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Frank ICSE Mathematics Class 8 Solutions Chapter 3 Squares and Square Roots
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 3, Squares and Square Roots. Here students can easily find step by step solutions of all the problems for Squares and Square Roots, Exercise 3.1, 3.2, 3.3 and 3.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here in this post all the solutions are based on latest Syllabus.
Squares and Square Roots Exercise 3.1 Solution :
Question no – (1)
Solution :
(a) (11)2 = 121
= 11 × 11 = 121
(101)2 = 10201
= 101 × 101 = 10201
(1001)2 = 1002001
= 1001 × 1001 = 1002001
Now, (1000000)2
= 1000000 × 1000000
= 100000020000001
(b) (111)2
= 111 × 111 = 12321
(1111)2
= 1111 × 1111 = 1234321
(11111)2 =
= 11111 × 11111 = 123454321
(111111)2
= 111111 × 111111 = 1234564321
(1111111)2
1111111 × 1111111 = 1234567654321
When we multiply the given number by itself according to the given pattern we find the number, thus, if we multiply the 11111, 111111 and 1111111 by itself then we get the missing number.
Question no – (2)
Solution :
In the given question we get the following numbers,
841, 144, 900, 169, 576, 225, 1210000, 729, 256
Before we Classify the following numbers first we need to know what is odd and what is even numbers, the numbers that can divided into two equal parts are called Even numbers and the numbers that cannot divided into two equal parts are called Odd numbers.
Even number are : 900, 576, 1210000, 256
Odd number are : 841, 169, 225, 729
Question no – (3)
Solution :
(a) The square of the number 47 has 9 in its units place.
(b) The square of the number 84 has the digit 6 in units place.
(c) The square of the number 612 has the digit 4 in units place.
Question no – (4)
Solution :
(a) (70 + 5)2 = (70)2
= (70)2 + 2.70.5 + (5)2
= 5625
(b) (90 + 5)2 = (95)2
= (90)2 + 2.90.5 + (5)2
= 9025
(c) (65)2 = (60 + 5)2
= (60)2 + 2.60.5 + (5)2
= 3600 + 600 + 25
= 4225
(d) (125)2 = (100 + 25)2
= (100)2 + 2.100.25 + (25)2
= 10000 + 5000 + 625
= 15625
(e) (31)2 = (30 + 1)2
= (30)2 + 2.30 + (1)2
= 900 + 60 + 1
= 961
(f) (41)2 = (40 + )2
= (40)2 + 2.401 + (1)2
= 1600 + 80 + 1
= 1681
(g) (61)2 = (60 + 1)2
= (60)2 + 2.60.1 + (1)2
= 3600 + 120 + 1
= 3721
(h) (91)2 = (90 + 1)2
= (90)2 + 2.90.1 + (1)2
= 8100 + 180 + 1
= 8281
Question no – (6)
Solution :
(a) 112
= 112/2
= 121/2 = 60.5
The numbers are = 60.5 ± 0.5
∴ 112 = 60 + 61
(b) 192
= 192/2
= 361/2 = 180.5
The numbers are = 180.5 ± 0.5
∴ 192 = 180 + 181
(c) 232
= 232/2
= 529/2 = 264.5
The numbers are 264.5 ± 0.5
∴ 232 = 264 + 265
(d) 152
= 152/2
= 225/2 = 112.5
The numbers are = 112.5 ± 0.5
∴ 152 = 112 + 113
Question no – (7)
Solution :
(a) 682
= 62 = 36
= 82 = 64
Now,
36 64
+ 9 60 → 2 × 6 × 8
——————
4624
(b) 732
= 72 = 49
= 32 = 9
Now,
= 49 09
+ 4 20
——————
5329
(c) 232
= 22 = 4
= 32 = 9
Now,
= 04 09
+ 1 20
——————
529
(d) 872
= 82 = 64
= 72 = 49
Now,
64 49
+ 11 20
——————
7569
Question no – (8)
Solution :
(a) 1 + 3 + 5 + 7
= 42
= 16
(b) 1 + 3 + 5 + 7 + 9 + 11
= 62
= 36
(c) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
= 92
= 81
Question no – (9)
Solution :
(a) 1375
Here, factors are not same
∴ 1375 is not a perfect square
(b) 625
Here, all the factors are same.
