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

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 $$81^2$$ = 1 [Unit’s place=1 i.e, $$1 \times 1$$ =1]
(ii) Unit digit of $$272^2$$ = 4 [Unit’s place=2 i.e, $$2 \times 2$$ =4]

(iii) Unit digit of $$799^2$$ = 1 [Unit’s place=1 i.e, $$9 \times 9$$ =1]

(iv) Unit digit of $$3853^2$$ = 9 [Unit’s place=3 i.e, $$3 \times 3$$ =9]

(v) Unit digit of $$1234^2$$ = 6 [Unit’s place=4 i.e, $$4 \times 4$$ =6]

(vi) Unit digit of $$26387^2$$ = 9 [Unit’s place=7 i.e, $$7 \times 7$$ =9]

(vii) Unit digit of $$52698^2$$= 4 [Unit’s place=8 i.e, $$8 \times 8$$ =4]

(viii) Unit digit of $$99880^2$$ = 0 [Unit’s place=0 i.e, $$0 \times 0$$ =0]

(ix) Unit digit of $$12796^2$$ = 6 [Unit’s place=6 i.e, $$6 \times 6$$ =6]

(x) Unit digit of $$55555^2$$ = 5 [Unit’s place=5 i.e, $$5 \times 5$$ =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) $$431^2$$ is an odd number as the unit's digit is odd.

(ii) $$2826^2$$ is an even number as the unit's digit is even.

(iii) $$7779^2$$ is an odd number as the unit's digit is odd.

(iv) $$82004^2$$ is an even number as the unit's digit is even.

4.Observe the following pattern and find the missing digits.
$$11^2$$ = 121

$$101^2$$ = 10201

$$1001^2$$ = 1002001

$$100001^2$$ = 1…2…1

$$10000001^2$$ = ………

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
$$100001^2$$ = 10000200001

$$10000001^2$$ = 100000020000001

5.Observe the following pattern and supply the missing numbers.
$$11^2$$ = 121

$$101^2$$ = 10201

$$10101^2$$ = 102030201

$$1010101^2$$ = ……….

$$……….^2$$ = 10203040504030201

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

Answer :

According to the above pattern, we have
$$1010101^2$$ = 1020304030201

$$101010101^2$$ = 10203040504030201

6.Using the given pattern, find the missing numbers.
$$1^2 + 2^2 + 2^2 = 3^2$$

$$2^2 + 3^2 + 6^2 = 7^2$$

$$3^2 + 4^2 + 12^2 = 13^2$$

$$4^2 + 5^2 + ….^2 = 21^2$$

$$5^2 + ….^2 + 30^2 = 31^2$$

$$6^2 + 7^2 + …..^2 = ……2$$

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.

$$4^2 + 5^2 + 20^2 = 21^2$$

$$5^2 + 6^2 + 30^2 = 31^2$$

$$6^2 + 7^2 + 42^2 = 43^2$$

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 =$$n^2$$.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 $$n^2$$ and $$(n + 1)^2$$ = 2n

(i) Numbers between $$12^2$$ and $$13^2$$ = (2n) = $$2 \times 12$$ = 24

(ii) Numbers between $$25^2$$ and $$26^2$$ = $$2 \times 25$$ = 50 [ n = 25]

(iii) Numbers between $$99^2$$ and $$100^2$$ = $$2 \times 99$$ = 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, $$32 = 30 + 2$$

