subject experts at BYJU’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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rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
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rd sharma class 8 maths chapter 3
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rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
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rd sharma class 8 maths chapter 3
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rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3
rd sharma class 8 maths chapter 3

 

Access the answers to Maths RD Sharma Solutions For Class 8 Chapter 3 – Squares and Square Roots

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 = 2×2×11×11

By grouping the prime factors in equal pairs we get,

= (2×2) × (11×11)

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

∴ 484 is a perfect square.

(ii) 625

First find the prime factors for 625

625 = 5×5×5×5

By grouping the prime factors in equal pairs we get,

= (5×5) × (5×5)

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

∴ 625 is a perfect square.

(iii) 576

First find the prime factors for 576

576 = 2×2×2×2×2×2×3×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (2×2) × (3×3)

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

∴ 576 is a perfect square.

(iv) 941

First find the prime factors for 941

941 = 941 × 1

We know that 941 itself is a prime factor.

∴ 941 is not a perfect square.

(v) 961

First find the prime factors for 961

961 = 31×31

By grouping the prime factors in equal pairs we get,

= (31×31)

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

∴ 961 is a perfect square.

(vi) 2500

First find the prime factors for 2500

2500 = 2×2×5×5×5×5

By grouping the prime factors in equal pairs we get,

= (2×2) × (5×5) × (5×5)

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

∴ 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 = 2×2×17×17

By grouping the prime factors in equal pairs we get,

= (2×2) × (17×17)

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

∴ 1156 is a perfect square.

To find the square of the given number

1156 = (2×17) × (2×17)

= 34 × 34

= (34)2

∴ 1156 is a square of 34.

(ii) 2025

First find the prime factors for 2025

2025 = 3×3×3×3×5×5

By grouping the prime factors in equal pairs we get,

= (3×3) × (3×3) × (5×5)

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

∴ 2025 is a perfect square.

To find the square of the given number

2025 = (3×3×5) × (3×3×5)

= 45 × 45

= (45)2

∴ 2025 is a square of 45.

(iii) 14641

First find the prime factors for 14641

14641 = 11×11×11×11

By grouping the prime factors in equal pairs we get,

= (11×11) × (11×11)

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

∴ 14641 is a perfect square.

To find the square of the given number

14641 = (11×11) × (11×11)

= 121 × 121

= (121)2

∴ 14641 is a square of 121.

(iv) 4761

First find the prime factors for 4761

4761 = 3×3×23×23

By grouping the prime factors in equal pairs we get,

= (3×3) × (23×23)

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

∴ 4761 is a perfect square.

To find the square of the given number

4761 = (3×23) × (3×23)

= 69 × 69

= (69)2

∴ 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 = 3×3×23×23×5

By grouping the prime factors in equal pairs we get,

= (3×3) × (23×23) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

23805 × 5 = (3×3) × (23×23) × (5×5)

= (3×5×23) × (3×5×23)

= 345 × 345

= (345)2

∴ Product is the square of 345.

(ii) 12150

First find the prime factors for 12150

12150 = 2×2×2×2×3×3×5×5×2

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (3×3) × (5×5) × 2

By observation, prime factor 2 is left out.

So, multiply by 2 we get,

12150 × 2 = (2×2) × (2×2) × (3×3) × (5×5) × (2×2)

= (2×2×3×5×2) × (2×2×3×5×2)

= 120 × 120

= (120)2

∴ Product is the square of 120.

(iii) 7688

First find the prime factors for 7688

7688 = 2×2×31×31×2

By grouping the prime factors in equal pairs we get,

= (2×2) × (31×31) × 2

By observation, prime factor 2 is left out.

So, multiply by 2 we get,

7688 × 2 = (2×2) × (31×31)× (2×2)

= (2×31×2) × (2×31×2)

= 124 × 124

= (124)2

∴ 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) 12283
(ii) 1800
(iii) 2904

Solution:

(i) 12283

First find the prime factors for 12283

12283 = 3×3×3×23×23

By grouping the prime factors in equal pairs we get,

= (3×3) × (23×23) × 3

By observation, prime factor 3 is left out.

So, divide by 3 to eliminate 3 we get,

12283/3 = (3×3) × (23×23)

= (3×23) × (3×23)

= 69 × 69

= (69)2

∴ Resultant is the square of 69.

(ii) 1800

First find the prime factors for 1800

1800 = 2×2×5×5×3×3×2

By grouping the prime factors in equal pairs we get,

= (2×2) × (5×5) × (3×3) × 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

1800/2 = (2×2) × (5×5) × (3×3)

= (2×5×3) × (2×5×3)

= 30 × 30

= (30)2

∴ Resultant is the square of 30.

(iii) 2904

First find the prime factors for 2904

2904 = 2×2×11×11×2×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (11×11) × 2 × 3

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

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

2904/6 = (2×2) × (11×11)

= (2×11) × (2×11)

= 22 × 22

= (22)2

∴ 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 = 32×3×7

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

225 prime factors are

225 = (5×5) × (3×3)

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

2048 prime factors are

2048 = (2×2) × (2×2) × (2×2) × (2×2) × (2×2) × 2

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

343 prime factors are

343 = (7×7) × 7

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

441 prime factors are

441 = (7×7) × (3×3)

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

2961 prime factors are

2961 = (3×3) × (3×3) × (3×3) × (2×2)

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

11025 prime factors are

11025 = (3×3) × (5×5) × (7×7)

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

3549 prime factors are

3549 = (13×13) × 3 × 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) 8820

(ii) 3675
(iii) 605

(iv) 2880
(v) 4056

(vi) 3468
(vii) 7776

Solution:

(i) 8820

First find the prime factors for 8820

8820 = 2×2×3×3×7×7×5

By grouping the prime factors in equal pairs we get,

= (2×2) × (3×3) × (7×7) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

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

= (2×3×7×5) × (2×3×7×5)

= 210 × 210

= (210)2

∴ Product is the square of 210.

