Exercise 3.3 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8 PDF Download

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

(i) 25

(ii) 37

(iii) 54

(iv) 71

(v) 96

Ans: 

(i) 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:

This matches with the result obtained by the column method.

(ii) 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 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:

37² = 1369

Using multiplication:

This matches with the result obtained using the column method.

(iii) 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 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:

54² = 2916

Using multiplication:

This matches with the result obtained using the column method.

(iv) 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 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:

71² = 5041

Using multiplication:

This matches with the result obtained using the column method.

(v) 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:

This matches with the result obtained using the column method.

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

(i) 98

(ii) 273

(iii) 348

(iv) 295

(v) 171

Ans: 

(i)

∴ 98² = 9604

(ii)

∴  273² = 74529

(iii)

∴  348² = 121104

(iv)

∴  295² = 87025

(v)

∴  171² = 29241

Q.3. Find the squares of the following numbers:

(i) 127

(ii) 503

(iii) 451

(iv) 862

(v) 265

Ans: We will use visual method as it is the most 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 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 side 265 units. Let us split it into 260 units and 5 units.

Hence, the square of 265 is 70225.

Q.4. Find the squares of the following numbers:

(i) 425

(ii) 575

(iii) 405

(iv) 205

(v) 95

(vi) 745

(vii) 512

(viii) 995

Ans: 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) Here, n = 42

∴ n(n + 1) = (42)(43) = 1806

∴ 425² = 180625

(ii) Here, n = 57

∴ n(n + 1) = (57)(58) = 3306

∴ 575² = 330625

(iii) Here n = 40

∴ n(n + 1) = (40)(41) = 1640

∴ 405² = 164025

(iv) Here n = 20

∴ n(n + 1) = (20)(21) = 420

∴ 205² =  42025

(v) Here n = 9

∴ n(n + 1) = (9)(10) = 90

∴ 95² = 9025

(vi) Here n = 74

∴ n(n + 1) = (74)(75) = 5550

∴ 745² = 555025

(vii) 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) Here, n = 99

∴ n(n + 1) = (99)(100) = 9900

∴ 995² = 990025

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

(i) 405

(ii) 510

(iii) 1001

(iv) 209

(v) 605

Ans: 

(i) On decomposing:

405 = 400 + 5

Here, a = 400 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) 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) 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) 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) 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

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

(i) 395

(ii) 995

(iii) 495

(iv) 498

(v) 99

(vi) 999

(vii) 599

Ans: 

(i) 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) 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) 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) Decomposing: 498 = 500 − 2

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) 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) 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) 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

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

(i) 52

(ii) 95

(iii) 505

(iv) 702

(v) 99

Ans: 

(i) 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) 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) 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) We have:

702 = 700 + 2

Let us draw a square 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) We have:

99 = 90 + 9

Let us draw a square 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.