$$(32)^2 = (30 + 2)^2$$

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

$$= 30^2 + 30 \times 2 + 2 \times 30 + 2^2$$

$$= 900 + 60 + 60 + 4$$

$$= 1024$$

So we have $$(32)^2 = 1024$$

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

$$(35)^2 = (30 + 5)^2$$

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

$$= (30)^2 + 30 \times 5 + 5 \times 30 + (5)^2$$

$$= 900 + 150 + 150 + 25$$

$$= 1225$$

So we have $$(35)^2 = 1225$$

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

$$86^2 = (80 + 6)^2$$

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

$$= (80)^2 + 80 \times 6 + 6 \times 80 + (6)^2$$

$$= 6400 + 480 + 480 + 36$$

$$= 7396$$

So we have, $$(86)^2 = 7396$$

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

$$93^2 = (90 + 3)^2$$

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

$$= (90)^2 + 90 \times 3 + 3 \times 90 + (3)^2$$

$$= 8100 + 270 + 270 + 9$$

$$= 8649$$

So we have, $$(93)^2 = 8649$$

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

$$71^2 = (70 + 1)^2$$

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

$$= (70)2 + 70 \times 1 + 1 \times 70 + (1)2$$

$$= 4900 + 70 + 70 + 1$$

$$= 5041$$

So we have, $$(71)^2 = 5041$$

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

$$46^2 = (40 + 6)^2$$

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

$$= (40)^2 + 40 \times 6 + 6 \times 40 + (6)^2$$

$$= 1600 + 240 + 240 + 36$$

$$= 2116$$

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 $$m^2 – 1 = 6$$

$$\Rightarrow m^2 = 6 + 1 = 7$$

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

Now, let us try for $$m^2 + 1 = 6$$

$$\Rightarrow m^2 = 6 – 1 = 5$$

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

Now, let 2m = 6

$$\Rightarrow m = 3\; which\; is \;an\; integer.$$

Other number are:

$$m^2 – 1 = 32 – 1 = 8$$ and $$m^2 + 1 = 32 + 1 = 10$$

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

(ii) Let $$m^2 – 1 = 14$$

$$\Rightarrow m^2 = 1 + 14 = 15$$

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

Now take 2m = 14

$$\Rightarrow$$ m = 7 which is an integer.

The other numbers of triplets are 2m =$$2 \times 7$$ = 14

$$m^2 – 1 = (7)2 – 1 = 49 – 1 = 48$$

and $$m^2 + 1 = (7)2 + 1 = 49 + 1 = 50$$

Hence the triplets are 14, 48 and 50.

(iii) Let, 2m = 16

$$\Rightarrow$$m = 8

The other numbers are

$$m^2 – 1 = (8)2 – 1 = 64 – 1 = 63$$

$$m^2 + 1 = (8)2 + 1 = 64 + 1 = 65$$

Hence the triplets are 16, 63 and 65

(iv) Let 2m = 18

$$\Rightarrow$$ m = 9

The other triplets are:

$$m^2 – 1 = (9)2 – 1 = 81 – 1 = 80$$

and $$m^2 + 1 = (9)2 + 1 = 81 + 1 = 82$$

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, $$\sqrt{100} = 10$$

(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, $$\sqrt{169} = 13$$

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:

$$\begin{array}{c|lcr} 3 & 729\\ \hline 3 & 243\\ \hline 3 & 81\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1 \end{array}$$

$$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^2 \times 3^2 \times 3^2$$

∴$$\sqrt{729}= 3 \times 3 \times 3 = 27$$

(ii) We have prime factors of 400 as

$$\begin{array}{c|lcr} 2 & 400\\ \hline 2 & 200\\ \hline 2 & 100\\ \hline 2 & 50\\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1 \end{array}$$

$$400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 = 2^2 \times 2^2 \times 5^2$$

∴$$\sqrt{400} = 2 \times 2 \times 5 = 20$$

(iii) We have prime factors of 1764 as:

$$\begin{array}{c|lcr} 2 & 1764\\ \hline 2 & 882\\ \hline 3 & 441\\ \hline 3 & 147\\ \hline 7 & 49\\ \hline 7 & 7\\ \hline & 1 \end{array}$$

$$1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7 = 2^2 \times 3^2 \times 7^2$$

∴$$\sqrt{1764 }= 2 \times 3 \times 7 = 42$$

(iv) We have prime factors of 4096 as:

$$\begin{array}{c|lcr} 2 & 4096\\ \hline 2 & 2048\\ \hline 2 & 1024\\ \hline 2 & 512\\ \hline 2 &256\\ \hline 2 & 128\\ \hline 2 & 64\\ \hline 2 & 32\\ \hline 2 & 16\\ \hline 2 & 8\\ \hline 2 & 4\\ \hline 2 & 2\\ \hline & 1 \end{array}$$

$$4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$$

$$= 2^2 \times 2^2 \times 2^2 \times 2^2 \times 2^2 \times 2^2$$

∴$$\sqrt{4096} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$$

(v)We have prime factors of 7744 as:

$$\begin{array}{c|lcr} 2 & 7744\\ \hline 2 & 3872\\ \hline 2 & 1936\\ \hline 2 & 968\\ \hline 2 & 484\\ \hline 2 & 242\\ \hline 11 & 121\\ \hline 11 & 11\\ \hline & 1\\ \end{array}$$

