Home » class 8 Maths » NCERT Solutions for Class 8 Maths- NCERT Solution Chapter 6 – Squares and Square Roots

NCERT Solutions for Class 8 Maths- NCERT Solution Chapter 6 – Squares and Square Roots


Exercise 6.1 : Solutions of Questions on Page Number : 96


Q1 : 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 :
We know that if a number has its unit’s place digit as a, then its square will end with the unit digit of the multiplication a x a.
(i) 81
Since the given number has its unit’s place digit as 1, its square will end with the unit digit of the multiplication (1 x 1 = 1) i.e., 1.
(ii) 272
Since the given number has its unit’s place digit as 2, its square will end with the unit digit of the multiplication (2 x 2 = 4) i.e., 4.
(iii) 799
Since the given number has its unit’s place digit as 9, its square will end with the unit digit of the multiplication (9 x 9 = 81) i.e., 1.
(iv) 3853
Since the given number has its unit’s place digit as 3, its square will end with the unit digit of the multiplication (3 x 3 = 9) i.e., 9.
(v) 1234
Since the given number has its unit’s place digit as 4, its square will end with the unit digit of the multiplication (4 x 4 = 16) i.e., 6.
(vi) 26387
Since the given number has its unit’s place digit as 7, its square will end with the unit digit of the multiplication (7 x 7 = 49) i.e., 9.
(vii) 52698
Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (8 x 8 = 64) i.e., 4.
(viii) 99880
Since the given number has its unit’s place digit as 0, its square will have two zeroes at the end. Therefore, the unit digit of the square of the given number is 0.
(xi) 12796
Since the given number has its unit’s place digit as 6, its square will end with the unit digit of the multiplication (6 x 6 = 36) i.e., 6.
(x) 55555
Since the given number has its unit’s place digit as 5, its square will end with the unit digit of the multiplication (5
x 5 = 2
5) i.e., 5


Q2 : 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 square of numbers may end with any one of the digits 0, 1, 5, 6, or 9. Also, a perfect square has even number of zeroes at the end of it.
(i) 1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.
(ii) 23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.
(iii) 7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.
(iv) 222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.
(v) 64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.
(vi) 89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.
(vii) 222000 has three zeroes at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.
(viii) 505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.


Q3 : 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. Here, 431 and 7779 are odd numbers.
Thus, the square of 431 and 7779 will be an odd number.


Q4 :  Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1…2…1
100000012 = …
Answer :
In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,
1000012 = 10000200001
100000012 = 100000020000001


Q5 : Observe the following pattern and supply the missing number.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = …
2 = 10203040504030201

Answer :
By following the given pattern, we obtain
10101012 = 1020304030201
1010101012 = 10203040504030201


Q6 : 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 :
From the given pattern, it can be observed that,
(i) The third number is the product of the first two numbers.
(ii) The fourth number can be obtained by adding 1 to the third number.
Thus, the missing numbers in the pattern will be as follows.
42 + 52 + = 212
52 + + 302 = 312
62 + 72 + =


Q7 : 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 :
We know that the sum of first n odd natural numbers is n2.
(i) Here, we have to find the sum of first five odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 = (5)2 = 25
(ii) Here, we have to find the sum of first ten odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100
(iii) Here, we have to find the sum of first twelve odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)2 = 144


Q8 : (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11odd numbers.
Answer :
We know that the sum of first n odd natural numbers is n2.
(i) 49 = (7)2
Therefore, 49 is the sum of first 7 odd natural numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = (11)2
Therefore, 121 is the sum of first 11 odd natural numbers.
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21


Q9 :  How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Answer :
We know that there will be 2n numbers in between the squares of the numbers n and (n + 1).
(i) Between 122 and 132, there will be 2 x 12 = 24 numbers
(ii) Between 252 and 262, there will be 2 x 25 = 50 numbers
(iii) Between 992 and 1002, there will be 2 x 99 = 198 numbers


Exercise 6.2 : Solutions of Questions on Page Number : 98


Q1 : Find the square of the following numbers
(i) 32 (ii) 35
(iii) 86 (iv) 93
(v) 71 (vi) 46
Answer :
(i) 322 = (30 + 2)2
= 30 (30 + 2) + 2 (30 + 2)
= 302 + 30 x 2 + 2 x 30 + 22
= 900 + 60 + 60 + 4
= 1024
(ii) The number 35 has 5 in its unit’s place. Therefore,
352 = (3) (3 + 1) hundreds + 25
= (3 x 4) hundreds + 25
= 1200 + 25 = 1225
(iii) 862 = (80 + 6)2
= 80 (80 + 6) + 6 (80 + 6)
= 802 + 80 x 6 + 6 x 80 + 62
= 6400 + 480 + 480 + 36
= 7396
(iv) 932 = (90 + 3)2
= 90 (90 + 3) + 3 (90 + 3)
= 902 + 90 x 3 + 3 x 90 + 32
= 8100 + 270 + 270 + 9
= 8649
(v) 712 = (70 + 1)2
= 70 (70 + 1) + 1 (70 + 1)
= 702 + 70 x 1 + 1 x 70 + 12
= 4900 + 70 + 70 + 1
= 5041
(vi) 462 = (40 + 6)2
= 40 (40 + 6) + 6 (40 + 6)
= 402 + 40 x 6 + 6 x 40 + 62
= 1600 + 240 + 240 + 36
= 2116


