NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots
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Page 1
NCERT
SOLUTIONS
CLASS - 8TH
aglase .co
SOLUTIONS
CLASS - 8TH
aglase .co
Page 2
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Class : 8th
Subject : Maths
Chapter : 6
Chapter Name : Squares and Square Roots
Exercise 6.1
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 O, 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
Page 1 of 39 Aglasem Schools
Class : 8th
Subject : Maths
Chapter : 6
Chapter Name : Squares and Square Roots
Exercise 6.1
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 O, 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
Page 1 of 39 Aglasem Schools
Page 3
Book : Mathematics Ncert Solutions | Chapter-6 Maths
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 = 25) i.e., 5.
Page : 96 , Block Name : Exercise 6.1
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 zeros 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 zeros 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.
Page : 96 , Block Name : Exercise 6.1
Q3 Find the smallest number by which each of the following numbers must be multiplied to
obtain a perfect cube.
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Answer. (i) 243 - 3 x 3 x 3 x 3 x 3
Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.
In that case, 243 x 3 =3 x 3 x 3 x 3 x 3 x 3 = 729 is a perfect cube.
Hence, the smallest natural number by which 243 should be multiplied to make it a perfect
cube is 3.
(ii) 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.
256 x 2 =2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 is a perfect cube.
Hence, the smallest natural number by which 256 should be multiplied to make it a perfect
cube is 2.
(iii) 72 = 2 x 2 x 2 x 3 x 3
Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.
Page 2 of 39 Aglasem Schools
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 = 25) i.e., 5.
Page : 96 , Block Name : Exercise 6.1
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 zeros 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 zeros 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.
Page : 96 , Block Name : Exercise 6.1
Q3 Find the smallest number by which each of the following numbers must be multiplied to
obtain a perfect cube.
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Answer. (i) 243 - 3 x 3 x 3 x 3 x 3
Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.
In that case, 243 x 3 =3 x 3 x 3 x 3 x 3 x 3 = 729 is a perfect cube.
Hence, the smallest natural number by which 243 should be multiplied to make it a perfect
cube is 3.
(ii) 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.
256 x 2 =2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 is a perfect cube.
Hence, the smallest natural number by which 256 should be multiplied to make it a perfect
cube is 2.
(iii) 72 = 2 x 2 x 2 x 3 x 3
Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.
Page 2 of 39 Aglasem Schools
Page 4
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Then, we obtain
72 x 3 = 2 x 2 x 2 x 3 x 3 x 3 = 216 is a perfect cube.
Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube
is 3.
(iv) 675 = 3 x 3 x 3 x 5 x 5
Here, two 5s are left which are not in a triplet. TO make 675 a cube, one more 5 is required.
Then, we obtain
675 x 5 = 3 x 3 x 3 x 5 x 5 x 5 = 3375 is a perfect cube.
Hence, the smallest natural number by which 675 should be multiplied to make it a perfect
cube is 5.
(v) 100 = 2 x 2 x 5 x 5
Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one
more 2 and one more 5.
Then, we obtain
100 x 2 x 5 = 2 x 2 x 2x 5 x 5 x 5 = 1000 is a perfect cube
Hence, the smallest natural number by which 100 should be multiplied to make it a perfect
cube is 2 x 5 = 10.
Page : 96 , Block Name : Exercise 6.1
Q4 Observe the following pattern and nd the missing digits.
2
11 = 121
2
101 = 10201
2
1001 = 1002001
2
100001 = 1………2………1
2
10000001 = …………………
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,
2
100001 = 10000200001
2
10000001 = 100000020000001
Page : 96 , Block Name : Exercise 6.1
Q5 Observe the following pattern and supply the missing numbers.
2
11 = 121
2
101 = 10201
2
10101 = 102030201
2
1010101 = …………………
2
……… = … 0203040504030201
Answer. By following the given pattern, we obtain
1010101 = 1020304030201
2
Page 3 of 39 Aglasem Schools
Then, we obtain
72 x 3 = 2 x 2 x 2 x 3 x 3 x 3 = 216 is a perfect cube.
Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube
is 3.
(iv) 675 = 3 x 3 x 3 x 5 x 5
Here, two 5s are left which are not in a triplet. TO make 675 a cube, one more 5 is required.
Then, we obtain
675 x 5 = 3 x 3 x 3 x 5 x 5 x 5 = 3375 is a perfect cube.
Hence, the smallest natural number by which 675 should be multiplied to make it a perfect
cube is 5.
(v) 100 = 2 x 2 x 5 x 5
Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one
more 2 and one more 5.
Then, we obtain
100 x 2 x 5 = 2 x 2 x 2x 5 x 5 x 5 = 1000 is a perfect cube
Hence, the smallest natural number by which 100 should be multiplied to make it a perfect
cube is 2 x 5 = 10.
Page : 96 , Block Name : Exercise 6.1
Q4 Observe the following pattern and nd the missing digits.
2
11 = 121
2
101 = 10201
2
1001 = 1002001
2
100001 = 1………2………1
2
10000001 = …………………
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,
2
100001 = 10000200001
2
10000001 = 100000020000001
Page : 96 , Block Name : Exercise 6.1
Q5 Observe the following pattern and supply the missing numbers.
2
11 = 121
2
101 = 10201
2
10101 = 102030201
2
1010101 = …………………
2
……… = … 0203040504030201
Answer. By following the given pattern, we obtain
1010101 = 1020304030201
2
Page 3 of 39 Aglasem Schools
Page 5
Book : Mathematics Ncert Solutions | Chapter-6 Maths
101010101
2
= 10203040504030201
Page : 96 , Block Name : Exercise 6.1
Q6 Using the given pattern, nd the missing numbers.
Answer.
2 2 2 2 2 2 2 2
1 + 2 + 2 = 3 4 + 5 + − = 21
2 2 2 2 2 2 2 2
2 + 3 + 6 = 7 5 + − + 30 = 31
2 2 2 2 2 2 2 2
3 + 4 + 12 = 13 6 + 7 + = _
−
Answer. From the given pattern, it can be observed that,
(i) The third number is the product of the rst 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.
