NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots

Question

Here you will find NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots, which is one of the most important study materials for students preparing for the CBSE board examination. Our NCERT Solutions are based on the most recent CBSE syllabus. These solutions, prepared and designed by subject matter experts, are to the point and explained in a point-by-point manner to make it easier for students to learn, study, and revise at the last minute. All of the shortcuts, equations, rules, theorems, and axioms have also been covered. We have explained all of the topics in detail in the NCERT Solution Class 8 that we have provided, which will undoubtedly help students clear all of the required concepts. Important topics covered here include:

• Perfect square
• Properties of Square numbers
• Finding the Square of a Number
• Square Roots
• Square Roots of Decimals
• Estimating Square Root

Our NCERT Solutions Class 8 Maths is the best resource for practicing and learning all of the key concepts. Regular practice of our solutions will assist students in developing subject-related problem-solving skills. We have answers to all types of questions, including exercise and in-text questions.

Answer 1

NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots

1. What will be the unit digit of the squares of the following numbers?

i. 81

ii. 272

iii. 799

iv. 3853

v. 1234

vi. 26387

vii. 52698

viii. 99880

ix. 12796

x. 55555

Solution:

The unit digit of square of a number having ‘a’ at its unit place ends with a×a.

i. The unit digit of the square of a number having digit 1 as unit’s place is 1.

∴ Unit digit of the square of number 81 is equal to 1.

ii. The unit digit of the square of a number having digit 2 as unit’s place is 4.

∴ Unit digit of the square of number 272 is equal to 4.

iii. The unit digit of the square of a number having digit 9 as unit’s place is 1.

∴ Unit digit of the square of number 799 is equal to 1.

iv. The unit digit of the square of a number having digit 3 as unit’s place is 9.

∴ Unit digit of the square of number 3853 is equal to 9.

v. The unit digit of the square of a number having digit 4 as unit’s place is 6.

∴ Unit digit of the square of number 1234 is equal to 6.

vi. The unit digit of the square of a number having digit 7 as unit’s place is 9.

∴ Unit digit of the square of number 26387 is equal to 9.

vii. The unit digit of the square of a number having digit 8 as unit’s place is 4.

∴ Unit digit of the square of number 52698 is equal to 4.

viii. The unit digit of the square of a number having digit 0 as unit’s place is 01.

∴ Unit digit of the square of number 99880 is equal to 0.

ix. The unit digit of the square of a number having digit 6 as unit’s place is 6.

∴ Unit digit of the square of number 12796 is equal to 6.

x. The unit digit of the square of a number having digit 5 as unit’s place is 5.

∴ Unit digit of the square of number 55555 is equal to 5.

2. The following numbers are obviously not perfect squares. Give reason.

i. 1057

ii. 23453

iii. 7928

iv. 222222

v. 64000

vi. 89722

vii. 222000

viii. 505050

Solution:

We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.

i. 1057 ⟹ Ends with 7

ii. 23453 ⟹ Ends with 3

iii. 7928 ⟹ Ends with 8

iv. 222222 ⟹ Ends with 2

v. 64000 ⟹ Ends with 0

vi. 89722 ⟹ Ends with 2

vii. 222000 ⟹ Ends with 0

viii. 505050 ⟹ Ends with 0

3. The squares of which of the following would be odd numbers?

i. 431

ii. 2826

iii. 7779

iv. 82004

Solution:

We know that the square of an odd number is odd and the square of an even number is even.

i. The square of 431 is an odd number.

ii. The square of 2826 is an even number.

iii. The square of 7779 is an odd number.

iv. The square of 82004 is an even number.

4. Observe the following pattern and find the missing numbers. 11² = 121

101² = 10201

1001² = 1002001

100001² = 1 …….2………1

10000001² = ……………………..

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.

∴ 100001² = 10000200001

10000001² = 100000020000001

5. Observe the following pattern and supply the missing numbers. 112 = 121

1012 = 10201

101012 = 102030201

10101012 = ………………………

…………2 = 10203040504030201

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And, the square is symmetric about the middle digit. If the middle digit is 4, then the number to be squared is 10101 and its square is 102030201.

