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Frank Learning Maths Class 8 Solutions Chapter 3 Squares and Square Roots
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 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. Here students will find solutions for Exercise 3.1, 3.2 and 3.3 and also solution for chapter Check-up problems. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Squares and Square Roots Exercise 3.1 Solution
Question no – (1)
Solution :
(a) 192
= 19 × 19
= 361
(b) 472
= 47 × 47
= 2209
(c) 622
= 62 × 62
= 3844
(d) 1012
= 10201
Question no – (2)
Solution :
Let, Pythagorean triplets are 2m, m2 – 1, m2 + 1
(a) 2 m = 6
M = 3
M2 – 1 = 9 – 1 = 8
M2 + 1 = 9 + 1 = 10
(b) 2 m = 14
M = 7
M2 = 1 = 48
M2 + 1 = 50
(c) 2m = 18
M = 9
= M2 + 1 = 82
(d) 2 m = 16
M = 8
M2 – 1 = 63
M2 + 1 = 65
Question no – (3)
Solution :
(a) (29)2 – (28)2
=(29 + 28) (29 – 28)
= 57
(b) (68)2 – (67)2
= (68 – 67) (68 +67)
= 135
(c) (121)2 – (120)2
= (121) (121 – 120) (121 + 120)2
= 241
(d) (278)2 – (277)2
= (278 – 277) (278 + 277)
= 555
Question no – (4)
Solution :
LCM of 8, 9, 10 = 360
∴ 360 = 2 × 2 × 2 × 3 × 3 × 3 × 5
∴ We should multiple the given number with (2 × 5) for perfect square.
∴ 360 × 10 = 36000
Therefore, 3600 is the perfect square.
Question no – (5)
Solution :
∴ LCM = 180
= 2 × 2 × 3 × 3 × 5
∴ We should multiple the number by 5
∴ 180 × 5 = 900
Thus, the least square number is 900
Question no – (8)
Solution :
(i) 92 – 82
= 81 – 64
= 17
(ii) 132 – 122
= 25
Therefore, the Natural numbers between then 16 and 24
Question no – (9)
Solution :
(a) 502 – 512
= (51 + 50) (51 – 50)
= 101
∴ Non-squares number between 512, 502 = (101 – 1)
= 100
(b) 902 – 912
= (91 + 90) (91 – 90)
= 181
∴ Non-square number (181 – 1) = 180
(c) 10002 – 10012
= (1001 + 1000) ( 1001 – 1000)
= 2001
∴ Non-square numbers (2001 – 1) = 2000
Question no – (10)
Solution :
(a) 13 × 15
= 13 (13 + 2)
= 132 + 13 x 2
= 169 + 26
= 195
(b) 28 × 30
= (30 – 2) x 30
= 302 – 2 x 30
= 900 – 60
= 840
(c) 53 × 55
= (55 – 2) x 55
= (55)2 – 2 x 55
= 3025 – 120
= 2905
(d) 100 × 102
= 100 (100 + 2)
= 1002 + 2 x 100
= 10000 + 200
= 10200
Squares and Square Roots Exercise 3.2 Solution
Question no – (3)
Solution :
(a) 121 – 1 → 120, 120 – 3 → 117, 117 – 5 → 112, 112 – 7 → 105, 105 – 9 → 96, 96 – 11 → 85, 85 – 13 → 72, 72 – 15 → 57, 57 – 17 → 40, 40 – 19 → 21, 21 – 21 → 0.
∴ √121 = 11
(b) 64 – 1 → 63, 63 – 3 → 60, 60 – 5 → 55, 55 – 7 → 48, 48 – 9 → 39, 39 – 11 → 29, 29 – 13 → 15, 15 – 15 = 0.
∴ √64 = 8
(c) 49 – 1 = 48, 48 – 3 = 45, 45 – 5 = 40, 40 – 7 = 33, 33 – 9 = 24, 24 – 11 = 13, 13 – 13 = 0.
∴ √49 = 7
(d) 144 – 1 = 143, 143 – 3 = 140, 140 – 5 = 135, 135 – 7 = 128, 128 – 9 = 119, 119 – 11 = 108, 108 – 13 = 95, 95 – 15 = 80, 80 – 17 = 63, 63 – 19 = 44, 44 – 21 = 23, 23 – 23 = 0.