(ii) 3675

First find the prime factors for 3675

3675 = 5×5×7×7×3

By grouping the prime factors in equal pairs we get,

= (5×5) × (7×7) × 3

By observation, prime factor 3 is left out.

So, multiply by 3 we get,

3675 × 3 = (5×5) × (7×7) × (3×3)

= (5×7×3) × (5×7×3)

= 105 × 105

= (105)2

∴ Product is the square of 105.

(iii) 605

First find the prime factors for 605

605 = 5×11×11

By grouping the prime factors in equal pairs we get,

= (11×11) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

605 × 5 = (11×11) × (5×5)

= (11×5) × (11×5)

= 55 × 55

= (55)2

∴ Product is the square of 55.

(iv) 2880

First find the prime factors for 2880

2880 = 5×3×3×2×2×2×2×2×2

By grouping the prime factors in equal pairs we get,

= (3×3) × (2×2) × (2×2) × (2×2) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

2880 × 5 = (3×3) × (2×2) × (2×2) × (2×2) × (5×5)

= (3×2×2×2×5) × (3×2×2×2×5)

= 120 × 120

= (120)2

∴ Product is the square of 120.

(v) 4056

First find the prime factors for 4056

4056 = 2×2×13×13×2×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (13×13) × 2 × 3

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

So, multiply by 6 we get,

4056 × 6 = (2×2) × (13×13) × (2×2) × (3×3)

= (2×13×2×3) × (2×13×2×3)

= 156 × 156

= (156)2

∴ Product is the square of 156.

(vi) 3468

First find the prime factors for 3468

3468 = 2×2×17×17×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (17×17) × 3

By observation, prime factor 3 is left out.

So, multiply by 3 we get,

3468 × 3 = (2×2) × (17×17) × (3×3)

= (2×17×3) × (2×17×3)

= 102 × 102

= (102)2

∴ Product is the square of 102.

(vii) 7776

First find the prime factors for 7776

7776 = 2×2×2×2×3×3×3×3×2×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (3×3) × (3×3) × 2 × 3

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

So, multiply by 6 we get,

7776 × 6 = (2×2) × (2×2) × (3×3) × (3×3) × (2×2) × (3×3)

= (2×2×3×3×2×3) × (2×2×3×3×2×3)

= 216 × 216

= (216)2

∴ 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 = 7×7×13×13×2

By grouping the prime factors in equal pairs we get,

= (7×7) × (13×13) × 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

16562/2 = (7×7) × (13×13)

= (7×13) × (7×13)

= 91 × 91

= (91)2

∴ Resultant is the square of 91.

(ii) 3698

First find the prime factors for 3698

3698 = 2×43×43

By grouping the prime factors in equal pairs we get,

= (43×43) × 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

3698/2 = (43×43)

= (43)2

∴ Resultant is the square of 43.

(iii) 5103

First find the prime factors for 5103

5103 = 3×3×3×3×3×3×7

By grouping the prime factors in equal pairs we get,

= (3×3) × (3×3) × (3×3) × 7

By observation, prime factor 7 is left out.

So, divide by 7 to eliminate 7 we get,

5103/7 = (3×3) × (3×3) × (3×3)

= (3×3×3) × (3×3×3)

= 27 × 27

= (27)2

∴ Resultant is the square of 27.

(iv) 3174

First find the prime factors for 3174

3174 = 2×3×23×23

By grouping the prime factors in equal pairs we get,

= (23×23) × 2 × 3

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

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

3174/6 = (23×23)

= (23)2

∴ Resultant is the square of 23.

(v) 1575

First find the prime factors for 1575

1575 = 3×3×5×5×7

By grouping the prime factors in equal pairs we get,

= (3×3) × (5×5) × 7

By observation, prime factor 7 is left out.

So, divide by 7 to eliminate 7 we get,

1575/7 = (3×3) × (5×5)

= (3×5) × (3×5)

= 15 × 15

= (15)2

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image1

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image2

∴ 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 = 3×3×7×7×11

By grouping the prime factors in equal pairs we get,

= (3×3) × (7×7) × 11

∴ 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 = 2×2×3×3×3×17×17

By grouping the prime factors in equal pairs we get,

= (2×2) × (3×3) × (17×17) × 3

∴ 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

28812 = 2×2×2×2×2×2×2×3×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (2×2) × (3×3) × 3

∴ 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 = 2×2×2×2×2×2×3×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (2×2) × (3×3)

= 26 × 32

= 242

∴ 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 unit’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 = (22) = 4

(ii) 977

Unit digit of (977)2 = (72) = 49 = 9

(iii) 4583

Unit digit of (4583)2 = (32) = 9

(iv) 78367

Unit digit of (78367)2 = (72) = 49 = 9

(v) 52698

Unit digit of (52698)2 = (82) = 64 = 4

(vi) 99880

Unit digit of (99880)2 = (02) = 0

(vii) 12796

Unit digit of (12796)2 = (62) = 36 = 6

(viii) 55555

Unit digit of (55555)2 = (52) = 25 = 5

(ix) 53924

Unit digit of (53924)2 = (42) = 16 = 6

5. Observe the following pattern
1+3 = 22

1+3+5 = 32

1+3+5+7 = 42
And write the value of 1+3+5+7+9+……… 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.

