NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

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Written by Team Trustudies
Updated at 2021-05-07


NCERT solutions for class 8 Maths Chapter 6 Squares And Square Roots 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) 20387

(vii) 52698

(viii) 99880

(ix) 12796

(x) 55555



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

As we know that unit's place is affected only by the unit’s place of the number of which square is seen. So we have: (i) Unit digit of 812 = 1 [Unit’s place=1 i.e, 1×1 =1]
(ii) Unit digit of 2722 = 4 [Unit’s place=2 i.e, 2×2 =4]

(iii) Unit digit of 7992 = 1 [Unit’s place=1 i.e, 9×9 =1]

(iv) Unit digit of 38532 = 9 [Unit’s place=3 i.e, 3×3 =9]

(v) Unit digit of 12342 = 6 [Unit’s place=4 i.e, 4×4 =6]

(vi) Unit digit of 263872 = 9 [Unit’s place=7 i.e, 7×7 =9]

(vii) Unit digit of 526982= 4 [Unit’s place=8 i.e, 8×8 =4]

(viii) Unit digit of 998802 = 0 [Unit’s place=0 i.e, 0×0 =0]

(ix) Unit digit of 127962 = 6 [Unit’s place=6 i.e, 6×6 =6]

(x) Unit digit of 555552 = 5 [Unit’s place=5 i.e, 5×5 =5]

2. The following numbers are not perfect squares. Give reason.
(i) 1057

(ii) 23453

(iii) 7928

(iv) 222222

(v) 64000

(vi) 89722

(vii) 222000

(viii) 505050



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

For checking a number to be perfect square we need to see the unit's digit or the number of digits repeating from the end of the number given.If the unit's digit is perfect square then it may be a perfect square.Also if the number of digits repeating are odd in number then it won't be a perfect square.So here we have:
(i) It is not a perfect square number because 1057 ends with 7 at unit place.

(ii) It is not a perfect square number because 23453 ends with 3 at unit place.

(iii) It is not a perfect square number because 7928 ends with 8 at unit place.

(iv) It is not a perfect square number because 222222 ends with 2 at unit place.

(v) It is not a perfect square number because 64000 ends with odd number of zeros i.e, 3.

(vi) It is not a perfect square number because 89722 ends with 2 at unit place.

(vii) It is not a perfect square number because 22000 ends with 3 zeros.

(viii) It is not a perfect square number because 505050 ends with 1 zero.

3.The squares of which of the following would be odd numbers?
(i) 431

(ii) 2826

(iii) 7779

(iv) 82004



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

The squares of a number would be odd number depending upon the unit's digit of the number.If the number has an even unit digit then it will have an even square number and if the unit's digit is odd then it will have an odd square number.Here we have:
(i) 4312 is an odd number as the unit's digit is odd.

(ii) 28262 is an even number as the unit's digit is even.

(iii) 77792 is an odd number as the unit's digit is odd.

(iv) 820042 is an even number as the unit's digit is even.

4.Observe the following pattern and find the missing digits.
112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1…2…1

100000012 = ………



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

According to the above pattern, we can observe that the squares of the given numbers have the same number of zeroes before and after the digit 2.Therefore, we have
1000012 = 10000200001

100000012 = 100000020000001

5.Observe the following pattern and supply the missing numbers.
112 = 121

1012 = 10201

101012 = 102030201

10101012 = ……….

.2 = 10203040504030201



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

According to the above pattern, we have
10101012 = 1020304030201

1010101012 = 10203040504030201

6.Using the given pattern, find the missing numbers.
12+22+22=32

22+32+62=72

32+42+122=132

42+52+.2=212

52+.2+302=312

62+72+..2=2



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

According to the given pattern, we have the third number as the product of the first two numbers and fourth number is obtained after adding 1 to the third number.

42+52+202=212

52+62+302=312

62+72+422=432

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

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

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



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

We know that the sum of n odd numbers =n2.So, we have:

(i) 1 + 3 + 5 + 7 + 9 = (5)2 = 25 [? n = 5]

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100 [? n = 10]

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12)2 = 144 [? n = 12]

8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

(i) 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)

(ii) 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)

9.How many numbers lie between squares of the following numbers?
(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100.



