NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots

Square means a number will be multiplied by itself. For example:- If 6 is multiplied by 6, then the result will be 36. 36 is said to be the square of 6 and conversely, 6 is said to be the square root of 36. Suppose you need to find the area of a square. How will you calculate it? You just need to take the square of the side of the square. There are many other applications where square and square root are needed, just like in Pythagoras' theorem, solving quadratic equations, etc. Here, we are going to discuss several approaches to find the square and roots ie. long division method, factorization, etc.

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1. Squares and Square Roots Class 8 Questions And Answers PDF Free Download
2. Squares and Square Roots Class 8 Solutions - Important Formulae
3. NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots - Topics
4. NCERT Solutions for Class 8 Maths: Chapter Wise
5. Importance of Solving NCERT Questions of Class 8 Maths Chapter 5

These NCERT Solutions are created by the expert team at craeers360 keeping the latest syllabus and pattern of CBSE 2025-26. The NCERT solutions for Class 8 Maths chapter 6 Squares and Square Roots cover questions related to the squares of numbers and the square roots of numbers. Practicing questions is important to score good marks in Mathematics. Here you will get the detailed NCERT Solutions for Class 8 by clicking on the link.

Squares and Square Roots Class 8 Questions And Answers PDF Free Download

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Squares and Square Roots Class 8 Solutions - Important Formulae

Square Root Formula: If q is a natural number such that p 2 = q, then √q = p or -p.

Properties of Squares and Square Roots:

• There are 2n non-perfect square numbers between n 2 and (n+1) 2 .
• If a perfect square has n digits, its square root will have n/2 digits if n is even, or (n+1)/2 digits if n is odd.

Question 1: Find the perfect square numbers between

(i) $30$ and $40$

(ii) $50$ and $60$

Answer:

(i) We know that

$5^2=25,6^2=36,7^2=49$

So clearly 36 is the perfect square number between 30 and 40.

(ii) We know

$7^2=49;8^2=64;9^2=81$

So clearly it can be seen that there does not exist any perfect square number between 50 and 60

Question 1: Can we say whether the following numbers are perfect squares? How do we know?

$(i)1057$

Answer:

We know that numbers that end with 0, 1, 4, 5, 6, or 9 at unit place may be a perfect square number, and all other numbers are not perfect square numbers.

Since the given number has 7 at the unit's place hence this number is not a perfect square.

Question 1(ii): Can we say whether the following numbers are perfect squares? How do we know?

$23453$

Answer:

We have 23453.

Since this number ends with digit 3 it cannot be a perfect square. (As we know a number must end with 0, 1, 4, 5, 6, or 9 for being a perfect square number.)

Question 1(iii): Can we say whether the following numbers are perfect squares? How do we know?

$7928$

Answer:

It is known that a number must end with 0, 1, 4, 5, 6, or 9 at the unit's place for being a perfect square.

The given number ends with 8 so it is not a perfect square.

Question 1(iv): Can we say whether the following numbers are perfect squares? How do we know?

$222222$

Answer:

The given number ends with the digit 2.

We know that only a number ending with 0, 1, 4, 5, 6, or 9 at unit place can be a perfect square.

Therefore 222222 is not a perfect square number.

Question 1(v): Can we say whether the following numbers are perfect squares? How do we know?

$1069$

Answer:

Since the unit place of a given number is 9, thus it may or may not be a perfect square number.

As we know a number ending with 0, 1, 4, 5, 6, or 9 at unit place can be a perfect square number.

Question 1(vi): Can we say whether the following numbers are perfect squares? How do we know?

$2061$

Answer:

It is known that the numbers that end with 0, 1, 4, 5, 6, or 9 at unit place may be a perfect square.

Given number has 1 as the last digit it may be a perfect square number.

Question 2: Write five numbers that you cannot decide just by looking at their unit digit (or unit place) and whether they are square numbers or not.

Answer:

The five numbers can be: 521, 655, 124, 729, 1940, etc.