So, 10101012 =1020304030201

1010101012 =10203040505030201

6. Using the given pattern, find the missing numbers. 1² + 2² + 2² = 3²

2² + 3² + 6² = 7²

3² + 4² + 12² = 13²

4² + 5² + _² = 21²

5 + _ ² + 30² = 31²

6 + 7 + _ ² = _ ²

Solution:

Given, 1² + 2² + 2² = 3²

i.e 1² + 2² + (1×2 )² = ( 1² + 2² -1 × 2 )²

2² + 3² + 6² = 7²

∴ 2² + 3² + (2×3 )² = (2² + 3² -2 × 3)²

3² + 4² + 12² = 13²

∴ 3² + 4² + (3×4 )² = (3² + 4² – 3 × 4)²

4² + 5² + (4×5 )² = (4² + 5² – 4 × 5)²

∴ 4² + 5² + 20² = 21²

5² + 6² + (5×6 )² = (5²+ 6² – 5 × 6)²

∴ 5² + 6² + 30² = 31²

6² + 7² + (6×7 )² = (6² + 7² – 6 × 7)²

∴ 6² + 7² + 42² = 43²

7. Without adding, find the sum.

i. 1 + 3 + 5 + 7 + 9

ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution:

i.  Sum of first five odd number = (5)² = 25

ii. Sum of first ten odd number = (10)² = 100

iii. Sum of first thirteen odd number = (12)² = 144

Answer 2

8. (i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

Solution:

(i) We know, sum of first n odd natural numbers is n².

Since,49 = 7²

∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Since, 121 = 11²

∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?

i. 12 and 13

ii. 25 and 26

iii. 99 and 100

Solution:

Between n² and (n + 1)², there are 2n non–perfect square numbers.

i. 122 and 132 there are 2 × 12 = 24 natural numbers.

ii. 252 and 262 there are 2 × 25 = 50 natural numbers.

iii. 992 and 1002 there are 2 × 99 =198 natural numbers.

10. Find the square of the following numbers.

i. 32

ii. 35

iii. 86

iv. 93

v. 71

vi. 46

Solution:

i. (32)²

= (30 + 2)²

= (30)² + (2)² + 2 × 30 × 2 [Since, (a + b)² = a² + b² + 2ab]

= 900 + 4 + 120

= 1024

ii. (35)²

= (30+5 )²

= (30)² + (5)² + 2 × 30 × 5 [Since, (a + b)² = a² + b² + 2ab]

= 900 + 25 + 300

= 1225

iii. (86)²

= (90 – 4)²

= (90)² + (4)² – 2 × 90 × 4 [Since, (a + b)² = a² + b² +2ab]

= 8100 + 16 – 720

= 8116 – 720

= 7396

iv. (93)²

= (90 + 3)²

= (90)² + (3)² + 2 × 90 × 3 [Since, (a + b)² = a² + b² + 2ab]

= 8100 + 9 + 540

= 8649

v. (71)²

= (70 + 1 )²

= (70)² + (1)² + 2 × 70 × 1 [Since, (a + b)² = a² + b² + 2ab]

= 4900 + 1 + 140

= 5041

vi. (46)²

= (50 - 4 )²

= (50)² + (4)² – 2 × 50 × 4 [Since, (a + b)² = a² + b² + 2ab]

= 2500 + 16 – 400

= 2116

11. Write a Pythagorean triplet whose one member is.

i. 6

ii. 14

iii. 16

iv. 18

Solution:

(i) Let m² – 1 = 6

[Triplets are in the form 2m, m² – 1, m² + 1]

m² = 6 + 1 = 7

So, the value of m will not be an integer.

Now, let us try for m² + 1 = 6

⇒ m² = 6 – 1 = 5

Also, the value of m will not be an integer.

Now we let 2m = 6 ⇒ m = 3 which is an integer.

Other members are:

m² – 1 = 3² – 1 = 8 and m² + 1 = 3² + 1 = 10

Hence, the required triplets are 6, 8 and 10

(ii) Let m² – 1 = 14

⇒ m² = 1 + 14 = 15

The value of m will not be an integer.

Now take 2m = 14 ⇒ m = 7 which is an integer.

The member of triplets are 2m = 2 × 7 = 14

m² – 1 = (7)² – 1 = 49 – 1 = 48

and m² + 1 = (7)² + 1 = 49 + 1 = 50

i.e., (14, 48, 50)

(iii) Let 2m = 16

m = 8

The required triplets are 2m = 2 × 8 = 16

m² – 1 = (8)² – 1 = 64 – 1 = 63

m² + 1 = (8)² + 1 = 64 + 1 = 65

i.e., (16, 63, 65)

(iv) Let 2m = 18

⇒ m = 9

Required triplets are:

2m = 2 × 9 = 18

m² – 1 = (9)² – 1 = 81 – 1 = 80

and m² + 1 = (9)² + 1 = 81 + 1 = 82

i.e., (18, 80, 82)

12. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

i. 9801

ii. 99856

iii. 998001

iv. 657666025

Solution:

i. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[9²=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 9801 is equal to 1 or 9.

ii. We know that the unit’s digit of the square of a number having digit as unit’s

place 6 is 6 and also 4 is 6 [6²=36 and 4²=16, both the squares have unit digit 6].