$$7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11$$

$$= 2^2 \times 2^2 \times 2^2 \times 11^2$$

∴$$\sqrt{7744} = 2 \times 2 \times 2 \times 11 = 88$$

(vi) We have prime factors of 9604 as:

$$\begin{array}{c|lcr} 2 & 9604\\ \hline 2 & 4802\\ \hline 7 & 2401\\ \hline 7 & 343\\ \hline 7 & 49\\ \hline 7 & 7\\ \hline & 1 \end{array}$$

$$9604 = 2 \times 2 \times 7 \times 7 \times 7 \times 7 = 2^2 \times 7^2 \times 7^2$$

∴$$\sqrt{9604} = 2 \times 7 \times 7 = 98$$

(vii) We have prime factors of 5929 as:

$$\begin{array}{c|lcr} 7 & 5929\\ \hline 7 & 847\\ \hline 11 & 121\\ \hline 11 & 11\\ \hline & 1 \end{array}$$

$$5929 = 7 \times 7 \times 11 \times 11 = 7^2 \times 11^2$$

∴$$\sqrt{5929} = 7 \times 11 = 77$$

(viii) We have prime factors of 9216 as:

$$\begin{array}{c|lcr} 2 & 9216\\ \hline 2 & 4608\\ \hline 2 & 2304\\ \hline 2 & 1152\\ \hline 2 & 576\\ \hline 2 & 288\\ \hline 2 & 144\\ \hline 2 & 72\\ \hline 2 & 36\\ \hline 2 & 18\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1\\ \end{array}$$

$$9216 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$$

$$= 2^2 \times 2^2 \times 2^2 \times 2^2 \times 2^2 \times 3^2$$

∴$$\sqrt{9216 }= 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 96$$

(ix) We have prime factors of 529 as:

$$\begin{array}{c|lcr} 23 & 529\\ \hline 23 & 23\\ \hline & 1 \end{array}$$

$$529 = 23 \times 23 = 23^2$$

∴$$\sqrt{529} = 23$$

(x)We have prime factors of 8100 as:

$$\begin{array}{c|lcr} 2 & 8100\\ \hline 2 & 4050\\ \hline 3 & 2025\\ \hline 3 & 675\\ \hline 3 & 225\\ \hline 3 & 75\\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1\\ \end{array}$$

$$8100 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 = 2^2 \times 3^2 \times 3^2 \times 5^2$$

∴$$\sqrt{8100} = 2 \times 3 \times 3 \times 5 = 90$$

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:
$$\qquad \begin{array}{c|lcr} 2 & 252\\ \hline 2 & 126\\ \hline 3 & 63\\ \hline 3 & 21\\ \hline 7 & 7\\ \hline & 1\\ \end{array}$$

$$252 = 2 \times 2 \times 3 \times 3 \times 7$$

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 $$252 \times 7 = 1764$$

∴Square root of 1764 is

$$\sqrt{1764} = 2 \times 3 \times 7 = 42$$

(ii)We have prime factors of180 as:

$$\begin{array}{c|lcr} 2 & 180\\ \hline 2 & 90\\ \hline 5 & 45\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1\\ \end{array}$$

$$180 = 2 \times 2 \times 3 \times 3 \times 5$$

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

New square number = $$180 \times 5 = 900$$

The square root of 900 is

∴$$\sqrt{900} = 2 \times 3 \times 5 = 30$$

(iii) We have prime factors of 1008 as:

$$\begin{array}{c|lcr} 2 & 1008\\ \hline 2 & 504\\ \hline 2 & 252\\ \hline 2 & 126\\ \hline 3 & 63\\ \hline 3 & 21\\ \hline 7 & 7\\ \hline & 1\\ \end{array}$$

$$1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7$$

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

New square number = $$1008 \times 7 = 7056$$

Square root of 7056 is

∴$$\sqrt{7056}= 2 \times 2 \times 3 \times 7 = 84$$

(iv) We have prime factors of 2028 as:

$$\begin{array}{c|lcr} 2 & 2028\\ \hline 2 & 1014\\ \hline 3& 507\\ \hline 13 & 169\\ \hline 13 & 13\\ \hline & 1\\ \end{array}$$

