NCERT Solutions for Class 8 Maths: Squares and Square Roots
In NCERT Solution of maths of class 8 of chapter 6 Squares and Square Roots get worksheet,solve, questions and answers.
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 1:
If there is m at the unit place of a number than its square has the digit in unit place of product of m × m
(i) Science, 81 contain digit 1 in its unit place then its square has digit in unit place is (1 × 1)=1
(ii) 272 contain 2 in its unit place, so square of 272 has digit in unit place (2. × 2)= 4
(iii) There is 9 at unit place of 799 and product of (9×9)=81. Therefore square of 799 has 1 digit at unit place.
(iv) since there is 3 digit in unit place of 3853, hence it square contain (3×3) =9 at unit place.
(v) there is 4 at unit place of 1234 ant product of (4×4)=16. Therefore the square of 1234 has 6 digit in its unit place
(vi) there is 7 at unit place of 26387 and product of (7×7)=49. therefore the square of the number 26387 has the digit in unit place is 9
(vii) there is a digit 8 at unit place of 52698 and product of (8×8)=64. Therefore the digit at unit place of 52698 is 4.
(viii) there is a digit 0 at unit place of 99880 and product of (0×0)=0. Therefore the digit at unit place of square of 9 9880 is zero.
(ix) there is 6 at unit place of the number 12796 and product of (6×6)=36. Therefore the digit at unit place of 12796 is 6.
(x) there is the digit 5 at unit place of the number 55555 and product of (5×5)=25. Therefore the digit at unit place of 55555 is 5.
Question 2:
The following numbers are obviously not perfect squares. Give reasons.
(i) 1057 (ii) 23453(iii) 7928 (iv) 222222(v) 64000 (vi) 89722(vii) 222000 (viii) 505050
Answer 2:
(i) a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 7 at unit place of 1057, so it is not a perfect square number
(ii) a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 3 at unit place of 23453, so it is not a perfect square number
(iii)a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 8 at unit place of 7928, so it is not a perfect square number
(iv) a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 2 at unit place of 222222, so it is not a perfect square number
(v)a perfect square number has digit at unit place are 1,4,5,6,9 and has even number of 0 digit. Since there is odd number of 0 digit at unit place of 64000, so it is not a perfect square number
(vi) a perfect square number has digit at unit place are 1,4,5,6,9 and even number of 0. Since there is 2 at unit place of 89722, so it is not a perfect square number
(vii)a perfect square number has digit at unit place are 1,4,5,6,9 and has even number of 0 digit. Since there is odd number of 0 digit at unit place of 222000, so it is not a perfect square.
(viii) a perfect square number has digit at unit place are 1,4,5,6,9 and has even number of 0 digit. Since there is odd number of 0 digit at unit place of 505050, so it is not a perfect square number
Question 3:
The squares of which of the following would be odd number:
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Answer 3:
Square of an odd number is also an odd number and square of an even number is also an even number.
Sales 431 and 7779 are odd number hence the square is also odd number.
visit for more information
Question 4:
Observe the following pattern and find the missing digits:
11² = 121
101² = 10201
1001² = 1002001
100001² = 1…….2…….1
1000000² = 1……………………
Answer 4:
11² = 121
101² = 10201
1001² = 1002001
100001² = 10000200001
1000000² = 100000020000001
Question 5:
Observe the following pattern and supply the missing numbers:
11² = 121
101²= 10201
10101² = 102030201
1010101²= ………………………
……………………….² = 10203040504030201
Answer 5:
11² = 121
101²= 10201
10101² = 102030201
1010101² = 1020304030201
101010101² = 10203040504030201
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 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
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 7:
We know that sum of first n natural odd number =n²
(i) therefore, sum of first 5 odd natural numbers = 5² = 25
1 + 3 + 5 + 7 + 9 = 52 = 25
(ii) sum of first 10 odd natural numbers = 10² = 100
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
(iii) sum of first 12 odd natural numbers = (12)² = 144.
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144
Question 8:
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer 8:
(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
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:
Since there is 2n numbers present between squares of n and (n+1)
(i) the numbers present in between 12² and 13² = (2×12) = 24.
