Ncert math solution of class 8 ch-6 (Square and Square Roots)
Answers
Answer:
What will be the unit digit of the squares of the following numbers?
i. 81
ii. 272
iii. 799
iv. 3853
v. 1234
vi. 26387
vii. 52698
viii. 99880
ix. 12796
x. 55555
Solution:
The unit digit of square of a number having ‘a’ at its unit place ends with a×a.
i. The unit digit of the square of a number having digit 1 as unit’s place is 1.
∴ Unit digit of the square of number 81 is equal to 1.
ii. The unit digit of the square of a number having digit 2 as unit’s place is 4.
∴ Unit digit of the square of number 272 is equal to 4.
iii. The unit digit of the square of a number having digit 9 as unit’s place is 1.
∴ Unit digit of the square of number 799 is equal to 1.
iv. The unit digit of the square of a number having digit 3 as unit’s place is 9.
∴ Unit digit of the square of number 3853 is equal to 9.
v. The unit digit of the square of a number having digit 4 as unit’s place is 6.
∴ Unit digit of the square of number 1234 is equal to 6.
vi. The unit digit of the square of a number having digit 7 as unit’s place is 9.
∴ Unit digit of the square of number 26387 is equal to 9.
vii. The unit digit of the square of a number having digit 8 as unit’s place is 4.
∴ Unit digit of the square of number 52698 is equal to 4.
viii. The unit digit of the square of a number having digit 0 as unit’s place is 01.
∴ Unit digit of the square of number 99880 is equal to 0.
ix. The unit digit of the square of a number having digit 6 as unit’s place is 6.
∴ Unit digit of the square of number 12796 is equal to 6.
x. The unit digit of the square of a number having digit 5 as unit’s place is 5.
∴ Unit digit of the square of number 55555 is equal to 5
hope it helps
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1. What will be the unit digit of the squares of the following numbers?
i. 81
ii. 272
iii. 799
iv. 3853
v. 1234
vi. 26387
vii. 52698
viii. 99880
ix. 12796
x. 55555
Solution:
The unit digit of square of a number having ‘a’ at its unit place ends with a×a.
i. The unit digit of the square of a number having digit 1 as unit’s place is 1.
∴ Unit digit of the square of number 81 is equal to 1.
ii. The unit digit of the square of a number having digit 2 as unit’s place is 4.
∴ Unit digit of the square of number 272 is equal to 4.
iii. The unit digit of the square of a number having digit 9 as unit’s place is 1.
∴ Unit digit of the square of number 799 is equal to 1.
iv. The unit digit of the square of a number having digit 3 as unit’s place is 9.
∴ Unit digit of the square of number 3853 is equal to 9.
v. The unit digit of the square of a number having digit 4 as unit’s place is 6.
∴ Unit digit of the square of number 1234 is equal to 6.
vi. The unit digit of the square of a number having digit 7 as unit’s place is 9.
∴ Unit digit of the square of number 26387 is equal to 9.
vii. The unit digit of the square of a number having digit 8 as unit’s place is 4.
∴ Unit digit of the square of number 52698 is equal to 4.
viii. The unit digit of the square of a number having digit 0 as unit’s place is 01.
∴ Unit digit of the square of number 99880 is equal to 0.
ix. The unit digit of the square of a number having digit 6 as unit’s place is 6.
∴ Unit digit of the square of number 12796 is equal to 6.
x. The unit digit of the square of a number having digit 5 as unit’s place is 5.
∴ Unit digit of the square of number 55555 is equal to 5.
2. The following numbers are obviously not perfect squares. Give reason.
i. 1057
ii. 23453
iii. 7928
iv. 222222
v. 64000
vi. 89722
vii. 222000
viii. 505050
Solution:
We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.
i. 1057 ⟹ Ends with 7
ii. 23453 ⟹ Ends with 3
iii. 7928 ⟹ Ends with 8
iv. 222222 ⟹ Ends with 2
v. 64000 ⟹ Ends with 0
vi. 89722 ⟹ Ends with 2
vii. 222000 ⟹ Ends with 0
viii. 505050 ⟹ Ends with 0
42 + 52 + _2 = 212
5 + _ 2 + 302 = 312
6 + 7 + _ 2 = _ 2
Solution:
Given, 12 + 22 + 22 = 32
i.e 12 + 22 + (1×2 )2 = ( 12 + 22 -1 × 2 )2
22 + 32 + 62 =72
∴ 22 + 32 + (2×3 )2 = (22 + 32 -2 × 3)2
32 + 42 + 122 = 132
∴ 32 + 42 + (3×4 )2 = (32 + 42 – 3 × 4)2
42 + 52 + (4×5 )2 = (42 + 52 – 4 × 5)2
∴ 42 + 52 + 202 = 212
52 + 62 + (5×6 )2 = (52+ 62 – 5 × 6)2
∴ 52 + 62 + 302 = 312
62 + 72 + (6×7 )2 = (62 + 72 – 6 × 7)2
∴ 62 + 72 + 422 = 432
7. Without adding, find the sum.
i. 1 + 3 + 5 + 7 + 9
Solution:
Sum of first five odd number = (5)2 = 25
ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
Solution:
Sum of first ten odd number = (10)2 = 100
iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Sum of first thirteen odd number = (12)2 = 144
8. (i) Express 49 as the sum of 7 odd numbers.
Solution:
We know, sum of first n odd natural numbers is n2 . Since,49 = 72
∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers. Solution:
Since, 121 = 112
∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. How many numbers lie between squares of the following numbers?
i. 12 and 13
ii. 25 and 26
iii. 99 and 100
Solution:
Between n2 and (n+1)2, there are 2n non–perfect square numbers.
i. 122 and 132 there are 2×12 = 24 natural numbers.
ii. 252 and 262 there are 2×25 = 50 natural numbers.
iii. 992 and 1002 there are 2×99 =198 natural numbers.