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

We know that numbers between n2 and (n+1)2 = 2n

(i) Numbers between 122 and 132 = (2n) = 2×12 = 24

(ii) Numbers between 252 and 262 = 2×25 = 50 [ n = 25]

(iii) Numbers between 992 and 1002 = 2×99 = 198 [ n = 99]

NCERT solutions for class 8 Maths Chapter 6 Squares And Square Roots Exercise 6.2

1.Find the square of the following numbers.
(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

(i) We have, 32=30+2

(32)2=(30+2)2

=30(30+2)+2(30+2)

=302+30×2+2×30+22

=900+60+60+4

=1024

So we have (32)2=1024

(ii)We have, 35=(30+5)

(35)2=(30+5)2

=30(30+5)+5(30+5)

=(30)2+30×5+5×30+(5)2

=900+150+150+25

=1225

So we have (35)2=1225

(iii) We have, 86=(80+6)

862=(80+6)2

=80(80+6)+6(80+6)

=(80)2+80×6+6×80+(6)2

=6400+480+480+36

=7396

So we have, (86)2=7396

(iv)We have, 93=(90+3)

932=(90+3)2

=90(90+3)+3(90+3)

=(90)2+90×3+3×90+(3)2

=8100+270+270+9

=8649

So we have, (93)2=8649

(v) We have, 71=(70+1)

712=(70+1)2

=70(70+1)+1(70+1)

=(70)2+70×1+1×70+(1)2

=4900+70+70+1

=5041

So we have, (71)2=5041

(vi)We have, 46=(40+6)

462=(40+6)2

=40(40+6)+6(40+6)

=(40)2+40×6+6×40+(6)2

=1600+240+240+36

=2116

So we have, (46)2=2116

2.Write a Pythagorean triplet whose one member is
(i) 6

(ii) 14

(iii) 16

(iv) 18



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

For a natural number m greater than 1,Triplets are in the form 2m, m2 – 1, m2 + 1
(i) Let m21=6

?m2=6+1=7

So,here we observe that value of m is not integer.

Now, let us try for m2+1=6

?m2=61=5

So, here too we have no integer value of m.

Now, let 2m = 6

?m=3whichisaninteger.

Other number are:

m21=321=8 and m2+1=32+1=10

Hence, the required triplets are 6, 8 and 10.

(ii) Let m21=14

?m2=1+14=15

So, we observe that the value of m will not be an integer.

Now take 2m = 14

? m = 7 which is an integer.

The other numbers of triplets are 2m =2×7 = 14

m21=(7)21=491=48

and m2+1=(7)2+1=49+1=50

Hence the triplets are 14, 48 and 50.

(iii) Let, 2m = 16

?m = 8

The other numbers are

m21=(8)21=641=63

m2+1=(8)2+1=64+1=65

Hence the triplets are 16, 63 and 65

(iv) Let 2m = 18

? m = 9

The other triplets are:

m21=(9)21=811=80

and m2+1=(9)2+1=81+1=82

Hence we have the triplets as 18, 80 and 82.

NCERT solutions for class 8 Maths Chapter 6 Squares And Square Roots 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



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

One’s digit of any number depends on the one’s digit of the number of which it is square of. Here we have:
(i) One’s digit in the square root of 9801 can be 1 or 9.

(ii) One’s digit in the square root of 99856 can be 4 or 6.

(iii) One’s digit in the square root of 998001 can be 1 or 9.

(iv) One’s digit in the square root of 657666025 can be 5.

2.Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153

(ii) 257

(iii) 408

(iv) 441



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares because they don't exist at the end of any square number.
(i) 153 is not a perfect square number because it ends with 3.

(ii) 257 is not a perfect square number because it ends with 7.

(iii) 408 is not a perfect square number because it ends with 8.

(iv) 441 is a perfect square number.

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



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

Using the method of repeated subtraction of consecutive odd numbers, we have

(i) 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

Thus it takes 10 times repetition to reach 0.

So, 100=10

(ii) 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

It takes thirteen times repetition to reach 0.

