RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

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Class 8 : RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

The document RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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QUESTION 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

ANSWER 1: (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. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
42025

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). 

Column I 
Column II
Column III 
a2 
2 x a x b 
b2 
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 a2 in Column I. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
4 + 2 
20 + 2 
25 
6 
22 

Step 4. Underline the number in Column I.

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
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:
252 = 625
Using multiplication:
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
9 
42 
49 

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). 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
942+449

46

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. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
9 + 4 
42 + 4 
49 
13 
46 

Step 4. Underline the number in Column I. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
9 + 4 
42 + 4 
49
13 46 

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: 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
25 
40 
16

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). 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
25 
40 + 1 
16 

41 

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. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
25 + 4 
40 + 1 
16 
29 
41 

Step 4. Underline the number in Column I. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
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:
542 = 2916
Using multiplication: 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
49141

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). 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
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 a2 in Column I. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
49 + 1 
14 + 0 
1
50 
14 

Step 4. Underline the number in Column I. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
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:
712 = 5041
Using multiplication: 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
81 
10836

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). 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
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 a2 in Column I. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
81 + 11 
108 + 3 
36
92 
111

Step 4. Underline the number in Column I. 

Column I 
Column II 
Column III 
a2 
2 x a x b 
b2 
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:
962 = 9216
Using multiplication: 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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

Answer 2:

 RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

∴ 982 = 9604 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

∴ 2732 = 74529 

(iii) 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

∴ 3482 = 121104 

(iv) 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

∴ 2952 = 87025 

(v) 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

∴ 1712 = 29241 

QUESTION 3: FIND THE SQUARES OF THE FOLLOWING NUMBERS:
(I) 127
(II) 503
(III) 451
(IV) 862
(V) 265

ANSWER 3: 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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

Hence, the square of 265 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

Answer 4: 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(+ 1) before 25. 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(+ 1) before 25. 

(i) Here, n = 42
n(+ 1) = (42)(43) = 1806
4252 = 180625
(ii) Here, n = 57
n(+ 1) = (57)(58) = 3306
5752 = 330625
(iii) Here n = 40
n(+ 1) = (40)(41) = 1640
4052 = 164025
(iv) Here n = 20
 n(+ 1) = (20)(21) = 420
2052 =  42025
(v) Here n = 9
n(+ 1) = (9)(10) = 90
952 = 9025
(vi) Here n = 74
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(+ 1) = (99)(100) = 9900
∴ 9952 = 990025 

Question 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

Answer 5: (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 

Question 6: Find the squares of the following numbers using the identity (ab)2 = a2 − 2ab + b2:
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599

Answer 6: (i) Decomposing: 395 = 400 − 5
Here, a = 400 and b = 5
Using the identity (ab)2 = a2 − 2ab + b2:

3952 = (400  5)2 = 400 2(400)(5) + 52 = 160000  4000 + 25 = 156025
(ii) Decomposing: 995 = 1000 − 5
Here, a = 1000 and b = 5
Using the identity (ab)2 = a2 − 2ab + b2:

9952 = (1000 − 5)2 = 1000 2(1000)(5) + 52 = 1000000  10000 + 25 = 990025
(iii) Decomposing: 495 = 500 − 5
Here, a = 500 and b = 5
Using the identity (ab)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 (ab)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 (ab)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 (ab)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 (ab)2 = a2 − 2ab + b2:

5992 = (600  1)2 = 600 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

Answer 7: (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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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. 

RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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.
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math Class 8 Notes | EduRev

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

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