∴ 625 is a perfect square.
(c) 1024
Here, all the factors are same.
∴ 1024 is perfect square.
(d) 729
Here, all the factors are same.
∴ 729 is a perfect square
Question no – (10)
Solution :
(a) 6
= m = 6,
= 2m = 12,
= m2 – 1 = 35,
= m2 + 1 = 37
So, here Pythagorean triplets are 12, 35 and 37
(b) 7
= m = 7,
= m = 14,
= m2 – 1 = 48,
= m2 + 1 = 50
So, here Pythagorean triplets are 14, 48 and 50
(c) 8
= m = 8,
= 2m = 10,
= m2 -1 = 63,
= m2 + 1 = 65
So, here Pythagorean triplets are 10, 63 and 65
(d) 9
= m = 9,
= 2m = 18,
= m2 – 1 = 80
= m2 + 1 = 82
So, here Pythagorean triplets are 18, 80 and 82
Question no – (11)
Solution :
(a) (29)2 – (28)2
= (29 + 28)
= 57
(b) (68)2 – (67)2
= (68 + 67)
= 135
(c) (121)2 – (120)2
= (121 + 120)
= 241
(d) (278)2 – (277)2
= (278)2 – (277)2
= (278 + 277)
= 55
Question no – (12)
Solution :
(a) 13 × 15
= (14 – 1) (14 + 1)
= 142 – 12
= 196 – 1
(b) 29 × 31
= (30 – 1) (30 + 1)
= 302 – 12
= 900 – 1
= 899
(c) 49 × 51
= (50 – 1) (50 + 1)
= 502-12
= 2500 – 1
= 2499
(d) 24 × 26
= (25 – 1) (25 + 1)
= 252 – 12
= 625 – 1
= 624
Squares and Square Roots Exercise 3.2 Solution :
Question no – (1)
Solution :
(a) 289
= √289 = 2 = 89
Now, 2 lies between 12 = 1 and 22 = 4
The tens digit will be 1
Now, 32 = 9 and 72 = 49
So, we have to choose between 13 and 17
Since 152 = 225 < 289
√289 = 17
∴ Square root of 289 is 17
(b) 361
= √361 = 3 61
3 lies between 12 = 1 and 22 = 4
The tens digit will be 1
Since 12 = 1, 92 = 81. we have to chose between 11 and 19
Since 152 = 225 < 361
Therefore = √361 = 19
∴ Square root of 361 is 19
(c) 529
= √529 = 5 29
Since 22 < 5 < 32The tens digit will be 2
Now,
= 32 = 9, 72 = 49 and 252 = 625 > 259
∴ √529 = 23
∴ Square root of 529 is 23
(d) 441
√441 = 4 41
Since 22 = 4 The tens digit will be 2
Now,
= 12 = 1, 92 and 252 = 625 > 441
∴ √441 = 21
∴ Square root of 441 is 21
(e) 2916
= √2916 = 29 16, Since 52 < 29 < 62
∴ Tens digit will be 5,
42 = 16, 62 = 36 and 552 = 3025 > 2916
∴ √2916 = 54
∴ Square root of 2916 is 54
(f) 3136
√3136 = 31 36
Since 52 < 31 < 62 = the tens digit will be 5
Now, 42 = 16, 62 = 36 and 552 = 3025 < 3136
Therefore √3136 = 56
∴ Square root of 3136 is 56
(g) 2704
= √2704 = 27 04
Since, 52 < 27 < 62 ∴The tens digit will be 5.