2
2 2 20 2
4 + 5 + = 21
5
2
2 6 2 2
5 + + 30 = 31
5
2 2 2 2
6 + 7 + 42 = 43
Page : 96 , Block Name : Exercise 6.1
Q7 Without adding, nd 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 rst n Odd natural numbers is n . 2
(i) Here, we have to nd the sum of rst ve odd natural numbers.
Therefore, 1 +3 5+7 + 9 = (5) = 25 2
(ii) Here, we have to nd the sum of rst ten odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10) = 100 2
(iii) Here, we have to nd the sum of rst twelve Odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 19 + 21 + 23 = (12) = 44 2
Page : 96 , Block Name : Exercise 6.1
Q8 (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer. We know that the sum of rst n odd natural numbers is (n) . 2
(i) 49 = (7) 2
Therefore, 49 is the sum of rst 7 odd natural numbers.
Page 4 of 39 Aglasem Schools
101010101
2
= 10203040504030201
Page : 96 , Block Name : Exercise 6.1
Q6 Using the given pattern, nd the missing numbers.
Answer.
2 2 2 2 2 2 2 2
1 + 2 + 2 = 3 4 + 5 + − = 21
2 2 2 2 2 2 2 2
2 + 3 + 6 = 7 5 + − + 30 = 31
2 2 2 2 2 2 2 2
3 + 4 + 12 = 13 6 + 7 + = _
−
Answer. From the given pattern, it can be observed that,
(i) The third number is the product of the rst 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.
2
2 2 20 2
4 + 5 + = 21
5
2
2 6 2 2
5 + + 30 = 31
5
2 2 2 2
6 + 7 + 42 = 43
Page : 96 , Block Name : Exercise 6.1
Q7 Without adding, nd 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 rst n Odd natural numbers is n . 2
(i) Here, we have to nd the sum of rst ve odd natural numbers.
Therefore, 1 +3 5+7 + 9 = (5) = 25 2
(ii) Here, we have to nd the sum of rst ten odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10) = 100 2
(iii) Here, we have to nd the sum of rst twelve Odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 19 + 21 + 23 = (12) = 44 2
Page : 96 , Block Name : Exercise 6.1
Q8 (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer. We know that the sum of rst n odd natural numbers is (n) . 2
(i) 49 = (7) 2
Therefore, 49 is the sum of rst 7 odd natural numbers.
Page 4 of 39 Aglasem Schools
Page 6
Book : Mathematics Ncert Solutions | Chapter-6 Maths
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = (11) 2
Therefore, 121 is the sum of rst 11 odd natural numbers.
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Page : 96 , Block Name : Exercise 6.1
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 (12) and (13) , there will be 2 x 12 = 24 numbers
2 2
(ii) Between (25) and (26) , there will be 2 x 25 = 50 numbers
2 2
(iii) Between (99) and (100) , there will be 2 x 99 = 198 numbers
2 2
Page : 96 , Block Name : Exercise 6.1
Exercise 6.2
Q1 Find the square of the following numbers
(i) 32 (ii) 35
(iii) 86 (iv) 93
(v) 71 (vi) 46
Answer.
2 2
(i) 32 = (30 + 2)
= 30(30 + 2) + 2(30 + 2)
2 2
= 30 + 30 × 2 + 2 × 30 + 2
= 900 + 60 + 60 + 4
= 1024
(ii) The number 35 has 5 in its unit's place. Therefore,
2
35 = (3)(3 + 1) hundreds + 25
= (3 × 4) hundreds + 25
= 1200 + 25 = 1225
Page 5 of 39 Aglasem Schools
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = (11) 2
Therefore, 121 is the sum of rst 11 odd natural numbers.
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Page : 96 , Block Name : Exercise 6.1
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 (12) and (13) , there will be 2 x 12 = 24 numbers
2 2
(ii) Between (25) and (26) , there will be 2 x 25 = 50 numbers
2 2
(iii) Between (99) and (100) , there will be 2 x 99 = 198 numbers
2 2
Page : 96 , Block Name : Exercise 6.1
Exercise 6.2
Q1 Find the square of the following numbers
(i) 32 (ii) 35
(iii) 86 (iv) 93
(v) 71 (vi) 46
Answer.
2 2
(i) 32 = (30 + 2)
= 30(30 + 2) + 2(30 + 2)
2 2
= 30 + 30 × 2 + 2 × 30 + 2
= 900 + 60 + 60 + 4
= 1024
(ii) The number 35 has 5 in its unit's place. Therefore,
2
35 = (3)(3 + 1) hundreds + 25
= (3 × 4) hundreds + 25
= 1200 + 25 = 1225
Page 5 of 39 Aglasem Schools
Page 7
Book : Mathematics Ncert Solutions | Chapter-6 Maths
2 2
(iii) 86 = (80 + 6)
= 80(80 + 6) + 6(80 + 6)
2 2
= 80 + 80 × 6 + 6 × 80 + 6
= 6400 + 480 + 480 + 36
= 7396
2 2
( iv) 93 = (90 + 3)
= 90(90 + 3) + 3(90 + 3)
2 2
= 90 + 90 × 3 + 3 × 90 + 3
= 8100 + 270 + 270 + 9
= 8649
2 2
(v)71 = (70 + 1)
= 70(70 + 1) + 1(70 + 1)
2 2
= 70 + 70 × 1 + 1 × 70 + 1
= 4900 + 70 + 70 + 1 = 5041
2 2
(vi) 46 = (40 + 6)
= 40(40 + 6) + 6(40 + 6)
2 2
= 40 + 40 × 6 + 6 × 40 + 6
= 1600 + 240 + 240 + 36
= 2116
Page : 98 , Block Name : Exercise 6.2
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, m − 1, m + 1 forms a Pythagorean triplet.
