Chapter: “Squares & Square Roots 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
(ii) 37
(iii) 54
(iv) 71
(v) 96
Solution:
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.
Solution:
(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 4: 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.
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.
(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.
(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.
2.) Find the squares of the following numbers using diagonal method:
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171
:
3.) Find the squares of the following numbers:
Solution:
(i) 127
(ii) 503
(iii) 451
(iv) 862
(v) 265
Solution:
We will use visual method as it is the efficient method to solve this problem.
4.) Find the squares of the following numbers:
(i) 425
(ii) 575
(iii) 405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995
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
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Solution:
(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
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599
Solution:
(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
(ii) 95
(iii) 505
(iv) 702
(v) 99
Solution: