NCERT Solutions for Class 8th: Ch 6 Squares and Square Roots Part-1

NCERT Solutions for Class 8th: Ch 6 Squares and Square Roots Maths Part-1

Page No: 96

Exercise 6.1

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

The unit digit of square of a number having 'a' at its unit place ends with a×a.

(i) The unit digit of the square of a number having digit 1 as unit’s place is 1.
∴ Unit digit of the square of number 81 is equal to 1.

(ii) The unit digit of the square of a number having digit 2 as unit’s place is 4.
∴ Unit digit of the square of number 272 is equal to 4.

(iii) The unit digit of the square of a number having digit 9 as unit’s place is 1.
∴ Unit digit of the square of number 799 is equal to 1.

(iv) The unit digit of the square of a number having digit 3 as unit’s place is 9.
∴ Unit digit of the square of number 3853 is equal to 9.

(v) The unit digit of the square of a number having digit 4 as unit’s place is 6.
∴ Unit digit of the square of number 1234 is equal to 6.

(vi) The unit digit of the square of a number having digit 7 as unit’s place is 9.
∴ Unit digit of the square of number 26387 is equal to 9.

(vii) The unit digit of the square of a number having digit 8 as unit’s place is 4.
∴ Unit digit of the square of number 52698 is equal to 4.

(viii) The unit digit of the square of a number having digit 0 as unit’s place is 01.
∴ Unit digit of the square of number 99880 is equal to 0.

(ix) The unit digit of the square of a number having digit 6 as unit’s place is 6.
∴ Unit digit of the square of number 12796 is equal to 6.

(x) The unit digit of the square of a number having digit 5 as unit’s place is 5.
∴ Unit digit of the square of number 55555 is equal to 5.

2. The following numbers are obviously not perfect squares. Give reason.
(i) 1057 
(ii) 23453 
(iii) 7928 
(iv) 222222
(v) 64000 
(vi) 89722 
(vii) 222000 
(viii) 505050

Solution

As we know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect square.
∴ 1057, 23453, 7928, 222222, 64000, 89722, 222000 and 505050 are not perfect squares.

3. The following numbers are obviously not perfect squares. Give reason.

Solution

As we know that the square of an odd number is odd and the square of an even number is even. Therefore,
(i) The square of 431 is an odd number.
(ii) The square of 2826 is an even number.
(iii) The square of 7779 is an odd number.
(iv)The square of 82004 is an even number.

4.The squares of which of the following would be odd numbers?
112 = 121 
1012 = 10201 
10012 = 1002001 
1000012 = 1 .......2.........1 
100000012 = ..........................

Solution

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.
∴ 100001= 10000200001
100000012 = 100000020000001

5. Observe the following pattern and supply the missing numbers. 
112 = 121 
1012 = 10201 
101012 = 102030201 
10101012 = ...........................
............2  = 10203040504030201

Solution

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And, the square is symmetric about the middle digit. If the middle digit is 4, then the number to be squared is 10101 and its square is 102030201.
So, 1010101=1020304030201
1010101012 =10203040505030201

6. Using the given pattern, find the missing numbers. 
12 + 22 + 22 = 32
22  + 32 + 62 = 72 
32  + 42 + 122 = 132 
42  + 52 + _ = 212
5  + _ 2 + 302 = 312
6  + 7 + _ 2 = __2

Solution

Given, 12  + 22 + 2= 32
i.e  1+ 2+ (1×2 )= ( 1+ 2-1 × 2 )2

• 2+ 3+ 6=72
∴ 2+ 3+ (2×3 )= (2+ 3-2 × 3)

• 3+ 4+ 12= 132
∴ 3+ 4+ (3×4 )= (3+ 4- 3 × 4)

 • 4+ 5+ (4×5 )= (4 + 5- 4 × 5)
∴ 4+ 5+ 20= 212

• 5+ 6+ (5×6 )= (52+ 6- 5 × 6)2
∴ 5+ 6 + 30 = 312

• 6+ 7+ (6×7 )= (6+ 7- 6 × 7)
∴ 6+ 72 + 42= 432

7. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9 

Solution

Sum of first five odd number = (5)= 25

(ii) 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

Solution

Sum of first ten odd number = (10)= 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution

Sum of first thirteen odd number = (12)= 144

8. (i) Express 49 as the sum of 7 odd numbers.

Solution

We know, sum of first n odd natural numbers is n.
Since, 49 = 7
∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers.

Solution

Since, 121 = 112
∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

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

Solution

Between n2 and (n+1)2, there are 2n non–perfect square numbers.
(i) 12and 132 there are 2×12 = 24 natural numbers.
(ii) 252 and 262 there are 2×25 = 50 natural numbers.
(iii) 992 and 100 there are 2×99 =198 natural numbers.

Exercise 6.2

1. Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46

Solution

(i) (32)2
= (30 +2)2
= (30)+ (2)+ 2×30×2 [Since, (a+b)= a2+b2 +2ab]
= 900 + 4 + 120
= 1024

(ii) (35)2
= (30+5 )2
= (30)+ (5)+ 2×30×5 [Since, (a+b)= a2+b+2ab]
= 900 + 25 + 300
= 1225

(iii) (86)2
= (90 - 4)2
= (90)+ (4)- 2×90×4 [Since, (a-b)= a+ b- 2ab]
= 8100 + 16 - 720
= 8116 - 720
= 7396

(iv)  (93)2
= (90+3 )2
= (90)+ (3)+ 2×90×3 [Since, (a+b)= a2+b+2ab]
= 8100 + 9 + 540
= 8649

(v) (71)2
= (70+1 )2
= (70)+ (1)+2×70×1 [Since, (a+b)= a2+b+2ab]
= 4900 + 1 + 140
= 5041

(v) (46)2
= (50 -4 )2
= (50)+ (4)- 2×50×4 [Since, (a-b)= a+ b- 2ab]
= 2500 + 16 - 400
= 2516 - 400
= 2116


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

Solution

For any natural number m, we know that
2m, m2–1, m2+1 is a Pythagorean triplet.

(i) 2m = 6
⇒ m = 6/2 = 3

m2–1= 32 – 1 = 9–1 = 8

m2+1= 32+1 = 9+1 = 10

∴ (6, 8, 10) is a Pythagorean triplet.

(ii) 2m = 14
⇒ m = 14/2 = 7

m2–1= 72–1 = 49–1 = 48

m2+1 = 72+1 = 49+1 = 50

∴ (14, 48, 50) is not a Pythagorean triplet.

(iii) 2m = 16
⇒ m = 16/2 = 8

m2–1 = 82–1 = 64–1 = 63

m2+ 1 = 82+1 = 64+1 = 65

∴ (16, 63, 65) is a Pythagorean triplet.

(iv) 2m = 18
⇒ m = 18/2 = 9

m2–1 = 92–1 = 81–1 = 80

m2+1 = 92+1 = 81+1 = 82

∴ (18, 80, 82) is a Pythagorean triplet.

Exercise 6.3

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

(i) As we know that the unit’s digit of the square of a number having digit as unit’s place 1 is 1.
∴ Unit’s digit of the square root of number 9801 is equal to 1.

(ii) As we know that the unit’s digit of the square of a number having digit as unit’s place 6 is 6.
∴ Unit’s digit of the square root of number 99856 is equal to 6.

(iii) As we know that the unit’s digit of the square of a number having digit as unit’s place 1 is 1.
∴ Unit’s digit of the square root of number 998001 is equal to 1.

(iv) As we know that the unit’s digit of the square of a number having digit as unit’s place 5 is 5.
∴ Unit’s digit of the square root of number 657666025 is equal to 5.

2. Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153
(ii) 257
(iii) 408 
(iv) 441

Solution

As we know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect square.
∴ 153, 257, 408 are not perfect squares and 441 is a perfect square.

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Solution

• 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
Therefore, we have performed subtraction tenth times.
∴ √100 = 10

• 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
Therefore, we have performed subtraction thirteenth times.
∴ √169 = 13

4.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

Solution

(i)
729 = 3×3×3×3×3×3×1
⇒ 729 = (3×3)×(3×3)×(3×3) 
⇒ 729 = (3×3×3)×(3×3×3)
⇒ 729 = (3×3×3)2
⇒ √729 = 3×3×3 = 27

(ii)
400 = 2×2×2×2×5×5×1
⇒ 400 = (2×2)×(2×2)×(5×5) 
⇒ 400 = (2×2×5)×(2×2×5)
⇒ 400 = (2×2×5)2
⇒ √400 = 2×2×5 = 20

(iii)
1764 = 2×2×3×3×7×7
⇒ 1764 = (2×2)×(3×3)×(7×7)
⇒ 1764 = (2×3×7)×(2×3×7) 
⇒ 1764 = (2×3×7)2
⇒ √1764 = 2 ×3×7 = 42

(iv) 
4096 = 2×2×2×2×2×2×2×2×2×2×2×2 
⇒ 4096 = (2×2)×(2×2)×(2×2)×(2×2)×(2×2)×(2×2)
⇒ 4096 = (2×2×2×2×2×2)×(2×2×2×2×2×2)
⇒ 4096 = (2×2×2×2×2×2)2
⇒ √4096 = 2×2×2 ×2×2×2 = 64