∴ The value of 1+3+5+7+9+……… up to n terms = n2 (as there are only n terms).

6. Observe the following pattern

22 -12 = 2 + 1

32 – 22 = 3 + 2

42 – 32 = 4 + 3

52 – 42 = 5 + 4

And find the value of

(i) 1002 -992

(ii)1112 – 1092

(iii) 992 – 962

Solution:

(i) 1002 -992

100 + 99 = 199

(ii) 1112 – 1092

(1112 – 1102) + (1102 – 1092)

(111 + 110) + (100 + 109)

440

(iii) 99 – 96

(992 – 982) + (982 – 972) + (972 – 962)

(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 = 82 + 152

= 289

RHS = 172

= 289

LHS = RHS

∴ The given triplet is a Pythagorean.

(ii) (18, 80, 82)

LHS = 182 + 802

= 6724

RHS = 822

= 6724

LHS = RHS

∴ The given triplet is a Pythagorean.

(iii) (14, 48, 51)

LHS = 142 + 482

= 2500

RHS = 512

= 2601

LHS ≠ RHS

∴ The given triplet is not a Pythagorean.

(iv) (10, 24, 26)

LHS = 102 + 242

= 676

RHS = 262

= 676

LHS = RHS

∴ The given triplet is a Pythagorean.

(v) (16, 63, 65)

LHS = 162 + 632

= 4225

RHS = 652

= 4225

LHS = RHS

∴ The given triplet is a Pythagorean.

(vi) (12, 35, 38)

LHS = 122 + 352

= 1369

RHS = 382

= 1444

LHS ≠ RHS

∴ The given triplet is not a Pythagorean.

8. Observe the following pattern

(1×2) + (2×3) = (2×3×4)/3

(1×2) + (2×3) + (3×4) = (3×4×5)/3

(1×2) + (2×3) + (3×4) + (4×5) = (4×5×6)/3

And find the value of

(1×2) + (2×3) + (3×4) + (4×5) + (5×6)

Solution:

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

9. Observe the following pattern

1 = 1/2 (1×(1+1))

1+2 = 1/2 (2×(2+1))

1+2+3 = 1/2 (3×(3+1))

1+2+3+4 = 1/2 (4×(4+1))

And find the values of each of the following:

(i) 1+2+3+4+5+…+50

(ii) 31+32+….+50

Solution:

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

(i) 1+2+3+4+5+…+50 = 1/2 (5×(5+1))

25 × 51 = 1275

(ii) 31+32+….+50 = (1+2+3+4+5+…+50) – (1+2+3+…+30)

1275 – 1/2 (30×(30+1))

1275 – 465

810

10. Observe the following pattern

12 = 1/6 (1×(1+1)×(2×1+1))

12+22 = 1/6 (2×(2+1)×(2×2+1)))

12+22+32 = 1/6 (3×(3+1)×(2×3+1)))

12+22+32+42 = 1/6 (4×(4+1)×(2×4+1)))

And find the values of each of the following:

(i) 12+22+32+42+…+102

(ii) 52+62+72+82+92+102+112+122

Solution:

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

(i) 12+22+32+42+…+102 = 1/6 (10×(10+1)×(2×10+1))

= 1/6 (2310)

= 385

(ii) 52+62+72+82+92+102+112+122 = 12+22+32+…+122 – (12+22+32+42)

1/6 (12×(12+1)×(2×12+1)) – 1/6 (4×(4+1)×(2×4+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

∴ 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.

∴ 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 unit’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 22 and 32 is 13 which is not square number.

(iv) False, because difference of 32 and 22 is 5, which is not square number.

(v) True, because the square of 22 and 32 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
a2

4

+2

6

2ab

20

+2

22

b2

25

6 2 5

∴ 252 = 625

Where, it can be expressed as

252 = 25× 25 = 625

(ii) 37

So here, a = 3 and b = 7

Column I Column II Column III
a2

9

+4

13

2ab

42

+4

46

b2

49

13 6 9

∴ 372 = 1369

Where, it can be expressed as

252 = 37× 37 = 1369

(iii) 54

So here, a = 5 and b = 4

Column I Column II Column III
a2

25

+4

29

2ab

40

+1

41

b2

16

29 1 6

∴ 542 = 2916

Where, it can be expressed as

542 = 54 × 54 = 2916

(iv) 71

So here, a = 7 and b = 1

Column I Column II Column III
a2

49

+1

49

2ab

14

+0

14

b2

01

49 4 1

∴ 712 = 4941

Where, it can be expressed as

712 = 71 × 71 = 4941

(v) 96

So here, a = 9 and b = 6

Column I Column II Column III
a2

81

+11

92

2ab

108

+3

111

b2

36

92 1 6

∴ 962 = 9216

Where, it can be expressed as

962 = 96 × 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 n2 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.

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image3

∴ 982 = 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 n2 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.

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image4

∴ 2732 = 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 n2 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.

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image5

∴ 3482 = 121104

(iv) 295

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 n2 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.

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image6

∴ 2952 = 87025

(v) 171

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 n2 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.