(ii) the numbers present in between 25² and 26² = (2×25) = 50
(III) the numbers present in between 99² and 100² = (2×99) = 198
For More NCERT Solutions for Class 8 Maths: Squares and Square Roots
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 1:
If there is m at the unit place of a number than its square has the digit in unit place of product of m × m
(i) Science, 81 contain digit 1 in its unit place then its square has digit in unit place is (1 × 1)=1
(ii) 272 contain 2 in its unit place, so square of 272 has digit in unit place (2. × 2)= 4
(iii) There is 9 at unit place of 799 and product of (9×9)=81. Therefore square of 799 has 1 digit at unit place.
(iv) since there is 3 digit in unit place of 3853, hence it square contain (3×3) =9 at unit place.
(v) there is 4 at unit place of 1234 ant product of (4×4)=16. Therefore the square of 1234 has 6 digit in its unit place
(vi) there is 7 at unit place of 26387 and product of (7×7)=49. therefore the square of the number 26387 has the digit in unit place is 9
(vii) there is a digit 8 at unit place of 52698 and product of (8×8)=64. Therefore the digit at unit place of 52698 is 4.
(viii) there is a digit 0 at unit place of 99880 and product of (0×0)=0. Therefore the digit at unit place of square of 9 9880 is zero.
(ix) there is 6 at unit place of the number 12796 and product of (6×6)=36. Therefore the digit at unit place of 12796 is 6.
(x) there is the digit 5 at unit place of the number 55555 and product of (5×5)=25. Therefore the digit at unit place of 55555 is 5.
Question 2:
The following numbers are obviously not perfect squares. Give reasons.
(i) 1057 (ii) 23453(iii) 7928 (iv) 222222(v) 64000 (vi) 89722(vii) 222000 (viii) 505050
Answer 2:
(i) a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 7 at unit place of 1057, so it is not a perfect square number
(ii) a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 3 at unit place of 23453, so it is not a perfect square number
(iii)a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 8 at unit place of 7928, so it is not a perfect square number
(iv) a perfect square number has digit at unit place are 1,4,5,6,9 and 0. Since there is 2 at unit place of 222222, so it is not a perfect square number
(v)a perfect square number has digit at unit place are 1,4,5,6,9 and has even number of 0 digit. Since there is odd number of 0 digit at unit place of 64000, so it is not a perfect square number
(vi) a perfect square number has digit at unit place are 1,4,5,6,9 and even number of 0. Since there is 2 at unit place of 89722, so it is not a perfect square number
(vii)a perfect square number has digit at unit place are 1,4,5,6,9 and has even number of 0 digit. Since there is odd number of 0 digit at unit place of 222000, so it is not a perfect square.
(viii) a perfect square number has digit at unit place are 1,4,5,6,9 and has even number of 0 digit. Since there is odd number of 0 digit at unit place of 505050, so it is not a perfect square number
Question 3:
The squares of which of the following would be odd number:
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Answer 3:
Square of an odd number is also an odd number and square of an even number is also an even number.
Sales 431 and 7779 are odd number hence the square is also odd number.
visit for more information
Question 4:
Observe the following pattern and find the missing digits:
11² = 121
101² = 10201
1001² = 1002001
100001² = 1…….2…….1
1000000² = 1……………………
Answer 4:
11² = 121
101² = 10201
1001² = 1002001
100001² = 10000200001
1000000² = 100000020000001
Question 5:
Observe the following pattern and supply the missing numbers:
11² = 121
101²= 10201
10101² = 102030201
1010101²= ………………………
……………………….² = 10203040504030201
Answer 5:
11² = 121
101²= 10201
10101² = 102030201
1010101² = 1020304030201
101010101² = 10203040504030201
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 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
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 7:
We know that sum of first n natural odd number =n²
(i) therefore, sum of first 5 odd natural numbers = 5² = 25
1 + 3 + 5 + 7 + 9 = 52 = 25
(ii) sum of first 10 odd natural numbers = 10² = 100
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
(iii) sum of first 12 odd natural numbers = (12)² = 144.
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144
Question 8:
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer 8:
(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
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:
Since there is 2n numbers present between squares of n and (n+1)
(i) the numbers present in between 12² and 13² = (2×12) = 24.
(ii) the numbers present in between 25² and 26² = (2×25) = 50
(III) the numbers present in between 99² and 100² = (2×99) = 198
For More NCERT Solutions for Class 8 Maths: Squares and Square Roots