∴ √144 = 12
Question no – (4)
Solution :
(a) 324
∴ The units digit of the square root may be 2 or 8
Tens digit – 1
∴ probably 12 , 18
∴ √324 = 18
(b) 4225
∴ The unit digit of the square root is – 5
∴ tens digit – 6
∴ √4225 = 65
(c) 9801
∴ The unit digit of the square root may be – 1 , 9
Tens digit – 9
∴ probably – 91 , 99
∴ √9801 = 99
(d) 5041
∴ The unit digits may be – 1, 9
∴ Tens digit – 7
∴ probably – 71, 79
∴ √5041 = 71
Question no – (5)
Solution :
(a) 4931
∴ We should subtracted by 31 ten perfect square,
= 4900 = (70)2
(b) 18255
∴ We should subtracted by 30 ten perfect square,
= 18225 = (135)2
(c) 27285
∴ We should subtracted by 160 ten perfect square,
27285 = (165)2
Question no – (6)
Solution :
(a) 3450
∴ We have 582 < 3450 < 592
∴ perfect square should be – 592
(b) 7895
∴ We have 892 < 1895 < 902
∴ Perfect square should be – 902
(c) 54725
∴ We have 2332 < 54725 < 2342
∴ Perfect square should be – 2342
Question no – (7)
Solution :
∴ We should multiplied by -11
∴ The square,
= (2 × 11 × 13)
= 286
Question no – (8)
Solution :
∴ We should divided by 6
∴ The square = 5 × 7 = 35
Question no – (9)
Solution :
∴ The least number of 6 digit that is a perfect square = (317)2
Question no – (10)
Solution :
∴ The perfect square is = (316)2
Question no – (11)
Solution :
(a) 6, 11, 12
∴ The smallest square,
= 2 × 2 × 3 × 11 × 3 × 11
= (2 × 3 × 11)2
= (66)2
(b) 5, 14, 21
∴ The smallest square,
= 2 × 2 × 3 × 3 × 5 × 5 × 7 × 7
= (2 × 3 × 5 × 7)2
= (210)2
Question no – (12)
Solution :
∴ There are 52 number of students.
Question no – (13)
Solution :
Therefore, 11 students are left out.
Question no – (14)
Solution :
∴ Area of the square park,
= 96 × 54
= 5184
Therefore, the Length of the side will be 72 m
Question no – (15)
Solution :
Let the number one x, 2x
∴ 2x × x = 2000000
X2 = 1000000
X = 1000
∴ The numbers are – 1000, 2000
Squares and Square Roots Exercise 3.3 Solution
Question no – (1)
Solution :
(a) √121/625
= √121/√625
= √11 × 11/ √25 × 25
= 11/25
(b) √225/729
= √225/729
= 15/27
(c) √64/√441
= 8/21
(d) √81/576
= √81/576
= 9/24
Question no – (2)
Solution :
(a) √31 × √98
= √32 98
= √4 × 4 × 2 × 2 × 7 × 7
= 4 × 2 × 7
= 56
(b) √12 × 27
= √3 × 4 × 3 × 9
= 3 × 2 × 3
= 18
(c) √48 × 108
= √4 × 4 × 3 × 2 × 2 × 3 × 9
= 4 × 3 × 2 × 3
= 72
(d) √104 × 234
= √2 × 2 × 2 × 13 × 2 × 13 × 9
= 2 × 2 × 13 × 3
= 156
Question no – (3)
Solution :
(a) Given, √5 19/25
= √144/25
= 12/5
(b) Given, √96/294
= √16/49
= 4/7
(c) Given, √84 37/121
= √10201/121
= 101/11
(d) Given, √6 30/289
= √1764/289
= 42/17
Question no – (4)
Solution :
(a) Given, 33.64
(b) Given, 0.002809
(c) Given, 15.625
(d) Given, 44.1
(e) Given, 3
Question no – (5)
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 – (6)
Solution :
(a) √21609.0 = 147.0
(b) √216.09 = 14.7
(c) √2.1609 = 1.47
(d) √0.021609 = 0.147
(e) √0.00021609 = 0.0147
(f) √2160900 = 1470
Question no – (7)
Solution :
∴ Length of the side 11.5 km
4 side of the square 4 × 11.5 = 46 km
He travelled 3 times = (46 × 3) = 138 km
Therefore, the distance he travelled is 138 km.