Basically, numbers ending with 0, 1, 4, 5, 6, or 9 at units place can be square numbers.

Question 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

Answer:

The unit digit of the squares of the following numbers will be:-

(i) 81:- 1

(ii) 272:- 4

(iii) 799:- 1

(iv) 3853 - 9

(v) 1234 - 6

(vi) 26387 - 9

(vii) 52698 - 4

(viii) 99880:- 0

(ix) 12796 - 6

(x) 55555 :- 5

Question 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

Answer:

We know that only the numbers that end with 0, 1, 4, 5, 6, or 9 at unit place can be perfectly square numbers.

Also, a perfectly square number has several zeros in multiple of 2.

Since these numbers have either odd no. of zeros or their unit place is 2, 3, 7, 8 thus they are not perfectly square numbers.

Question 3: The squares of which of the following would be odd numbers?

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Answer:

It is known that the square of an odd number is always an odd number.

Therefore the squares of 431 and 7779 will also be an odd number.

Question 4: Observe the following pattern and find the missing digits.

${11}^2=121$

${101}^2=10201$

${1001}^2=1002001$

${100001}^2=1...........2................1$

${10000001}^2=............................$

Answer:

By observation, it is clear that the no. of zeros between 1 and 1 in LHS are equal to the no. of zeros between 1-2 and 2-1 in the RHS.

So, ${100001}^2=10000200001$

and ${10000001}^2=100000020000001$

Question 5: Observe the following pattern and supply the missing numbers.

${11}^2$ $=121$

${101}^2=10201$

${10101}^2=102030201$

${1010101}^2=..................................$

$......{.}^2=10203040504030201$

Answer:

The solution to the above-written question is as follows

By observation we get,

${1010101}^2=1020304030201$

and ${101010101}^2=10203040504030201$

Question 6: Using the given pattern, find the missing numbers.

$1^2+2^2+2^2=3^2$

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

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

$4^2+5^2+-{}^2={21}^2$

$5^2+{-}^2+{30}^2={31}^2$

$6^2+7^2+{-}^2={-}^2$

Answer:

The pattern is clearly visible.

The first two numbers and the last two numbers are the consecutive numbers.

Moreover, the third number is obtained when the first is multiplied by the second number.

So required numbers can be found.

i.e., 4 $\times$ 5 = 20 and 6 $\times$ 7 = 42

Hence $4^2+5^2+{20}^2={21}^2$

and $5^2+6^2+{30}^2={31}^2$

and $6^2+7^2+{42}^2={43}^2$

Question 7: Without adding, find the sum.

(i) $1+3+5+7+9$

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

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

Answer:

It is known that the sum of the odd consecutive number starting from 1 is $n^2$.

(i) n = 5 i.e., $5^2=25$

(ii) n = 10 i.e., ${10}^2=100$

(iii) n = 12 i.e., ${12}^2=144$

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

Answer:

The solution to the above-mentioned question is as follows:-

The splitter form of 49 (In increasing odd numbers):- 1 + 3 + 5 + 7 + 9 + 11 + 13

Question 8(ii): Express 121 as the sum of 11 odd numbers.

Answer:

The splitter form of number 121 (starting with odd numbers in increasing orders) = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Question 9(i): How many numbers lie between squares of the following numbers?

$12and13$

Answer:

We know that there are 2n non-perfect square numbers between the squares of the numbers n and (n + 1).

So for this question, n = 12

So total numbers that lie between squares 12 and 13 are = 2(12) = 24.

Question 9(ii): How many numbers lie between squares of the following numbers?

$25and26$

Answer:

It is known that there are 2n non-perfect square numbers between the squares of the numbers n and (n + 1).

So the total number that lies between 25 and 26 will be = 2(25) = 50

Question 9(iii): How many numbers lie between squares of the following numbers?

$99and100$

Answer:

We know that there are 2n non-perfect square numbers between the squares of the numbers n and (n + 1).

In this question, we have n = 99

Thus total number that lies between 99 and 100 = 2(99) = 198

Question 1: Find the square of the following numbers.

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

Answer:

(i) ${32}^2$ = $(30+2{)}^2$ = 30(30 + 2) + 2(30 + 2) = 30(32) + 2(32) = 960 + 64 = 1024

(ii) ${35}^2$ = $(30+5{)}^2$ = 30(30 + 5) + 5(30 + 5) = 30(35) + 5(35) = 1050 + 175 = 1225

(iii) ${86}^2$ = $(80+6{)}^2$ = 80(80 + 6) + 6(80 + 6) = 80(86) + 6(86) = 6880 + 516 = 7396

(iv) ${93}^2$ = $(90+3{)}^2$ = 90(90 + 3) + 3(90 +3) = 90(93) + 3(93) = 8370 + 279 = 8649

(v) ${71}^2$ = $(70+1{)}^2$ = 70(70 + 1) + 1(70 + 1) = 70(71) + 1(71) = 4970 + 71 = 5041

(vi) ${46}^2$ = $(40+6{)}^2$ = 40(40 + 6) + 6(40 + 6) = 40(46) + 6(46) = 1840 + 276 = 2110

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

Answer:

For any natural number m > 1, 2m, $m^2-1$ and $m^2+1$ forms a Pythagorean triplet.

So if we take, $m^2-1=6$

$m^2=6+1=7$

But the value of m will not be an integer.

Now we take, $m^2+1=6$

$m^2=6-1=5$

But the value of m will not be an integer.

If we take 2m = 6

then m = 3

Then $m^2-1=$ 9 - 1 = 8 and $m^2+1=$ 9 + 1 = 10.

Therefore, the required triplet is 6, 8, and 10

Question 2(ii): Write a Pythagorean triplet whose one member is 14.

Answer:

For any natural number m > 1, 2m, $m^2-1$ and $m^2+1$ forms a Pythagorean triplet.

So if we take, $m^2-1=14$

$m^2=14+1=15$

But then the value of m will not be an integer.

We take, $m^2+1=14$

$m^2=14-1=13$

but the value of m will not be an integer.

If we take 2m = 14

or m = 7

Then $m^2-1=$ 49 - 1 = 48 and $m^2+1=$ 49 + 1 = 50.

Therefore, the combination of numbers is 14, 48, and 50.

Question 2(iii): Write a Pythagorean triplet whose one member is 16.

Answer:

For any natural number m > 1, 2m, $m^2-1$ and $m^2+1$ forms a Pythagorean triplet.

So if we take, $m^2-1=16$

$m^2=16+1=17$

But the value of m will not be an integer.

Now we take, $m^2+1=16$

$m^2=16-1=15$

But the value of m will not be an integer.

If we take 2m = 16

then m = 8

Then $m^2-1=$ 64 - 1 = 63 and $m^2+1=$ 64 + 1 = 65.

Therefore, the required numbers are 16, 63, and 65.

Question 2(iv): Write a Pythagorean triplet whose one member is.

18

Answer:

For any natural number m > 1, 2m, $m^2-1$ and $m^2+1$ forms a Pythagorean triplet.

So if we take, $m^2-1=18$

$m^2=18+1=19$

But the value of m will not be an integer.

Now we take, $m^2+1=18$

$m^2=18-1=17$

but the value of m will not be an integer.

If we take 2m = 18

then m = 9

Then $m^2-1=$ 81 - 1 = 80 and $m^2+1=$ 81 + 1 = 82.
Therefore, the required combination is 18, 8,0, and 82

1 (i). What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

9801

Answer:

We know that the square of digits ending with 1 and 9 gives 1 at the unit's place.

So the number whose square ends in 1 = 1 & 9

So, the possible unit digit of the square root of 9801 = 1 and 9.

1 (ii). What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

99856

Answer:

We know that the square of digits ending with 4 and 6 gives 6 at its unit place.

So possible ‘ones’ digits of the square root of 99856 are 4 and 6.

Q1 (iii): What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

998001

Answer:

We know that the square of digits ending with 1 and 9 gives 1 at the unit's place.

So the possible ‘one’s’ digits of the square root of 998001 are 1 and 9.

Q1 (iv): What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

657666025

Answer:

We know that the square of a number ending with 5 gives 5 at its unit place.

So the possible ‘one’s’ digits of the square root of 657666025 are 5.

Q.2 Without doing any calculation, find the numbers that are surely not perfect squares.

(i) 153

(ii) 257

(iii) 408

(iv) 441

Answer:

As we know the unit place of a perfect square cannot be 2, 3, 7, and 8.

So 153, 257, and 408 are surely not perfect squares.

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

Answer:

(i) For 100 :- 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.

We obtain zero at 10th step so $\sqrt{100}=10$

(ii) For 169 :- 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.

We obtain Zero at the 13th step so $\sqrt{169}=13$

Q.4 (i) Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729

Answer:

By prime factorization, we know that

$729=3\times 3\times 3\times 3\times 3\times 3$

or $729=(3\times 3\times 3{)}^2={27}^2$

Thus, the square root of 729 is 27.

Q.4 (ii) Find the square roots of the following numbers by the Prime Factorisation Method.

400

Answer:

By prime factorization, we get

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

or $400=(2\times 2\times 5{)}^2={20}^2$

Thus, the square root of 400 is 20

Q4 (iii). Find the square roots of the following numbers by the Prime Factorisation Method.

1764

Answer:

We have 1764, by prime factorization we get

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

or $1764=(2\times 3\times 7{)}^2={42}^2$

Thus, the square root of 1764 is 42.

Q.4 (iv) Find the square roots of the following numbers by the Prime Factorisation Method.

4096

Answer:

We have 4096, by prime factorization:

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

or $4096=(2\times 2\times 2\times 2\times 2\times 2{)}^2={64}^2$ .

So the square root of 4096 is 64.

Q.4 (v) Find the square roots of the following numbers by the Prime Factorisation Method.

(v) 7744

Answer:

We have in 7744. By prime factorization, we get

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

or $7744=(2\times 2\times 11{)}^2={44}^2$

Thus, the square root of 7744 is 44.

Q.4 (vi) Find the square roots of the following numbers by the Prime Factorisation Method.

(vi) 9604

Answer:

We have in 9604. By prime factorization, we get,

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

or $9604=(2\times 7\times 7{)}^2={98}^2$

Hence, the square root of 9604 is 98.

Q.4 (vii) Find the square roots of the following numbers by the Prime Factorisation Method.

5929

Answer:

The solution to the above question is as follows

Prime factorization of number 5929,

$5929=7\times 7\times 11\times 11$

or $5929=(7\times 11{)}^2={77}^2$ .

Thus, the square root of 5929 is 77.

Q4 (viii). Find the square roots of the following numbers by the Prime Factorisation Method.

(viii) 9216

Answer:

The solution to the above-written question is as follows

prime factorization of 9216,

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

or $9216=(2\times 2\times 2\times 2\times 2\times 3{)}^2={96}^2$ .

Thus, the square root of 9216 is 96.

Q.4 Find the square roots of the following numbers by the Prime Factorisation Method.

(ix) 529

Answer:

The solution to the above-written question is as follows

We have 529.

Prime factorization gives $529=23\times 23$

So square root of 529 is 23.

Q.4 Find the square roots of the following numbers by the Prime Factorisation Method.

(x) 8100

Answer:

The solution to the above-written question is as follows

We have in 8100.

By prime factorization, we get: $8100=2\times 2\times 3\times 3\times 3\times 3\times 5\times 5$

or $8100=(2\times 3\times 3\times 5{)}^2={90}^2$ .

So the square root of 8100 is 90.