(v)
7744 = 2×2×2×2×2×2×11×11×1
⇒ 7744 = (2×2)×(2×2)×(2×2)×(11×11)
⇒ 7744 = (2×2×2×11)×(2×2×2×11)
⇒ 7744 = (2×2×2×11)2
⇒ √7744 = 2×2×2×11 = 88

(vi) 

9604 = 62 × 2 × 7 × 7 × 7 × 7 
⇒ 9604 = ( 2 × 2 ) × ( 7 × 7 ) × ( 7 × 7 ) 
⇒ 9604 = ( 2 × 7 ×7 ) × ( 2 × 7 ×7 )
⇒ 9604 = ( 2×7×7 )2
⇒ √9604 = 2×7×7
               = 98

(vii) 
5929 = 7×7×11×11 
⇒ 5929 = (7×7)×(11×11) 
⇒ 5929 = (7×11)×(7×11) 
⇒ 5929 = (7×11)2
⇒ √5929 = 7×11 = 77

(viii)
9216 = 2×2×2×2×2×2×2×2×2×2×3×3×1
⇒ 9216 = (2×2)×(2×2) × ( 2 × 2 ) × ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 )  
⇒ 9216  =  ( 2 × 2 × 2 × 2 × 2 × 3)  × ( 2 × 2 × 2 × 2 × 2 × 3)  
⇒ 9216 =   96 × 96
⇒ 9216 = ( 96 )2
⇒ √9216 = 96

(ix) 
529 = 23×23
⇒ 529 = (23)2
⇒ √529 = 23

(x) 
8100 = 2×2×3×3×3×3×5×5×1
⇒ 8100 = (2×2) ×(3×3)×(3×3)×(5×5)  
⇒ 8100 = (2×3×3×5)×(2×3×3×5)  
⇒ 8100 = 90×90
⇒ 8100 = (90)2
⇒ √8100 = 90

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 = 2×2×3×3×7
= (2×2)×(3×3)×7
Here, 7 cannot be paired.
∴ We will multiply 252 by 7 to get perfect square.
New number = 252 × 7 = 1764
1764 = 2×2×3×3×7×7
⇒ 1764 = (2×2)×(3×3)×(7×7)
⇒ 1764 = 22×32×72
⇒ 1764 = (2×3×7)2
⇒ √1764 = 2×3×7 = 42

(ii) 
180 = 2×2×3×3×5
= (2×2)×(3×3)×5
Here, 5 cannot be paired.
∴ We will multiply 180 by 5 to get perfect square.
New number = 180 × 5 = 900
900 = 2×2×3×3×5×5×1  
⇒ 900 = (2×2)×(3×3)×(5×5) 
⇒ 900 = 22×32×52  
⇒ 900 = (2×3×5)2
⇒ √900 = 2×3×5 = 30

(iii) 
1008 =  2×2×2×2×3×3×7
= (2×2)×(2×2)×(3×3)×7
Here, 7 cannot be paired.
∴ We will multiply 1008 by 7 to get perfect square.
New number = 1008×7 = 7056

7056 = 2×2×2×2×3×3×7×7  
⇒ 7056 = (2×2)×(2×2)×(3×3)×(7×7)
⇒ 7056 = 22×22×32×72  
⇒ 7056 = (2×2×3×7)2
⇒ √7056 = 2×2×3×7 = 84

(iv) 
2028 = 2×2×3×13×13
= (2×2)×(13×13)×3
Here, 3 cannot be paired.
∴ We will multiply 2028 by 3 to get perfect square.
New number = 2028×3 = 6084


6084 = 2×2×3×3×13×13 
⇒ 6084 = (2×2)×(3×3)×(13×13)
⇒ 6084 = 22×32×132  
⇒ 6084 = (2×3×13)2
⇒ √6084 = 2×3×13 = 78

(v) 
1458 = 2×3×3×3×3×3×3
= (3×3)×(3×3)×(3×3)×2 
Here, 3 cannot be paired.
∴ We will multiply 1458 by  2 to get perfect square.
New number = 1458 × 2 = 2916
2916 = 2×2×3×3×3×3×3×3
⇒ 2916 = (3×3)×(3×3)×(3×3)×(2×2)
⇒ 2916 =  32×32×32×22  
⇒ 2916 = (3×3×3×2)2
⇒ √2916 = 3×3×3×2 = 54