Q2 :  Write a Pythagorean triplet whose one member is
(i) 6 (ii) 14
(iii) 16 (iv) 18

Answer :
For any natural number m > 1, 2m, m2 – 1, m2 + 1 forms a Pythagorean triplet.
(i) If we take m2 + 1 = 6, then m2 = 5
The value of m will not be an integer.
If we take m2 – 1 = 6, then m2 = 7
Again the value of m is not an integer.
Let 2m = 6
m = 3
Therefore, the Pythagorean triplets are 2 x 3, 32 – 1, 32 + 1 or 6, 8, and 10.
(ii) If we take m2 + 1 = 14, then m2 = 13
The value of m will not be an integer.
If we take m2 – 1 = 14, then m2 = 15
Again the value of m is not an integer.
Let 2m = 14
m = 7
Thus, m2 – 1 = 49 – 1 = 48 and m2 + 1 = 49 + 1 = 50
Therefore, the required triplet is 14, 48, and 50.
(iii) If we take m2 + 1 = 16, then m2 = 15
The value of m will not be an integer.
If we take m2 – 1= 16, then m2 = 17
Again the value of m is not an integer.
Let 2m = 16
m = 8
Thus, m2 – 1 = 64 – 1 = 63 and m2 + 1 = 64 + 1 = 65
Therefore, the Pythagorean triplet is 16, 63, and 65.
(iv) If we take m2 + 1 = 18,
m2 = 17
The value of m will not be an integer.
If we take m2 – 1 = 18, then m2 = 19
Again the value of m is not an integer.
Let 2m =18
m = 9
Thus, m2 – 1 = 81 – 1 = 80 and m2 + 1 = 81 + 1 = 82
Therefore, the Pythagorean triplet is 18, 80, and 82.


Exercise 6.3 : Solutions of Questions on Page Number : 102


Q1 : 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
Answer :
(i) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.
(ii) If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.
(iii) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.
(iv) If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.


Q2 : Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153 (ii) 257
(iii) 408 (iv) 441
Answer :
The perfect squares of a number can end with any of the digits 0, 1, 4, 5, 6, or 9 at unit’s place. Also, a perfect square will end with even number of zeroes, if any.
(i) Since the number 153 has its unit’s place digit as 3, it is not a perfect square.
(ii) Since the number 257 has its unit’s place digit as 7, it is not a perfect square.
(iii) Since the number 408 has its unit’s place digit as 8, it is not a perfect square.
(iv) Since the number 441 has its unit’s place digit as 1, it is a perfect square.


Q3 : Find the square roots of 100 and 169 by the method of repeated subtraction.
Answer :
We know that the sum of the first n odd natural numbers is n2.
Consider.
(i) 100 – 1 = 99 (ii) 99 – 3 = 96 (iii) 96 – 5 = 91
(iv) 91 – 7 = 84 (v) 84 – 9 = 75 (vi) 75 – 11= 64
(vii) 64 – 13 = 51 (viii) 51 – 15 = 36 (ix) 36 – 17 = 19
(x) 19 – 19 = 0
We have subtracted successive odd numbers starting from 1 to 100, and obtained 0 at 10th step.
Therefore,
The square root of 169 can be obtained by the method of repeated subtraction as follows.
(i) 169 – 1 = 168 (ii) 168 – 3 = 165 (iii) 165 – 5 = 160
(iv) 160 – 7 = 153 (v) 153 – 9 = 144 (vi) 144 – 11 = 133
(vii) 133 – 13 = 120 (viii) 120 – 15 = 105 (ix) 105 – 17 = 88
(x) 88 – 19 = 69 (xi) 69 – 21 = 48 (xii) 48 – 23 = 25
(xiii)25 – 25 = 0
We have subtracted successive odd numbers starting from 1 to 169, and obtained 0 at 13th step
Therefore,


Q4 : Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729 (ii) 400
(iii) 1764 (iv) 4096
(v) 7744 (vi) 9604
(vii) 5929 (viii) 9216
(ix) 529 (x) 8100
Answer :




Q5 : 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
Answer:




Q6 : For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square number. Also find the square root of the square number so obtained.
(i) 252 (ii) 2925
(iii) 396 (iv) 2645
(v) 2800 (vi) 1620
Answer :



Q7 : The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Answer :
It is given that each student donated as many rupees as the number of students of the class. Number of students in the class will be the square root of the amount donated by the students of the class.
The total amount of donation is Rs 2401.
Number of students in the class =


Hence, the number of students in the class is 49.


