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
Solution :
(i) The number 81 contains its unit’s place digit 1. So, square of 1 is 1.
Hence, unit’s digit of square of 81 is 1.
(ii) The number 272 contains its unit’s place digit 2. So, square of 2 is 4.
Hence, unit’s digit of square of 272 is 4.
(iii) The number 799 contains its unit’s place digit 9. So, square of 9 is 81.
Hence, unit’s digit of square of 799 is 1.
(iv) The number 3853 contains its unit’s place digit 3. So, square of 3 is 9.
Hence, unit’s digit of square of 3853 is 9.
(v) The number 1234 contains its unit’s place digit 4. So, square of 4 is 16.
Hence, unit’s digit of square of 1234 is 6.
(vi) The number 26387 contains its unit’s place digit 7. So, square of 7 is 49.
Hence, unit’s digit of square of 26387 is 9.
(vii) The number 52698 contains its unit’s place digit 8. So, square of 8 is 64.
Hence, unit’s digit of square of 52698 is 4.
(viii) The number 99880 contains its unit’s place digit 0. So, square of 0 is 0.
Hence, unit’s digit of square of 99880 is 0.
(ix) The number 12796 contains its unit’s place digit 6. So, square of 6 is 36.
Hence, unit’s digit of square of 12796 is 6.
(x) The number 55555 contains its unit’s place digit 5. So, square of 5 is 25.
Hence, unit’s digit of square 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
Solution :
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.
Therefore 1057 is not a perfect square because its unit’s place digit is 7.
(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 23453 is not a perfect square because its unit’s place digit is 3.
(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 7928 is not a perfect square because its unit’s place digit is 8.
(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222222 is not a perfect square because its unit’s place digit is 2.
(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 64000 is not a perfect square because its unit’s place digit is single 0.
(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 89722 is not a perfect square because its unit’s place digit is 2.
(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.
Question 3.
The squares of which of the following would be odd number:
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution :
(i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number.
(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number.
(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81. Therefore, square of 7779 would be an odd number.
(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16. Therefore, square of 82004 would not be an odd number.
Question 4.
Observe the following pattern and find the missing digits:
11² = 121
101²= 10201
1001²= 1002001
10001²= 1…….2…….1
100001²= 1……………………
Solution :
11²= 121
101²= 10201
1001²= 1002001
10001²= 10000200001
10000001²= 100000020000001
Question 5.
Observe the following pattern and supply the missing numbers:
11²= 121
101²= 10201
10101²= 102030201
1010101²= ………………………
10101011²= 10203040504030201
Solution :
11² = 121
101² = 10201
10101² = 102030201
1010101² = 1020304030201
10101010²= 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 + _² + _² = 7²
Solution :
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21²
5² + _² + 30² = 31²
6² + 7² + 42² = 7²
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
Solution :
(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
Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution :
(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
Solution :
(i) Since, non-perfect square numbers between n² and (n + 1)² vare 2n
Here, n= 12
Therefore, non-perfect square numbers between 12 and 13 =n2 =(n + 1)² = 24
(ii) Since, non-perfect square numbers between and are
Here, n= 25
Therefore, non-perfect square numbers between 25 and 26 =2n =2× 12= 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 = 2×99 = 198
NCERT Solutions for Class 8 Maths Exercise 6.2
Question 1.
Find the squares of the following numbers:
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Solution :
= 900 + 300 + 25 = 1225
= 4900 + 140 + 1 = 5041
= 1600 + 480 + 36 = 2116
Question 2.
Write a Pythagoras triplet whose one member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution :
NCERT Solutions for Class 8 Maths Exercise 6.3
Question 1.
What could be the possible ‘one’s’ digits of the square root of each of the following numbers:
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution :
Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:
(i) 1 (ii) 6 (iii) 1 (iv) 5
Question 2.
Without doing any calculation, find the numbers which are surely not perfect squares:
(i) 153
(ii) 257
(iii) 408
(iv) 441
Solution :
Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) But given number 153 has its unit digit 3. So it is not a perfect square number.
(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.
(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.
(iv) Given number 441 has its unit digit 1. So it would be a perfect square number
Question 3.
Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution :
By successive subtracting odd natural numbers from 100,
100 – 1 = 99
99 – 3 = 96
96 – 5 = 91
91 – 7 = 84
84 – 9 = 75
75 – 11 = 64
64 – 13 = 51
51 – 15 = 36
36 – 17 = 19
19 – 19 = 0
This successive subtraction is completed in 10 steps.
Therefore √100 =10
By successive subtracting odd natural numbers from 169,
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0
This successive subtraction is completed in 13 steps.
Therefore √169 =13
Question 4.
Find the square roots of the following numbers by the Prime Factorization method:
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
Solution :
(i) 729
=3x3x3
= 27
(ii) 400
(iii) 1764
(iv) 4096
= 64
(v) 7744
= 88
(vi) 9604
= 98
(vii) 5929
(viii) 9216
= 96
(ix) 529
= 23
(x) 8100
= 90
Question 5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained:
(i) 252 (ii) 180
(iii) 1008 (iv) 2028
(v) 1458 (vi) 768
Solution :
(i) 252 =
Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.
= 1764
And (i)= 2×3×7= 42
(ii) 180 = 2×2×3×3×5
Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.
∴180 x 5 = 900
And
= 30
(iii) 1008 =
Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.
∴ 1008 × 7 = 7056
And
= 84
(iv) 2028 =
Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.
∴ 2028 × 3= 6084
And
= 78
(v) 1458 =
Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.
= 2916
And
= 54
(vi) 768 =
Here, prime factor 3 has no pair. Therefore 768 must be multiplied by 3 to make it a perfect square.