Q.5 For each of the following numbers, find the smallest whole number by which it should be multiplied to get a perfect square number. Also, find the square root of the square number so obtained

(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

Answer:

(i) 252: Prime factorization of 252 = $2\times 2\times 3\times 3\times 7$.

To make pairs we will multiply 252 by 7.

So the number is 1764 and its square root is 42.

(ii) 180: Prime factorization of 180 = $2\times 2\times 3\times 3\times 5$.

To make it a perfect square, multiply by 5.

So the number is 900 and its square root is 30.

(iii) 1008: Prime factorization of 1008 gives = $2\times 2\times 2\times 2\times 3\times 3\times 7$.

To make pairs we need to multiply it by 7.

So the number we get is 7056 and its square root is 84.

(iv) 2028: Prime factorization of 2028 = $2\times 2\times 3\times 13\times 13$.

To make pairs we multiply the number by 3.

So the number obtained is 6084 and its square root is 78.

(v) 1458: Prime factorization of 1458 gives = $2\times 3\times 3\times 3\times 3\times 3\times 3$

To make pairs we need to multiply the number by 2.

So the number obtained is 2916 and its square root is 54.

(vi) 768: Prime factorization of 768 gives = $2\times 2\times 2\times 2\times 2\times 2\times 2\times 3$

To make pairs we need to multiply the given number by 6.

So the required number is 4608 and its square root is 48.

Q.6 For each of the following numbers, find the smallest whole number by which it should be divided 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

Answer:

(i) 252: Prime factorization of 252 gives = $2\times 2\times 3\times 3\times 7$.

For making pairs we will divide the given number by 7.

The obtained number is 36 and its square root is 6.

(ii) 2925: Prime factorization of 2925 gives = $3\times 3\times 5\times 5\times 13$

To make pairs divide the given number by 13.

So the obtained number is 225 and its square root is 15.

(iii) 396: Prime factorization if 396 = $2\times 2\times 3\times 3\times 11$

To obtain a perfect square number we need to divide the given number by 11.

So the required number is 36 and its square root is 6.

(iv) 2645: Prime factorization of 2645 = $5\times 23\times 23$

We need to divide the given number by 5 to obtain the perfect square number.

So the obtained number is 529 and its square root is 23.

(v) 2800: Prime factorization of 2800 = $2\times 2\times 2\times 2\times 5\times 5\times 7$

To make pairs we need to divide 2800 by 7.

So the required number is 400 and its square root is 20.

(vi) 1620: Prime factorization of 1620 gives = $2\times 2\times 3\times 3\times 3\times 3\times 5$

To make pairs divide the given number by 5.

We get the number = 324 and its square root = 18.

Q.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.

Answer:

Let the number of students in a class be x.

According to the question,

Number of students = money donated by each of the students

So total money donated = $x^2$

or $x^2=2401$

Prime factorization of $2401=7\times 7\times 7\times 7=(7\times 7{)}^2={49}^2=x^2$

So the number of students in the class = 49.

Q.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.

Answer:

The total number of plants = No. of rows $\times$ No. of plants in 1 row.

Since in this case no.of rows = no. of plants in each row.

Thus, let us assume the number of rows to be x.

Then the equation becomes: $x^2=2025$

Prime factorization of 2025 gives = $3\times 3\times 3\times 3\times 5\times 5$

So the value of x is = 45.

Hence no. of rows = 45; and no. of plants in each row = 45.

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

Answer:

This has to be done in two steps. First, we will find the LCM of the given numbers, then we will make it a perfect square.

So the LCM of 4, 9, 10 is 180. 4 = 2 $\times$ 2; 9 = 3 $\times$ 3; 10 = 2 $\times$ 5

Prime factorization of 180 gives = $2\times 2\times 3\times 3\times 5$.

To make it a perfect square we need to multiply it with 5.

So, the smallest square number which is divisible by each of the numbers 4, 9, and 10 = 900.

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

Answer:

This has to be done in two steps. First, we will find the LCM of the given numbers, then we will make it a perfect square.