2 2
(i) If we take m + 1 = 6, then m = 5
2 2
The value of m will not be an integer.
If we take m − 1 = 6, then m = 7
2 2
Again the value of m is not an integer .
Let 2m = 6
m=3
Therefore, The Pythagorean triplets are 2 x 3, 3 − 1, (3^{2} + 1\) or 6, 8, and 10.
2
(ii) If we take m + 1 = 14, then m = 13
2 2
The value of m will not be an integer.
If we take m - 1 = 14, then m = 15
2 2
Again the value of m is not an integer.
Let 2m = 14
m=7
Thus, m - 1 = 49 -1 = 48 and m + 1 = 49 + 1 = 50
2 2
Therefore, the required triplet is 14, 48, and 50.
Page 6 of 39 Aglasem Schools
2 2
(iii) 86 = (80 + 6)
= 80(80 + 6) + 6(80 + 6)
2 2
= 80 + 80 × 6 + 6 × 80 + 6
= 6400 + 480 + 480 + 36
= 7396
2 2
( iv) 93 = (90 + 3)
= 90(90 + 3) + 3(90 + 3)
2 2
= 90 + 90 × 3 + 3 × 90 + 3
= 8100 + 270 + 270 + 9
= 8649
2 2
(v)71 = (70 + 1)
= 70(70 + 1) + 1(70 + 1)
2 2
= 70 + 70 × 1 + 1 × 70 + 1
= 4900 + 70 + 70 + 1 = 5041
2 2
(vi) 46 = (40 + 6)
= 40(40 + 6) + 6(40 + 6)
2 2
= 40 + 40 × 6 + 6 × 40 + 6
= 1600 + 240 + 240 + 36
= 2116
Page : 98 , Block Name : Exercise 6.2
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, m − 1, m + 1 forms a Pythagorean triplet.
2 2
(i) If we take m + 1 = 6, then m = 5
2 2
The value of m will not be an integer.
If we take m − 1 = 6, then m = 7
2 2
Again the value of m is not an integer .
Let 2m = 6
m=3
Therefore, The Pythagorean triplets are 2 x 3, 3 − 1, (3^{2} + 1\) or 6, 8, and 10.
2
(ii) If we take m + 1 = 14, then m = 13
2 2
The value of m will not be an integer.
If we take m - 1 = 14, then m = 15
2 2
Again the value of m is not an integer.
Let 2m = 14
m=7
Thus, m - 1 = 49 -1 = 48 and m + 1 = 49 + 1 = 50
2 2
Therefore, the required triplet is 14, 48, and 50.
Page 6 of 39 Aglasem Schools
Page 8
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(iii) If we take m + 1 = 16, then m = 15
2 2
The value of m not be an integer.
If we take m - 1 = 16, then m = 17
2 2
Again the value of m is not an integer.
Let 2m = 16
m=8
Thus, m - 1 = 64 -1 = 63 and m + 1 = 64 + 1 = 65
2 2
Therefore, the Pythagorean triplet is 16, 63, and 65.
(iv) If we take m + 1 = 18
2
m = 17
2
The value of m will not be an integer.
Let 2m = 18
m=9
Thus m - 1 = 81 - 1 = 80 and m + 1 = 81 + 1 = 82
2 2
Therefore, the Pythagorean triplet is 18, 80, and 82.
Page : 98 , Block Name : Exercise 6.2
Exercise 6.3
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.
Page : 102 , Block Name : Exercise 6.3
Q2 Without doing any calculation, nd 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.
Page 7 of 39 Aglasem Schools
(iii) If we take m + 1 = 16, then m = 15
2 2
The value of m not be an integer.
If we take m - 1 = 16, then m = 17
2 2
Again the value of m is not an integer.
Let 2m = 16
m=8
Thus, m - 1 = 64 -1 = 63 and m + 1 = 64 + 1 = 65
2 2
Therefore, the Pythagorean triplet is 16, 63, and 65.
(iv) If we take m + 1 = 18
2
m = 17
2
The value of m will not be an integer.
Let 2m = 18
m=9
Thus m - 1 = 81 - 1 = 80 and m + 1 = 81 + 1 = 82
2 2
Therefore, the Pythagorean triplet is 18, 80, and 82.
Page : 98 , Block Name : Exercise 6.2
Exercise 6.3
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.
Page : 102 , Block Name : Exercise 6.3
Q2 Without doing any calculation, nd 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.
Page 7 of 39 Aglasem Schools
Page 9
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(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.
Page : 102 , Block Name : Exercise 6.3
Q3 Find the square roots of 100 and 169 by the method of repeated subtraction.
Answer. We know that the sum of the rst n odd natural numbers is n . Consider √100
2
(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, √100 = 10
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
(xii) 25 - 25 = 0
We have subtracted successive odd numbers starting from 1 to 169, and obtained O at Step.
Therefore, √169 = 13
Page : 102 , Block Name : Exercise 6.3
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. (i) 729 can be factorised as follows.
Page 8 of 39 Aglasem Schools
(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.
Page : 102 , Block Name : Exercise 6.3
Q3 Find the square roots of 100 and 169 by the method of repeated subtraction.
Answer. We know that the sum of the rst n odd natural numbers is n . Consider √100
2
(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, √100 = 10
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
(xii) 25 - 25 = 0
We have subtracted successive odd numbers starting from 1 to 169, and obtained O at Step.
Therefore, √169 = 13
Page : 102 , Block Name : Exercise 6.3
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. (i) 729 can be factorised as follows.
Page 8 of 39 Aglasem Schools
Page 10
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(ii)400 can be factorised as follows.
(iii) 1764 can be factorised as follows.
(iv) 4096 can be factorised as follows.
Page 9 of 39 Aglasem Schools
(ii)400 can be factorised as follows.
(iii) 1764 can be factorised as follows.
(iv) 4096 can be factorised as follows.
Page 9 of 39 Aglasem Schools
Page 11
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(v) 7744 can be factorised as follows.