Hence, 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



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

(i) We have prime factors of 729 as:

3729324338132739331

729=3×3×3×3×3×3=32×32×32

?729=3×3×3=27

(ii) We have prime factors of 400 as

240022002100250525551

400=2×2×2×2×5×5=22×22×52

?400=2×2×5=20

(iii) We have prime factors of 1764 as:

21764288234413147749771

1764=2×2×3×3×7×7=22×32×72

?1764=2×3×7=42

(iv) We have prime factors of 4096 as:

2409622048210242512225621282642322162824221

4096=2×2×2×2×2×2×2×2×2×2×2×2

=22×22×22×22×22×22

?4096=2×2×2×2×2×2=64

(v)We have prime factors of 7744 as:

2774423872219362968248422421112111111

7744=2×2×2×2×2×2×11×11

=22×22×22×112

?7744=2×2×2×11=88

(vi) We have prime factors of 9604 as:

2960424802724017343749771

9604=2×2×7×7×7×7=22×72×72

?9604=2×7×7=98

(vii) We have prime factors of 5929 as:

7592978471112111111

5929=7×7×11×11=72×112

?5929=7×11=77

(viii) We have prime factors of 9216 as:

2921624608223042115225762288214427223621839331

9216=2×2×2×2×2×2×2×2×2×2×3×3

=22×22×22×22×22×32

?9216=2×2×2×2×2×3=96

(ix) We have prime factors of 529 as:

2352923231

529=23×23=232

?529=23

(x)We have prime factors of 8100 as:

28100240503202536753225375525551

8100=2×2×3×3×3×3×5×5=22×32×32×52

?8100=2×3×3×5=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



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

(i) We have prime factors of 252 as:
22522126363321771

252=2×2×3×3×7

Here, the prime factors are not in pair as 7 has no pair. Thus, the smallest whole number by which the given number is to be multiplied to get a perfect square number is 7.

The new square number is 252×7=1764

?Square root of 1764 is

1764=2×3×7=42

(ii)We have prime factors of180 as:

218029054539331

180=2×2×3×3×5

Here, 5 has no pair.So the least number to be multiplied to get perfect square is 5

New square number = 180×5=900

The square root of 900 is

?900=2×3×5=30

(iii) We have prime factors of 1008 as:

21008250422522126363321771

1008=2×2×2×2×3×3×7

Here, 7 has no pair.So the least number to be multiplied to get perfect square is 7

New square number = 1008×7=7056

Square root of 7056 is

?7056=2×2×3×7=84

(iv) We have prime factors of 2028 as:

220282101435071316913131

2028=2×2×3×13×13

Here, 3 is not in pair.So the least number to be multiplied to get perfect square is 3

New square number =2028×3=6084

Square root of 6084 is

?6084=2×13×3=78

(v)We have prime factors of 1458 as:

214582729324338132739331

1458=2×3×3×3×3×3×3

Here, 2 is not in pair.So the least number to be multiplied to get perfect square is 2

New square number =1458×2=2916

Square root of 2916 is

?2916=3×3×3×2=54

(vi) We have prime factors of 768 as:

27682384219229624822421226331

768=2×2×2×2×2×2×2×2×3

Here, 3 is not in pair.So the least number to be multiplied to get perfect square is 3

New square number = 768×3=2304

Square root of 2304 is

?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



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

(i) We have prime factors of 252 as:
22522126363321771

252=2×2×3×3×7

Here, the prime factors are not in pair as 7 has no pair. Thus, the smallest whole number by which the given number is to be divided to get a perfect square number is 7.

The new square number is 252÷7=36

?Square root of 36 is 36=6

(ii) We have prime factors of 2925 as:
329253975532556513131

2925=3×3×5×5×13

Here, 13 has no pair.So, the smallest whole number by which 2925 is divided to get a square number is 13.

New square number =2925÷13=225

? 225=15

(iii)We have prime factors of 396 as:

2396219839933311111

396=2×2×3×3×11

Here 11 is not in pair.So,the required smallest whole number by which 396 is divided to get a square number is 11.

New square number = 396÷11=36

? 36=6

(iv)We have prime factors of 2645 as:

526452352923231

2645=5×23×23

Here, 5 is not in pair.So, 5 is the required smallest whole number by which 2645 is divided to get a square number

New square number = 2645÷5=529

? 529=23

(v) We have prime factors of 2800 as:

2280021400270023505175535771

2800=2×2×2×2×5×5×7

Here, 7 is not in pair.So, 7 is the required smallest number by which 2800 is divided to get a square number.

New square number =2800÷7=400

? 400=20

(vi)We have prime factors of 1620 as:

21620281034053135345315551

1620=2×2×3×3×3×3×5

Here, 5 is not in pair.So, 5 is the required smallest prime number by which 1620 is divided to get a square number = 1620÷5=324

? 324=18

7.The students of class VIII of a school donated ? 2401 in all, for the 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.



NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots


Answer :

Let x be the number of students in the class. Therefore, total money donated by students:x×x=?x2
Here we have, x2=2401