∴ Unit’s digit of the square root of number 99856 is equal to 6.

iii. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[9²=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 998001 is equal to 1 or 9.

iv. We know that the unit’s digit of the square of a number having digit as unit’s

place 5 is 5.

∴ Unit’s digit of the square root of number 657666025 is equal to 5.

13. Without doing any calculation, find the numbers which are surely not perfect squares.

i. 153

ii. 257

iii. 408

iv. 441

Solution:

We know that natural numbers ending with the digits 0, 2, 3, 7 and 8 are not perfect square.

i. 153⟹ Ends with 3.

∴ 153 is not a perfect square

ii. 257⟹ Ends with 7

∴ 257 is not a perfect square

iii. 408⟹ Ends with 8

∴ 408 is not a perfect square

iv. 441⟹ Ends with 1

∴ 441 is a perfect square.

Answer 3

14. Find the square roots of 100 and 169 by the method of repeated subtraction.

Solution:

100

100 – 1 = 99

99 – 3 = 96

96 – 5 = 91

91 – 7 = 84

84 – 9 = 75

75 – 11 = 64

64 – 13 = 51

51 – 15 = 36

36 – 17 = 19

19 – 19 = 0

Here, we have performed subtraction ten times.

∴ √100 = 10

169

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

Here, we have performed subtraction thirteen times.

∴ √169 = 13

15. Find the square roots of the following numbers by the Prime Factorisation Method.

i. 729

ii. 400

iii. 1764

iv. 4096

v. 7744

vi. 9604

vii. 5929

viii. 9216

ix. 529

x. 8100

Solution:

(i) 729

Prime factors of 729

729 = 3 × 3 × 3 × 3 × 3 × 3 = 3² × 3² × 3²

√729 = 3 × 3 × 3 = 27

(ii) 400

Prime factors of 400

400 = 2 × 2 × 2 × 2 × 5 × 5 = 2² × 2² × 5²

√400 = 2 × 2 × 5 = 20

(iii) 1764

1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7²

√1764 = 2 × 3 × 7 = 42

(iv) 4096

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

= 2² × 2² × 2² × 2² × 2² × 2²

√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

(v)  7744

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

= 2² × 2² × 2² × 11²

√7744 = 2 × 2 × 2 × 11 = 88

(vi) 9604

9604 = 2 × 2 × 7 × 7 × 7 × 7 = 2² × 7² × 7²

√9604 = 2 × 7 × 7 = 98

(vii) 5929

5929 = 7 × 7 × 11 × 11 = 7² × 11²

√5929 = 7 × 11 = 77

(viii) 9216

9216 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3

= 2² × 2² × 2² × 2² × 2² × 3²

√9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96

(ix) 529

529 = 23 × 23 = 23²

√529 = 23

(x) 8100

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2² × 3² × 3² × 5²

√8100 = 2 × 3 × 3 × 5 = 90

Answer 4

16. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.

(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

Solution:

(i) Prime factorisation of 252 is

252 = 2 × 2 × 3 × 3 × 7

Here, the prime factorisation is not in pair. 7 has no pair.

Thus, 7 is the smallest whole number by which the given number is multiplied to get a perfect square number.

The new square number is 252 × 7 = 1764

Square root of 1764 is

√1764 = 2 × 3 × 7 = 42

(ii) Primp factorisation of 180 is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

New square number = 180 × 5 = 900

The square root of 900 is

√900 = 2 × 3 × 5 = 30

Thus, 5 is the smallest whole number by which the given number is multiplied to get a square number.

(iii) Prime factorisation of 1008 is

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here, 7 has no pair.

New square number = 1008 × 7 = 7056

Thus, 7 is the required number.

Square root of 7056 is

√7056 = 2 × 2 × 3 × 7 = 84

(iv) Prime factorisation of 2028 is

2028 = 2 × 2 × 3 × 13 × 13

Here, 3 is not in pair.

Thus, 3 is the required smallest whole number.

New square number = 2028 × 3 = 6084

Square root of 6084 is

√6084 = 2 × 13 × 3 = 78

(v) Prime factorisation of 1458 is

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, 2 is not in pair.

