# ML Aggarwal Solutions for Class 8 Maths Chapter 3: Squares and Square Roots

ML Aggarwal Solutions for Class 8 Maths Chapter 3 Squares and Square Roots help students understand the types of methods to be followed in solving problems effortlessly. The solutions can be used by the students, irrespective of time constraint to boost their annual exam preparation. Students can download ML Aggarwal Solutions for Class 8 Maths Chapter 3 Squares and Square Roots free PDF, from the links which are given below.

Chapter 3 has problems on determining the squares and square roots of fraction, natural number and decimals. The various steps are explained in simple language to make it easy for the students during revision. The solutions are designed with the aim of reducing fear among students and improving confidence in answering complex problems in a shorter duration.

## Access ML Aggarwal Solutions for Class 8 Maths Chapter 3: Squares and Square Roots

Exercise 3.1

1. Which of the following natural numbers are perfect squares? Give reasons in support of your answer.

(i) 729

(ii) 5488

(iii) 1024

(iv) 243

Solution:

(i) 729

We know that

It can be written as

729 = 3 × 3 × 3 × 3 × 3 × 3

Here

729 is the product of pairs of equal prime factors

Therefore, 729 is a perfect square.

(ii) 5488

We know that

It can be written as

5488 = 2 × 2 × 2 × 2 × 7 × 7 × 7

Here

After pairing the same prime factors, one factor 7 is left unpaired.

Therefore, 5488 is not a perfect square.

(iii) 1024

We know that

It can be written as

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Here

After pairing the same prime factors, there is no factor left.

Therefore, 1024 is a perfect square.

(iv) 243

We know that

It can be written as

243 = 3 × 3 × 3 × 3 × 3

Here

After pairing the same prime factors, factor 3 is left unpaired.

Therefore, 243 is not 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.

(i) 1296

(ii) 1764

(iii) 3025

(iv) 3969

Solution:

(i) 1296

We know that

It can be written as

1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

Here

After pairing the same prime factors, no factor is left.

Therefore, 1296 is a perfect square of 2 × 2 × 3 × 3 = 36.

(ii) 1764

We know that

It can be written as

1764 = 2 × 2 × 3 × 3 × 7 × 7

Here

After pairing the same factors, no factor is left.

Therefore, 1764 is a perfect square of 2 × 3 × 7 = 42.

(iii) 3025

We know that

It can be written as

3025 = 5 × 5 × 11 × 11

Here

After pairing the same prime factors, no factor is left.

Therefore, 3025 is a perfect square of 5 × 11 = 55.

(iv) 3969

We know that

It can be written as

3969 = 3 × 3 × 3 × 3 × 7 × 7

Here

After pairing the same prime factors, no factor is left.

Therefore, 3969 is a perfect square of 3 × 3 × 7 = 63.

3. Find the smallest natural number by which 1008 should be multiplied to make it a perfect square.

Solution:

We know that

It can be written as

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here

After pairing the same kind of prime factors, one factor 7 is left.

Now multiplying 1008 by 7

We get a perfect square

Therefore, the required smallest number is 7.

4. Find the smallest natural number by which 5808 should be divided to make it a perfect square. Also, find the number whose square is the resulting number.

Solution:

We know that

It can be written as

5808 = 2 × 2 × 2 × 2 × 3 × 11 × 11

Here

After pairing the same kind of prime factors, factor 3 is left.

Now dividing the number by 3, we get a perfect square.

Therefore, the square root of the resulting number is 2 × 2 × 11 = 44.

Exercise 3.2

1. Write five numbers which you can decide by looking at their one’s digit that they are not square numbers. Solution:

We know that

A number which ends with the digits 2, 3, 7 or 8 at its unit places is not a perfect square.

Example – 111, 372, 563, 978, 1282 are not square numbers.

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

(i) 951

(ii) 502

(iii) 329

(iv) 643

(v) 5124

(vi) 7625

(vii) 68327

(viii) 95628

(ix) 99880

(x) 12796

Solution:

(i) 951

The unit digit of the square is 1.

(ii) 502

The unit digit of the square is 4.

(iii) 329

The unit digit of the square is 1.

(iv) 643

The unit digit of the square is 9.