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image7

∴ 1712 = 29241

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

(ii) 503
(iii) 450

(iv) 862
(v) 265

Solution:

(i) 127

1272 = 127 × 127 = 16129

(ii) 503

5032 = 503 × 503 = 253009


(iii)
450

4502 = 450 × 450 = 203401

(iv) 862

8622 = 862 × 862 = 743044


(v)
265

2652 = 265 × 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

4252 = 425 × 425 = 180625

(ii) 575

5752 = 575 × 575 = 330625


(iii)405

4052 = 405 × 405 = 164025

(iv) 205

2052 = 205 × 205 = 42025


(v) 95

952 = 95 × 95 = 9025

(vi) 745

7452 = 745 × 745 = 555025


(vii) 512

5122 = 512 × 512 = 262144

(viii) 995

9952 = 995 × 995 = 990025

5. Find the squares of the following numbers using the identify (a+b) 2= a2+2ab+b2:
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605

Solution:

(i) 405

We know, (a+b) 2= a2+2ab+b2

405 = (400+5)2

= (400)2 + 52 + 2 (400) (5)

= 160000 + 25 + 4000

= 164025

(ii) 510

We know, (a+b) 2= a2+2ab+b2

510 = (500+10)2

= (500)2 + 102 + 2 (500) (10)

= 250000 + 100 + 10000

= 260100


(iii)
1001

We know, (a+b) 2= a2+2ab+b2

1001 = (1000+1)2

= (1000)2 + 12 + 2 (1000) (1)

= 1000000 + 1 + 2000

= 1002001


(iv)
209

We know, (a+b) 2= a2+2ab+b2

209 = (200+9)2

= (200)2 + 92 + 2 (200) (9)

= 40000 + 81 + 3600

= 43681


(v)
605

We know, (a+b) 2= a2+2ab+b2

605 = (600+5)2

= (600)2 + 52 + 2 (600) (5)

= 360000 + 25 + 6000

= 366025

6. Find the squares of the following numbers using the identity (a-b) 2= a2-2ab+b2
(i) 395

(ii) 995
(iii)495

(iv) 498
(v) 99

(vi) 999
(vii)599

Solution:

(i) 395

We know, (a-b) 2= a2-2ab+b2

395 = (400-5)2

= (400)2 + 52 – 2 (400) (5)

= 160000 + 25 – 4000

= 156025

(ii) 995

We know, (a-b) 2= a2-2ab+b2

995 = (1000-5)2

= (1000)2 + 52 – 2 (1000) (5)

= 1000000 + 25 – 10000

= 990025


(iii)
495

We know, (a-b) 2= a2-2ab+b2

495 = (500-5)2

= (500)2 + 52 – 2 (500) (5)

= 250000 + 25 – 5000

= 245025

(iv) 498

We know, (a-b) 2= a2-2ab+b2

498 = (500-2)2

= (500)2 + 22 – 2 (500) (2)

= 250000 + 4 – 2000

= 248004


(v)
99

We know, (a-b) 2= a2-2ab+b2

99 = (100-1)2

= (100)2 + 12 – 2 (100) (1)

= 10000 + 1 – 200

= 9801

(vi) 999

We know, (a-b) 2= a2-2ab+b2

999 = (1000-1)2

= (1000)2 + 12 – 2 (1000) (1)

= 1000000 + 1 – 2000

= 998001


(vii)
599

We know, (a-b) 2= a2-2ab+b2

599 = (600-1)2

= (600)2 + 12 – 2 (600) (1)

= 360000 + 1 – 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= a2+2ab+b2

52 = (50+2)2

= (50)2 + 22 + 2 (50) (2)

= 2500 + 4 + 200

= 2704

(ii) 95

We know, (a-b) 2= a2-2ab+b2

95 = (100-5)2

= (100)2 + 52 – 2 (100) (5)

= 10000 + 25 – 1000

= 9025


(iii)
505

We know, (a+b) 2= a2+2ab+b2

505 = (500+5)2

= (500)2 + 52 + 2 (500) (5)

= 250000 + 25 + 5000

= 255025

(iv) 702

We know, (a+b) 2= a2+2ab+b2

702 = (700+2)2

= (700)2 + 22 + 2 (700) (2)

= 490000 + 4 + 2800

= 492804


(v)
99

We know, (a-b) 2= a2-2ab+b2

99 = (100-1)2

= (100)2 + 12 – 2 (100) (1)

= 10000 + 1 – 200

= 9801


EXERCISE 3.4 PAGE NO: 3.38

1.Write the possible unit’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) 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) 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) 998001

We know that unit digit of 998001 = 10

Unit digit of square root = 10 or 9

Since the number is odd, square root is also odd

(iv) 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 let’s find the prime factors for