Page 10 of 39 Aglasem Schools
(v) 7744 can be factorised as follows.
Page 10 of 39 Aglasem Schools
Page 12
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(vi) 9604 can be factorised as follows.
(vii) 5929 can be factorised as follows.
(viii) 9216 can be factorised as follows.
Page 11 of 39 Aglasem Schools
(vi) 9604 can be factorised as follows.
(vii) 5929 can be factorised as follows.
(viii) 9216 can be factorised as follows.
Page 11 of 39 Aglasem Schools
Page 13
Book : Mathematics Ncert Solutions | Chapter-6 Maths
9216 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2x3 x 3
∴ √9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96
(ix) 529 can be factorised as follows.
(x) 8100 can be factorised as follows.
Page 12 of 39 Aglasem Schools
9216 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2x3 x 3
∴ √9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96
(ix) 529 can be factorised as follows.
(x) 8100 can be factorised as follows.
Page 12 of 39 Aglasem Schools
Page 14
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Page : 102 , Block Name : Exercise 6.3
Q5 For each of the following numbers, nd the smallest whole number by which it should be
multiplied so as to get a perfect square number. Also nd the square root of the square
number so obtained.
(i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768
Answer. (i)252 can be factorised as follows. .
Here, prime factor 7 does not have its pair.
If 7 gets a pair, then the number will become a perfect square. Therefore, 252 has to be
multiplied with 7 to obtain a perfect square.
225 x 7 = 2 x 2 x 3 x 3 x 7 x 7
Therefore, 252 x 7 = 1764 is a perfect square.
∴ √1764 = 2 × 3 × 7 = 42
Page 13 of 39 Aglasem Schools
Page : 102 , Block Name : Exercise 6.3
Q5 For each of the following numbers, nd the smallest whole number by which it should be
multiplied so as to get a perfect square number. Also nd the square root of the square
number so obtained.
(i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768
Answer. (i)252 can be factorised as follows. .
Here, prime factor 7 does not have its pair.
If 7 gets a pair, then the number will become a perfect square. Therefore, 252 has to be
multiplied with 7 to obtain a perfect square.
225 x 7 = 2 x 2 x 3 x 3 x 7 x 7
Therefore, 252 x 7 = 1764 is a perfect square.
∴ √1764 = 2 × 3 × 7 = 42
Page 13 of 39 Aglasem Schools
Page 15
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(ii)180 can be factorised as follows.
Here, prime factor 5 does not have its pair. If 5 gets a pair, then the number will become a
perfect square. Therefore, 180 has to be multiplied with 5 to obtain a
perfect square.
180 x 5 = 900 = 2 x 2 x 3 x 3 x 5 x 5
Therefore, 180 x 5 = 900 is a perfect square.
∴ √900 = 2 x 3 x 5 = 30
(iii)1008 can be factorised as follows.
Here, prime factor 7 does not have its pair. If 7 gets a pair, then the number will become a
perfect square. Therefore, 1008 can be multiplied with 7 to obtain a
Page 14 of 39 Aglasem Schools
(ii)180 can be factorised as follows.
Here, prime factor 5 does not have its pair. If 5 gets a pair, then the number will become a
perfect square. Therefore, 180 has to be multiplied with 5 to obtain a
perfect square.
180 x 5 = 900 = 2 x 2 x 3 x 3 x 5 x 5
Therefore, 180 x 5 = 900 is a perfect square.
∴ √900 = 2 x 3 x 5 = 30
(iii)1008 can be factorised as follows.
Here, prime factor 7 does not have its pair. If 7 gets a pair, then the number will become a
perfect square. Therefore, 1008 can be multiplied with 7 to obtain a
Page 14 of 39 Aglasem Schools
Page 16
Book : Mathematics Ncert Solutions | Chapter-6 Maths
perfect square.
1008 x 7 = 7056 = 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7
Therefore, 1008 x 7 = 7056 is a perfect square.
∴ √7056 = 2x2x3x7 = 84
(iv) 2028 can be factorised as follows.
Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a
perfect square. Therefore, 2028 has to be multiplied with 3 to obtain a
perfect square.
Therefore, 2028 x 3 = 6084 is a perfect square.
2028 x 3 = 6084 = 2 x 2 x 3 x 3 x 13 x 13
∴ √6084 = 2 x 3 x 13 = 78
(v) 1458 can be factorised as follows.
Page 15 of 39 Aglasem Schools
perfect square.
1008 x 7 = 7056 = 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7
Therefore, 1008 x 7 = 7056 is a perfect square.
∴ √7056 = 2x2x3x7 = 84
(iv) 2028 can be factorised as follows.
Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a
perfect square. Therefore, 2028 has to be multiplied with 3 to obtain a
perfect square.
Therefore, 2028 x 3 = 6084 is a perfect square.
2028 x 3 = 6084 = 2 x 2 x 3 x 3 x 13 x 13
∴ √6084 = 2 x 3 x 13 = 78
(v) 1458 can be factorised as follows.
Page 15 of 39 Aglasem Schools
Page 17
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Here, prime factor 2 does not have its pair. If 2 gets a pair, then the number will become a
perfect square. Therefore, 1458 has to be multiplied with 2 to obtain a
perfect square.
Therefore, 1458 x 2 = 2916 is a perfect square.
1458 x 2 = 2916 = 2 x 2 x 3 x 3 x 3 x 3
∴ √2916 = 2 x 3 x 3 x 3 = 54
(vi) 768 can be factorised as follows.
768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a
perfect square. Therefore, 768 has to be multiplied with 3 to obtain a
perfect square.
Therefore, 768 x 3 = = 2304 is a perfect square.
768 x 3 = 2304= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
√2304 = 2 × 2 × 2 × 2 × 3 = 48
Page : 102 , Block Name : Exercise 6.3
Page 16 of 39 Aglasem Schools
Here, prime factor 2 does not have its pair. If 2 gets a pair, then the number will become a
perfect square. Therefore, 1458 has to be multiplied with 2 to obtain a
perfect square.
