# Brilliant’s Composite Mathematics Class 8 Solutions Chapter 1

## Brilliant’s Composite Mathematics Class 8 Solutions Chapter 1 Squares and Square Roots

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 1, Squares and Square Roots. Here students can easily find step by step solutions of all the problems for Squares and Square Roots, Exercise 1.1, 1.2 and 1.3 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

#### Squares and Square Roots Exercise 1.1 Solution :

Question no – (1)

(a) 2

= 22

= 2 × 2

= 4

(b) 21

= 212

= 21 × 21

= 441

(c) 21

= 212

= (20 + 1)2

= 202 + 20 × 12

= 400 + 40 + 1

= 441

(d) 29

= 292

= (30 – 1)2

= 302 – 2 × 30 × 1 + 11

= 900 – 60 + 1

= 641

(e) 111

= 1002 + 2 ×100 × 11 × 112

= 10000 + 2200 + 121

= 12,321

(f) 138

= 1382

= (140 – 2)2

=1402 – 2 × 140 × 2 + 22

= 19600 – 560 + 4

= 19044

(g) 37

= 372

= (40 – 3)2

= 402 – 2 × 40 × 3 + 32

= 1600 – 240 + 9

= 1369

(h) 48

= 482

= (50 – 2)

= 502 – 2 × 50 × 2 + 22

= 2500 – 200 + 4

= 2304

(i) 2/5

= 2/52

= 2/5 × 2/5

= 4/25

(j) 102/300

= 102/3002

= 102/300 × 102/300

= 289/2500

= 0.1156

(k) 0.7

= (0.7)2

= 7/10 × 7/10

= 49/100

= 0.49

Question no – (2)

(a) Given number, 16

= 16 = 2 × 2 × 2 × 2  (Perfect square)

(b) 12

= 12 = 2 × 2 × 3  (Not perfect square)

(c) 20

20 = 2 × 2 × 5 (Not perfect square)

(d) 3125

= 3125 = 25 × 25 × 5  (Not perfect square)

(e) 2156

= 2156 = 14 × 14 × 11 (Not perfect square)

(f) 118

= 188 = 2 × 59  (Not perfect square)

(g) 350

= 350 = 5 × 7 × 5 × 2  (Not perfect square)

(h) 1600

= 1600 = 4 × 4 × 10 × 10  (perfect square)

(i) 252

= 252 = 2 × 2 × 3 × 3 × 7  (Not perfect square)

Question no – (3)

Solution :

(a) 25

= √25

= 5  (odd Number)

(b) 121

= √121

= 11  (odd Number)

(c) 64

= √64

= 8

(d) 100

= √100

= 10

(e) 256

= √256

= 16

(f) 1296

= √1296

= 36

(g) 36

= √36

= 6

(h) 6561

= √6561

= 81

Question no – (4)

Solution :

(a) 9

= √9

= 3

(b) 49

= √49

= 7

(c) 36

= √36

= 6

(d) 169

= √169

= 13

(e) 225

= √225

= 15

(f) 900

= √900

= 30

∴ They are the Squares of odd number.

Question no – (7)

Solution :

Given, 22050

= 2 × 5 × 5 × 21 × 21

Since there are only 2 left.

So, if multiplied by 2, it become perfect square.

Question no – (8)

Solution :

Given, 6000

= 10 × 10 × 2 × 2 × 15

∴ If we divide by 15 it become perfect square.