Q8 : 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Answer :
It is given that in the garden, each row contains as many plants as the number of rows.
Hence,
Number of rows = Number of plants in each row
Total number of plants = Number of rows × Number of plants in each row
Number of rows × Number of plants in each row = 2025
(Number of rows)2 = 2025


Thus, the number of rows and the number of plants in each row is 45.


Q9 :
Find the smallest square number that is divisible by each of the numbers 4, 9, and 10.
Answer :
The number that will be perfectly divisible by each one of 4, 9, and 10 is their LCM. The LCM of these numbers is as follows.

LCM of 4, 9, 10 = 2 × 2 × 3 × 3 × 5 =180
Here, prime factor 5 does not have its pair. Therefore, 180 is not a perfect square. If we multiply 180 with 5, then the number will become a perfect square. Therefore, 180 should be multiplied with 5 to obtain a perfect square.
Hence, the required square number is 180 × 5 = 900


Q10 :
Find the smallest square number that is divisible by each of the numbers 8, 15,
and 20.
Answer :
The number that is perfectly divisible by each of the numbers 8, 15, and 20 is their LCM.

LCM of 8, 15, and 20 = 2 × 2 × 2 × 3 × 5 =120
Here, prime factors 2, 3, and 5 do not have their respective pairs. Therefore, 120 is not a perfect square.
Therefore, 120 should be multiplied by 2 × 3 × 5, i.e. 30, to obtain a perfect square.
Hence, the required square number is 120 × 2 × 3 × 5 = 3600


Exercise 6.4 : Solutions of Questions on Page Number : 107


Q1 : Find the square root 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
Answer :
(i) The square root of 2304 can be calculated as follows.



Q2 : 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
Answer :
(i) By placing bars, we obtain

Since there is only one bar, the square root of 64 will have only one digit in it.
(ii) By placing bars, we obtain

Since there are two bars, the square root of 144 will have 2 digits in it.
(iii) By placing bars, we obtain

Since there are two bars, the square root of 4489 will have 2 digits in it.
(iv) By placing bars, we obtain

Since there are three bars, the square root of 27225 will have three digits in it.
(v) By placing the bars, we obtain

Since there are three bars, the square root of 390625 will have 3 digits in it.


Q3 : Find the square root of the following decimal numbers.
(i) 2.56 (ii) 7.29
(iii) 51.84 (iv) 42.25
(v) 31.36
Answer :
(i) The square root of 2.56 can be calculated as follows.


Q4 : 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
Answer :
(i) The square root of 402 can be calculated by long division method as follows.


Q5 : 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
Answer :
(i) The square root of 525 can be calculated by long division method as follows.

The remainder is 69.
It represents that the square of 41 is less than 1750.
The next number is 42 and 422 = 1764
Hence, number to be added to 1750 = 422 – 1750 = 1764 – 1750 = 14
The required perfect square is 1764 and
(iii) The square root of 252 can be calculated by long division method as follows.


Q6 : Find the length of the side of a square whose area is 441 m2.
Answer :
Let the length of the side of the square be x m.
Area of square = (x)2 = 441 m2

The square root of 441 can be calculated as follows.


Q7 : In a right triangle ABC, ∠ B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC = 13 cm, BC = 5 cm, find AB
Answer :
(a) ΔABC is right-angled at B.
Therefore, by applying Pythagoras theorem, we obtain
AC2 = AB2 + BC2
AC2 = (6 cm)2 + (8 cm)2
AC2 = (36 + 64) cm2 =100 cm2
AC=
AC = 10 cm
(b) ΔABC is right-angled at B.
Therefore, by applying Pythagoras theorem, we obtain
AC2 = AB2 + BC2
(13 cm)2 = (AB)2 + (5 cm)2
AB2 = (13 cm)2 – (5 cm)2 = (169 – 25) cm2 = 144 cm2
AB=
AB = 12 cm


Q8 : A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Answer :
It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same.
We have to find the number of more plants that should be there, so that when the gardener plants them, the number of rows and columns are same.
That is, the number which should be added to 1000 to make it a perfect square has to be calculated.
The square root of 1000 can be calculated by long division method as follows.

The remainder is 39. It represents that the square of 31 is less than 1000.
The next number is 32 and 322 = 1024
Hence, number to be added to 1000 to make it a perfect square
= 322 – 1000 = 1024 – 1000 = 24
Thus, the required number of plants is 24.


Q9 : These 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?
Answer :
It is given that there are 500 children in the school. They have to stand for a P.T. drill such that the number of rows is equal to the number of columns.
The number of children who will be left out in this arrangement has to be calculated. That is, the number which should be subtracted from 500 to make it a perfect square has to be calculated.
The square root of 500 can be calculated by long division method as follows.

The remainder is 16.
It shows that the square of 22 is less than 500 by 16. Therefore, if we subtract 16 from 500, we will obtain a perfect square.
Required perfect square = 500 – 16 = 484
Thus, the number of children who will be left out is 16.


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