(v) 5124

The unit digit of the square is 6.

(vi) 7625

The unit digit of the square is 5.

(vii) 68327

The unit digit of the square is 9.

(viii) 95628

The unit digit of the square is 4.

(ix) 99880

The unit digit of the square is 0.

(x) 12796

The unit digit of the square is 6.

3. The following numbers are obviously not perfect. Give reason.

(i) 567

(ii) 2453

(iii) 5298

(iv) 46292

(v) 74000

Solution:

In the given numbers

If the square of a number does not have 2, 3, 7, 8 or 0 as its unit digit, the squares 567, 2453, 5208, 46292 and 74000 cannot be the perfect squares as they have 7, 2, 8, 2 digits at the unit place.

4. The square of which of the following numbers would be an odd number or an even number? Why?

(i) 573

(ii) 4096

(iii) 8267

(iv) 37916

Solution:

We know that

The square of an odd number is odd and a square of an even number is even.

So 573 and 8262 are odd numbers and their squares will be an odd number.

4096 and 37916 are even numbers and their square will also be even number.

5. How many natural numbers lie between square of the following numbers?

(i) 12 and 13

(ii) 90 and 91

Solution:

(i) We know that

No. of natural numbers between the squares of 12 and 13 = (132 – 122) – 1

By further calculation

= (13 + 12 – 1)

So we get

= 25 – 1

= 24

(ii) We know that

No. of natural numbers between the squares of 90 and 91 = (912 – 902) – 1

By further calculation

= (91 + 90 – 1)

So we get

= 181 – 1

= 180

6. Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

Solution:

(i) We know that

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = n2

Here n = 8

So the sum = 82 = 64

(ii) We know that

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = n2

Here n = 15

So the sum = 152 = 225

7. (i) Express 64 as the sum of 8 odd numbers.

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

Solution:

(i) We know that

64 as the sum of 8 odd numbers = 82 = n2

Here n = 8

It can be written as

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

(ii) We know that

121 as the sum of 11 odd numbers = 112 = n2

Here n = 11

It can be written as

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

8. Express the following as the sum of two consecutive integers.

(i) 192

(ii) 332

(iii) 472

Solution:

We know that

n2 = (n2 – 1)/ 2 + (n2 + 1)/ 2 is the sum of two consecutive integers when n is odd

(i) 192 = (192 – 1)/ 2 + (192 + 1)/ 2

We know that

192 = 361

By further calculation

= (361 – 1)/ 2 + (361 + 1)/ 2

So we get

= 180 + 181

(ii) 332 = (332 – 1)/ 2 + (332 + 1)/ 2

We know that

332 = 1089

By further calculation

= (1089 – 1)/ 2 + (1089 + 1)/ 1

So we get

= 1088/2 + 1090/2

= 544 + 545

(iii) 472 = (472 – 1)/ 2 + (472 + 1)/ 2

We know that

472 = 2209

By further calculation

= (2209 – 1)/ 2 + (2209 + 1)/ 1

So we get

= 2208/2 + 2210/2

= 1104 + 1105

9. Find the squares of the following numbers without actual multiplication:

(i) 31

(ii) 42

(iii) 86

(iv) 94

Solution:

We know that

(a + b)2 = a2 + 2ab + b2

(i) 312 = (30 + 1)2

We can write it as

= 302 + 2 × 30 × 1 + 12

By further calculation

= 900 + 60 + 1

= 961

(ii) 422 = (40 + 2)2

We can write it as

= 402 + 2 × 40 × 2 + 22

By further calculation

= 1600 + 160 + 4

= 1764

(iii) 862 = (80 + 6)2

We can write it as

= 802 + 2 × 80 × 6 + 62

By further calculation

= 6400 + 960 + 36

= 7396

(iv) 942 = (90 + 4)2

We can write it as

= 902 + 2 × 90 × 4 + 42

By further calculation

= 8100 + 720 + 16

= 8836

10. Find the squares of the following numbers containing 5 in unit’s place:

(i) 45

(ii) 305

(iii) 525

Solution:

(i) 452 = n52

It can be written as

= n (n + 1) hundred + 52

Substituting the values

= 4 × 5 hundred + 25

By further calculation

= 2000 + 25

= 2025

(ii) 3052 = (30 × 31) hundred + 25

By further calculation

= 93000 + 25

= 93025

(iii) 5252 = (52 × 53) hundred + 25

By further calculation

= 275600 + 25

= 275625

11. Write a Pythagorean triplet whose one number is

(i) 8

(ii) 15

(iii) 63

(iv) 80

Solution:

(i) 8

Take n = 8

So the triplet will be 2n, n2 – 1, n2 + 1

Here

If 2n = 8, then n = 8/2 = 4

Substituting the values

n2 – 1 = 42 – 1 = 16 – 1 = 15

n2 + 1 = 42 + 1 = 16 + 1 = 17

Therefore, the triplets are 8, 15 and 17.

(ii) 15

Take 2n = 15

So n = n/2 is not possible

n2 – 1 = 15

By further calculation

n2 = 15 + 1 = 16 = 42

n = 4

So we get

2n = 2 × 4 = 8

n2 – 1 = 15

n2 + 1 = 42 + 1 = 16 + 1 = 17

Therefore, the triplets are 8, 15 and 17.

(iii) 63

Take n2 – 1 = 63

By further calculation

n2 = 63 + 1 = 64 = 82

So n = 8

Here

2n = 2 × 8 = 16

n2 – 1 = 63

n2 + 1 = 82 + 1 = 64 + 1 = 65

Therefore, the triplets are 16, 63 and 65.

(iv) 80

Take 2n = 80

n = 80/2 = 40

Here

n2 – 1 = 402 – 1 = 1600 – 1 = 1599

n2 + 1 = 402 + 1 = 1600 + 1 = 1601

Therefore, the triplets are 80, 1599 and 1601.

12. Observe the following pattern and find the missing digits:

212 = 441

2012 = 40401

20012 = 4004001

200012 = 4—4—1

2000012 = ——

Solution:

212 = 441

2012 = 40401

20012 = 4004001

200012 = 400040001

2000012 = 40000400001

13. Observe the following pattern and find the missing digits:

92 = 81

992 = 9801

9992 = 998001

99992 = 99980001

999992 = 9—8—01

9999992 = 9—0—1

Solution:

92 = 81

992 = 9801

9992 = 998001

99992 = 99980001

999992 = 9999800001

9999992 = 9999998000001

14. Observe the following pattern and find the missing digits:

72 = 49

672 = 4489

6672 = 444889

66672 = 44448889

666672 = 4—-8—–9

6666672 = 4—-8—-8-

Solution:

72 = 49

672 = 4489

6672 = 444889

66672 = 44448889

666672 = 4444488889

6666672 = 4444448888889

Exercise 3.3

1. By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root:

(i) 121

(ii) 55

(iii) 36

(iv) 90

Solution:

(i) We know that

Square root of 121

121 – 1 = 120

120 – 3 = 117

117 – 5 = 112

112 – 7 = 105

105 – 9 = 96

96 – 11 = 85

85 – 13 = 72

72 – 15 = 57

57 – 17 = 40

40 – 19 = 21

21 – 21 = 0

So the square root of 121 is 11

Hence, 121 is a perfect square.

(ii) We know that

Square root of 55

55 – 1 = 54

54 – 3 = 51

51 – 5 = 46

46 – 7 = 39

39 – 9 = 30

30 – 11 = 19

19 – 13 = 6

6 – 15 = – 9 is not possible

Hence, 55 is not a perfect square.

(iii) We know that

Square root of 36

36 – 1 = 35

35 – 3 = 32

32 – 5 = 27

27 – 7 = 20

20 – 9 = 11

11 – 11 = 0

Hence, 36 is a perfect square and its square root is 6.

(iv) We know that

Square root of 90

90 – 1 = 89

89 – 3 = 86

86 – 5 = 81

81 – 7 = 74

74 – 9 = 65

65 – 11 = 54

54 – 13 = 41

41 – 15 = 26

26 – 17 = 9

9 – 19 = – 10 which is not possible

Hence, 90 is not a perfect square.