$$2028 = 2 \times 2 \times 3 \times 13 \times 13$$

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

New square number =$$2028 \times 3 = 6084$$

Square root of 6084 is

∴$$\sqrt{6084 }= 2 \times 13 \times 3 = 78$$

(v)We have prime factors of 1458 as:

$$\begin{array}{c|lcr} 2 & 1458\\ \hline 2 & 729\\ \hline 3 & 243\\ \hline 3 & 81\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1\\ \end{array}$$

$$1458 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$$

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

New square number =$$1458 \times 2 = 2916$$

Square root of 2916 is

∴$$\sqrt{2916 }= 3 \times 3 \times 3 \times 2 = 54$$

(vi) We have prime factors of 768 as:

$$\begin{array}{c|lcr} 2 & 768\\ \hline 2 & 384\\ \hline 2 & 192\\ \hline 2 & 96\\ \hline 2 & 48\\ \hline 2 & 24\\ \hline 2 & 12\\ \hline 2 & 6\\ \hline 3 & 3\\ \hline & 1\\ \end{array}$$

$$768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$$

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

New square number = $$768 \times 3 = 2304$$

Square root of 2304 is

∴$$\sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3 = 48$$

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:
$$\qquad \begin{array}{c|lcr} 2 & 252\\ \hline 2 & 126\\ \hline 3 & 63\\ \hline 3 & 21\\ \hline 7 & 7\\ \hline & 1\\ \end{array}$$

$$252 = 2 \times 2 \times 3 \times 3 \times 7$$

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 $$\sqrt{36} = 6$$

(ii) We have prime factors of 2925 as:
$$\qquad \begin{array}{c|lcr} 3 & 2925\\ \hline 3 & 975\\ \hline 5 & 325\\ \hline 5 & 65\\ \hline 13 & 13\\ \hline & 1\\ \end{array}$$

$$2925 = 3 \times 3 \times 5 \times 5 \times 13$$

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 =$$2925 ÷ 13 = 225$$

∴ $$\sqrt{225} = 15$$

(iii)We have prime factors of 396 as:

$$\qquad \begin{array}{c|lcr} 2 & 396\\ \hline 2 & 198\\ \hline 3 & 99\\ \hline 3 & 33\\ \hline 11 & 11\\ \hline & 1\\ \end{array}$$

$$396 = 2 \times 2 \times 3 \times 3 \times 11$$

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 = $$396 ÷ 11 = 36$$

∴ $$\sqrt{36} = 6$$

(iv)We have prime factors of 2645 as:

$$\qquad \begin{array}{c|lcr} 5 & 2645\\ \hline 23 & 529\\ \hline 23 & 23\\ \hline & 1\\ \end{array}$$

$$2645 = 5 \times 23 \times 23$$

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 = $$2645 ÷ 5 = 529$$

∴ $$\sqrt{529} = 23$$

(v) We have prime factors of 2800 as:

$$\qquad \begin{array}{c|lcr} 2 & 2800\\ \hline 2 & 1400\\ \hline 2 & 700\\ \hline 2 & 350\\ \hline 5 & 175\\ \hline 5 & 35\\ \hline 7 & 7\\ \hline & 1\\ \end{array}$$

$$2800 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7$$

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 =$$2800 ÷ 7 = 400$$

∴ $$\sqrt{400} = 20$$

(vi)We have prime factors of 1620 as:

$$\qquad \begin{array}{c|lcr} 2 & 1620\\ \hline 2 & 810\\ \hline 3 & 405\\ \hline 3 & 135\\ \hline 3 & 45\\ \hline 3 & 15\\ \hline 5 & 5\\ \hline & 1\\ \end{array}$$

$$1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5$$

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

∴ $$\sqrt{324} = 18$$

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:$$x \times x=₹ x^2$$
Here we have, $$x^2 =2401$$

$$\Rightarrow x=\sqrt{2401}$$

$$\qquad \begin{array}{c|lcr} 7 & 2401\\ \hline 7 & 343\\ \hline 7 & 49\\ \hline 7 & 7\\ \hline & 1\\ \end{array}$$

$$\Rightarrow x=\sqrt{7 \times 7 \times 7 \times 7}$$

$$\Rightarrow x=7 \times 7$$

$$\Rightarrow x=49$$

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

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

Answer :

