**Chapter 3: Squares and Square Roots Exercise – 3.3**

**Question: 1**

Find the squares of the following numbers using column method. Verify the result finding the square using the usual multiplication.

(i) 25

(ii) 37

(iii) 54

(iv) 71

(v) 96

**Solution:**

Here a = 2, b = 5

Step: 1 Make 3 columns and write the values of a^{2}, 2 x a x b, and b^{2} in these columns.

Sol.

(i) 25

Here a = 2, b = 5

Step: 1 Make 3 columns and write the values of a^{2}, 2 x a x b, and b^{2} in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

4 | 20 | 25 |

Step: 2 Underline the unit digit of b^{2} (in Column III) and add its tens digit, if any, with 2 x a x b (in column II)

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

4 | 20 + 2 | 25 |

22 |

Step: 3 Underline the unit digit in Column II and add the number formed by the tens and other digits if any, with a^{2} in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b |

4 + 2 | 20 + 2 | 25 |

6 | 22 |

Step 4: Underline the number in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

4 + 2 | 20 + 2 | 25 |

6 | 22 |

Step: 5 write the underlined digits at the bottom of each column to obtain the square of the given number.

In this case, we have:

25^{2} = 625

Using Multiplication:

25 x 25 = 625

This matches with the result obtained by the column method:

(ii) 37

Here, a = 3, b = 7

Step: 1 Make 3 columns and write the values of a^{2}, 2 x a x b, and b^{2} in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

9 | 42 | 49 |

Step: 2 Underline the unit digit of b^{2} (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II)

Column I | Column II | Column III |

A^{2} |
2 x a x b | B^{2} |

9 + 4 | 42 + 4 | 49 |

13 | 46 |

Step: 3 Underline the unit digit in Column II and add the number formed by tens and others digits if any, with a^{2} in Column I.

Column I | Column II | Column III |

A^{2} |
2 x a x b | B^{2} |

9 + 4 | 42 + 4 | 49 |

13 |
46 |

Step 4: Write the underlined digits at the bottom of each column to obtain the square of the given number.

In this case, we have:

37^{2} = 1369

Using multiplication:

37 x 37 = 1369

This matches with the result obtained using the column method.

(iii) 54

Here, a = 5, b = 4

Step 1: make 3 columns and write the values of a^{2}, 2 x a x b and b^{2} in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

25 | 40 |

Step: 2 Underline the unit digit of b^{2} (in Column III) and add its tens digit, if any, with 2x a x b (in Column II)

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

25 | 40 +1 | 16 |

41 |

Step: 3 Underline the digit in Column II and add the number formed by the tens and other digits if any, with a^{2} in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

25 + 4 | 40 + 1 | 16 |

29 | 41 |

Step: 4 underline the number in Column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

25 + 4 | 40 + 1 | 16 |

29 |
41 |

Step: 5 write the underlined digits at the bottom of each column to obtain the square of the given number.

In this case, we have:

54^{2} = 2916

Using multiplication:

54 x 54 = 2916

This matches with the result obtained using the column method.

(iv) 71

Here, a = 7, b = 1

Step: 1 Make 3 columns and write the values of a^{2}, 2 x a x b and b^{2} in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

49 | 14 | 1 |

Step: 2 Underline the unit digit of b^{2} (in column III) and add its ten digit, if any with 2 x a x b (in column II)

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

49 | 14 + 0 | 1 |

14 |

Step: 3 Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a^{2} in column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

49 + 1 | 14 + 0 | 1 |

50 | 14 |

Step: 4 underline the number in column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

49 + 1 | 14 + 0 | 1 |

50 |
14 |

Step: 5 write the underlined digits at the bottom of each column to obtain the square of the given number:

In this case, we have:

71^{2} = 5041

Using multiplication:

71 x 71 = 5041

This matches with the result obtained using the column method.

(v) 96

Here, a = 9, b = 6

Step: 1 Make 3 columns and write the values of a^{2}, 2 x a x b and b^{2} in these columns.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

81 | 108 | 36 |

Step: 2 Underline the unit digit of b^{2} (in column III) and add its tens digit, if any with 2 x a x b (in column II)

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

81 | 108 + 3 | 36 |

111 |

Step: 3 Underline the unit digit in Column II and add the number formed by the tens and other digits if any, with a^{2} in column I.