441 = 3×3×7×7

= 32×72

√441 = 3×7

= 21

(ii) 196

Firstly let’s find the prime factors for

196 = 2×2×7×7

= 22×72

√196 = 2×7

= 14


(iii)
529

Firstly let’s find the prime factors for

529 = 23×23

= 232

√529 = 23

(iv) 1764

Firstly let’s find the prime factors for

1764 = 2×2×3×3×7×7

= 22×32×72

√1764 = 2×3×7

= 42


(v)
1156

Firstly let’s find the prime factors for

1156 = 2×2×17×17

= 22×172

√1156 = 2×17

= 34

(vi) 4096

Firstly let’s find the prime factors for

4096 = 2×2×2×2×2×2×2×2×2×2×2×2

= 212

√4096 = 26

= 64

(vii) 7056

Firstly let’s find the prime factors for

7056 = 2×2×2×2×21×21

= 22×22×212

√7056 = 2×2×21

= 84

(viii) 8281

Firstly let’s find the prime factors for

8281 = 91×91

= 912

√8281 = 91


(ix)
11664

Firstly let’s find the prime factors for

11664 = 2×2×2×2×3×3×3×3×3×3

= 22×22×32×32×32

√11664 = 2×2×3×3×3

= 108

(x) 47089

Firstly let’s find the prime factors for

47089 = 217×217

= 2172

√47089 = 217


(xi)
24336

Firstly let’s find the prime factors for

24336 = 2×2×2×2×3×3×13×13

= 22×22×32×132

√24336 = 2×2×3×13

= 156

(xii) 190969

Firstly let’s find the prime factors for

190969 = 23×23×19×19

= 232×192

√190969 = 23×19

= 437


(xiii)
586756

Firstly let’s find the prime factors for

586756 = 2×2×383×383

= 22×3832

√586756 = 2×383

= 766

(xiv) 27225

Firstly let’s find the prime factors for

27225 = 5×5×3×3×11×11

= 52×32×112

√27225 = 5×3×11

= 165


(xv)
3013696

Firstly let’s find the prime factors for

3013696 = 2×2×2×2×2×2×217×217

= 26×2172

√3013696 = 23×217

= 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 let’s find the prime factors for

180 = (2 × 2) × (3 × 3) × 5

=22 × 32 × 5

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

180 × 5 = 22 × 32 × 52

∴ Square root of √ (180 × 5) = 2 × 3 × 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 let’s find the prime factors for

147 = (7 × 7) × 3

=72 × 3

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

147 × 3 = 72 × 32

∴ Square root of √ (147 × 3) = 7 × 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 let’s find the prime factors for

3645 = (3 × 3) × (3 × 3) × 5 × 3

=32 × 32 × 5 × 3

To make the unpaired 5 and 3 into paired, the number 3645 has to be divided by 5 × 3=15

3645 ÷ 15 = 32 × 32

∴ Square root of √ (3645 ÷ 15) = 3 × 3

= 9

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 let’s find the prime factors for

1152 = (2 × 2) × (2 × 2) × (2 × 2) × 2 × (3 × 3)

=22 × 22 × 22 × 32 × 2

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

1152 ÷ 2 = 22 × 22 × 22 × 32

∴ Square root of √ (1152 ÷ 2) = 2 × 2 × 2 × 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 × b = 1296

16b × b = 1296

16b2 = 1296

b2 = 1296/16 = 81

b = 9

a = 16b

= 16(9)

= 144

∴ 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) = a2

We know that the total Collection = 202500

a = √ 202500

a = √a(2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 5)

= 2 × 3 × 3 × 5 × 5a

= 450

∴ 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

a2 = 9216

a = √9216

= 2 × 2 × 2 × 12

= 96

∴ 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 = a2paise

= 230400 paise

a = √230400

= √2304 × √100

= 10√2304

a = 10 × 2 × 2 × 12

a = 480

∴ There were 480 students in the school

11. The area of a square field is 5184 m2. 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

a2 = 5184 m2

a = √5184m

a = 2 × 2 ×2 × 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 × 48 m2

= 4608 m2

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 = 22 × 32 × 5

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

180 × 5 = 22 × 32 × 52

∴ 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 = 22 × 32 ×2 × 5

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

360 × 10 = 22 × 32 × 52 × 22

∴ 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 – 1 = 120

120 – 3 = 117

117 – 5 = 112

112 – 7 = 115

115 – 9 = 106

106 – 11 = 95

95 – 13 = 82

82 – 15 = 67

67 – 17 = 50

50 – 19 = 31

31 – 21 = 10

Clearly, we have performed operation 11 times

∴ √121 = 11

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 88

88 – 19 = 69

69 – 21 = 48

48 – 23 = 25

25 – 25 = 0

Clearly, we have performed subtraction 13 times

∴ √169 = 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) 7744

Prime factors of 7744 is

7744 = 22 × 22 × 22 × 112

∴ The square root of 7744 is

√7744 = 2 × 2 × 2 × 11

= 88

(ii) 9604

Prime factors of 9604 is

9604 = 22 × 72 × 72

∴ The square root of 9604 is

√9604 = 2 × 7 × 7

= 98

(iii) 5929

Prime factors of 5929 is

5929 = 112 × 72

∴ The square root of 5929 is

√5929 = 11 × 7

= 77

(iv) 7056

Prime factors of 7056 is

7056 = 22 × 22 × 72 × 32

∴ The square root of 7056 is

√7056 = 2 × 2 × 7 × 3

= 84

15. The students of class VIII of a school donated Rs 2401 for PM’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 × a rupees = 2401