Let the number of rows be x and x be the number of plants in each row.
∴we have total number of plants=$$x^2$$ = $$\sqrt{2025}$$

$$\Rightarrow x=\sqrt{2401}$$

$$\qquad \begin{array}{c|lcr} 3 & 2025\\ \hline 3 & 675\\ \hline 3 & 225\\ \hline 3 & 75\\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1\\ \end{array}$$

$$\Rightarrow \sqrt { 3\times 3\times 3\times 3\times 5\times 5 }$$

$$\Rightarrow \sqrt { { 3 }^{ 2 }\times { 3 }^{ 2 }\times { 5 }^{ 2 } }$$

$$\Rightarrow 3 \times 3 \times 5$$

$$\Rightarrow 45$$
Hence, the number of rows is 45 and the number of plants in each row is also 45

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

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

Answer :

We have, LCM of 4, 9, 10 = 180

$$\qquad \begin{array}{c|lcr} 2 & 4 \;,\;9\;,\;10\\ \hline 2 & 2 \;,\;9\;,\;5\\ \hline 3 & 1 \;,\;9\;,\;5\\ \hline 3 & 1 \;,\;3\;,\;5\\ \hline 5 & 1 \;,\;1\;,\;5\\ \hline & 1\;,\;1\;,\;1\\ \end{array}$$

The least number divisible by 4, 9 and 10 = 180

Now we have, prime factors of 180 as:

$$180 = 2 \times 2 \times 3 \times 3 \times 5$$

Here, 5 has no pair.So the smallest number to be multiplied to make it a perfect square is 5.

The required smallest square number = $$180 \times 5 = 900$$

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

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

Answer :

The smallest number divisible by 8, 15 and 20 is equal to their LCM.
$$\qquad \begin{array}{c|lcr} 2 & 8 \;,\;15\;,\;20\\ \hline 2 & 4 \;,\;15\;,\;10\\ \hline 2 & 2 \;,\;15\;,\;5\\ \hline 3 & 1 \;,\;15\;,\;5\\ \hline 5 & 1 \;,\;5\;,\;5\\ \hline & 1\;,\;1\;,\;1\\ \end{array}$$

LCM =$$2 \times 2 \times 2 \times 3 \times 5 = 120$$

Here, 2, 3 and 5 have no pair.So they are multiplied to get the least number divisible by 8,15 and 20.

The required smallest square number =$$120 \times 2 \times 3 \times 5 = 120 \times 30 = 3600$$

## NCERT solutions for class 8 Maths Chapter 6 Squares And Square Roots Exercise 6.4

1.Find the square root of each of the following numbers by Long 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

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

Answer :

(i)
$$\qquad \begin{array}{c|lcr} &48\\ \hline 4 &\;\; \;\bar{23}\bar{04}\\ &-16\\ \hline 88 &\;\;\;\;\; 704\\ &\;\;\;\;\; 704\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{2304}=48$$

(ii)
$$\qquad \begin{array}{c|lcr} &67\\ \hline 6 &\;\; \;\bar{44}\bar{89}\\ &-36\\ \hline 127 &\;\;\;\;\; 889\\ &\;\;\;\;\; 889\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{4489}=67$$

(iii)
$$\qquad \begin{array}{c|lcr} &59\\ \hline 5 &\;\; \;\bar{34}\bar{81}\\ &-25\\ \hline 109 &\;\;\;\;\; 981\\ &\;\;\;\;\; 981\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{3481}=59$$

(iv)
$$\qquad \begin{array}{c|lcr} &23\\ \hline 2 &\;\; \;{5}\bar{29}\\ &-4\\ \hline 43 &\;\;\;\;\; 129\\ &\;\;\;\;\; 129\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{529}=23$$

(v)
$$\qquad \begin{array}{c|lcr} &57\\ \hline 5 &\;\; \;\bar{32}\bar{49}\\ &-25\\ \hline 107 &\;\;\;\;\; 749\\ &\;\;\;\;\; 749\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{3249}=57$$

(vi)
$$\qquad \begin{array}{c|lcr} &37\\ \hline 3 &\;\; \;\bar{13}\bar{69}\\ &\;-9\\ \hline 67 &\;\;\;\;\;469\\ &\;\;\;\;\; 469\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{1369}=37$$

(vii)
$$\qquad \begin{array}{c|lcr} &76\\ \hline 7 &\;\; \;\bar{57}\bar{76}\\ &-49\\ \hline 146 &\;\;\;\;\; 876\\ &\;\;\;\;\; 876\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{5776}=76$$