2. Find the square roots of the following numbers by prime factorization method:

(i) 784

(ii) 441

(iii) 1849

(iv) 4356

(v) 6241

(vi) 8836

(vii) 8281

(viii) 9025

Solution:

(i) We know that

Square root of 784

It can be written as

So we get

= 2 × 2 × 7

= 28

(ii) We know that

Square root of 441

It can be written as

So we get

= 3 × 7

= 21

(iii) We know that

Square root of 1849

It can be written as

= 43

(iv) We know that

Square root of 4356

It can be written as

So we get

= 2 × 3 × 11

= 66

(v) We know that

Square root of 6241

It can be written as

= 79

(vi) We know that

Square root of 8836

It can be written as

So we get

= 2 × 47

= 94

(vii) We know that

Square root of 8281

It can be written as

So we get

= 7 × 13

= 91

(viii) We know that

Square root of 9025

It can be written as

So we get

= 5 × 19

= 95

3. Find the square roots of the following numbers by prime factorization method:

(i) 9 67/121

(ii) 17 13/36

(iii) 1.96

(iv) 0.0064

Solution:

(i) 9 67/121 = (9 × 121 + 67)/ 121

By further calculation

= (1089 + 67)/ 121

= 1156/121

By squaring we get

We know that

It can be written as

So we get

= (2 × 17)/ 11

= 34/11

= 3 1/11

(ii) 17 13/36 = (17 × 36 + 13)/ 36

By further calculation

= (612 + 13)/ 36

= 625/36

By squaring we get

We know that

It can be written as

So we get

= (5 × 5)/ (2 × 3)

= 25/6

= 4 1/6

(iii) 1.96 = 196/100

By squaring we get

We know that

It can be written as

So we get

= (2 × 7)/ (2 × 5)

= 14/10

= 1.4

(iv) 0.0064 = 64/10000

By squaring we get

We know that

It can be written as

So we get

= (2 × 2 × 2)/ (2 × 2 × 5 × 5)

= 8/100

= 0.08

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

(i) 588

(ii) 720

(iii) 2178

(iv) 3042

(v) 6300

Solution:

(i) 588 = 2 × 2 × 3 × 7 × 7

We know that

By pairing the same kind of factors, one factor 3 is left unpaired.

So to make it a pair we must multiply it by 3

Required least number = 3

Square root of 588 × 3 = 1764

Here

2 × 3 × 7 = 42

(ii) 720 = 2 × 2 × 2 × 2 × 3 × 3 × 5

We know that

By pairing the same kind of factors, one factor 5 is left unpaired.

So to make it a pair we must multiply it by 5

Required least number = 5

Square root of 720 × 5 = 3600

Here

2 × 2 × 3 × 5 = 60

(iii) 2178 = 2 × 3 × 3 × 11 × 11

We know that

By pairing the same kind of factors, one factor 2 is left unpaired.

So to make it a pair we must multiply it by 2

Required least number = 2

Square root of 2178 × 2 = 4356

Here

2 × 3 × 11 = 66

(iv) 3042 = 2 × 3 × 3 × 13 × 13

We know that

By pairing the same kind of factors, one factor 2 is left unpaired

So to make it a pair we must multiply it by 2

Required least number = 2

Square root of 3042 × 2 = 6084

Here

2 × 3 × 13 = 78

(v) 6300 = 2 × 2 × 3 × 3 × 5 × 5 × 7

We know that

By pairing the same kind of factors, one factor 7 is left unpaired

So to make it a pair we must multiply it by 7

Required least number = 7

Square root of 6300 × 7 = 44100

Here

2 × 3 × 5 × 7 = 210

5. For each of the following numbers, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also, find the square root of the square number so obtained:

(i) 1872

(ii) 2592

(iii) 3380

(iv) 16224

(v) 61347

Solution:

(i) 1872 = 2 × 2 × 2 × 2 × 3 × 3 × 13

It can be written as

By pairing the same kind of factors, one factor 13 is left unpaired

Required least number = 13

The number 1872 should be divided by 13 so that the resultant number will be a perfect square

Resultant number = 1872 ÷ 13 = 144

Square root = 2 × 2 × 3 = 12

(ii) 2592 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

It can be written as

By pairing the same kind of factors, one factor 2 is left unpaired

Required least number = 2

The number 2592 should be divided by 2 so that the resultant number will be a perfect square