Now, 32 = 9 and 72 = 49 and 652 = 4225 > 3969
∴ √2704 = 52
∴ Square root of 2704 is 52
Question no – (2)
Solution :
(a) 484
= 2 × 2 × 2 × 11 × 11 × 11
√484 = 2 × 11 × 22
(b) 1089
= 1089 = 3 × 3 × 3 × 11 × 11 × 11
√1089 = 3 × 11 = 33
(c) 784
= 784 = 2 × 2 × 2 × 2 × 7 × 7 × 7
√784 = 2 × 2 × 2 × 7 = 28
(d) 1764
= 1764 = 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7
= √1764 = 2 × 3 × 7 = 42
Question no – (3)
Solution :
(a) 5929
∴ √5929 = 77
(b) 2809
∴ √2809 = 53
(c) 12544
∴ √12544 = 112
(d) 53361
∴ √53361 = 231
(e) 60516
∴ √60516 = 246
(f) 33856
∴ √33850 = 184
(g) 77841
∴ √77841 = 279
(h) 276676
∴ √276676 = 526
The square root of 810000 is 900.
Squares and Square Roots Exercise 3.3 Solution :
Question no – (1)
Solution :
(a) 121/625
= √121/625
= √121/√ 625
= √11 × 11/√5 × 5 × 5 × 5
= 11/5 × 5
= 11/25
∴ Square root of 121/625 is 11/25
(b) 196/225
= √196/225
= √2 × 2 × 7 × 7/√3 × 3 × 5 × 5
= 2 × 7/3 × 5
= 14/15
∴ Square root of 196/225 is 14/15
(c) 64/441
= √64/441
= √2 × 2 × 2 × 2 × 2 × 2/√3 × 3 × 7 × 7
= 2 × 2 × 2/3 × 7
= 8/21
∴ Square root of 64/441 is 8/21
(d) 49/169
= √49/169
= √7 × 7/√13 × 13
= 7/13
∴ Square root of 49/169 is 7/13
Question no – (2)
Solution :
(a) √32 × √98
Here,
Prime factors of 32 = 2 × 2 × 2 × 2 × 2
Prime factors of 98 = 2 × 7 × 7
∴ √2 × 2 × 2 × 2 × 2 × 2 × 7 × 7
= 2 × 2 × 2 × 7
= 56
Therefore, the value is 56
(b) √12 × √27
= √12 × 27
= √2 × 2 × 3 × 3 × 3 × 3
= 2 × 3 × 3
= 18
(c) √48 × √108
= √48 × 108
= √2 × 2 × 2 × 2 × 3 × 2 × 2 × 3 × 3 × 3
= 2 × 2 × 2 × 3 × 3
= 72
(d) √104 × √234
= √104 × 234
= √2 × 2 × 2 × 13 × 2 × 3 × 3 × 13
= 2 × 2 × 3 × 13
= 156
Question no – (3)
Solution :
(a) √5 19/25
= √(25 × 5) + 19/25
= √144/25
= √2 × 2 × 3 × 3 × 2 × 2/√5 × 5
= 12/5
(b) √96/294
= √2 × 2 × 2 × 2 × 2 × 3/√2 × 3 × 7 × 7
= √2 × 2 × 2 × 2/√7 × 7
= 4/7
(c) √1 13/36
(First we converted it into mixed fraction to proper fraction)
We get, = √49/36
Now, √49/36
Prime factors of 49 = 7 × 7
Prime factors of 36 = 6 × 6
= √7 × 7/√6 × 6
= 7/6
(d) √1 5/49
= √64/49
= √8 × 8/√7 × 7
= 8/7
Question no – (4)
Solution :
(a) 350
182 < 350 < 192
√350 to the nearest integer = 19
(b) 150
122 < 150 < 132
∴ √150 to the nearest integer = 13
(c) 5000
702 < 5000 < 72
∴ √5000 to the nearest integer 71
(d) 720
∴ 262 < 720 < 272
∴ √720 to the nearest integer 27
Question no – (5)
Solution :
(a) √21609 = 147
(b) √2160.09 = 14,7
(c) √2.1609 = 1.47
(d) √0.021609 = 0.147
(e) √0.00021609 = 0.0147
(f) √2160900 = 1470
Question no – (6)
Solution :
The area of a square field is 132.25 km2. find the perimeter of the field.
We have to find : the perimeter of the field?
Area of the square field = 132.25 km2
∴ Side of the square field √132,25 = 11.5 km
As we know that,
Perimeter of square = 4 × side
∴ Perimeter of the square field,
= 11.5 × 4
= 46 km
Therefore, the perimeter of the square field will be 46 km.
