RD Sharma Solutions Class 8 Squares And Square Roots Exercise 3.3

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 a², 2 x a x b, and b² in these columns.

Step: 2 Underline the unit digit of b² (in Column III) and add its tens digit, if any, with 2 x a x b (in column II)

Step: 3 Underline the unit digit in Column II and add the number formed by the tens and other digits if any, with a² in Column I.

Step 4: Underline the number in Column I.

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:

25² = 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 a², 2 x a x b, and b² in these columns.

Step: 2 Underline the unit digit of b² (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II)

Step: 3 Underline the unit digit in Column II and add the number formed by tens and others digits if any, with a² in Column I.

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:

37² = 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 a², 2 x a x b and b² in these columns.

Step: 2 Underline the unit digit of b² (in Column III ) and add its tens digit, if any, with 2x a x b (in Column II)

Step: 3 Underline the digit in Column II and add the number formed by the tens and other digits if any, with a² in Column I.

Step: 4 underline the number in Column I.

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:

54² = 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 a², 2 x a x b and b² in these columns.

Step: 2 Underline the unit digit of b² (in column III) and add its ten digit, if any with 2 x a x b (in column II)

Step: 3 Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a² in column I.

Step: 4 underline the number in column I.

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:

71² = 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 a² , 2 x a x b and b² in these columns.

Step: 2 Underline the unit digit of b² (in column III) and add its tens digit, if any with 2 x a x b (in column II)

Step: 3 Underline the unit digit in Column II and add the number formed by the tens and other digits if any, with a² in column I.

Step: 4 underline the number in Column I

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:

96² = 9216

Using multiplication:

96 x 96 = 2916

This matches with the result obtained using the column method.

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

(i) 98

\(∴ 98^{2}= 9604\)

(ii) 273

\(∴ 273^{2}= 74529\)

(iii) 348

\(∴ 348^{2}= 121104\)

(iv) 295

\(∴ 295^{2}= 87025\)

(v) 171

\(∴ 171^{2}= 29241\)

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.

(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.

4.) Find the squares of the following numbers:

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.

(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

\(∴\) 512² = (250+12)1000 + (12)2 = 262000 + 144 = 262144

(viii)  995

Here, n = 99

\(∴\)  n(n + 1) = (99)(100) = 9900

\(∴\) 995² = 990025

5.) Find the squares of the following numbers using the identity (a +b)² = a² + 2ab + b² :

(i) 405

On decomposing:

405 = 400 + 5

Here, a = 400x and b = 5

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

405² = (400 + 5)²= 400² + 2(400)(5) + 5²= 160000 + 4000 + 25 = 164025

(ii) 510

On decomposing:

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

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

510² = (500 + 10)² = 500² + 2(500)(10) + 10² = 250000 + 10000 + 100 = 260100

(iii) 1001

On decomposing:

1001 = 1000 + 1

Here, a = 1000 and b = 1

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

1001² = (1000 + 1)²= 1000² + 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 = a²+ 2ab + b²:

209² = (200 + 9)² = 200² + 2(200)(9) + 9² = 40000 + 3600 + 81 = 43681

(v) 605

On decomposing:

605 = 600 + 5

Here, a = 600 and b = 5

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

605² = (600 + 5)² = 6002 + 2(600)(5) + 5² = 360000 + 6000 + 25 = 366025

6.)Find the squares of the following numbers using the identity (a  –  b)² = a²  – 2ab + b² :

 (i) 395

Decomposing: 395 = 400 — 5

Here, a= 400 and b = 5

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

395² = (400 — 5)² = 400² — 2(400)(5) + 5² = 160000 — 4000 + 25 = 156025

(ii) 995

Decomposing:

995 = 1000 — 5

Here, a =1000 and b = 5

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

995² = (1000 — 5)² = 1000² — 2(1000)(5) + 5²= 1000000 —10000 + 25 = 990025

(iii) 495

Decomposing: 495 = 500 — 5 Here, a = 500 and b = 5 Using the identity (a — b) ² = a² — 2ab + b²:

495² = (500 — 5)² = 500² — 2(500) (5) + 5² = 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 = a² — 2ab + b²:

4982 = (500 — 2)2 = 5002 — 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 = a2 — 2ab + b2: 992 = (100 — 1)2 = 1002 — 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  = a² — 2ab + b²:

999² = (1000 – 1)² – 1000² – 2(1000) (1)+1² = 1000000 – 2000 +1 = 998001

(vii) 599

Decomposing: 599 = 600 — 1

Here, a = 600 and b = 1

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

5992 = (600 —1)² = 600²— 2(600) (1) + 12 = 360000 —1200 + 1 = 358801

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

(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.