a2 = 2401

a = √2401

a = 49

∴ 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 = a2

Total students a2 + 71 = 6000

a2 = 5929

a = √5929

a = 77

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image8

∴ the square root of 12544

√12544 = 112

(ii) 97344

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image9

∴ the square root of 97344

√97344 = 312


(iii)
286225

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image10

∴ the square root of 286225

√286225 = 535

(iv) 390625

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image11

∴ the square root of 390625

√390625 = 625


(v)
363609

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image12

∴ the square root of 363609

√36369 = 603

(vi) 974169

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image13

∴ the square root of 974169

√974169 = 987

(vii) 120409

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image14

∴ the square root of 120409

√120409 = 347

(viii) 1471369

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image15

∴ the square root of 1471369

√1471369 = 1213


(ix)
291600

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image16

∴ the square root of 291600

√291600 = 540

(x) 9653449

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image17

∴ the square root of 9653449

√9653449 = 3107


(xi)
1745041

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image18

∴ the square root of 1745041

√1745041 = 1321

(xii) 4008004

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image19

∴ the square root of 4008004

√4008004 = 2002


(xiii)
20657025

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image20

∴ the square root of 20657025

√20657025 = 4545

(xiv) 152547201

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image21

∴ the square root of 152547201

√152547201 = 12351


(xv)
20421361

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image22

∴ the square root of 20421361

√20421361 = 4519

(xvi) 62504836

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image23

∴ the square root of 62504836

√62504836 = 7906


(xvii)
82264900

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image24

∴ the square root of 82264900

√82264900 = 9070

(xviii) 3226694416

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image25

∴ the square root of 3226694416

√3226694416 = 56804


(xix)
6407522209

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image26

∴ the square root of 6407522209

√6407522209 = 80047

(xx) 3915380329

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image27

∴ the square root of 3915380329

√3915380329 = 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image28

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


(ii)
194491

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image29

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


(iii)
26535

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image30

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


(iv)
161605

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image31

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


(v)
4401624

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image32

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image33

The remainder is 131

Since, (74)2 < 5607

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

(75)2 = 5625 > 5607

So, the number to be added = 5625 – 5607 = 18


(ii)
4931

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image34

The remainder is 31

Since, (70)2 < 4931

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

(71)2 = 5041 > 4931

So, the number to be added = 5041 – 4931 = 110


(iii)
4515600

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image35

The remainder is 4224

Since, (2124)2 < 4515600

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

(2125)2 = 4515625 > 4515600

So, the number to be added = 4515625 – 4515600 = 25


(iv)
37460

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image36

The remainder is 211

Since, (193)2 < 37460

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

(194)2 = 37636 > 37460

So, the number to be added = 37636 – 37460 = 176


(v)
506900

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image37

The remainder is 1379

Since, (711)2 < 506900

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

(712)2 = 506944 > 506900

So, the number to be added = 506944 – 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image38

The remainder is 143

So, the greatest 5 digit perfect square number is:

99999 – 143 = 99856

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image39

The remainder is 39

Since, (31)2 < 1000

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

(32)2 = 1024 > 1000

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image40

The remainder is 144

Since, (316)2 < 100000

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

(317)2 = 100489 > 100000

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image41

The remainder is 198

So, the greatest 4 digit perfect square number is:

9999 – 198 = 9801

∴ 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 – 60 = 8100

∴ number of soldiers in each row = √8100

= √ (9×9×10×10)

= 9×10

= 90

9. The area of a square field is 60025m2. 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 m2

Speed of cyclist = 18 km/h

= 18 × (1000/60×60)

= 5 m/s2

Area = 60025 m2

Side2 = 60025

Side = √60025

= 245

We know, Total length of boundary = 4 × Side

= 4 × 245

= 980 m

∴ Time taken to return to the starting point = 980/5

= 196 seconds

= 3 minutes 16 seconds

10. The cost of levelling and turning a square lawn at Rs 2.50 per m2 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 m2

Total cost of levelling and turning = Rs. 13322.50

Total area of square lawn = 13322.50/2.50

= 5329 m2

Side2 = 5329

Side of square lawn = √5329

= 73 m

So, total length of lawn = 4 × 73

= 292 m

∴ Cost of fencing the lawn at Rs 5 per metre = 292 × 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image42

The remainder is 38

So, the greatest 3 digit perfect square number is:

999 – 38 = 961

∴ 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 let’s find the square root of 2300

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image43

The remainder is 91

Since, (47)2 < 2300

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

(48)2 = 2304 > 2300

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

2304 – 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

√441/961 = 21/31

(ii) 324/841

The square root of

√324/841= 18/29

(iii) 4 29/29

The square root of

√(4 29/29) = √(225/49) = 15/7

(iv) 2 14/25

The square root of

√(2 14/25) = √(64/25) = 8/5

(v) 2 137/196

The square root of

√2 137/196 = √ (529/196) = 23/14

(vi) 23 26/121

The square root of

√(23 26/121) = √(2809/121) = 53/11

(vii) 25 544/729

The square root of

√(25 544/729) = √(18769/729) = 137/27

(viii) 75 46/49

The square root of

√(75 46/49) = √(3721/49) = 61/7

(ix) 3 942/2209

The square root of

√(3 942/2209) = √(7569/2209) = 87/47

(x) 3 334/3025

The square root of

√(3 334/3025) = √(9409/3025) = 97/55

(xi) 21 2797/3364

The square root of

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

(xii) 38 11/25

The square root of

√(38 11/25) = √(961/25) = 31/5

(xiii) 23 394/729

The square root of

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

(xiv) 21 51/169

The square root of

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

(xv) 10 151/225

The square root of

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

2. Find the value of:

(i) √80/√405

(ii) √441/√625

(iii) √1587/√1728

(iv) √72 ×√338

(v) √45 × √20

Solution:

(i) √80/√405

√80/√405 = √16/√81 = 4/9

(ii) √441/√625

√441/√625 = 21/25

(iii) √1587/√1728

√1587/√1728 = √529/√576 = 23/24

(iv) √72 ×√338

√72 ×√338 = √(2×2×2×3×3) ×√(2×13×13)

By using the formula √a × √b = √(a×b)

= √(2×2×2×3×3×2×13×13)

