(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
(i) Since, non-perfect square numbers between and are n² and (n+1)² are 2n Here, = 12 n Therefore, non-perfect square numbers between 12 and 13 = 2n = 2x12=24 (ii) Since, non-perfect square numbers between n² and (n+1)² are 2n. Here, n = 25 Therefore, non-perfect square numbers between 25 and 26 = 2Read more
(i) Since, non-perfect square numbers between and are n² and (n+1)² are 2n
Here, = 12 n
Therefore, non-perfect square numbers between 12 and 13 = 2n = 2×12=24
(ii) Since, non-perfect square numbers between n² and (n+1)² are 2n.
Here, n = 25
Therefore, non-perfect square numbers between 25 and 26 = 2n = 2 x 25 = 50
(iii) Since, non-perfect square numbers between n² and (n+1)² are 2n.
Here, n = 99
Therefore, non-perfect square numbers between 99 and 100 = 2n = 2 x 99 = 198
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
(i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers. 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video for more answersRead more
(i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
(i) Here, there are five odd numbers. Therefore square of 5 is 25. ∴ 1 + 3 + 5 + 7 + 9 = 5² = 25 (ii) Here, there are ten odd numbers. Therefore square of 10 is 100. ∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10² = 100 (iii) Here, there are twelve odd numbers. Therefore square of 12 is 144. ∴ 1Read more
(i) Here, there are five odd numbers. Therefore square of 5 is 25.
∴ 1 + 3 + 5 + 7 + 9 = 5² = 25
(ii) Here, there are ten odd numbers. Therefore square of 10 is 100.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10² = 100
(iii) Here, there are twelve odd numbers. Therefore square of 12 is 144.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12² = 144
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
11² = 121 101² = 10201 10101² = 102030201 1010101² = 1020304030201 101010101² = 10203040504030201 Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
11² = 121 101² = 10201 1001² = 1002001 100001² = 10000200001 1000000² = 100000020000001 Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
(i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number. (ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number. (iii) 7779 – Unit’s digit of given number is 9 and square of 9Read more
(i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number.
(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number.
(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81. Therefore, square of 7779 would be an odd number.
(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16. Therefore, square of 82004 would not be an odd number.
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0. Therefore 1057 is not a perfect square because its unit’s place digit is 7. (ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 234Read more
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.
Therefore 1057 is not a perfect square because its unit’s place digit is 7.
(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 23453 is not a perfect square because its unit’s place digit is 3.
(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 7928 is not a perfect square because its unit’s place digit is 8.
(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222222 is not a perfect square because its unit’s place digit is 2.
(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 64000 is not a perfect square because its unit’s place digit is single 0.
(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 89722 is not a perfect square because its unit’s place digit is 2.
(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.
(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 505050 is not a perfect square because its unit’s place digit is 0.
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
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
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Find the squares of the following numbers: (i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
(i) (32)² = (30+2)² = (30)² +2x30x2+(2)² [∵ (a+b)²=a²+2ab+b²] = 900+120+4=1024 (ii) (35)² =(30+5)² =(30)² +2x30x5+(5)² [∵ (a+b)²=a²+2ab+b²] = 900+300+25=1225 (iii) (86)²= (80+6)² =(80)² +2x80x6+(6)² [∵ (a+b)²=a²+2ab+b²] =1600+960+36=1386 (iv) (93)² =(90+3)² =(90)² +2x90x3+(3)² [∵ (a+b)²=a²+2ab+b²] =Read more
(i) (32)² = (30+2)² = (30)² +2x30x2+(2)² [∵ (a+b)²=a²+2ab+b²]
= 900+120+4=1024
(ii) (35)² =(30+5)² =(30)² +2x30x5+(5)² [∵ (a+b)²=a²+2ab+b²]
= 900+300+25=1225
(iii) (86)²= (80+6)² =(80)² +2x80x6+(6)² [∵ (a+b)²=a²+2ab+b²]
=1600+960+36=1386
(iv) (93)² =(90+3)² =(90)² +2x90x3+(3)² [∵ (a+b)²=a²+2ab+b²]
= 8100+540+9=8649
(v) (71)² = (70+1)² =(70)² +2x70x1+(1)² [∵ (a+b)²=a²+2ab+b²]
= 4900+140+1=5041
(vi) (46)²=(40+6)² =(40)² +2x40x6+(6)² [∵ (a+b)²=a²+2ab+b²]
=1600+480+36=2216
Class 8 Maths Chapter 6 Exercise 6.2 Solution in Video
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How many numbers lie between squares of the following numbers: (i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
(i) Since, non-perfect square numbers between and are n² and (n+1)² are 2n Here, = 12 n Therefore, non-perfect square numbers between 12 and 13 = 2n = 2x12=24 (ii) Since, non-perfect square numbers between n² and (n+1)² are 2n. Here, n = 25 Therefore, non-perfect square numbers between 25 and 26 = 2Read more
(i) Since, non-perfect square numbers between and are n² and (n+1)² are 2n
Here, = 12 n
Therefore, non-perfect square numbers between 12 and 13 = 2n = 2×12=24
(ii) Since, non-perfect square numbers between n² and (n+1)² are 2n.