Thus, 2 is the required smallest whole number.

New square number = 1458 × 2 = 2916

Square root of 1458 is

√2916 = 3 × 3 × 3 × 2 = 54

(vi) Prime factorisation of 768 is

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, 3 is not in pair.

Thus, 3 is the required whole number.

New square number = 768 × 3 = 2304

Square root of 2304 is

√2304 = 2 × 2 × 2 × 2 × 3 = 48

17. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained.

(i) 252

(ii) 2925

(iii) 396

(iv) 2645

(v) 2800

(vi) 1620

Solution:

(i) Prime factorisation of 252 is

252 = 2 × 2 × 3 × 3 × 7

Here 7 has no pair.

7 is the smallest whole number by which 252 is divided to get a square number.

New square number = 252 ÷ 7 = 36

Thus, √36 = 6

(ii) Prime factorisation of 2925 is

2925 = 3 × 3 × 5 × 5 × 13

Here, 13 has no pair.

13 is the smallest whole number by which 2925 is divided to get a square number.

New square number = 2925 ÷ 13 = 225

Thus √225 = 15

(iii) Prime factorisation of 396 is

396 = 2 × 2 × 3 × 3 × 11

Here 11 is not in pair.

11 is the required smallest whole number by which 396 is divided to get a square number.

New square number = 396 ÷ 11 = 36

Thus √36 = 6

(iv) Prime factorisation of 2645 is

2645 = 5 × 23 × 23

Here, 5 is not in pair.

5 is the required smallest whole number.

By which 2645 is multiplied to get a square number

New square number = 2645 ÷ 5 = 529

Thus, √529 = 23

(v) Prime factorisation of 2800 is

2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

Here, 7 is not in pair.

7 is the required smallest number.

By which 2800 is multiplied to get a square number.

New square number = 2800 ÷ 7 = 400

Thus √400 = 20

(vi) Prime factorisation of 1620 is

1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

Here, 5 is not in pair.

5 is the required smallest prime number.

By which 1620 is multiplied to get a square number = 1620 ÷ 5 = 324

Thus √324 = 18

Answer 5

18. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’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 the number of students in the school be, x.

∴ Each student donate Rs.x .

Total many contributed by all the students= x × x = x²

Given, x² = Rs.2401

x² = 7 × 7 × 7 × 7

⇒ x² = (7 × 7) × (7 × 7)

⇒ x² = 49 × 49

⇒ x = √(49 × 49)

⇒ x = 49

∴ The number of students = 49

19. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Solution:

Let the number of rows be, x.

∴ the number of plants in each rows = x.

Total many contributed by all the students = x × x = x²

Given,

x₂ = Rs.2025

x² = 3 × 3 × 3 × 3 × 5 × 5

⇒ x² = (3 × 3) × (3 × 3) × (5 × 5)

⇒ x² = (3 × 3 × 5) × (3 × 3 × 5)

⇒ x² = 45 × 45

⇒ x = √45 × 45

⇒ x = 45

∴ The number of rows = 45 and the number of plants in each rows = 45.

20. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution:

LCM of 4, 9, 10 = 180

The least number divisible by 4, 9 and 10 = 180

Now prime factorisation of 180 is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

The required smallest square number = 180 × 5 = 900

21. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution:

L.C.M of 8, 15 and 20 is (2 × 2 × 5 × 2 × 3) 120.

120 = 2 × 2 × 3 × 5 × 2

= (2 × 2) × 3 × 5 × 2

Here, 3, 5 and 2 cannot be paired.

∴ We will multiply 120 by (3 × 5 × 2) 30 to get perfect square.

Hence, the smallest square number divisible by 8, 15 and 20 = 120 × 30 = 3600

22. Find the square root of each of the following numbers by Long Division method.

(i) 2304

(ii) 4489

(iii) 3481

(iv) 529

(v) 3249

(vi) 1369

(vii) 5776

(viii) 7921

(ix) 576

(x) 1024

(xi) 3136

(xii) 900

Solution:

Answer 6

23. Find the number of digits in the square root of each of the following numbers (without any calculation).

i. 64

ii. 144

iii. 4489

iv. 27225

v. 390625

Solution:

We know that if n is number of digits in a square number then

Number of digits in the square root = \(\frac n2\) if n is even and \(\frac{n+1}2\) if n is odd.