Therefore, 1458 x 2 = 2916 is a perfect square.
1458 x 2 = 2916 = 2 x 2 x 3 x 3 x 3 x 3
∴ √2916 = 2 x 3 x 3 x 3 = 54
(vi) 768 can be factorised as follows.
768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a
perfect square. Therefore, 768 has to be multiplied with 3 to obtain a
perfect square.
Therefore, 768 x 3 = = 2304 is a perfect square.
768 x 3 = 2304= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
√2304 = 2 × 2 × 2 × 2 × 3 = 48
Page : 102 , Block Name : Exercise 6.3
Page 16 of 39 Aglasem Schools
Page 18
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Q6 For each of the following numbers, nd the smallest whole number by which it should be
divided so as to get a perfect square. Also nd the square root of the square number so
obtained.
(i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620
Answer. (i) 252 can be factorised as follows.
Here, prime factor 7 does not have its pair.
If we divide this number by 7, then the number will become a perfect square.
Therefore, 252 has to be divided by 7 to obtain a perfect square.
252 ÷ 7 = 36 is a perfect square.
36 = 2 x 2 x 3 x 3
∴ √36 = 2 × 3 = 6
(ii) 2925 can be factorised as follows.
Page 17 of 39 Aglasem Schools
Q6 For each of the following numbers, nd the smallest whole number by which it should be
divided so as to get a perfect square. Also nd the square root of the square number so
obtained.
(i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620
Answer. (i) 252 can be factorised as follows.
Here, prime factor 7 does not have its pair.
If we divide this number by 7, then the number will become a perfect square.
Therefore, 252 has to be divided by 7 to obtain a perfect square.
252 ÷ 7 = 36 is a perfect square.
36 = 2 x 2 x 3 x 3
∴ √36 = 2 × 3 = 6
(ii) 2925 can be factorised as follows.
Page 17 of 39 Aglasem Schools
Page 19
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Here, prime factor 13 does not have its pair.
If we divide this number by 13, then the number will become a perfect square.
Therefore, 2925 has to be divided by 13 to obtain a perfect square.
2925 ÷ 13 = 225 is a perfect square.
225 = 3 x 3 x 5 x 5
∴ √225 = 3 × 5 = 15
(iii)396 can be factorised as follows.
Here, prime factor 11 does not have its pair.
If we divide this number by 11, then the number will become a perfect square.
Therefore, 396 has to be divided bv 11 to obtain a perfect square.
396 ÷ 11 = 36 is a perfect square.
36 = 2 x 2x 3 x 3
∴ √36 = 2 × 3 = 6
(iv) 2645 can be factorised as follows.
Page 18 of 39 Aglasem Schools
Here, prime factor 13 does not have its pair.
If we divide this number by 13, then the number will become a perfect square.
Therefore, 2925 has to be divided by 13 to obtain a perfect square.
2925 ÷ 13 = 225 is a perfect square.
225 = 3 x 3 x 5 x 5
∴ √225 = 3 × 5 = 15
(iii)396 can be factorised as follows.
Here, prime factor 11 does not have its pair.
If we divide this number by 11, then the number will become a perfect square.
Therefore, 396 has to be divided bv 11 to obtain a perfect square.
396 ÷ 11 = 36 is a perfect square.
36 = 2 x 2x 3 x 3
∴ √36 = 2 × 3 = 6
(iv) 2645 can be factorised as follows.
Page 18 of 39 Aglasem Schools
Page 20
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Here, prime factor 5 does not have its pair.
If we divide this number by 5, then the number will become a perfect square.
Therefore, 2645 has to be divided by 5 to obtain a perfect square.
2645 ÷ 5 = 529 is a perfect square.
529 = 23 × 23
∴ √529 = 23
(v)2800 can be factorised as follows.
Here, prime factor 7 does not have its pair.
If we divide this number by 7, then the number will become a perfect square.
Therefore, 2800 has to be divided by 7 to obtain a perfect square.
2800 ÷ 7 = 400 is a perfect square.
Page 19 of 39 Aglasem Schools
Here, prime factor 5 does not have its pair.
If we divide this number by 5, then the number will become a perfect square.
Therefore, 2645 has to be divided by 5 to obtain a perfect square.
2645 ÷ 5 = 529 is a perfect square.
529 = 23 × 23
∴ √529 = 23
(v)2800 can be factorised as follows.
Here, prime factor 7 does not have its pair.
If we divide this number by 7, then the number will become a perfect square.
Therefore, 2800 has to be divided by 7 to obtain a perfect square.
2800 ÷ 7 = 400 is a perfect square.
Page 19 of 39 Aglasem Schools
Page 21
Book : Mathematics Ncert Solutions | Chapter-6 Maths
400 = 2 x 2 x 2 x 2 x 5 x 5
√400 = 2 × 2 × 5 = 20
(vi) 1620 can be factorised as follows.
Here, prime factor 5 does not have its pair.
If we divide this number by 5, then the number will become a perfect square.
Therefore, 1620 has to be divided by 5 to obtain a perfect Square.
\(\1620 \div 5=324) is a perfect square.
324 = 2 x 2 x3 x 3 x 3 x 3
√324 = 2 × 3 × 3 = 18
Page : 102 , Block Name : Exercise 6.3
Q7 The students of Class VIII of a school donated ` 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 = √2401
2401 = 7 x 7 x7 x 7
√2401 = 7 × 7 = 49
Hence, the number of students in the class is 49.
Page : 102 , Block Name : Exercise 6.3
Page 20 of 39 Aglasem Schools
400 = 2 x 2 x 2 x 2 x 5 x 5
√400 = 2 × 2 × 5 = 20
(vi) 1620 can be factorised as follows.
Here, prime factor 5 does not have its pair.
If we divide this number by 5, then the number will become a perfect square.
Therefore, 1620 has to be divided by 5 to obtain a perfect Square.
\(\1620 \div 5=324) is a perfect square.
