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.
Column I | Column II | Column III |
a2 | 2 x a x b | b2 |
4 | 20 | 25 |
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 | b |
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:
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.
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 |
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 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 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:
372 = 1369
Using multiplication:
37 x 37 = 1369
This matches with the result obtained using the column method.
Â
(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.
Column I | Column II | Column III |
a2 | 2 x a x b | b2 |
25 | 40 |
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)
Column I | Column II | Column III |
a2 | 2 x a x b | b2 |
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 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:
54 x 54 = 2916
This matches with the result obtained using the column method.
Â
(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.
Column I | Column II | Column III |
a2 | 2 x a x b | b2 |
49 | 14 | 1 |
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)
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:
71 x 71 = 5041
This matches with the result obtained using the column method.
Â
(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.
Column I | Column II | Column III |
a2 | 2 x a x b | b2 |
81 | 108 | 36 |
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:
96 x 96 = 2916
This matches with the result obtained using the column method.
Â
2.) Find the squares of the following numbers using diagonal method:
(i) 98
\(∴ 98^{2}= 9604\)
(ii) 273
\(∴ 273^{2}= 74529\)
(iii) 348
\(∴ 348^{2}= 121104\)
Â
(iv) 295
\(∴ 295^{2}= 87025\)
 Â
(v) 171
\(∴ 171^{2}= 29241\)
Â
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.
Â
(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.
Â
(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.
Â
(i) Â 425
Here, n = 42
\(∴\)
\(∴\)
(ii) 575
Here, n = 57
\(∴\)
\(∴\)
(iii) 405
Here n = 40
\(∴\)
\(∴\)
Â
(iv) 205
Here n = 20
\(∴\)
\(∴\)
Â
(v) 95
Here n = 9
\(∴\)
\(∴\)
(vi) 745
Here n = 74
\(∴\)
\(∴\)
(vii) 512
We know: The square of a three-digit number of the form 5ab = (250 + ab) 1000 + (ab)2
\(∴\)
Â
(viii) Â 995
Here, n = 99
\(∴\)
\(∴\)
Â
5.) Find the squares of the following numbers using the identity (a +b)2 = a2 + 2ab + b2 :
Â
(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
Â
(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
Â
(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
Â
(iv) Â 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
Â
(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
Â
6.)Find the squares of the following numbers using the identity (a – b)2 = a2 – 2ab + b2 :
Â
 (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
Â
(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
Â
(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
Â
(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:
Â
(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.
Â
(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.
Â
(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.
Â
(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.
Â
(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.