= 22 × 3 × 13

= 156

(v) √45 × √20

√45 × √20 = √(5×9×9) × √(5×2×2)

By using the formula √a × √b = √(a×b)

= √(5×9×9×5×2×2)

= 5 × 9 × 2

= 90

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

Solution:

We know that the given area = 80 244/729 m2

= 58564/729 m2

If L is length of each side

L2 = 58564/729

L = √ (58564/729) = √58564/√729

= 242/27

= 8 26/27

∴ Length is 8 26/27

4. The area of a square field is 30 1/4m2. Calculate the length of the side of the square.

Solution:

We know that the given area = 30 1/4 m2

= 121/4 m2

If L is length of each side then,

L2 = 121/4

L = √(121/4)  = √121/√4

= 11/2

∴ Length is 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 × b

= 72 × 338 m2

= 24336 m2

Area of square, L2 = 24336 m2

L = √24336

= 156 m

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image44

∴ the square root of 84.8241

√84.8241 = 9.21

2. 0.7225

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image45

∴ the square root of 0.7225

√0.7225 = 0.85

3. 0.813604

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image46

∴ the square root of 0.813604

√0.813604 = 0.902

4. 0.00002025

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image47

∴ the square root of 0.00002025

√0.00002025 = 0.0045

5. 150.0625

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image48

∴ the square root of 150.0625

√150.0625 = 12.25

6. 225.6004

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image49

∴ the square root of 225.6004

√225.6004 = 15.02

7. 3600.720036

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image50

∴ the square root of 3600.720036

√3600.720036 = 60.006

8. 236.144689

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image51

∴ the square root of 236.144689

√236.144689 = 15.367

9. 0.00059049

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image52

∴ the square root of 0.00059049

√0.00059049 = 0.0243

10. 176.252176

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image53

∴ the square root of 176.252176

√176.252176 = 13.276

11. 9998.0001

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image54

∴ the square root of 9998.0001

√9998.0001 = 99.99

12. 0.00038809

Solution:

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image55

∴ the square root of 0.00038809

√0.00038809 = 0.0197

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

Solution:

Let us consider a number a

Where, a = √227.798649

= 15.093

By using long division method let us verify

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image56

∴ 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., L2 = 256.6404 m2

L = √256.6404

= 16.02m

By using long division method let us verify

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image57

∴ 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 = √0.00053361

= 0.0231

By using long division method let us verify

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image58

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

16. Simplify:

(i) (√59.29 – √5.29)/ (√59.29 + √5.29)

(ii) (√0.2304 + √0.1764)/ (√0.2304 – √0.1764)

Solution:

(i) (√59.29 – √5.29)/ (√59.29 + √5.29)

Firstly let us find the square root √59.29 and √5.29

√59.29 = √5929/ √100

= 77/10

= 7.7

√5.29 = √5.29/ √100

= 23/10

= 2.3

So, (7.7 – 2.3)/ (7.7 + 2.3)

= 54/10

= 0.54

(ii) (√0.2304 + √0.1764)/ (√0.2304 – √0.1764)

Firstly let us find the square root √0.2304 and √0.1764

√0.2304 = √2304/ √10000

= 48/100

= 0.48

√0.1764 = √1764/ √10000

= 42/100

= 0.42

So, (0.48 + 0.42)/ (0.48 – 0.42)

= 0.9/0.06

= 15

17. Evaluate √50625 and hence find the value of √506.25 + √5.0625

Solution:

By using long division method let us find the √50625 

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image59

So now, 506.25 = √50625/ √100

= 225/10

= 22.5

√5.0625 = √50625/ √10000

= 225/100

= 2.25

So equating in the above equation we get,

√506.25 + √5.0625 = 22.5 + 2.25

= 24.75

18. Find the value of √103.0225 and hence find the value of
(i) √10302.25
(ii) √1.030225

Solution:

By using long division method let us find the √103.0225

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image60

So now, (i)10302.25 = √(10302.25× 100)

= 10 × 10.15

= 101.5

(ii)√1.030225 = √1.030225/ √1000

= 10.15/10

= 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) 7/8 (xviii) 5/12
(xix) 2 1/2 (xx) 287 5/8

Solution:

(i) 5

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image61

∴ the square root of 5 is 2.236

(ii) 7

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image62

∴ the square root of 7 is 2.646


(iii)
17

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image63

∴ the square root of 17 is 4.123

(iv) 20

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image64

∴ the square root of 20 is 4.472


(v)
66

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image65

∴ the square root of 66 is 8.124

(vi) 427

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image66

∴ the square root of 427 is 20.664


(vii)
1.7

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image67

∴ the square root of 1.7 is 1.304

(viii) 23.1

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image68

∴ the square root of 23.1 is 4.806


(ix)
2.5

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image69

∴ the square root of 2.5 is 1.581

(x) 237.615

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image70

∴ the square root of 237.615 is 15.415


(xi)
15.3215

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image71

∴ the square root of 15.3215 is 3.914

(xii) 0.9

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image72

∴ the square root of 0.9 is 0.949


(xiii)
0.1

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image73

∴ the square root of 0.1 is 0.316

(xiv) 0.016

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image74

∴ the square root of 0.016 is 0.126


(xv)
0.00064

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image75

∴ the square root of 0.00064 is 0.025

(xvi) 0.019

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image76

∴ the square root of 0.019 is 0.138


(xvii) 
7/8

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image77

∴ the square root of 7/8 is 0.935

(xviii) 5/12

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image78

∴ the square root of 5/12 is 0.645


(xix) 
2 1/2 

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image79

∴ the square root of 5/2 is 1.581

(xx) 287 5/8

By using long division method

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image80

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image81

∴ 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

RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots- image82

∴ the square root of 11 is 3.31662

4. Give that: √2 = 1.414, √3 = 1.732, √5 = 2.236 and √7 = 2.646, evaluate each of the following:
(i) √ (144/7)
(ii) √ (2500/3)