Resultant number = 2592 ÷ 2 = 1296

Square root = 2 × 2 × 3 × 3 = 36

(iii) 3380 = 2 × 2 × 5 × 13 × 13

It can be written as

By pairing the same kind of factors, one factor 5 is left unpaired

Required least number = 5

The number 3380 should be divided by 5 so that the resultant number will be a perfect square

Resultant number = 3380 ÷ 5 = 676

Square root = 2 × 13 = 26

(iv ) 16224 = 2 × 2 × 2 × 2 × 2 × 3 × 13 × 13

It can be written as

By pairing the same kind of factors, two factors 2 and 3 is left unpaired

Required least number = 2 × 3 = 6

The number 16224 should be divided by 6 so that the resultant number will be a perfect square

Resultant number = 16224 ÷ 6 = 2704

Square root = 2 × 2 × 13 = 52

(v) 61347 = 3 × 11 × 11 × 13 × 13

It can be written as

By pairing the same kind of factors, one factor 3 is left unpaired

Required least number = 3

The number 61347 should be divided by 3 so that the resultant number will be a perfect square

Resultant number = 61347 ÷ 3 = 20449

Square root = 11 × 13 = 143

6. Find the smallest square number that is divisible by each of the following numbers:

(i) 3, 6, 10, 15

(ii) 6, 9, 27, 36

(iii) 4, 7, 8, 16

Solution:

(i) 3, 6, 10, 15

Number which is divisible by

3, 6, 10, 15 = LCM of 3, 6, 10, 15

It can be written as

So we get

= 2 × 3 × 5

= 30

(ii) 6, 9, 27, 36

Number which is divisible by

6, 9, 27, 36 = LCM of 6, 9, 27, 36

It can be written as

So we get

= 3 × 3 × 2 × 2 × 3

= 108

Here the smallest square

= 108 × 3

= 324

(iii) 4, 7, 8, 16

Number which is divisible by

4, 7, 8, 16 = LCM of 4, 7, 8, 16

It can be written as

So we get

= 2 × 2 × 2 × 2 × 7

= 112

Here the smallest square

= 112 × 7

= 784

7. 4225 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:

It is given that

Total number of plants = 4225

Here

Number of rows = Number of plant in each row

So the number of rows = square root of 4225

It can be written as

So we get

= 5 × 13

= 65

Hence, the number of rows is 65 and the number of plants in each row is 65.

8. The area of rectangle is 1936 sq. m. If the length of the rectangle is 4 times its breadth, find the dimensions of the rectangle.

Solution:

It is given that

Area of rectangle = 1936 sq. m

Length = 4x m

So we get

4x2 = 1936

By further calculation

x2 = 1936/4 = 484

We know that

It can be written as

By further calculation

= 2 × 11

= 22

Here

Length = 4x = 4 × 22 = 88 m

Breadth = x = 22 m

9. In a school a P.T. teacher wants to arrange 2000 students in the form of rows and columns for P.T. display. If the number of rows is equal to number of columns and 64 students could not be accommodated in this arrangement. Find the number of rows.

Solution:

It is given that

Total number of students in a school = 2000

The P.T. teacher arranges in such a way that

No. of rows = no. of students in each row

So 64 students are left

Required number of students = 2000 – 64 = 1936

No. of rows = √1936

We know that

It can be written as

By further calculation

= 2 × 2 × 11

= 44

10. In a school, the students of class VIII collected ₹2304 for a picnic. Each student contributed as many rupees as the number of students in the class. Find the number of students in the class.

Solution:

It is given that

Amount collected for picnic = ₹2304

We know that

No. of students = no. of rupees contributed by each student = √2304

Here

It can be written as

By further calculation

= 2 × 2 × 2 × 2 × 34= 48

Therefore, the number of students in class VIII is 4811.

