(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
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/
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
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
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
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
(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
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
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
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/