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