11. The product of two numbers is 7260. If one number is 15 times the other number, find the numbers.

Solution:

It is given that

Product of two numbers = 7260

Consider one number = x

Second number = 15x

It can be written as

15x × x = 7260

15x2 = 7260

By further calculation

x2 = 7260/15 = 484

So we get

= 2 × 11

= 22

Here

One number = 22

Second number = 22 × 15 = 330

12. Find three positive numbers in the ratio 2: 3: 5, the sum of whose squares is 950.

Solution:

It is given that

Ratio of three positive numbers = 2: 3: 5

Sum of their squares = 950

Consider

First number = 2x

Second number = 3x

Third number = 5x

It can be written as

(2x)2+ (3x)2 + (5x)2 = 950

By further calculation

4x2 + 9x2 + 25x2 = 950

38x2 = 950

So we get

X2 = 950/38 = 25

x = √25 = 5

Here

First number = 2 × 5 = 10

Second number = 3 × 5 = 15

Third number = 5 × 5 = 25

13. The perimeter of two squares is 60 metres and 144 metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.

Solution:

It is given that

Perimeter of first square = 60 m

Side = 60/4 = 15 m

Perimeter of second square = 144 m

Side = 144/4 = 36 m

So the sum of perimeters of two squares = 60 + 144 = 204 m

Sum of areas of these two squares = 152 + 362

= 225 + 1296

= 1521 m2

Here

Area of third square = 1521 m2

We know that

So we get

Side = √Area = √1521

It can be written as

= 3 × 13

= 39 m

Here

Perimeter = 4 × side

Substituting the values

= 4 × 39

= 156 m

Exercise 3.4

1. Find the square root of each of the following by division method:

(i) 2401

(ii) 4489

(iii) 106929

(iv) 167281

(v) 53824

(vi) 213444

Solution:

(i) √2401 = 49

By division method

(ii) √4489 = 67

By division method

(iii) √106929 = 327

By division method

(iv) √167281 = 409

By division method

(v) √53824 = 232

By division method

(vi) √213444 = 462

By division method

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

(i) 81

(ii) 169

(iii) 4761

(iv) 27889

(v) 525625

Solution:

(i) 81

We know that

In 81, a group of two’s is 1

Therefore, its square root has one digit.

(ii) 169

We know that

In 169, group of two’s are 2

Therefore, its square root has two digits.

(iii) 4761

We know that

In 4761, group of two’s are 2

Therefore, its square root has two digits.

(iv) 27889

We know that

In 27889, groups of two’s are 3

Therefore, its square root has three digits.

(v) 525625

We know that

In 525625, groups of two’s are 3

Therefore, its square root has three digits.

3. Find the square root of the following decimal numbers by division method:

(i) 51.84

(ii) 42.25

(iii) 18.4041

(iv) 5.774409

Solution:

(i) √51.84 = 7.2

By division method

(ii) √42.25 = 6.5

By division method

(iii) √18.4041 = 4.29

By division method

(iv) √5.774409 = 2.403

By division method

4. Find the square root of the following numbers correct to two decimal places:

(i) 645.8

(ii) 107.45

(iii) 5.462

(iv) 2

(v) 3

Solution:

(i) √645.8 = 25.41

It can be written as

(ii) √107.45 = 10.36

It can be written as

(iii) √5.462 = 2.337 = 2.34

It can be written as

(iv) √2 = 1.41

It can be written as

(v) √3 = 1.73

It can be written as

5. Find the square root of the following fractions by division method:

(i) 841/1521

(ii) 8 257/529

(iii) 16 169/441

Solution:

(i) 841/1521

By squaring

It can be written as

(ii) 8 257/529

By squaring

It can be written as

(iii) 16 169/441

By squaring

It can be written as

6. Find the least number which must be subtracted from each of the following numbers to make them a perfect square. Also find the square root of the perfect square number so obtained:

(i) 2000

(ii) 984

(iii) 8934

(iv) 11021

Solution:

(i) 2000

We know that

By taking square root, 64 is left as remainder

Subtracting 64 from 2000

We get 1936 which is a perfect square and its square root is 44.

(ii) 984

We know that

By taking square root, 23 is left as remainder

Subtracting 23 from 984

We get 961 which is a perfect square and its square root is 31.

(iii) 8934

We know that

By taking square root, 98 is left as remainder

Subtracting 98 from 894

We get 8934 – 98 = 8836 which is a perfect square and its square root is 94.

(iv) 11021

We know that

By taking square root, 205 is left as remainder

Subtracting 205 from 11021

We get 11021 – 205 = 10816 which is a perfect square and its square root is 104.