(vi) 
768 = 2×2×2×2×2×2×2×2×3
= (2×2)×(2×2)×(2×2)×(2×2)×3
Here, 3 cannot be paired.
∴ We will multiply 768 by 3 to get perfect square.
New number = 768×3 = 2304

2304 = 2×2×2×2×2×2×2×2×3×3
⇒ 2304 = (2×2)×(2×2)×(2×2)×(2×2)×(3×3)
⇒ 2304 = 22×22×22×22×32
⇒ 2304 = (2×2×2×2×3)2
⇒ √2304 = 2×2×2×2×3 = 48

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

Solution

(i)
252 = 2×2×3×3×7
= (2×2)×(3×3)×7
Here, 7 cannot be paired.
∴ We will divide 252 by 7 to get perfect square.
New number = 252 ÷ 7 = 36
36 = 2×2×3×3 
⇒ 36 = (2×2)×(3×3)
⇒ 36 = 22×32
⇒ 36 = (2×3)2
⇒ √36 = 2×3 = 6

(ii) 
2925 = 3×3×5×5×13
= (3×3)×(5×5)×13
Here, 13 cannot be paired.
∴ We will divide 2925 by 13 to get perfect square.
New number = 2925 ÷ 13 = 225

225 = 3×3×5×5 
⇒ 225 = (3×3)×(5×5) 
⇒ 225 = 32×52
⇒ 225 = (3×5)2
⇒ √36 = 3×5 = 15

(iii)
396 = 2×2×3×3×11
= (2×2)×(3×3)×11
Here, 11 cannot be paired.
∴ We will divide 396 by 11 to get perfect square.
New number = 396 ÷ 11 = 36
36 = 2×2×3×3 
⇒ 36 = (2×2)×(3×3)
⇒ 36 = 22×32
⇒ 36 = (2×3)2
⇒ √36 = 2×3 = 6

(iv)
2645 = 5×23×23
⇒ 2645 = (23×23)×5 
Here, 5 cannot be paired.
∴ We will divide 2645 by 5 to get perfect square.
New number = 2645 ÷ 5 = 529
529 = 23×23
⇒ 529 = (23)2
⇒ √529 = 23

(v)
2800 = 2×2×2×2×5×5×7
= (2×2)×(2×2)×(5×5)×7
Here, 7 cannot be paired.
∴ We will divide 2800 by 7 to get perfect square.
New number = 2800 ÷ 7 = 400
400 = 2×2×2×2×5×5 
⇒ 400 = (2×2)×(2×2)×(5×5) 
⇒ 400 = (2×2×5)2
⇒ √400 = 20

(vi)
1620 =  2×2×3×3×3×3×5
= (2×2)×(3×3)×(3×3)×5
Here, 5 cannot be paired.
∴ We will divide 1620 by 5 to get perfect square.
New number = 1620 ÷ 5 = 324

324 = 2×2×3×3×3×3 
⇒ 324 = (2×2)×(3×3)×(3×3)
⇒ 324 = (2×3×3)2
⇒ √324 = 18

7. 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.

Solution

Let the number of students in the school be, x.
∴ Each student donate Rs.x .
Total many contributed by all the students = x×x = x2
Given, x2 = Rs.2401
x2 = 7×7×7×7
⇒ x2  = (7×7)×(7×7)  
⇒ x2 = 49×49
⇒ x =
⇒ x = 49
∴ The number of students = 49

8. 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.

Solution

Let the number of rows be, x.
∴ the number of plants in each rows =  x.
Total many contributed by all the students = x×x = x2
Given,
x2 = Rs.2025
x2 = 3×3×3×3×5×5
⇒ x2 = (3×3)×(3×3)×(5×5)  
⇒ x2 = (3×3×5)×(3×3×5)
⇒ x2 = 45×45
⇒ x =
⇒ x = 45
∴ The number of rows = 45 and the number of plants in each rows = 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution


L.C.M of 4, 9 and 10 is (2×2×9×5) 180.

180 = 2×2×9×5
= (2×2)×3×3×5 
= (2×2)×(3×3)×5
Here, 5 cannot be paired. 
∴ we will multiply 180 by 5 to get perfect square.
Hence, the smallest square number divisible by 4, 9 and 10 = 180×5 = 900

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution

L.C.M of 8, 15 and 20 is (2×2×5×2×3) 120.
120 = 2×2×3×5×2
= (2×2)×3×5×2
Here, 3, 5 and 2 cannot be paired. 
∴ We will multiply 120 by (3×5×2) 30  to get perfect square.
Hence, the smallest square number divisible by 8, 15 and 20 =120×30 = 3600


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