So the LCM of 8, 15, 20 is 120 . 8 = 2 $\times$ 2 $\times$ 2; 15 = 3 $\times$; 5 20 = 2 $\times$ 2 $\times$ 5

Prime factorization of 120 gives = $2\times 2\times 2\times 3\times 5$.

To make it a perfect square we need to multiply it with 30.

So the smallest square number that is divisible by each of the numbers 4, 9, and 10 is 3600.

Q.1 (i) Find the square root of each of the following numbers by the Division method.

2304

Answer:

The detailed explanation for the above question is as follows,

We will find the square root using the division method.

Q.1 (ii) Find the square root of each of the following numbers by the Division method.

4489

Answer:

The square root of 4489 is 67.

Q.1 (iii ) Find the square root of each of the following numbers by Division method.

3481

Answer:

The square root of 3481 is obtained as 59.

Q1 (iv). Find the square root of each of the following numbers by Division method.

529

Answer:

The detailed solution for the above-written question is as follows

The square root of 529 is 23.

Q1 (v). Find the square root of each of the following numbers by the Division method.

3249

Answer:

The detailed solution for the above question is as follows,

The square root of 3249 is 57.

Q1 (vi). Find the square root of each of the following numbers by the Division method.

1369

Answer:

The detailed solution for the above question is as follows,

The square root of 1369 is 37.

Q1 (vii). Find the square root of each of the following numbers by Division method.

5776

Answer:

The solution to the above question is as follows,

The square root of 5776 is 76.

Q1 (viii). Find the square root of each of the following numbers by the Division method.

7921

Answer:

The detailed solution for the above-written question is as follows,

The square root of 7921 is 89.

Q1 (ix). Find the square root of each of the following numbers by Division method.

576

Answer:

The detailed solution for the above-written question is as follows,

The square root of 576 is 24.

Q1 (x). Find the square root of each of the following numbers by Division method.

1024

Answer:

The detailed solution for the above-written question is as follows,

The square root of 1024 is 32.

Q.1(xi) Find the square root of each of the following numbers by the Division method.

3136

Answer:

The detailed solution for the above-written question is as follows,

The square root of 3136 is 56.

Q.1 (xii) Find the square root of each of the following numbers by the Division method.

900

Answer:

The detailed solution for the above-written question is as follows

The square root of 900 is 30.

Question 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

Answer:

(i) 64:- The number of digits in the square root will be

$\frac{n}{2}=\frac{2}{2}=1$

(ii) 144:- The number of digits in the square root will be

$\frac{n+1}{2}=\frac{4}{2}=2$

(iii) 4489:- The number of digits in the square root will be

$\frac{n}{2}=\frac{4}{2}=2$

(iv) 27225:- The number of digits in the square root will be

$\frac{n+1}{2}=\frac{6}{2}=3$

(v) 390625:- The number of digits in the square root will be

$\frac{n}{2}=\frac{6}{2}=3$

Question 3: Find the square root of the following decimal numbers.

(i) $2.56$

(ii) $7.29$

(iii) $51.84$

(iv) $42.25$

(v) $31.36$

Answer:

The detailed solution for the given questions is as follows

(i) Square root of 2.56 using the division method

(ii) The square root of 7.29 using the division method

(iii) The square root of 51.84 using the division method

(iv) The square root of 42.25 using the division method

(v) The square root of 31.36 using the division method

Question 4: Find the least number that must be subtracted from each of the following numbers to get a perfect square. Also, find the square root of the perfect square so obtained.

(i) 402

(ii) 1989

(iii) 3250

(iv) 825

(v) 4000

Answer:

(i) 402:- It can be seen that 2 is the remainder. So we will subtract 2 from 402.

The required number is 400 and its square root is 20.

(ii) 1989:- It can be seen that 53 is the remainder here. So we will subtract 53 from 1989 to make it a perfect square.

The required number is 1936, and its square root is 44.