7. Find the least number which must be added to each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:

(i) 1750

(ii) 6412

(iii) 6598

(iv) 8000

Solution:

(i) 1750

We know that

By taking square root

412 is less than 1750

So by taking 422

164 – 150 = 14 less

Adding 14 we get a square of 42 which is 1764.

(ii) 6412

We know that

By taking square root

802 is less than 6412

So by taking 812

161 – 12 = 14 less

Adding 149 we get a square of 81 which is 6561.

(iii) 6598

We know that

By taking square root

812 is less than 6598

So by taking 822

324 – 198 = 126 less

Adding 126 we get a square of 82 which is 6724.

(iv) 8000

We know that

By taking square root

892 is less than 8000

So by taking 902

8100 – 8000 = 100 less

Adding 100 we get a square of 90 which is 8100.

8. Find the smallest four-digit number which is a perfect square.

Solution:

It is given that

Smallest four – digit number = 1000

We know that

By taking square root, we find that 39 is left.

If we subtract any number from 1000 we get 3 digit number

Take 322 = 1024

Here 1024 – 1000 = 24 is to be added to get a perfect square of least 4 digit number

Therefore, the required 4 digit smallest number is 1024.

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

Solution:

It is given that

Greatest six digit number = 999999

We know that

By taking square root, we find that 1998 is left

If we subtract 1998 from 999999 we get 998001 which is a perfect square.

Therefore, required six digit greatest number is 998001.

10. In a right triangle ABC, ∠B = 900.

(i) If AB = 14 cm, BC = 48 cm, find AC.

(ii) If AC = 37 cm, BC = 35 cm, find AB.

Solution:

(i) In a right angled triangle ABC

It is given that

AB = 14 cm and BC = 48 cm

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

= 142 + 482

By further calculation

= 196 + 2304

= 2500

So we get

AC = √2500 = 50 cm

(ii) In right triangle ABC

B = 900, AC = 37 cm, BC = 35 cm

Using Pythagoras Theorem

AC2 = AB2 + BC2

Substituting the values

372 = AB2 + 352

By further calculation

1369 = AB2 + 1225

AB2 = 1369 – 1225 = 144

So we get

AB = √144 = 12 cm

11. A gardener has 1400 plants. He wants to plant these in such a way that the number of rows and number of columns remains same. Find the minimum number of plants he needs more for this.

Solution:

It is given that

Total number of plants = 1400

We know that

Here

Number of columns = Number of rows

By taking the square root of 1400

372 < 1400

So take 382 = 1444

We need 1444 – 1400 = 44 plants more

Therefore, the minimum number of plants he needs more for this is 44.

12. There are 1000 children in a school. For a P.T. drill they have to stand in such a way that the number of rows is equal to number of columns. How many children would be left out in this arrangement?

Solution:

It is given that

No. of total children in a school = 1000

For a P.T. drill, children have to stand in such a way that

No. of rows = No. of columns

Take the square root of 1000

39 is left as remainder

Left out children = 39

Hence, 39 children would be left out in this arrangement.

13. Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?

Solution:

It is given that

Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend’s house

Consider O as the house and A and B as the places

OA = 16 m, AO = 63 m

Using Pythagoras theorem

OB2 = OA2 + AB2

Substituting the values

= 162 + 632

By further calculation

= 256 + 3969

= 4225

So we get

OB = √4225 = 65

Therefore, Amit has to walk 65 m to reach his house.

14. A ladder 6 m long leaned against a wall. The ladder reaches the wall to a height of 4.8 m. Find the distance between the wall and the foot of the ladder.

Solution:

It is given that

Length of ladder = 6 m

Ladder reaches the wall to a height of 4.8 m

Consider AB as the ladder and AC as the height of the wall

AB = 6 m and AC = 4.8 m

Distance between the foot of ladder and wall is BC

Using Pythagoras theorem,

AB2 = AC2 + BC2

Substituting the values

62 = 4.82 + BC2

By further calculation

BC2 = 62 – 4.82

BC2 = 36 – 23.04 = 12.96

So we get

BC = √12.96 = 3.6 m

Hence, the distance between the wall and the foot of the ladder is 3.6 m.