Question no – (8)
Solution :
Given, √59.29 – √5.29 / √59.29 + √5.29
∴ (√59.29 – √5.29) (√59.29 + √5.29) / (√59.29 + √5.29)2
= 59.29 – 5.29 / 59.29 + 2 √59.29 – 5.29 + 5.29
= 54.00 / 64. 58 + 2√5929/100 × 529/100
= 54/ 6458/100 + 2√77 × 77 × 23 × 23/100
= 5400/6458 + 2 × 77 × 23
= 5400/6458 + 3542
= 5400/10000
= 0.54…(Simplified)
Question no – (9)
Solution :
(a) As per the question,
AB = 6cm
BC = 8cm
∴ AC = √62 + 82
= √36 + 64
= √100
= 10 cm
(b) As per the question,
AC = 13 cm,
BC = 5 cm
∴ AB = √132 – 52
= √169 – 25
= √144
= 12 cm
Squares and Square Roots Chapter Check-up Solution :
Question no – (1)
Solution :
(a) The value of √176 + v2401 = 15
Explanation :
= √176 + √2401
= √176 + 49
= √225
= 15
(b) There are 10 natural numbers lying between 52 and 62
(c) One’s digit in the square of 14 and 36 is 6.
(d) If one number of a Pythagorean triplet is 2 m, the other two one (m2 – 1) and (m2 + 1).
(e) The sum of first n odd natural numbers is n2.
(f) The perfect square will never end with 2, 3, 7, and 8.
(g) The value of √248 + √52 + √144 is 16.
Explanation :
= √248 + √52 + √144
= √248 + √52 + 12
= √248 + √64
= √248 + 8
= √256
= 16
(h) A number having 7 at its units place will have 9 at the unit place of its square.
Question no – (2)
Solution :
∴ 176 = 2 × 2 × 4 × 11
Therefore, we should multiplied by 11 for perfect square.
Question no – (3)
Solution :
Given, √(35)2 + (12)2
= √1225 + 144
= √1369
= 37 m
Therefore, he walk 37 m while returning.
Question no – (4)
Solution :
Two Pythagorean triplets each having one of the numbers as 5 are 3, 4, 5 and 5, 5.25, 6.25
Question no – (5)
Solution :
(a) 1369
(b) 5625
(c) 27.04
(d) 1.44
Question no – (6)
Solution :
Given number, 1385
∴ Subtracted by 16
∴ Square root be 37
Question no – (7)
Solution :
Given number, 6200
∴ 782 < 6200 < 792
∴ 792 = 6241
Therefore, we must be added 41 for a perfect square.
Question no – (8)
Solution :
We know, The greatest three digits is 999
∴ The greatest number of three digit is (31)2 = 961
Question no – (9)
Solution :
∴ LCM = 120
= 120 = 4 × 2 × 3 × 5
Therefore, we need 4 × 2 × 2 × 3 × 3 × 5 × 5 for divisible by the given number 3600
Question no – (10)
Solution :
As per the given question,
A decimal number is multiplied by itself.
The product is = 51.84,
The number = ?
Hence, the number will be 7.2
Question no – (11)
Solution :
4a = 96 m
a = 24
∴ Length = 24 m
∴ Area = (24)2 = 576 m2
Thus, the area of a square field will be 576 m2
Question no – (12)
Solution :
Given, √(35)2 + (12)2
= √1225 + 144
= √1369
= 37 m
Therefore, he walk 37 m while returning.
Question no – (13)
Solution :
(2x)2 + (3x)2 + (5x)2 = 608
4x2 + 9x2 + 25x2 = 608
38x2 = 608
x2 = 16
x = 4
Therefore, the numbers are 8, 12 and 20
Question no – (14)
Solution :
√101 1/400 m2
= 201/20 m
Therefore, Length of one side will be 10 1/20 m.
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