Chapter 6 Squares and Square Roots Exercise 6.1
Question 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
Answer:
(The unit digit of square of a number having ‘a’ at its unit place ends with a × a.)
i. 1 × 1 = 1
ii. 2 × 2 = 4
iii. 9 × 9 = 81 = 1
iv. 3 × 3 = 9
v. 4 × 4 = 16 = 6
vi. 7 × 7 = 49 = 9
vii. 8 × 8 = 64 = 4
viii. 0 × 0 = 0
ix. 6 × 6 = 36 = 6
x. 5 × 5 = 25 = 5
Question 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
Answer:
(The natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.)
i. 1057 is not a perfect square because the unit digits of 1057 is 7 which is not a perfect square.
ii. 23453 is not a perfect square because the unit digits of 23453 is 3 which is not a perfect square.
iii. 7928 is not a perfect square because the unit digits of 7928 is 8 which is not a perfect square.
iv. 222222 is not a perfect square because the unit digits of 222222 is 2 which is not a perfect square.
v. 64000 is not a perfect square because the number contains three zeros which is not a perfect square.
vi. 89722 is not a perfect square because the unit digits of 89722 is 2 which is not a perfect square.
vii. 222000 is not a perfect square because the number contains three zeros which is not a perfect square.
viii. 505050 is not a perfect square because the unit digits of 505050 is 0 which is not a perfect square.
Question 3: The squares of which of the following would be odd numbers?
i. 431
ii. 2826
iii. 7779
iv. 82004
Answer:
(The square of an odd number is odd and the square of an even number is even.)
i. 431² is an odd number as unit digits of 431 is 1 which is 1 × 1 = 1 which is an odd number.
ii. 2826² is an even number as unit digits of 2826 is 6 which is 6 × 6 = 36 which is an even number.
iii. 7779² is an odd number as unit digits of 7779 is 9 which is 9 × 9 = 81 which is an odd number.
iv. 82004² is an even number as unit digits of 82004 is 4 which is 4 × 4 = 16 which is an even number.
Question 4: Observe the following pattern and find the missing numbers.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1 …….2………1
10000001² = ……………………..
Answer:
The pattern is if there is 1 zero, the one zero will appear on the left side of 2 and another one on the right side of the zero. So if there are 2 zeros in a number, 2 zeros will appear on the left side of 2 while the other 2 will appear on the right side of the 2. So,
100001² = 10000200001
10000001² = 100000020000001
Question 5: Observe the following pattern and supply the missing numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………………………
…………2 = 10203040504030201
Answer:
We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And, the square is symmetric about the middle digit. If the middle digit is 4, then the number to be squared is 10101 and its square is 102030201.
So,
10101012 =1020304030201
1010101012 =10203040505030201
Question 6: Using the given pattern, find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 122 = 13²
4² + 5² + _² = 21²
5² + _² + 30² = 31²
6² + 7² + _² = _²
Answer:
Given,
1² + 2² + 2² = 3²
= 1 + 4 + 4 = 9
= 9 = 9
2² + 3² + 6² = 7²
= 4 + 9 + 36 = 49
= 49 = 49
3² + 4² + 12² = 13²
= 9 + 16 + 144 = 169
= 169 = 169
4² + 5² + _² = 21²
= 16 + 25 + x² = 441
= 441 - 41 = 400
= √400 = 20²
5² + _² + 30² = 31²
= 25 + x² + 900 = 961
= 961 - 925 = 36
= √36 = 6
6² + 7² + _² = _²
= 36 + 49 + x² = y²
= 36 + 49 + (6 × 7)² = y²
= 36 + 49 + 1764 = y²
= 36 + 49 + 1764 = 1849
= √1849 = 43
Question 7: 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
Answer:
(The sum of the first n odd natural numbers is n².)
i.
Total numbers = 5
= 5² = 25
Therefore, the sum of 1 + 3 + 5 + 7 + 9 = 25.
ii.
Total numbers = 10
= 10² = 100
Therefore, the sum of 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100.
iii.
Total numbers = 12
= 12² = 144
Therefore, the sum of 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 144.
Question 8:
i) Express 49 as the sum of 7 odd numbers.
ii) Express 121 as the sum of 11 odd numbers.
Answer:
i)
= 49 - 1 - 3 - 5 - 7 - 9 - 11 - 13
= 0
ii)
= 121 - 1 - 3 - 5 - 7 - 9 - 11 - 13 - 15 - 17 - 19 - 21
= 0
Question 9: How many numbers lie between squares of the following numbers?
i. 12 and 13
ii. 25 and 26
iii. 99 and 100
Answer:
(Between any 2 perfect squares there are 2n non-perfect squares.)
i.
To find: non-perfect squares between 12 and 13
= 12 × 2
= 24
Therefore, there are 24 non-perfect squares between 12 and 13.
ii.
To find: non-perfect squares between 25 and 26
= 25 × 2
= 50
Therefore, there are 50 non-perfect squares between 25 and 26.
iii.
To find: non-perfect squares between 99 and 100
= 99 × 2
= 198
Therefore, there are 198 non-perfect squares between 99 and 100.
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