Question no – (7)
Solution :
(a) 90.25
= √90.25
= √9025/√100
Now,√9025/√100
= 95/10
= 9.5
∴ Square root of 90.25 is 9.5
(b) 11.56
= √1156/√100
= 34/10
= 3.4
∴ Square root of 11.56 is 3.4
(c) 16129
= 127
∴ Square root of 16129 is 127
(d) 9604
= √9604 = 98
∴ Square root of 9604 is 98
(e) 8.0656
= √8.0656 = 2.84
∴ Square root of 8.0656 is 2.84
(f) 6.1504
= √6.1504 = 2.48
∴ Square root of 6.1504 is 2.48
(g) 13.9876
= √13.9876 = 3.74
∴ Square root of 13.9876 is 3.74
(h) 1049.76
= √1049.76 = 32.4
∴ Square root of 1049.76 is 32.4
(i) 3340.84
= √3340.84 = 57.8
∴ Square root of 3340.84 is 57.8
(j) 4596.84
= √4596.84 = 67.8
∴ Square root of 4596.84 is 67.8
(k) 5715.36
= √5715.36 = 75.6
∴ Square root of 5715.36 is 75.6
(l) 51.6961
= √51.6961 = 7.19
∴ Square root of 51.6961 is 7.19
(m) 81796
= √81796 = 286
∴ Square root of 81796 is 286
(n) 56169
= √56169 = 237
∴ Square root of 56169 is 237
(o) 32.2624
= √32.2624 = 6.68
∴ Square root of 32.2624 is 6.68
Question number – (8)
Solution :
We have to find : How many rows are there?
We can solve this problem using long division method
Therefore, There were total 64 rows.
Squares and Square Roots Exercise 3.4 Solution :
Question no – (1)
Solution :
(a) 3789
= 3800
∴ The round off of 3789 is 3800.
(b) 2304
= 2300
∴ The round off of 2304 is 2300
(c) 79.83
= 80
∴ The round off of 79.83 is 80
(d) 0.01304
= 0
∴ The round off of 0.01304 is 0
(e) 4.967
= 5
∴ The round off of 4.967 is 5
Question no – (2)
Solution :
(a) 34.47
= 35
∴ The round off of 34.47 is 35
(b) 0.01234
= 0.01
∴ The round off of 0.01234 is 0.01
(c) 0.07057
= 0.07
∴ The round off of 0.07057 is 0.07
(d) 0.4003
= 0.4
∴ The round off of 0.4003 is 0.4
(e) 0.2996
= 0.3
∴ The round off of 0.2996 is 0.3
Question no – (3)
Solution :
(a) 17.835
= 17.83
∴ The round off of 17.835 is 17.83
(b) 46.796
= 46.8
∴ The round off of 46.796 is 46.8
(c) 0.610467
= 0.61
∴ The round off of 0.610467 is 0.61
(d) 67895
= 6789
∴ The round off of 67895 is 6789
(e) 35.7449
= 35.74
∴ The round off of 35.7449 is 35.74
Revision Exercise Questions Solution :
Question no – (1)
Solution :
(a) 2225
Since, 89 does not have pair.
Therefore, 2225 is not a perfect square.
(b) 4356
Clearly, all the prime factors appear in pairs that’s why this is a perfect square
Question no – (2)
Solution :
We can solve this problem by long division method.
Given number is 62575
So now,
∴ 75 has to be subtracted
and 250 would be the square root.
Question no – (3)
Solution :
Since,
18496 > 10000
= (100)2,
Therefore, the √18496 would be a three digit number.
Question no – (4)
Solution :
The units digit of the square of 74 is 6
Since, 42 = 16,
The unit digit of square of 74 will be 6.
Question no – (5)
Solution :
As per the given question here we have to find the smallest number by which 1323 must be divided to make it a perfect square.
We can solve this problem using Prime factorisation method
Clearly, 1323 must be divided by 3 to get a perfect square.
Next Chapter Solution :
👉 Chapter 4 👈