Solution:

(i) √ (144/7)

Now let us simplify the given equation

√ (144/7) = √ (12×12)/ √7

= 12/ 2.646

= 4.535

(ii) √ (2500/3)

Now let us simplify the given equation

√ (2500/3) = √ (5×5×10×10) /√3

= 5×10/ 1.732

= 50/1.732

= 28.867

5. Given that √2 = 1.414, √3 = 1.732, √5 = 2.236 and √7 = 2.646 find the square roots of the following:
(i) 196/75
(ii) 400/63
(iii) 150/7
(iv) 256/5
(v) 27/50

Solution:

(i) 196/75

Let us find the square root for196/75

(196/75) = (196)/ (75)

= (14×14)/ (5×5×3)

= 14/ (53)

= 14/ (5×1.732)

= 14/8.66

= 1.617


(ii) 
400/63

Let us find the square root for400/63

(400/63) = (400)/ (63)

= (20×20)/ (3×3×7)

= 20/ (37)

= 20/ (3×2.646)

= 20/7.938

= 2.520

(iii) 150/7

Let us find the square root for150/7

(150/7) = (150)/ (7)

= (3×5×5×2)/ (7)

= (52)/ (7)

= 5×1.732×1.414/ (2.646)

= 12.245/2.646

= 4.628

(iv) 256/5

Let us find the square root for256/5

(256/5) = (256)/ (5)

= (16×16)/ (5)

= 16/ (5)

= 16/2.236

= 7.155


(v) 
27/50

Let us find the square root for27/50

(27/50) = (27)/ (50)

= (3×3×3)/ (5×5×2)

= (33)/ (52)

= (3×1.732)/ (5×1.414)

= 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:

√7 = 2.645

∴ The square root of 7 is 2.645

2. 15

Solution:

From square root table we know,

Square root of 7 is:

15 = 3.8729

∴ The square root of 15 is 3.873

3. 74

Solution:

From square root table we know,

Square root of 74 is:

74 = 8.6023

∴ The square root of 74 is 8.602

4. 82

Solution:

From square root table we know,

Square root of 82 is:

82 = 9.0553

∴ The square root of 82 is 9.055

5. 198

Solution:

From square root table we know,

Square root of 198 is:

198 = 14.0712

∴ The square root of 198 is 14.071

6. 540

Solution:

From square root table we know,

Square root of 540 is:

540 = 23.2379

∴ The square root of 540 is 23.24

7. 8700

Solution:

From square root table we know,

Square root of 8700 is:

8700 = 93.2737

∴ The square root of 8700 is 93.27

8. 3509

Solution:

From square root table we know,

Square root of 3509 is:

3509 = 59.2368

∴ The square root of 3509 is 59.235

9. 6929

Solution:

From square root table we know,

Square root of 6929 is:

6929 = 83.2406

∴ The square root of 6929 is 83.239

10. 25725

Solution:

From square root table we know,

Square root of 25725 is:

25725 = 160.3901

∴ The square root of 25725 is 160.41

11. 1312.

Solution:

From square root table we know,

Square root of 1312 is:

1312 = 36.2215

∴ The square root of 1312 is 36.22

12. 4192

Solution:

From square root table we know,

Square root of 4192 is:

4192 = 64.7456

∴ The square root of 4192 is 64.75

13. 4955

Solution:

From square root table we know,

Square root of 4955 is:

4955 = 70.3917

∴ The square root of 4955 is 70.39

14. 99/144

Solution:

From square root table we know,

Square root of 99/144 is:

(99/144) = 0.82915

∴ 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:

√(57/169) = 0.58207

∴ 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:

√(101/169) = 0.77306

∴ 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:

√13.21 = 3.6345

∴ 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:

√21.97 = 4.6872

∴ The square root of 21.97 is 4.6872

19. 110

Solution:

From square root table we know,

Square root of 110 is:

√110 = 10.4880

∴ The square root of 110 is 10.488

20. 1110

Solution:

From square root table we know,

Square root of 1110 is:

√1110 = 33.3166

∴ The square root of 1110 is 33.317

21. 11.11

Solution:

From square root table we know,

Square root of 11.11 is:

√11.11 = 3.33316

∴ The square root of 11.11 is 3.333

22. The area of a square field is 325m2. Find the approximate length of one side of the field.

Solution:

We know that the given area of the field = 325 m2

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

√325 = 18.027 m

∴ 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

Side2 = 240 × 70

Side = √(240 × 70)

= √(10×10×2×2×2×3×7)

= 20√(42)

= 20 × 6.48

= 129.60 m

∴ 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.

Also, Access RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots

Exercise 3.1 Solutions 13 Questions

Exercise 3.2 Solutions 15 Questions

Exercise 3.3 Solutions 7 Questions

Exercise 3.4 Solutions 16 Questions

Exercise 3.5 Solutions 12 Questions

Exercise 3.6 Solutions 5 Questions

Exercise 3.7 Solutions 18 Questions

Exercise 3.8 Solutions 5 Questions

Exercise 3.9 Solutions 23 Questions

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 – 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.

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