Squares and Cube Roots Class 8 RS Aggarwal Exe-3B Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3. We provides step by step solutions of council presceibe publication / textbook. In this article you would learn properties and interesting pattern of perfect squires. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Squares and Cube Roots Class 8 RS Aggarwal Exe-3B Goyal Brothers ICSE Maths Solutions Ch-3
| Board | ICSE |
| Publications | Goyal Brothers Prakshan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Chapter-3 | Squires and Squire Roots, Cubes and Cube Roots |
| Exe-3B | properties and interesting pattern of perfect squires |
| Edition | 2024-2025 |
properties and interesting pattern of perfect squires
(Squares and Cube Roots Class 8 RS Aggarwal Exe-3B Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3)
Page- 44,45
Exercise- 3B
(Squares and Cube Roots Class 8 RS Aggarwal Exe-3B Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3)
Que-1: Give reason to show that none of the numbers given below is a perfect square :
(i) 2162 (ii) 6843 (iii) 9637 (iv) 6598 (v) 640 (vi) 49000 (vii) 3600000
Solution- (i) 2162
√2162 = 46.49
As perfect square never ends with 2.
(ii) 6843
√6843 = 82.72
As perfect square never ends with 3.
(iii) 9637
√9637 = 98.16
As perfect square never ends with 7.
(iv) 6593
√6593 = 81.19
As perfect square never ends with 3.
(v) 640
√640 = 25.29
As perfect square never ends with odd number of zero.
(vi) 49000
√49000 = 221.35
As perfect square never ends with odd number of zero.
(vii) 3600000
√3600000 = 1897.36
Que-2: Which of the following are squares of even numbers ?
(i) 676 (ii) 729 (iii) 1089 (iv) 2304 (v) 5625 (vi) 9216
Solution- (i) 26 x 26 = 676 26 is even number.
(ii) 27 x 27 = 729 27 is a odd number.
(iii) 33 x 33 = 1089 33 is a odd number.
(iv) 48 x 48 = 2304 48 is an even number.
(v) 75 x 75 = 5625 75 is a odd number.
(vi) 96 x 96 = 9216 96 is an even number.
Que-3: Which of the following are squares of odd numbers ?
(i) 484 (ii) 961 (iii) 4225 (iv) 7396 (v) 6241 (vi) 8649
Solution- (i) 484 is not odd Therefore it is not square of odd number.
(ii) 961 is odd Therefore it is square of odd number.
(iii) 4225 is odd Therefore it is square of odd number.
(iv) 7396 is not odd Therefore it is not square of odd number.
(v) 6241 is odd Therefore it is square of odd number.
(vi) 8649 is odd Therefore it is square of odd number.
Que-4: Without adding find the sum :
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
(ii) 1 + 3 + 5 + 7 + ………………….. + 47 + 49
(iii) 1+ 3 + 5 + 7 + …………………… + 97 + 99.
Solution- (i) 121
(ii) 625
(iii) 2500
Que-5: (i) Express 144 as the sum of 12 odd numbers.
(ii) Express 225 as the sum of 15 odd numbers.
Solution- (i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29
Que-6: Write a Pythagorean triplet whose smallest member is :
(i) 6 (ii) 14 (iii) 16 (iv) 20
Solution- (i) For any natural number m>1, the Pythagorean triplet adheres to
(2m)^2 + (m^2 − 1)^2 = (m^2 + 1) 2.
∵ smallest number is 6, 2m = 6 or m = 3.
m^2 + 1 = 3^2 + 1 = 9 + 1 = 10
m^2 − 1 = 3^2 − 1 = 9 − 1 = 8
So, the Pythagorean triplet is 6, 8, 10.
(ii) smallest number =14
means 2m =14m
= 14/2 = 7
m²-1= 7²-1
= 49-1= 48
m²+1=7²+1
= 49+1=50
hence the required triplet is 14, 48 and 50.
(iii) Here, 2m = 16
⇒ m = 16/2 = 8
∴Second number(m^2 − 1) = (8)^2 − 1 = 64−1 = 63
And third number (m^2 + 1) = (8)^2 + 1 = 64+1 = 65
Hence, Pythagorean triplet is (16, 63, 65).
(iv) Let 2m = 20
⇒ m = 10
∴ m^2 − 1 = (10)^2 − 1 = 100−1 = 99
and m^2 + 1 = (10)^2 + 1 = 100+1 = 101
∴ Pythagorean triplet = 20, 99, 101
Que-7: Evaluate :
(i) (29)² – (28)² (ii) (56)² – (55)² (iii) (82)² – (81)² (iv) (108) – (107)² (v) (221)² – (220)²
Solution- (i) 841 – 784 = 57
(ii) 3136 – 3025 = 111
(iii) 6724 – 6561 = 163
(iv) 11664 – 11449 = 215
(v) 48841 – 48400 = 441
Que-8: Using the identity (a + b)² = (a² + 2ab + b²), evaluate :
(i) (204)² (ii) (307)² (iii) (430)²
Solution- (i) 204²
= (200+4)²
= (200)² + 2(200)(4) + (4)²
= 40000 + 1600 + 16 = 41616 Ans.
(ii) 307²
= (300+7)²
= (300)² + 2(300)(7) + (7)²
= 90000 + 4200 + 49 = 94249 Ans.
(iii) 430²
= [400+30]²
= (400)² + 2(400)(30) + (30)²
= 160000 + 24000 + 900 = 184900 Ans.
Que-9: Using the identity (a – b)² = (a² – 2ab + b²), evaluate :
(i) (94)² (ii) (192)² (iii) (389)²
Solution- (i) (100-6)² = 100² – 2×100×6 + 6²
= 10000 – 1200 + 36
= 8836 Ans.
(ii) (200-8)² = 200² + 8² – 2(200)(8)
= 40000 + 64 – 3200
= 36864 Ans.
(iii) (400-11)² = 400² + 11² – 2(400)(11)
= 151321 Ans.
Que-10- Evaluate :
(i) 59 x 61 (ii) 88 x 92 (iii) 94 x 106 (iv) 48 x 52
Solution- (i) 3599
(ii) 8096
(iii) 9964
(iv) 2496
Que-11: State whether which of the following statements is true or false :
(i) A perfect square number always has an even number of digits.
(ii) Every number ending in an even number of zeroes is a perfect square.
(iii) The square of a prime number is a prime.
(iv) There are 10 perfect square numbers from 1 to 100.
(v) The sum of two perfect squares is a perfect square.
(vi) The difference of two perfect squares is a perfect square.
(vii) The product of two perfect squares is a perfect square.
Solution- (i) False
(ii) False
(iii) False
(iv) True
(v) False
(vi) False
(vii) True
— : End of Squares and Cube Roots Class 8 RS Aggarwal Exe-3B Goyal ICSE Maths Solutions Ch-3 :–
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