Question no – (9)

Solution :

(a) 182 – 172 = ____

= 182 – 172

= 324 – 289

= 35

(b) 242 – 232 = ____

= 242 – 232

= (24 + 23) (24 – 23)

= 47

(c) 322 – 312 = ____

= 322 – 312

= (32 + 31) (32 – 31)

= 63

(d) 5512 – 5502 = ____

= 5512 – 5502

= (551 + 550) (551 – 550)

= 1101

(e) 1012 – 1002 = ____

= 1012 – 1002

= (101 + 100) (101 – 100)

= 201

(f) 2012 – 2002 = ____

= 2012 – 2002

= (201 + 200) (201 – 200)

= 401

Question no – (10)

Solution :

(a) 552 = ____

= 552

= 55 × 55

= 3025

(b) 752 = ____

= 75 × 75

= 5625

(c) 952 = ____

= (100 – 5)2

= 1002 – 2 × 100 × 5 + 52

= 10000 – 100 + 25

= 9025

Question no – (11)

Solution :

(a) (1, 2, 3)

= 12 + 22

= 1 + 4

= 5 ≠ 32

∴ Not a Pythagorean triplet.

(b) (3, 4, 5)

= 32 + 42

= 9 + 16

= 25 = 52

∴ It is a Pythagorean triplet.

(c) (6, 8, 10)

= 62 + 82

= 36 + 64

= 100 = 102

∴ It is a Pythagorean triplet.

(d) (2, 2, 3)

= 22 + 22

= 4 + 4

= 8 ≠ 32

∴ Not a Pythagorean triplet.

(e) (1, 1, 1)

12 + 12

= 1 + 1

= 2 ≠ 12

∴ Not a Pythagorean triplet.

(f) (6, 7, 8)

= 62 + 72

= 36 + 49

= 85 ≠ 82

∴ Not a Pythagorean triplet.

#### Squares and Square Roots Exercise 1.2 Solution :

Question no – (1)

Solution :

(a) 100

100 = 2 × 2 × 5 × 5

∴ = √100 = 2 × 5

= 100

(b) 1225

1225 = 5 × 5 × 7 × 7

1225 = 5 × 5 × 7 × 7

∴ = √1225 = 5 × 7

= 35

(c) 625

625 = 5 × 5 × 5 × 5

∴ = √625 = 5 × 5

= 25

(d) 1296

1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

∴ = √1296 = 2 × 2 × 3 × 3

= 36

(e) 144

144 = 2 × 2 × 2× 2 × 3 × 3

∴ = √144 = 2 × 2 × 3

= 12

(f) 9/16

= √9/16

= = √3 × 3/2 × 2 × 2 × 2

= 3/2 × 2

= 3/4

(g) 4096

= 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ = √4096 = 2 × 2 × 2 × 2 × 2× 2

= 64

(h) 529

= 529 = 23 × 23

∴ = √629

= 23

(i) 144/169

= √144/169

= = √2 × 2 × 2 × 2 × 3 × 3/13× 13

= 2 × 2 × 3/13

= 12/13

(j) 3844/196

= √3844/196

= √2 × 2 × 31 × 31/2 × 2 × 7 × 7

= 2 × 31/2 × 7

= 31/7

Question no – (3)

Solution :

(a) √45 + 4

= √45 + 4

= √45

= √ 7 × 7

= 7 …(Simplified)

(b)  √520 + 9

= √520 + 9

= √524

= √23 × 23

= 23 …(Simplified)

(c) √49 + √169/√64 + √49

= √49 + √169/√64 + √49

= 7 + 13/8 + 7

= 20/15

= 4/3 …(Simplified)

(d) √1024 + √529/√4096 + √196

√1024 + √529/√4096 + √196

= 32 + 23/64 + 14

= 55/78 …(Simplified)

(e) √1764 + √400/√16 + √256

= 42 + 20/4 + 16

= 62/20

= 31/10 …(Simplified)

Question no – (4)

Solution :

So, 1100 = 2 × 2 × 5 × 5 × 11

We can see that 2 and 5 have pairs fret 11 doesn’t. So in order to get a perfect square either we have to divide 11.00 by 11 or multiply with 11.

In, first case,

= 1100 ÷ 11 = 100

and √100  = 10

In, second case,

1100 × 11 = 12100

and  √12100 = 110

Question no – (5)

Solution :

∴ 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7

Clearly, only 3 doesn’t pair.

Therefore, to get a perfect square, we have to divide 9408 by 3.

