Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9. (i) But given number 153 has its unit digit 3. So it is not a perfect square number. (ii) Given number 257 has its unit digit 7. So it is not a perfect square number. (iii) Given number 408 has its unit digit 8.Read more
Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) But given number 153 has its unit digit 3. So it is not a perfect square number.
(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.
(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.
(iv) Given number 441 has its unit digit 1. So it would be a perfect square number
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are: (i) 1 (ii) 6 (iii) 1 (iv) 5 Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/Read more
Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible
unit’s digits of the given numbers are:
(i) 1 (ii) 6 (iii) 1 (iv) 5
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
(i) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet. Here, 2m=6 ⇒ m=6/2=3 Therefore, Second number (m²-1)=(3)²-1=9-1=8 Third number m²+1=(3)²+1=9+1=10 Hence, Pythagorean triplet is (6,8,10). (ii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet. Here, 2m=14 ⇒ m=14/Read more
(i) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=6 ⇒ m=6/2=3
Therefore,
Second number (m²-1)=(3)²-1=9-1=8
Third number m²+1=(3)²+1=9+1=10
Hence, Pythagorean triplet is (6,8,10).
(ii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=14
⇒ m=14/2=7
Therefore,
Second number (m²-1)=(7)²-1=49-1=48
Third number m²+1=(7)²+1=49+1=50
Hence, Pythagorean triplet is (14,48,50).
(iii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=16
⇒ m=16/2=8
Therefore,
Second number (m²-1)=(8)²-1=64-1=63
Third number m²+1 =(8)²+1=64+1=65
Hence, Pythagorean triplet is (16,63,65)..
(iv) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=18
⇒ m=18/2=9
Therefore,
Second number (m²-1)=(9)²-1=81-1=80
Third number m²+1 =(9)²+1=81+1=82
Hence, Pythagorean triplet is (18, 80, 82).
Class 8 Maths Chapter 6 Exercise 6.2 Solution in Video
Find the square roots of the following numbers by the Prime Factorization method: (i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100
(i) 729 √729 = √3x3x3x3x3x3x = 3x3x3 = 27 (ii) 400 √400 = √2x2x2x2x5x5 = 2x2x5 = 20 (iii) 1764 √1764 = √2x2x3x3x7x7 = 2x3x7 = 42 (iv) 4096 √4096 = √2x2x2x2x2x2x2x2x2x2x2x2 = 2x2x2x2x2x2x = 64 (v) 7744 √7744 = √2x2x2x2x2x2x11x11 = 2x2x2x11 = 88 (vi) 9604 √9604 = √2x2x7x7x7x7 = 2x7x7 = 98 (vii) 5929 √Read more
(i) 729
√729 = √3x3x3x3x3x3x
= 3x3x3
= 27
(ii) 400
√400 = √2x2x2x2x5x5
= 2x2x5
= 20
(iii) 1764
√1764 = √2x2x3x3x7x7
= 2x3x7
= 42
(iv) 4096
√4096 = √2x2x2x2x2x2x2x2x2x2x2x2
= 2x2x2x2x2x2x
= 64
(v) 7744
√7744 = √2x2x2x2x2x2x11x11
= 2x2x2x11
= 88
(vi) 9604
√9604 = √2x2x7x7x7x7
= 2x7x7
= 98
(vii) 5929
√5929 = √7x7x11x11
= √7×11
= 77
(viii) 9216
√9216 = √2x2x2x2x2x2x2x2x2x2x3x3
= 2x2x2x2x2x3
= 96
(ix) 529
√529 = √23×23
= 23
(x) 8100
√8100 =√2x2x3x3x3x3x5x5
= 2x3x3x5
= 90
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
Find the square roots of 100 and 169 by the method of repeated subtraction.
By successive subtracting odd natural numbers from 100, 100 – 1 = 99 99 – 3 = 96 96 – 5 = 91 91 – 7 = 84 84 – 9 = 75 75 – 11 = 64 64 – 13 = 51 51 – 15 = 36 36 – 17 = 19 19 – 19 = 0 This successive subtraction is completed in 10 steps. Therefore √100 =10 By successive subtracting odd natural numbersRead more
By successive subtracting odd natural numbers from 100,
100 – 1 = 99 99 – 3 = 96 96 – 5 = 91 91 – 7 = 84
84 – 9 = 75 75 – 11 = 64 64 – 13 = 51 51 – 15 = 36
36 – 17 = 19 19 – 19 = 0
This successive subtraction is completed in 10 steps. Therefore √100 =10
By successive subtracting odd natural numbers from 169,
169 – 1 = 168 168 – 3 = 165 165 – 5 = 160 160 – 7 = 153
153 – 9 = 144 144 – 11 = 133 133 – 13 = 120 120 – 15 = 105
105 – 17 = 88 88 – 19 = 69 69 – 21 = 48 48 – 23 = 25
25 – 25 = 0
This successive subtraction is completed in 13 steps. Therefore √169 = 13
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
Without doing any calculation, find the numbers which are surely not perfect squares: (i) 153 (ii) 257 (iii) 408 (iv) 441
Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9. (i) But given number 153 has its unit digit 3. So it is not a perfect square number. (ii) Given number 257 has its unit digit 7. So it is not a perfect square number. (iii) Given number 408 has its unit digit 8.Read more
Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) But given number 153 has its unit digit 3. So it is not a perfect square number.
(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.
(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.
(iv) Given number 441 has its unit digit 1. So it would be a perfect square number
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
What could be the possible ‘one’s’ digits of the square root of each of the following numbers: (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025
Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are: (i) 1 (ii) 6 (iii) 1 (iv) 5 Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/Read more
Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible
unit’s digits of the given numbers are:
(i) 1 (ii) 6 (iii) 1 (iv) 5
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
Write a Pythagoras triplet whose one member is: (i) 6 (ii) 14 (iii) 16 (iv) 18
(i) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet. Here, 2m=6 ⇒ m=6/2=3 Therefore, Second number (m²-1)=(3)²-1=9-1=8 Third number m²+1=(3)²+1=9+1=10 Hence, Pythagorean triplet is (6,8,10). (ii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet. Here, 2m=14 ⇒ m=14/Read more
(i) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=6 ⇒ m=6/2=3
Therefore,
Second number (m²-1)=(3)²-1=9-1=8
Third number m²+1=(3)²+1=9+1=10
Hence, Pythagorean triplet is (6,8,10).
(ii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=14
⇒ m=14/2=7
Therefore,
Second number (m²-1)=(7)²-1=49-1=48
Third number m²+1=(7)²+1=49+1=50
Hence, Pythagorean triplet is (14,48,50).
(iii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=16
⇒ m=16/2=8
Therefore,
Second number (m²-1)=(8)²-1=64-1=63
Third number m²+1 =(8)²+1=64+1=65
Hence, Pythagorean triplet is (16,63,65)..
(iv) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=18
⇒ m=18/2=9
Therefore,
Second number (m²-1)=(9)²-1=81-1=80
Third number m²+1 =(9)²+1=81+1=82
Hence, Pythagorean triplet is (18, 80, 82).
Class 8 Maths Chapter 6 Exercise 6.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/