(viii)
$$\qquad \begin{array}{c|lcr} &89\\ \hline 8 &\;\; \;\bar{79}\bar{21}\\ &-64\\ \hline 169 &\;\;\;\;\; 1521\\ &\;\;\;\;\; 1521\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{7921}=89$$

(ix)
$$\qquad \begin{array}{c|lcr} &24\\ \hline 2 &\;\; \;5\bar{76}\\ &-4\\ \hline 44 &\;\;\;\;\; 176\\ &\;\;\;\;\; 176\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{576}=24$$

(x)
$$\qquad \begin{array}{c|lcr} &32\\ \hline 3 &\;\; \;\bar{10}\bar{24}\\ &-9\\ \hline 62 &\;\;\;\;\; 124\\ &\;\;\;\;\; 124\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{1024}=32$$

(xi)
$$\qquad \begin{array}{c|lcr} &56\\ \hline 5 &\;\; \;\bar{31}\bar{36}\\ &-25\\ \hline 106 &\;\;\;\;\; 636\\ &\;\;\;\;\; 636\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{3136}=56$$

(xii)
$$\qquad \begin{array}{c|lcr} &30\\ \hline 3 &\;\; \;9\bar{00}\\ &-9\\ \hline &\;\;\;\;\; 00\\ \hline \end{array}$$

Hence we have, $$\sqrt{900}=30$$

2.Find the number of digits in the square root of each of the following numbers (without any calculation)
(i) 64

(ii) 144

(iii) 4489

(iv) 27225

(v) 390625

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

Answer :

We know that if n is number of digits in a perfect square number, then it's square root will have $$\frac{n}2$$ digits if n is even and \{\frac{(n+1)}2\) digits if n is odd.
(i) 64

Here n is even i.e, 2.

So, number of digits in $$\sqrt{64} = \frac { 2 }{ 2 } = 1$$

(ii) 144

Here n is odd i.e, 3.

So, number of digits in square root = $$\frac { 3+1 }{ 2 } = 2$$

(iii) 4489

Here n is even i.e, 4.

Number of digits in square root =$$\frac { 4 }{ 2 } = 2$$

(iv) 27225

Here n is odd i.e, 5.

Number of digits in square root =$$\frac { 5+1 }{ 2 } = 3$$

(iv) 390625

Here n is even i.e, 6.

So, number of digits in square root =$$\frac { 6 }{ 2 } = 3$$

3.Find the square root of the following decimal numbers.
(i) 2.56

(ii) 7.29

(iii) 51.84

(iv) 42.25

(v) 31.36

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

Answer :

(i)
$$\qquad \begin{array}{c|lcr} &1.6\\ \hline 1 &\;\; \;\bar{2.}\bar{56}\\ &-1\\ \hline 26 &\;\;\; 156\\ &\;\;\; 156\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{2.56}=1.6$$

(ii)
$$\qquad \begin{array}{c|lcr} &2.7\\ \hline 2 &\;\; \;\bar{7.}\bar{29}\\ &-4\\ \hline 47 &\;\;\; 329\\ &\;\;\; 329\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{7.29}=2.7$$

(iii)
$$\qquad \begin{array}{c|lcr} &7.2\\ \hline 7 &\;\; \;\bar{51.}\bar{84}\\ &-49\\ \hline 142 &\;\;\; 284\\ &\;\;\; 284\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{51.84}=7.2$$

(iv)
$$\qquad \begin{array}{c|lcr} &6.5\\ \hline 6 &\;\; \;\bar{42.}\bar{25}\\ &-36\\ \hline 125 &\;\;\; 625\\ &\;\;\; 625\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{42.25}=6.5$$

(v)
$$\qquad \begin{array}{c|lcr} &5.6\\ \hline 5 &\;\; \;\bar{31.}\bar{36}\\ &-25\\ \hline 106 &\;\;\;\;\; 636\\ &\;\;\;\;\; 636\\ \hline &\;\;\;\;\; 0\\ \hline \end{array}$$

Hence we have, $$\sqrt{31.36}=5.6$$

4.Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402

(ii) 1989

(iii) 3250

(iv) 825

(v) 4000

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

Answer :