324 = 2 x 2 x3 x 3 x 3 x 3
√324 = 2 × 3 × 3 = 18
Page : 102 , Block Name : Exercise 6.3
Q7 The students of Class VIII of a school donated ` 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 = √2401
2401 = 7 x 7 x7 x 7
√2401 = 7 × 7 = 49
Hence, the number of students in the class is 49.
Page : 102 , Block Name : Exercise 6.3
Page 20 of 39 Aglasem Schools
Page 22
Book : Mathematics Ncert Solutions | Chapter-6 Maths
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 x Number of plants in each row
Number of rows x Number of plants in each row = 2025
2
(Number of rows) = 2025
Number of rows = √2025
2025 = 5 x 5 x 3 x 3 x 3 x 3
√2025 = 5 × 3 × 3 = 45
Page : 103 , Block Name : Exercise 6.3
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 x 2 x 3 x 3 x 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 x 5 = 900
Page : 103 , Block Name : Exercise 6.3
Q10 Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Page 21 of 39 Aglasem Schools
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 x Number of plants in each row
Number of rows x Number of plants in each row = 2025
2
(Number of rows) = 2025
Number of rows = √2025
2025 = 5 x 5 x 3 x 3 x 3 x 3
√2025 = 5 × 3 × 3 = 45
Page : 103 , Block Name : Exercise 6.3
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 x 2 x 3 x 3 x 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 x 5 = 900
Page : 103 , Block Name : Exercise 6.3
Q10 Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Page 21 of 39 Aglasem Schools
Page 23
Book : Mathematics Ncert Solutions | Chapter-6 Maths
Answer. The number that is perfectly divisible by each of the numbers 8, 15, and 20 is their
LCM.
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 x 3 x 5, i.e. 30, to obtain a perfect square. Hence, the
required square number is 120 x 2 x 3 x 5 = 3600
Page : 103 , Block Name : Exercise 6.3
Exercise 6.4
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.
Page 22 of 39 Aglasem Schools
Answer. The number that is perfectly divisible by each of the numbers 8, 15, and 20 is their
LCM.
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 x 3 x 5, i.e. 30, to obtain a perfect square. Hence, the
required square number is 120 x 2 x 3 x 5 = 3600
Page : 103 , Block Name : Exercise 6.3
Exercise 6.4
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.
Page 22 of 39 Aglasem Schools
Page 24
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(ii) The square root of 4489 can be calculated as follows.
(iii) The square root of 3481 can be calculated as follows.
Page 23 of 39 Aglasem Schools
(ii) The square root of 4489 can be calculated as follows.
(iii) The square root of 3481 can be calculated as follows.
Page 23 of 39 Aglasem Schools
Page 25
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(iv) The square root of 529 can be calculated as follows.
(v) The square root of 3249 can be calculated as follows.
Page 24 of 39 Aglasem Schools
(iv) The square root of 529 can be calculated as follows.
(v) The square root of 3249 can be calculated as follows.
Page 24 of 39 Aglasem Schools
Page 26
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(vi) The square root Of 1369 can be calculated as follows.
(vii) The square root of 5776 can be calculated as follows.
Page 25 of 39 Aglasem Schools
(vi) The square root Of 1369 can be calculated as follows.
(vii) The square root of 5776 can be calculated as follows.
Page 25 of 39 Aglasem Schools
Page 27
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(viii) The square root of 7921 can be calculated as follows.
(ix) The square root of 576 can be calculated as follows.
Page 26 of 39 Aglasem Schools
(viii) The square root of 7921 can be calculated as follows.
(ix) The square root of 576 can be calculated as follows.
Page 26 of 39 Aglasem Schools
Page 28
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(x) The square root of 1024 can be calculated as follows.
(xi) The square root of 3136 can be calculated as follows.
Page 27 of 39 Aglasem Schools
(x) The square root of 1024 can be calculated as follows.
(xi) The square root of 3136 can be calculated as follows.
Page 27 of 39 Aglasem Schools
Page 29
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(xii) The square root of 900 can be calculated as follows.
Page : 107 , Block Name : Exercise 6.4
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
¯
¯¯¯
¯¯
64 = 64
Since there is only one bar, the square root of 64 will have only one digit in it.
Page 28 of 39 Aglasem Schools
(xii) The square root of 900 can be calculated as follows.
Page : 107 , Block Name : Exercise 6.4
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
¯
¯¯¯
¯¯
64 = 64
Since there is only one bar, the square root of 64 will have only one digit in it.
Page 28 of 39 Aglasem Schools
Page 30
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(ii) ay placing bars, we obtain
¯
¯¯¯
¯¯¯
¯¯
144 = 1 44
Since there are two bars, the square root of 144 will have 2 digits in it.
(iii) By placing bars, we obtain
¯
¯¯¯
¯¯¯
¯¯¯
¯¯
4489 = 44 89
Since there are two bars, the square root Of 4489 will have 2 digits in it.
(iv) By placing bars, we obtain
¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯
27225 = 2 72 25
Since there are three bars, the square root Of 27225 will have three digits in it.
(v) By placing the bars, we obtain
¯
¯¯¯
¯¯¯¯
¯¯¯
¯¯¯
¯¯¯
390625 = 39 0625
Since there are three bars, the square root of 390625 will have 3 digits in it.
Page : 107 , Block Name : Exercise 6.4
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.
(ii) The square root Of 7.29 can be calculated as follows.
Page 29 of 39 Aglasem Schools
(ii) ay placing bars, we obtain
¯
¯¯¯
¯¯¯
¯¯
144 = 1 44
Since there are two bars, the square root of 144 will have 2 digits in it.
(iii) By placing bars, we obtain
¯
¯¯¯
¯¯¯
¯¯¯
¯¯
4489 = 44 89
Since there are two bars, the square root Of 4489 will have 2 digits in it.
(iv) By placing bars, we obtain
¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯
27225 = 2 72 25
Since there are three bars, the square root Of 27225 will have three digits in it.