Here, n = 25
Therefore, non-perfect square numbers between 25 and 26 = 2n = 2 x 25 = 50
(iii) Since, non-perfect square numbers between n² and (n+1)² are 2n.
Here, n = 99
Therefore, non-perfect square numbers between 99 and 100 = 2n = 2 x 99 = 198
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
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(i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers.
(i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers. 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video for more answersRead more
(i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
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Without adding, find the sum: (i) 1 + 3 + 5 + 7 + 9 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(i) Here, there are five odd numbers. Therefore square of 5 is 25. ∴ 1 + 3 + 5 + 7 + 9 = 5² = 25 (ii) Here, there are ten odd numbers. Therefore square of 10 is 100. ∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10² = 100 (iii) Here, there are twelve odd numbers. Therefore square of 12 is 144. ∴ 1Read more
(i) Here, there are five odd numbers. Therefore square of 5 is 25.
∴ 1 + 3 + 5 + 7 + 9 = 5² = 25
(ii) Here, there are ten odd numbers. Therefore square of 10 is 100.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10² = 100
(iii) Here, there are twelve odd numbers. Therefore square of 12 is 144.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12² = 144
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
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Using the given pattern, find the missing numbers:
1² + 2² + 2² = 3² 2² + 3² + 6² = 7² 3² + 4² + 12² = 13² 4² + 5² + 20² = 21² 5² + 6² + 30² = 31² 6² + 7² + 42² = 43² Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
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Observe the following pattern and supply the missing numbers:
11² = 121 101² = 10201 10101² = 102030201 1010101² = 1020304030201 101010101² = 10203040504030201 Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
11² = 121
101² = 10201
10101² = 102030201
1010101² = 1020304030201
101010101² = 10203040504030201
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
Observe the following pattern and find the missing digits:
11² = 121 101² = 10201 1001² = 1002001 100001² = 10000200001 1000000² = 100000020000001 Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
11² = 121
101² = 10201
1001² = 1002001
100001² = 10000200001
1000000² = 100000020000001
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
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The squares of which of the following would be odd number: (i) 431 (ii) 2826 (iii) 7779 (iv) 82004
(i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number. (ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number. (iii) 7779 – Unit’s digit of given number is 9 and square of 9Read more
(i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number.
(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number.
(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81. Therefore, square of 7779 would be an odd number.
(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16. Therefore, square of 82004 would not be an odd number.
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
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The following numbers are obviously not perfect squares. Give reasons. (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722(vii) 222000 (viii) 505050
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0. Therefore 1057 is not a perfect square because its unit’s place digit is 7. (ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 234Read more
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.
Therefore 1057 is not a perfect square because its unit’s place digit is 7.
(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 23453 is not a perfect square because its unit’s place digit is 3.
(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 7928 is not a perfect square because its unit’s place digit is 8.
(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222222 is not a perfect square because its unit’s place digit is 2.
(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 64000 is not a perfect square because its unit’s place digit is single 0.
(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 89722 is not a perfect square because its unit’s place digit is 2.
(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.
(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 505050 is not a perfect square because its unit’s place digit is 0.
Class 8 Maths Chapter 6 Exercise 6.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/