(i)
$$\qquad \begin{array}{c|lcr} &20\\ \hline 2 &\;\; \;\bar{4}\bar{02}\\ &-4\\ \hline 4 &\;\;\; 02\\ \hline \end{array}$$

So we have got remainder as 2.So we can say that 2 is the smallest required number to be subtracted from 402 to get a perfect square.
Therefore, we have new number as :402-2=400

So, $$\sqrt{400}=20$$

(ii)
$$\qquad \begin{array}{c|lcr} &44\\ \hline 4 &\;\; \;\bar{19}\bar{89}\\ &-16\\ \hline 84 &\;\;\; 389\\ &\;\;\; 336\\ \hline &\;\;\;53\\ \hline \end{array}$$

So we have got remainder as 53.So we can say that 53 is the smallest required number to be subtracted from 1989 to get a perfect square.
Therefore, we have new number as :1989-53=1936

So, $$\sqrt{1936}=44$$

(iii)
$$\qquad \begin{array}{c|lcr} &57\\ \hline 5 &\;\; \;\bar{32}\bar{50}\\ &-25\\ \hline 107 &\;\;\; 750\\ &\;\;\; 749\\ \hline &\;\;\;1\\ \hline \end{array}$$

So we have got remainder as 1.So we can say that 1 is the smallest required number to be subtracted from 3250 to get a perfect square.
Therefore, we have new number as :3250-1=3249

So, $$\sqrt{3249}=57$$

(iv)
$$\qquad \begin{array}{c|lcr} &28\\ \hline 2 &\;\; \;\bar{8}\bar{25}\\ &-4\\ \hline 48 &\;\;\; 425\\ &\;\;\; 384\\ \hline &\;\;\;41\\ \hline \end{array}$$

So we have got remainder as 41.So we can say that 41 is the smallest required number to be subtracted from 825 to get a perfect square.
Therefore, we have new number as :825-41=784

So, $$\sqrt{784}=28$$

(v)
$$\qquad \begin{array}{c|lcr} &63\\ \hline 6 &\;\; \;\bar{40}\bar{00}\\ &-36\\ \hline 123 &\;\;\; 400\\ &\;\;\; 369\\ \hline &\;\;\;31\\ \hline \end{array}$$

So we have got remainder as 31.So we can say that 31 is the smallest required number to be subtracted from 4000 to get a perfect square.
Therefore, we have new number as :4000-31=3969

So, $$\sqrt{3969}=63$$

5.Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.
(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

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

Answer :

(i)
$$\qquad \begin{array}{c|lcr} &22\\ \hline 2 &\;\; \;\bar{5}\bar{25}\\ &-4\\ \hline 42 &\;\;\; 125\\ &\;\;\; 84\\ \hline &\;\;\;41\\ \hline \end{array}$$

So we got remainder as 41.It represents that the square of 22 is less than 525.So, the next number is 23 and $$23^2 = 529$$.

Hence, the number to be added = 529 – 525 = 4

∴Required perfect square = 529

Thus we have, $$\sqrt{529}=23$$

(ii)
$$\qquad \begin{array}{c|lcr} &41\\ \hline 4 &\;\; \;\bar{17}\bar{50}\\ &-16\\ \hline 81 &\;\;\; 150\\ &\;\;\; 81\\ \hline &\;\;\;69\\ \hline \end{array}$$

So we got remainder as 69.It represents that the square of 41 is less than 1750.So, the next number is 42 and $$42^2 = 1764$$.

Hence, the number to be added = 1764– 1750 =14

∴Required perfect square = 1764

Thus we have, $$\sqrt{1764}=42$$

(iii)
$$\qquad \begin{array}{c|lcr} &15\\ \hline 1 &\;\; \;\bar{2}\bar{52}\\ &-16\\ \hline 25 &\;\;\; 152\\ &\;\;\; 125\\ \hline &\;\;\;27\\ \hline \end{array}$$

So we got remainder as 27.It represents that the square of 15 is less than 252.So, the next number is 16 and $$16^2 = 256$$.

Hence, the number to be added = 256 – 252 = 4

∴Required perfect square = 256

Thus we have, $$\sqrt{256}=16$$

(iv)
$$\qquad \begin{array}{c|lcr} &42\\ \hline 4 &\;\; \;\bar{18}\bar{25}\\ &-16\\ \hline 82 &\;\;\; 225\\ &\;\;\; 164\\ \hline &\;\;\;61\\ \hline \end{array}$$

So we got remainder as 61.It represents that the square of 42 is less than 1825.So, the next number is 43 and $$43^2 = 1849$$.