(v) By placing the bars, we obtain
¯
¯¯¯
¯¯¯¯
¯¯¯
¯¯¯
¯¯¯
390625 = 39 0625
Since there are three bars, the square root of 390625 will have 3 digits in it.
Page : 107 , Block Name : Exercise 6.4
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.
(ii) The square root Of 7.29 can be calculated as follows.
Page 29 of 39 Aglasem Schools
Page 31
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(iii) The square root of 51.84 can be calculated as follows.
(iv) The square root of 42.25 can be calculated as follows.
Page 30 of 39 Aglasem Schools
(iii) The square root of 51.84 can be calculated as follows.
(iv) The square root of 42.25 can be calculated as follows.
Page 30 of 39 Aglasem Schools
Page 32
Book : Mathematics Ncert Solutions | Chapter-6 Maths
(v) The square root of 31.36 can be calculated as follows.
Page : 108 , Block Name : Exercise 6.4
Q4 Find the least number which must be subtracted from each of the following numbers so as
to get a perfect square. Also nd 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.
Page 31 of 39 Aglasem Schools
(v) The square root of 31.36 can be calculated as follows.
Page : 108 , Block Name : Exercise 6.4
Q4 Find the least number which must be subtracted from each of the following numbers so as
to get a perfect square. Also nd 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.
Page 31 of 39 Aglasem Schools
Page 33
Book : Mathematics Ncert Solutions | Chapter-6 Maths
The remainder is 2. It represents that the square Of 20 is less than 402 by 2.
Therefore, a perfect square will be obtained by subtracting 2 from the given number 402.
Therefore, required perfect square 402 — 2 = 400
And, √400 = 20
(ii) The square root of 1989 can be calculated by long division method as follows.
The remainder is 53. It represents that the square of 44 is less than 1989 by 53.
Therefore, a perfect square will be obtained by subtracting 53 from the given number 1989.
Therefore, required perfect square = 1989 — 53 = 1936
And, √1936 = 44
(iii) The square root of 3250 can be calculated by long division method as follows.
Page 32 of 39 Aglasem Schools
The remainder is 2. It represents that the square Of 20 is less than 402 by 2.
Therefore, a perfect square will be obtained by subtracting 2 from the given number 402.
Therefore, required perfect square 402 — 2 = 400
And, √400 = 20
(ii) The square root of 1989 can be calculated by long division method as follows.
The remainder is 53. It represents that the square of 44 is less than 1989 by 53.
Therefore, a perfect square will be obtained by subtracting 53 from the given number 1989.
Therefore, required perfect square = 1989 — 53 = 1936
And, √1936 = 44
(iii) The square root of 3250 can be calculated by long division method as follows.
Page 32 of 39 Aglasem Schools
Page 34
Book : Mathematics Ncert Solutions | Chapter-6 Maths
The remainder is 1. It represents that the square of 57 is less than 3250 by 1.
Therefore, a perfect square can be obtained by subtracting 1 from the given number 3250.
Therefore, required perfect square = 3250 — 1 = 3249
And,√3249 = 57
(iv) The square root Of 825 can be calculated by long division method as follows.
The remainder is 41. It represents that the square of 28 is less than 825 by 41. Therefore, a
perfect square can be calculated by subtracting 41 from the given
number 825.
Therefore, required perfect square = 825 — 41 = 784
And,√784 = 28
Page 33 of 39 Aglasem Schools
The remainder is 1. It represents that the square of 57 is less than 3250 by 1.
Therefore, a perfect square can be obtained by subtracting 1 from the given number 3250.
Therefore, required perfect square = 3250 — 1 = 3249
And,√3249 = 57
(iv) The square root Of 825 can be calculated by long division method as follows.
The remainder is 41. It represents that the square of 28 is less than 825 by 41. Therefore, a
perfect square can be calculated by subtracting 41 from the given
number 825.
Therefore, required perfect square = 825 — 41 = 784
And,√784 = 28
Page 33 of 39 Aglasem Schools
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Book : Mathematics Ncert Solutions | Chapter-6 Maths
(v) The square root Of 4000 can be calculated by long division method as follows.
The remainder is 31. It represents that the square of 63 is less than 4000 by 31.
Therefore, a perfect square can be obtained by subtracting 31 from the given number 4000.
Therefore, required perfect square = 4000 — 31 = 3969
And, √3969= 63
Page : 108 , Block Name : Exercise 6.4
Q5 Find the least number which must be added to each of the following numbers so as to get a
perfect square. Also nd 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.
Page 34 of 39 Aglasem Schools
(v) The square root Of 4000 can be calculated by long division method as follows.
The remainder is 31. It represents that the square of 63 is less than 4000 by 31.
Therefore, a perfect square can be obtained by subtracting 31 from the given number 4000.
Therefore, required perfect square = 4000 — 31 = 3969
And, √3969= 63
Page : 108 , Block Name : Exercise 6.4
Q5 Find the least number which must be added to each of the following numbers so as to get a
perfect square. Also nd 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.
Page 34 of 39 Aglasem Schools
Page 36
Book : Mathematics Ncert Solutions | Chapter-6 Maths
The remainder is 41.
It represents that the square of 22 is less than 525.
2
Next number is 23 and 23 = 529
2
Hence, number to be added to 525 = 23 − 525 = 529 − 525 = 4
(ii) The square root of 1750 can be calculated by long division method as follows.
The remainder is 69.
It represents that the square of 41 is less than 1750.
2
The next number is 42 and 42 = 1764
2
Hence, number to be added to 1750 = 42 − 1750 = 1764 − 1750 = 14
The required perfect square is 1764 and √1764= 42
(iii) The square root of 252 can be calculated by long division method as follows.
Page 35 of 39 Aglasem Schools
The remainder is 41.
It represents that the square of 22 is less than 525.
2
Next number is 23 and 23 = 529
2
Hence, number to be added to 525 = 23 − 525 = 529 − 525 = 4
(ii) The square root of 1750 can be calculated by long division method as follows.