(i) 64

Here n = 2 (even)

Number of digits in √64 = \(\frac22\) = 1

(ii) 144

Here n = 3 (odd)

Number of digits in square root = \(\frac{3+1}2\) = 2

(iii) 4489

Here n = 4 (even)

Number of digits in square root = \(\frac42\) = 2

(iv) 27225

Here n = 5 (odd)

Number of digits in square root = \(\frac{5+1}2\) = 3

(v) 390625

Here n = 6 (even)

Number of digits in square root = \(\frac 62\) = 3

24. Find the square root of the following decimal numbers.

(i) 2.56

(ii) 7.29

(iii) 51.84

(iv) 42.25

(v) 31.36

Solution:

25. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 402

(ii) 1989

(iii) 3250

(iv) 825

(v) 4000

Solution:

(i) 

Here remainder is 2

2 is the least required number to be subtracted from 402 to get a perfect square

New number = 402 – 2 = 400

Thus, √400 = 20

(ii)

Here remainder is 53

53 is the least required number to be subtracted from 1989.

New number = 1989 – 53 = 1936

Thus, √1936 = 44

(iii)

Here remainder is 1

1 is the least required number to be subtracted from 3250 to get a perfect square.

New number = 3250 – 1 = 3249

Thus, √3249 = 57

(iv)

Here, the remainder is 41

41 is the least required number which can be subtracted from 825 to get a perfect square.

New number = 825 – 41 = 784

Thus, √784 = 28

(v)

Here, the remainder is 31

31 is the least required number which should be subtracted from 4000 to get a perfect square.

New number = 4000 – 31 = 3969

Thus, √3969 = 63

26. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.

(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

Solution:

(i)

Here remainder is 41

It represents that square of 22 is less than 525.

Next number is 23 an 23² = 529

Hence, the number to be added = 529 – 525 = 4

New number = 529

Thus, √529 = 23

(ii)

Here the remainder is 69

It represents that square of 41 is less than in 1750.

The next number is 42 and 42² = 1764

Hence, number to be added to 1750 = 1764 – 1750 = 14

Require perfect square = 1764

√1764 = 42

(iii)

Here the remainder is 27.

It represents that a square of 15 is less than 252.

The next number is 16 and 16² = 256

Hence, number to be added to 252 = 256 – 252 = 4

New number = 252 + 4 = 256

Required perfect square = 256

and √256 = 16

(iv)

The remainder is 61.

It represents that square of 42 is less than in 1825.

Next number is 43 and 43² = 1849

Hence, number to be added to 1825 = 1849 – 1825 = 24

The required perfect square is 1848 and √1849 =43

(v)

Here, the remainder is 12.

It represents that a square of 80 is less than in 6412.

The next number is 81 and 81² = 6561

Hence the number to be added = 6561 – 6412 = 149

The require perfect square is 6561 and √6561 = 8

Answer 7

27. Find the length of the side of a square whose area = 441 m²

Solution:

Let the length of the side of the square be x m.

Area of the square = (side)² = x² m²

x² = 441 ⇒ x = √441 = 21

Thus, x = 21 m.

Hence the length of the side of square = 21 m.

28. In a right triangle ABC, ∠B = 90°.

(a) If AB = 6 cm, BC = 8 cm, find AC

(b) If AC = 13 cm, BC = 5 cm, find AB

Solution:

(a) In right triangle ABC

AC² = AB² + BC² [By Pythagoras Theorem]

⇒ AC² = (6)² + (8)² = 36 + 64 = 100

⇒ AC = √100 = 10

Thus, AC = 10 cm.

(b) In right triangle ABC

AC² = AB² + BC² [By Pythagoras Theorem]

⇒ (13)² = AB² + (5)²

⇒ 169 = AB² + 25

⇒ 169 – 25 = AB²

⇒ 144 = AB²

AB = √144 = 12 cm

Thus, AB = 12 cm.

29. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs more for this.

Solution:

Let the number of rows be x.

And the number of columns also be x.

Total number of plants = x × x = x²

x² = 1000

⇒ x = √1000

Here the remainder is 39

So the square of 31 is less than 1000.

Next number is 32 and 32² = 1024

Hence the number to be added = 1024 – 1000 = 24

Thus the minimum number of plants required by him = 24.

Alternative method:

The minimum number of plants required by him = 24.

30. There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?

Solution:

Let the number of children in a row be x. And also that of in a column be x.

Total number of students = x × x = x²

x² = 500

⇒ x = √500

Here the remainder is 16

New Number 500 – 16 = 484

and, √484 = 22

Thus, 16 students will be left out in this arrangement.