Hence, the number to be added = 1849 – 1825 = 24

∴New number = 1849

Thus we have,$$\sqrt{1849}=43$$

(v)
$$\qquad \begin{array}{c|lcr} &80\\ \hline 8 &\;\; \;\bar{64}\bar{12}\\ &-64\\ \hline 160 &\;\;\; 12\\ &\;\;\; 0\\ \hline &\;\;\;12\\ \hline \end{array}$$

So we got remainder as 12.It represents that the square of 80 is less than 6412.So, the next number is 81 and $$81^2 = 6561$$.

Hence, the number to be added = 6561 – 6412 = 149

∴New number = 6561

Thus, we have $$\sqrt{6561}=81$$

6.Find the length of the side of a square whose area is $$441 m^2$$

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

Answer :

Let the length of the side of the square be x m.
So we have, Area of the square = $$(side)^2 = x^2\; m^2$$

$$x^2 = 441$$

$$\qquad \begin{array}{c|lcr} &21\\ \hline 2 &\;\; \;\bar{4}\bar{41}\\ &-4\\ \hline 41 &\;\;\; 41\\ &\;\;\;41\\ \hline &\;\;\;0\\ \hline \end{array}$$

$$\Rightarrow x =\sqrt{441} = 21$$

Hence the length of the side of square = 21 m.

7.In a right triangle ABC, ∠B =$$90^{\circ}$$.
(a) If AB = 6 cm, BC = 8 cm, find AC

(b) If AC = 13 cm, BC = 5 cm, find AB

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

Answer :

(a) In right triangle ABC

$$AC^2 = AB^2 + BC^2\quad$$ [By Pythagoras Theorem]

$$\Rightarrow AC^2 = (6)^2 + (8)^2 = 36 + 64 = 100$$

$$\Rightarrow AC = \sqrt{100} = 10$$

So we have, AC = 10 cm.

(b) In right triangle ABC

$$AC^2 = AB^2 + BC^2\quad$$ [By Pythagoras Theorem]

$$\Rightarrow (13)^2 = AB^2 + (5)^2$$

$$\Rightarrow 169 = AB^2 + 25$$

$$\Rightarrow 169 – 25 = AB^2$$

$$\Rightarrow 144 = AB^2$$

$$AB = \sqrt{144} = 12 cm$$

So we have, AB = 12 cm.

8.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 the same. Find the minimum number of plants he needs more for this.

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

Answer :

Let the number of rows and columns be x each.
So,the total number of plants =$$x \times x = x^2$$

So we can have, $$x^2 = 1000$$

$$\Rightarrow x = \sqrt{1000}$$

$$\qquad \begin{array}{c|lcr} &31\\ \hline 3 &\;\; \;\bar{10}\bar{00}\\ &\;-9\\ \hline 61 &\;\;\; \;\;100\\ &\;\;\;\;\;61\\ \hline &\;\;\;\;\;39\\ \hline \end{array}$$

We got the remainder as 39. This says that the square of 31 is less than 1000.So, the next number is 32 and $$32^2 = 1024$$

Hence the number to be added = 1024 – 1000 = 24

Thus, the minimum number of plants required by him = 24.

9.There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?

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

Answer :

Let the number of children in a row and column each be x.
Total number of students = $$x \times x = x^2$$

$$x^2 = 500$$

$$\Rightarrow x = \sqrt{500}$$

$$\qquad \begin{array}{c|lcr} &22\\ \hline 2 &\;\; \;\bar{5}\bar{00}\\ &\;-4\\ \hline 42 &\;\;\; \;\;100\\ &\;\;\;\;\;84\\ \hline &\;\;\;\;\;16\\ \hline \end{array}$$

We got the remainder as 16.

So the new number is = 500 – 16 = 484

and, $$\sqrt{484} = 22$$

Thus, 16 students will be left out in this arrangement.

##### FAQs Related to NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots
There are total 30 questions present in ncert solutions for class 8 maths chapter 6 squares and square roots
There are total 9 long question/answers in ncert solutions for class 8 maths chapter 6 squares and square roots
There are total 4 exercise present in ncert solutions for class 8 maths chapter 6 squares and square roots