The remainder is 69.
It represents that the square of 41 is less than 1750.
2
The next number is 42 and 42 = 1764
2
Hence, number to be added to 1750 = 42 − 1750 = 1764 − 1750 = 14
The required perfect square is 1764 and √1764= 42
(iii) The square root of 252 can be calculated by long division method as follows.
Page 35 of 39 Aglasem Schools
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Book : Mathematics Ncert Solutions | Chapter-6 Maths
The remainder is 27. It represents that the square of 15 is less than 252.
The next number is 16 and 16 = 256
2
Hence, number to be added to 252 = 16 — 252 = 256 - 252 = 4
2
The required perfect square is 256 and √256= 16
(iv) The square root of 1825 can be calculated by long division method as follows.
The remainder is 61. It represents that the square of 42 is less than 1825.
The next number is 43 and 43 = 1849
2
Hence, number to be added to 1825 = 43 - 1825 = 1849 - 1825 = 24
2
The required perfect square is 1849 and √1849= 43
(v) The square root of 6412 can be calculated by long division method as follows.
Page 36 of 39 Aglasem Schools
The remainder is 27. It represents that the square of 15 is less than 252.
The next number is 16 and 16 = 256
2
Hence, number to be added to 252 = 16 — 252 = 256 - 252 = 4
2
The required perfect square is 256 and √256= 16
(iv) The square root of 1825 can be calculated by long division method as follows.
The remainder is 61. It represents that the square of 42 is less than 1825.
The next number is 43 and 43 = 1849
2
Hence, number to be added to 1825 = 43 - 1825 = 1849 - 1825 = 24
2
The required perfect square is 1849 and √1849= 43
(v) The square root of 6412 can be calculated by long division method as follows.
Page 36 of 39 Aglasem Schools
Page 38
Book : Mathematics Ncert Solutions | Chapter-6 Maths
The remainder is 12.
It represents that the square of 80 is less than 6412.
The next number is 81 and 81 = 6561 2
Hence, number to be added to 6412 = 81 — 6412 = 6561 — 6412 = 149
2
The required perfect square is 6561 and √6561= 81
Page : 108 , Block Name : Exercise 6.4
Q6 Find the length of the side of a square whose area is 441 m 2
Answer. Let the length of the side of the square be x m.
2 2
Area of square = (x) = 441m
x = √441
Page : 108 , Block Name : Exercise 6.4
Q7 In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, nd AC (b) If AC = 13 cm, BC = 5 cm, nd AB
(a) △ABC is right-angled at B .
Therefore, by applying Pythagoras theorem, we obtain
2 2 2
AC = AB + BC
Answer. AC
2 2 2
= (6cm) + (8cm)
2 2 2
AC = (36 + 64)cm = 100cm
AC = (√100)cm = (√10 × 10)cm
AC = 10cm
Page 37 of 39 Aglasem Schools
The remainder is 12.
It represents that the square of 80 is less than 6412.
The next number is 81 and 81 = 6561 2
Hence, number to be added to 6412 = 81 — 6412 = 6561 — 6412 = 149
2
The required perfect square is 6561 and √6561= 81
Page : 108 , Block Name : Exercise 6.4
Q6 Find the length of the side of a square whose area is 441 m 2
Answer. Let the length of the side of the square be x m.
2 2
Area of square = (x) = 441m
x = √441
Page : 108 , Block Name : Exercise 6.4
Q7 In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, nd AC (b) If AC = 13 cm, BC = 5 cm, nd AB
(a) △ABC is right-angled at B .
Therefore, by applying Pythagoras theorem, we obtain
2 2 2
AC = AB + BC
Answer. AC
2 2 2
= (6cm) + (8cm)
2 2 2
AC = (36 + 64)cm = 100cm
AC = (√100)cm = (√10 × 10)cm
AC = 10cm
Page 37 of 39 Aglasem Schools
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Book : Mathematics Ncert Solutions | Chapter-6 Maths
(b) △ABC is right-angled at B .
Therefore, by applying Pythagoras theorem, we obtain
2 2 2
AC = AB + BC
2 2 2
(13cm) = (AB) + (5cm)
2 2 2 2 2
AB = (13cm) − (5cm) = (169 − 25)cm = 144cm
2 2
AB = (√144)cm = (√12 × 12)cm = 144cm
AB = 12cm
Page : 108 , Block Name : Exercise 6.4
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 nd 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 32 = 1024
2
Hence, number to be added to 1000 to make it a perfect square
= 32 - 1000 = 1024 - 1000 = 24
2
Thus, the required number of plants is 24.
Page 38 of 39 Aglasem Schools
(b) △ABC is right-angled at B .
Therefore, by applying Pythagoras theorem, we obtain
2 2 2
AC = AB + BC
2 2 2
(13cm) = (AB) + (5cm)
2 2 2 2 2
AB = (13cm) − (5cm) = (169 − 25)cm = 144cm
2 2
AB = (√144)cm = (√12 × 12)cm = 144cm
AB = 12cm
Page : 108 , Block Name : Exercise 6.4
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 nd 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 32 = 1024
2
Hence, number to be added to 1000 to make it a perfect square
= 32 - 1000 = 1024 - 1000 = 24
2
Thus, the required number of plants is 24.
Page 38 of 39 Aglasem Schools
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Book : Mathematics Ncert Solutions | Chapter-6 Maths
Page : 108 , Block Name : Exercise 6.4
Q9 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.
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.
Page : 108 , Block Name : Exercise 6.4
Page 39 of 39 Aglasem Schools
Page : 108 , Block Name : Exercise 6.4
Q9 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.
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.
Page : 108 , Block Name : Exercise 6.4
Page 39 of 39 Aglasem Schools
Document Details
| Board / Org | NCERT |
| Exam | Class 8 |
| Type | Solution |
| Pages | 40 |
| Language | english |
| Updated | 22 Jul 2026 |