(iii) 3250:- It can be seen that 1 is the remainder. So we will subtract 1 from 3250.

The required number is 3249, and its square root is 57.

(iv) 825:- It can be seen that 41 is the remainder. So we will subtract 41 from 825 to make it a perfect square number.

The required number is 784, and its square root is 28.

(v) 4000:- It can be seen that 31 is remainder here. So we will subtract 31 from 4000.

The required number is 3969, and its square root is 63.

Question 5: Find the least number that must be added to each of the following numbers to get a perfect square. Also, find the square root of the perfect square so obtained.

(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

Answer:

(i) 525:- It is visible that if we add 4 to the given number, the remainder will become zero.

So obtained number is 529 and its square root is 23.

(ii) 1750:- It is visible that if we add 14 to the given number, the remainder will become zero.

So the obtained number is 1764 and its square root is 42.

(iii) 252:- It is visible that if we add 4 to the given number, the remainder will become zero.

So the obtained number is 256 and its square root is 16.

(iv) 1825:- It is visible that if we add 24 to the given number, the remainder will become zero.

So the obtained number is 1849, and its square root is 43.

(v) 6412:- It is visible that if we add 149 to the given number, the remainder will become zero.

So the obtained number is 6561 and its square root is 81.

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

Answer:

Let the length of the side of a square be x m.

Area of square = $x^2$

So the equation becomes: $x^2=441$

By prime factorization of 441.

441 = $3\times 3\times 7\times 7$

Thus x = 21.

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

Question 7(a): In a right triangle $ABC,$ $\mathrm{\angle }B={90}^{\circ }$

(a) If $AB=6cm$ , $BC=8cm$ , find $AC$

Answer:

Using Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2=6^2+8^2=100$

By prime factorization of 100:- $2\times 2\times 5\times 5$

We get, AC = 10cm

Question 7(b): In a right triangle $ABC,\mathrm{\angle }B={90}^{\circ }$

If $AC=13cm,$ $BC=5cm,$ find $AB$

Answer:

Using Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

or ${13}^2=A{B}^2+5^2$

or $169=A{B}^2+25$

or $A{B}^2=169-25=144$

Prime factorization of 144 gives:- $2\times 2\times 2\times 2\times 3\times 3$

Hence, AB = 12 cm

Question 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 for this.

Answer:

It is given that the number of rows and the number of columns are the same.

Let the number of rows or a number of columns be x.

The number of plants required = $x^2$

The gardener has 1000 plants.

We need to find a perfect square just greater than 1000.

We know, ${31}^2=961$ and ${32}^2=1024$

So the minimum number of plants needed by the gardener = 1024 - 1000 = 24 plants.

Question 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 several columns. How many children would be left out in this arrangement?

Answer:

Given that the number of rows is equal to the number of columns. i.e., in the form of $x^2$

So the number of students that can stand in this order will be a perfect square number just less than 500.

We know that ${22}^2=484$ and ${23}^2=529$

So the number of students that would be left out in this arrangement = 500 - 484 = 16 students.

NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots - Topics

• Properties of Square Numbers
• Some More Interesting Patterns
• Finding the Square of a Number
• Square Roots
• Square Roots of Decimals
• Estimating Square Root

NCERT Solutions for Class 8 Maths: Chapter Wise

Importance of Solving NCERT Questions of Class 8 Maths Chapter 5

• NCERT questions help students understand core concepts like perfect squares, properties of square numbers, and patterns of squares.
• Builds a strong foundation in square roots and estimating square roots, which is useful for higher-level math and competitive exams.
• Regular practice boosts mental calculation skills, especially in identifying perfect squares and solving without calculators.
• Detailed step-by-step explanations for each problem make it easy for students to understand and apply mathematical concepts.
• NCERT exercises are structured from basic to advanced level, helping students gradually master the topic with confidence.

NCERT Solutions for Class 8 - Subject Wise

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

Also, check NCERT Books and NCERT Syllabus here:

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8