## RD Sharma Solutions Class 8 Chapter 3 Exercise 3.3

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

(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 xÂ 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 5:Â 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.

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(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.

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(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.

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(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.

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2.) Find the squares of the following numbers using diagonal method:

(i) 98

\(âˆ´ 98^{2}= 9604\)

(ii) 273

\(âˆ´ 273^{2}= 74529\)

(iii) 348

\(âˆ´ 348^{2}= 121104\)

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(iv) 295

\(âˆ´ 295^{2}= 87025\)

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(v) 171

\(âˆ´ 171^{2}= 29241\)

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3.) Find the squares of the following numbers:

Answer:

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.

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(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 Â and 2 units.

Hence, the square of 862 is 743044.

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(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.

4.) Find the squares of the following numbers:

Â

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.

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(i) Â 425

Here, n = 42

\(âˆ´\)

\(âˆ´\)

(ii) 575

Here, n = 57

\(âˆ´\)

\(âˆ´\)

(iii) 405

Here n = 40

\(âˆ´\)

\(âˆ´\)

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(iv) 205

Here n = 20

\(âˆ´\)

\(âˆ´\)

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(v) 95

Here n = 9

\(âˆ´\)

\(âˆ´\)

(vi) 745

Here n = 74

\(âˆ´\)

\(âˆ´\)

(vii) 512

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

\(âˆ´\)^{2} = (250+12)1000 + (12)2 = 262000 + 144 = 262144

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(viii) Â 995

Here, n = 99

\(âˆ´\)

\(âˆ´\)^{2} = 990025

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

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(i) 405

On decomposing:

405 = 400 + 5

Here, a = 400x and b = 5

Using the identity (a + b)2 = a2 + 2ab + b2:

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

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(ii) 510

On decomposing:

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

Using the identity (a + b)2 = a2 + 2ab + b2:

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

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(iii) 1001

On decomposing:

1001 = 1000 + 1

Here, a = 1000 and b = 1

Using the identity (a + b)2 = a2 + 2ab + b2:

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

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(iv) Â 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

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(v) 605

On decomposing:

605 = 600 + 5

Here, a = 600 and b = 5

Using the identity (a + b)^{ 2}= a2 + 2ab + b2:

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

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

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Â (i) 395

Decomposing: 395 = 400 â€” 5

Here, a= 400 and b = 5

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

395^{2} = (400 â€” 5)^{2} = 400^{2} â€” 2(400)(5) + 5^{2} = 160000 â€” 4000 + 25 = 156025

(ii) 995

Decomposing:

995 = 1000 â€” 5

Here, a =1000 and b = 5

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

995^{2} = (1000 â€” 5)^{2} = 1000^{2} â€” 2(1000)(5) + 5^{2}= 1000000 â€”10000 + 25 = 990025

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(iii) 495

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

495^{2} = (500 â€” 5)^{2} = 500^{2} â€” 2(500) (5) + 5^{2} = 250000 â€” 5000 + 25 = 245025

(iv) 498

Decomposing: 498 = 500 â€” 2 250000 â€” 5000 + 25 = 245025

Here, a = 500 and b = 2

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

4982 = (500 â€” 2)2 = 5002 â€” 2(500)(2) + 22 = 250000 â€” 2000 + 4 = 248004

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(v) 99

Decomposing: 99 = 100 â€” 1 Here, a = 100 and b = 1 Using the identity (a â€” b)2 = a2 â€” 2ab + b2: 992 = (100 â€” 1)2 = 1002 â€” 2(100)(1) + 12 = 10000 â€” 200 + 1 = 9801

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(vi) 999

Decomposing: 999 = 1000 â€“ 1

Here, a = 1000 and b = 1

Using the identity (a â€” b)^{2}Â = a^{2} â€” 2ab + b^{2}:

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

(vii) 599

Decomposing: 599 = 600 â€” 1

Here, a = 600 and b = 1

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

5992 = (600 â€”1)^{2} = 600^{2}â€” 2(600) (1) + 12 = 360000 â€”1200 + 1 = 358801

7.) Find the squares of the following numbers by visual method:

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(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.

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(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.

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(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.

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(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.

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(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.