∴ 768 × 3= 2304
And
= 48
NCERT Solutions for Class 8 Maths Exercise 6.4
Question 1.
Find the square roots of each of the following numbers by Division method:
(i) 2304 (ii) 4489
(iii) 3481 (iv) 529
(v) 3249 (vi) 1369
(vii) 5776 (viii) 7921
(ix) 576 (x) 1024
(xi) 3136 (xii) 900
Solution :
(i) 2304
Hence, the square root of 2304 is 48.
(ii) 4489
Hence, the square root of 4489 is 67.
(iii) 3481
Hence, the square root of 3481 is 59.
(iv) 529
Hence, the square root of 529 is 23.
(v) 3249
Hence, the square root of 3249 is 57.
(vi) 1369
Hence, the square root of 1369 is 37.
(vii) 5776
Hence, the square root of 5776 is 76.
(viii) 7921
Hence, the square root of 7921 is 89.
(ix) 576
Hence, the square root of 576 is 24.
(x) 1024
Hence, the square root of 1024 is 32.
(xi) 3136
Hence, the square root of 3136 is 56.
(xii) 900
Hence, the square root of 900 is 30.
Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation):
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625
Solution :
(i) Here, 64 contains two digits which is even.
Therefore, number of digits in square root = 
(ii) Here, 144 contains three digits which is odd.
Therefore, number of digits in square root = 
(iii) Here, 4489 contains four digits which is even.
Therefore, number of digits in square root = 
(iv) Here, 27225 contains fivr digits which is odd.
Therefore, number of digits in square root = 
(v) Here, 390625 contains six digits which is even.
Therefore, number of digits in square root = 
Question 3.
Find the square root of the following decimal numbers:
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution :
(i) 2.56
Hence, the square root of 2.56 is 1.6.
(ii) 7.29
Hence, the square root of 7.29 is 2.7.
(iii) 51.84
Hence, the square root of 51.84 is 7.2.
(iv) 42.25
Hence, the square root of 42.25 is 6.5.
(v) 31.36
Hence, the square root of 31.36 is 5.6.
Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained:
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000
Solution :
(i) 402
We know that, if we subtract the remainder from the number, we get a perfect square.
Here, we get remainder 2. Therefore 2 must be subtracted from 402 to get a perfect square.
∴ 402 – 2 = 400
Hence, the square root of 400 is 20.
(ii) 1989
We know that, if we subtract the remainder from the number, we get a perfect square.
Here, we get remainder 53. Therefore 53must be subtracted from 1989 to get a perfect square.
∴ 1989 – 53 = 1936
Hence, the square root of 1936 is 44.
(iii) 3250
We know that, if we subtract the remainder from the number, we get a perfect square.
Here, we get remainder 1. Therefore 1 must be subtracted from 3250 to get a perfect square.
∴3250 – 1 = 3249
Hence, the square root of 3249 is 57.
(iv) 825
We know that, if we subtract the remainder from the number, we get a perfect square.
Here, we get remainder 41. Therefore 41 must be subtracted from 825 to get a perfect square.
∴ 825 – 41 = 784
Hence, the square root of 784 is 28.
(v) 4000
We know that, if we subtract the remainder from the number, we get a perfect square.
Here, we get remainder 31. Therefore 31 must be subtracted from 4000 to get a perfect square.
∴ 4000 – 31 = 3969
Hence, the square root of 3969 is 63.
Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained:
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412
Solution :
(i) 525
Since remainder is 41.
Therefore 
Next perfect square number 23²= 529
Hence, number to be added
= 529 – 525 = 4
∴ 525 + 4 = 529
Hence, the square root of 529 is 23.
(ii) 1750
Since remainder is 69.
Therefore 
Next perfect square number 42² = 1764
Hence, number to be added
= 1764 – 1750 = 14
∴ 1750 + 14 = 1764
Hence, the square root of 1764 is 42.
(iii) 252
Since remainder is 27.
Therefore 
Next perfect square number 16² = 256
Hence, number to be added
= 256 – 252 = 4
∴ 252 + 4 = 256
Hence, the square root of 256 is 16.
(iv) 1825
Since remainder is 61.
Therefore 
Next perfect square number 43² = 1849
Hence, number to be added = 1849 – 1825 = 24
∴ 1825 + 24 = 1849
Hence, the square root of 1849 is 43.
(v) 6412
Since remainder is 12.
Therefore 80² < 6412
Next perfect square number 81²= 6561
Hence, number to be added
= 6561 – 6412 = 149
∴ 6412 + 149 = 6561
Hence, the square root of 6561 is 81.
Question 6.
Find the length of the side of a square whose area is
?
Solution :
Let the length of side of a square be x meter.
Area of square 
According to question,
Hence, the length of side of a square is 21 m.
Question 7.
In a right triangle ABC,∠B = 90º
(i) If AB = 6 cm, BC = 8 cm, find AC.
(ii) If AC = 13 cm, BC = 5 cm, find AB.
Solution :
(a) Using Pythagoras theorem,
(b) Using Pythagoras theorem,
Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and number of columns remain same. Find the minimum number of plants he needs more for this.
Solution :
Here, plants = 1000
Since remainder is 39.
Therefore 31² < 1000
Next perfect square number 32² = 1024
Hence, number to be added
= 1024 – 1000 = 24
∴ 1000 + 24 = 1024
Hence, the gardener required 24 more plants.
Question 9.
There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?
Solution :
Here, Number of children = 500
By getting the square root of this number, we get,
In each row, the number of children is 22.
And left out children are 16.