Question no – (6)

Solution :

Let, the number of students be x

∴ x2 = 4096

∴ x = √4096

∴ x = 64

Now,

∴ 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 2

∴ √4096 = 2 × 2 × 2 × 2 × 2 × 2

= 64

∴ There were 64 students in the school.

#### Squares and Square Roots Exercise 1.3 Solution :

Question no – (1)

Solution :

Given numbers, 64, 289, 169, 625, 4096.

√64

= √2 × 2 × 2 × 2 × 2 × 2

= 2 × 2× 2

= 8

√289

= √17 × 17

= 17

√169

= √13 × 13

= 13

√625

= √5 × 5 × 5 × 5

= 5 × 5

= 25

√4096

= √ 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2 × 2 × 2 × 2 × 2× 2

= 64

Question no – (2)

Solution :

(i) 3481

∴ √3481

= 59

(ii) 119025

∴ √119025

= 345

(iii) 152881

∴ √152881

= 391

(iv) 5776

∴ √5776

= 76

Question no – (3)

Solution :

(i) 363609

∴ √363609

= 603

(ii) 531441

∴ √ 531441

= 729

(iii) 22801

∴ √22801

= 151

(iv) 998001

∴ √998001

= 999

(v) 18662400

∴ √18662400

= 4320

(vi) 5803281

∴ √5803281

= 2409

(vii) 480249

∴ √480249

= 693

(viii) 63409369

∴ √63409369

= 7977

(ix) 9653449

∴ √9653449

= 3107

(x) 62504836

∴ √62504836

= 7906

Question no – (4)

Solution :

∴ To get a perfect square we have to subtracted 31 and the square root will be 127.

Question no – (5)

Solution :

Question no – (6)

Solution :

∴ To get a perfect square we have to subtracted 31 and the square root will be 127.

Question no – (7)

Solution :

∴ 489 should be added with 100000 to get a perfect square.

∴ 100489 is the least square number of 6 digits.

Question no – (8)

Solution :

∴ We have to subtract 198

∴ 9999 – 198

= 9801

Question no – (9)

Solution :

∴ There are 85 rows

Question no – (10)

Solution :

(i)  2 23/49

√2 23/49

= √(49 × 2) + 23/49

= √121/49

= √11 × 11/7 × 7

= 11/7

= 1 4/7

(ii) 256/625

= √256/625

= √2 × 2 × 2 × 2 × 2 × 2 × 2 × 2/5 × 5

= 2 × 2 × 2 × 2/5 × 5

= 16/5

(iii) 113 7/9

= √133 7/9

= √1024/9

= √2 v 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2/3 × 3

= 2 × 2 × 2 × 2 × 2/3

= 32/3

(iv) 29 4/25

= √ 29 4/25

= √729/25

= √3 × 3 × 3 × 3 × 3 × 3/ 5 × 5

= 3 × 3 × 3/5

= 27/5

(v) 21 51/169

= √21 51/169

= √3600/169

= √2 × 2 × 2 × 2 × 3 × 3 × 5 × 5/13 × 13

= 2 × 2 v 3 × 5/13

= 60/13

(vi) 21 2797/3364

= √73441/3364

= √ 271 × 271/58 × 58

= 271/58

Question no – (11)

Solution :

(i) 37.0881

∴ √37.0881

= 6.09

(ii) 0.1681

∴ √0.1681

= 0.41

(iii) 7260.7441

∴ √7260.7441

= 85.21

(iv) 1225.6004

Question no – (12)

Solution :

(i) 7

∴ √7

= 2.6457 ≈ 2.646

(ii) 14.01

∴ √14.01

= 3.7429 ≈ 3.743

(iii) 3.14

∴ √3.14

= 1.7708 ≈ 1.771

(iv) 25.6/52.9

= √256 × 10/529 × 10

= √256/529

= 16/23

(v) 71971.145

∴ √71971.145

= 268.2741 ≈ 268.274

Next Chapter Solutions :

Updated: May 29, 2023 — 2:44 pm