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

81 + 11 | 108 + 3 | 36 |

92 | 111 |

Step: 4 underline the number in Column I

Column I | Column II | Column III |

a^{2} |
2 x a x b | b^{2} |

81 + 11 | 108 + 3 | 36 |

92 |
111 |

Step: 5 write the underlined digits at the bottom of each column to obtain the square of the given number.

In this case, we have:

96^{2} = 9216

Using multiplication:

96 x 96 = 2916

This matches with the result obtained using the column method.

**Question: 2**

Find the squares of the following numbers using diagonal method:

(i) 98

(ii) 273

(iii) 348

(iv) 295

(v) 171

**Solution:**

(i) 98

∴ 98^{2} = 9604

(ii) 273

∴ 273^{2 }= 74529

(iii) 348

∴ 348^{2} = 121104

(iv) 295

∴ 295^{2} = 87025

(v) 171

∴ 171^{2} = 29241

**Question: 3**

Find the squares of the following numbers:

(i) 127

(ii) 503

(iii) 451

(iv) 862

(v) 265

**Solution:**

We will use visual method as it is the efficient method to solve this problem.

(i) We have:

127 = 120 + 7

Hence, let us draw a square having side 127 units. Let us split it into 120 units and 7 units.

Hence, the square of 127 is 16129.

(ii) We have:

503 = 500 + 3

Hence, let us draw a square having side 503 units. Let us split it into 500 units and 3 units.

Hence, the square of 503 is 253009.

(iii) We have:

451 = 450 + 1

Hence, let us draw a square of having side 451 units. Let us split it into 450 units and 1 units.

Hence, the square of 451 is 203401.

(iv) We have:

862 = 860 + 2

Hence, let us draw a square having side 862 units. Let us split it into 860 units and 2 units.

Hence, the square of 862 is 743044.

(v) We have:

265 = 260 +5

Hence, let us draw a square having 265 units. Let us split it into 260 units and 5 units.

Hence, the square of 265 units is 70225.

**Question: 4**

Find the squares of the following numbers:

(i) 425

(ii) 575

(iii) 405

(iv) 205

(v) 95

(vi) 745

(vii) 512

(viii) 995

Notice that all numbers except the one in question (vii) has 5 as their respective unit digits. We know that the square of a number with the form n5 is a number ending with 25 and has the number n(n + 1) before 25.

**Solution:**

Sol.

(i) 425

Here, n = 42

∴ n(n + 1) = (42)(43) = 1806

∴ 4252 = 180625

(ii) 575

Here, n = 57

∴n(n + 1) = (57)(58) = 3306

∴ 5752 = 330625

(iii) 405

Here n = 40

∴n(n + 1) = (40)(41) = 1640

∴ 4052 = 164025

(iv) 205

Here n = 20

∴n(n + 1) = (20)(21) = 420

∴ 2052 = 42025

(v) 95

Here n = 9

∴ n(n + 1) = (9)(10) = 90

∴ 952 = 9025

(vi) 745

Here n = 74

∴ n(n + 1) = (74)(75) = 5550

∴ 7452 = 555025

(vii) 512

We know: The square of a three-digit number of the form 5ab = (250 + ab) 1000 + (ab)^{2}

∴ 512^{2} = (250+12)1000 + (12)^{2} = 262000 + 144 = 262144

(viii) 995

Here, n = 99

∴ n(n + 1) = (99)(100) = 9900

∴ 995^{2} = 990025

**Question: 5**

Find the squares of the following numbers using the identity (a +b)^{2} = a^{2} + 2ab + b^{2}:

(i) 405

(ii) 510

(iii) 1001

(iv) 209

(v) 605

**Solution:**

(i) 405

On decomposing:

405 = 400 + 5

Here, a = 400x and b = 5

Using the identity (a + b)^{2} = a^{2} + 2ab + b^{2}:

405^{2} = (400 + 5)^{2}= 400^{2} + 2(400)(5) + 5^{2 }= 160000 + 4000 + 25 = 164025

(ii) 510

On decomposing:

510 = 500 + 10 Here, a = 500 and b = 10

Using the identity (a + b)^{2} = a^{2} + 2ab + b^{2}:

510^{2} = (500 + 10)^{2} = 500^{2} + 2(500)(10) + 10^{2} = 250000 + 10000 + 100 = 260100

(iii) 1001

On decomposing:

1001 = 1000 + 1

Here, a = 1000 and b = 1

Using the identity (a + b)^{2} = a^{2} + 2ab + b^{2}:

1001^{2} = (1000 +1)^{2 }= 1000^{2} + 2(1000) (1) + 12 = 1000000 + 2000 + 1 = 1002001

(iv) 209

On decomposing:

209 = 200 + 9

Here, a = 200 and b = 9

Using the identity (a + b)^{ 2} = a^{2}+ 2ab + b^{2}:

209^{2} = (200 + 9)^{2} = 200^{2} + 2(200)(9) + 9^{2} = 40000 + 3600 + 81 = 43681

(v) 605

On decomposing:

605 = 600 + 5

Here, a = 600 and b = 5

Using the identity (a + b)^{2 }= a^{2} + 2ab + b^{2}:

605^{2} = (600 + 5)^{2} = 6002 + 2(600)(5) + 5^{2} = 360000 + 6000 + 25 = 366025

**Question: 6**

Find the squares of the following numbers using the identity (a – b)^{2} = a^{2} – 2ab + b^{2}:

(i) 395

(ii) 995

(iii) 495

(iv) 498

(v) 99

(vi) 999

(vii) 599

**Solution:**

(i) 395

Decomposing: 395 = 400 - 5

Here, a = 400 and b = 5

Using the identity (a — b)^{ 2} = a^{2} – 2ab + b^{2}:

395^{2} = (400 - 5)^{2} = 400^{2 }- 2(400)(5) + 5^{2} = 160000 - 4000 + 25 = 156025

(ii) 995

Decomposing:

995 = 1000 - 5

Here, a =1000 and b = 5

Using the identity (a -b)^{ 2} = a^{2} - 2ab + b^{2}:

995^{2} = (1000 - 5)^{2} = 1000^{2} -2(1000)(5) + 5^{2 }= 1000000 -10000 + 25 = 990025

(iii) 495

Decomposing: 495 = 500 - 5 Here, a = 500 and b = 5 Using the identity (a - b)^{ 2} = a^{2} - 2ab + b^{2}:

495^{2} = (500 - 5)^{2} = 500^{2} - 2(500) (5) + 5^{2} = 250000 - 5000 + 25 = 245025

(iv) 498

Decomposing: 498 = 500 - 2 250000 - 5000 + 25 = 245025

Here, a = 500 and b = 2

Using the identity (a - b)^{2} = a^{2} - 2ab + b^{2}:

4982 = (500 -2)^{2} = 500^{2} -2(500)(2) + 22 = 250000 - 2000 + 4 = 248004

(v) 99

Decomposing: 99 = 100 -1 Here, a = 100 and b = 1 Using the identity (a - b)^{2} = a^{2} - 2ab + b^{2}: 992 = (100-1)^{2} = 100^{2}- 2(100)(1) + 12 = 10000 -200 + 1 = 9801

(vi) 999

Decomposing: 999 = 1000 – 1

Here, a = 1000 and b = 1

Using the identity (a - b)^{2} = a^{2} - 2ab + b^{2}:

999^{2} = (1000 – 1)^{2} – 1000^{2} – 2(1000) (1) + 1^{2} = 1000000 – 2000 +1 = 998001

(vii) 599

Decomposing: 599 = 600 -1

Here, a = 600 and b = 1

Using the identity (a - b)^{ 2} = a^{2} -2ab + b^{2}:

5992 = (600 -1)^{2} = 600^{2}-2(600) (1) + 12 = 360000 -1200 + 1 = 358801

**Question: 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 have:

52 = 50 + 2

Let us draw a square having side 52 units. Let us split it into 50 units and 2 units.

The sum of the areas of these four parts is the square of 52. Thus, the square of 52 is 2704.

(ii) 95

We have:

95 = 90 + 5

Let us draw a square having side 95 units. Let us split it into 90 units and 5 units.

The sum of the areas of these four parts is the square of 95. Thus, the square of 95 is 9025.

(iii) 505

We have:

505 = 500 + 5

Let us draw a square having side 505 units. Let us split it into 500 units and 5 units.

The sum of the areas of these four parts is the square of 505. Thus, the square of 505 is 255025.

(iv) 702

We have:

702 = 700 + 2

Let us draw a square of having side 702 units. Let us split it into 700 units and 2 units.

The sum of the areas of these four parts is the square of 702. Thus, the square of 702 is 492804.

(v) 99

We have:

99 = 90 + 9

Let us draw a square of having side 99 units. Let us split it into 90 units and 9 units.

The sum of the areas of these four parts is the square of 99. Thus, the square of 99 is 9801.