NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Exercise 5.1
1. What will be the unit digit of the squares of the following numbers?
(i) 81
Ans: `\because` `1^2`=1
`\therefore` unit digit of `81^2` will be 1
(ii) 272
Ans: `\because` `2^2`=4
`\therefore` unit digit of `272^2` will be 4
(iii) 799
Ans: `\because` `9^2`=81
& unit digit of 81 is 1
`\therefore` unit digit of `799^2` will be 1
(iv) 3853
Ans: `\because` `3^2`=9
`\therefore` unit digit of `3853^2` will be 9
(v) 1234
Ans: `\because` `4^2`=16
& unit digit of 16 is 6
`\therefore` unit digit of `1234^2` will be 6
(vi) 26387
Ans: `\because` `7^2`=49
& unit digit of 49 is 9
`\therefore` unit digit of `26387^2` will be 9
(vii) 52698
Ans: `\because` `8^2`=64
& unit digit of 64 is 4
`\therefore` unit digit of `52698^2` will be 4
(viii) 99880
Ans: `\because` `0^2`=0
`\therefore` unit digit of `99880^2` will be 0
(ix) 12796
Ans: `\because` `6^2`=36
& unit digit of 36 is 6
`\therefore` unit digit of `12796^2` will be 6
(x) 55555
Ans: `\because` `5^2`=25
& unit digit of 25 is 5
`\therefore` unit digit of `55555^2` will be 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
Ans: A number having 2,3,7 or 8 at unit's place is never a perfect square.
`\therefore` (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 and (vi) 89722 are not perfect squares.
Also numbers ending in an odd number of zeros is never a perfect square
`\therefore` (v) 64000 (vii) 222000 and (viii) 505050 having 3,3 and 1 zeroes at the end are not perfect squares.
3. The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Ans: Squares of (i) 431 and (iii) 7779 would be odd number because square of only odd numbers are odd.
4. Observe the following pattern and find the missing digits.
`11^2` = 121
`101^2` = 10201
`1001^2` = 1002001
`100001^2` = 1 ......... 2 ......... 1
`10000001^2` = ...........................
Ans: `100001^2` = 10000200001
`10000001^2` = 100000020000001
5. Observe the following pattern and supply the missing numbers.
`11^2` = 1 2 1
`101^2` = 1 0 2 0 1
`10101^2` = 102030201
`1010101^2` = ...........................
`............^2` = 10203040504030201
Ans: `1010101^2` = 1020304030201
`101010101^2` = 10203040504030201
6. Using the given pattern, find the missing numbers.
`1^2` + `2^2` + `2^2` = `3^2`
`2^2` + `3^2` + `6^2` = `7^2`
`3^2` + `4^2` + `12^2` = `13^2`
`4^2` + `5^2` + `_^2` = `21^2`
`5^2` + `_^2` + `30^2` = `31^2`
`6^2` + `7^2` + `_^2` = `_^2`
Ans: `4^2` + `5^2` + `20^2` = `21^2`
`5^2` + `6^2` + `30^2` = `31^2`
`6^2` + `7^2` + `42^2` = `43^2`
7. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
Ans: sum of first n odd natural numbers is `n^2`
`\therefore` sum of first 5 odd natural numbers = `5^2` = 25
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
Ans: sum of first 10 odd natural numbers = `10^2` = 100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Ans: sum of first 12 odd natural numbers = `12^2` = 144
8. (i) Express 49 as the sum of 7 odd numbers.
Ans: 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers.
Ans: 121 = 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
Ans: between `n^2` and `(n + 1)^2` there are 2n numbers
`\therefore` between `12^2` and `13^2` there are 2×12 = 24 numbers
(ii) 25 and 26
Ans: between `25^2` and `26^2` there are 2×25 = 50 numbers
(iii) 99 and 100
Ans: between `12^2` and `13^2` there are 2×99 = 198 numbers
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