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**NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6**.**1**

**Question 1.** What will be the unit digit of the squares of the following numbers?

- 81
- 272
- 799
- 3853
- 1234
- 20387
- 52698
- 99880
- 12796
- 55555

**Solution:**

- Unit digit of 81
^{2}= 1 - Unit digit of 272
^{2}= 4 - Unit digit of 799
^{2}= 1 - Unit digit of 3853
^{2}= 9 - Unit digit of 1234
^{2}= 6 - Unit digit of 26387
^{2}= 9 - Unit digit of 52698
^{2}= 4 - Unit digit of 99880
^{2}= 0 - Unit digit of 12796
^{2}= 6 - Unit digit of 55555
^{2}= 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

**Solution:**

(i) 1057 ends with 7 at unit place. So it is not a perfect square number.

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

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

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

(v) 64000 ends with 3 zeros. So it cannot a perfect square number.

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

(vii) 22000 ends with 3 zeros. So it can not be a perfect square number.

(viii) 505050 ends with 1 zero. So it is not a perfect square number.

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

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

**Solution:**

(i) 431^{2} is an odd number.

(ii) 2826^{2} is an even number.

(iii) 7779^{2} is an odd number.

(iv) 82004^{2} is an even 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} = …………………….

**Solution:**

According to the above pattern, we have

100001^{2} = 10000200001

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

**Solution:**

According to the above pattern, we have

1010101^{2} = 1020304030201

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}

**Solution:**

According to the given pattern, we have

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

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

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

Solution:

We know that the sum of n odd numbers = n^{2}

(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]

**Question 8.**

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

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

**Solution:**

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

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

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

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

**Question 9. **How many numbers lie between squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100.

**Solution:**

(i) 12 and 13

We know that numbers between n^{2} and (n + 1)^{2} = 2n

Numbers between 12^{2} and 13^{2} = (2n) = 2 × 12 = 24

(ii) 25 and 26

Numbers between 25^{2} and 26^{2} = 2 × 25 = 50 (∵ n = 25)

(iii) 99 and 100

Numbers between 99^{2} and 100^{2} = 2 × 99 = 198 (∵ n = 99)

**Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.2**

**Question 1.** Find the square of the following numbers.

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

**Solution:**

**(i) 32 = 30 + 2**

(32)^{2} = (30 + 2)^{2}

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

= 30^{2} + 30 × 2 + 2 × 30 + 2^{2}

= 900 + 60 + 60 + 4

= 1024

Thus (32)^{2} = 1024

**(ii) 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

Thus (35)^{2} = 1225

**(iii) 86 = (80 + 6)**

86^{2} = (80 + 6)^{2}

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

= (80)^{2} + 80 × 6 + 6 × 80 + (6)^{2}

= 6400 + 480 + 480 + 36

= 7396

Thus (86)^{2} = 7396

**(iv) 93 = (90+ 3)**

93^{2} = (90 + 3)^{2}

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

= (90)^{2} + 90 × 3 + 3 × 90 + (3)^{2}

= 8100 + 270 + 270 + 9

= 8649

Thus (93)^{2} = 8649

**(v) 71 = (70 + 1)**

71^{2} = (70 + 1)^{2}

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

= (70)^{2} + 70 × 1 + 1 × 70 + (1)^{2}

= 4900 + 70 + 70 + 1

= 5041

Thus (71)^{2} = 5041

**(vi) 46 = (40+ 6)**

46^{2} = (40 + 6)^{2}

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

= (40)^{2} + 40 × 6 + 6 × 40 + (6)^{2}

= 1600 + 240 + 240 + 36

= 2116

Thus (46)^{2} = 2116

**Question 2. **Write a Pythagorean triplet whose one member is.

(i) 6

(ii) 14

(iii) 16

(iv) 18

**Solution:**

**(i) Let m ^{2} – 1 = 6**

[Triplets are in the form 2m, m

^{2}– 1, m

^{2}+ 1]

m

^{2}= 6 + 1 = 7

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

Now, let us try for m

^{2}+ 1 = 6

⇒ m

^{2}= 6 – 1 = 5

Also, the value of m will not be an integer.

Now we let 2m = 6 ⇒ m = 3 which is an integer.

Other members are:

m

^{2}– 1 = 3

^{2}– 1 = 8 and m

^{2}+ 1 = 3

^{2}+ 1 = 10

Hence, the required triplets are 6, 8 and 10

**(ii) Let m ^{2} – 1 = 14 ⇒ m^{2} = 1 + 14 = 15**

The value of m will not be an integer.

Now take 2m = 14 ⇒ m = 7 which is an integer.

The member of triplets are 2m = 2 × 7 = 14

m

^{2}– 1 = (7)

^{2}– 1 = 49 – 1 = 48

and m

^{2}+ 1 = (7)

^{2}+ 1 = 49 + 1 = 50

i.e., (14, 48, 50)

**(iii) Let 2m = 16 m = 8**

The required triplets are 2m = 2 × 8 = 16

m^{2} – 1 = (8)^{2} – 1 = 64 – 1 = 63

m^{2} + 1 = (8)^{2} + 1 = 64 + 1 = 65

i.e., (16, 63, 65)

**(iv) Let 2m = 18 ⇒ m = 9**

Required triplets are:

2m = 2 × 9 = 18

m^{2} – 1 = (9)^{2} – 1 = 81 – 1 = 80

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

i.e., (18, 80, 82)

**NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.3**

**Question 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) One’s digit in the square root of 9801 maybe 1 or 9.

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

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

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

**Question 2. **Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153

(ii) 257

(iii) 408

(iv) 441

**Solution:**

We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares.

(i) 153 is not a perfect square number. (ending with 3)

(ii) 257 is not a perfect square number. (ending with 7)

(iii) 408 is not a perfect square number. (ending with 8)

(iv) 441 is a perfect square number.

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

**Solution:**

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

(Ten times repetition)

Thus √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

(Thirteen times repetition)

Thus √169 = 13

**Question 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) We have 729**The Prime factors of 729

729 = 3 × 3 × 3 × 3 × 3 × 3 = 3

^{2}× 3

^{2}× 3

^{2}

√729 = 3 × 3 × 3 = 27

**(ii) We have 400**The Prime factors of 400

400 = 2 × 2 × 2 × 2 × 5 × 5 = 2

^{2}× 2

^{2}× 5

^{2}

√400 = 2 × 2 × 5 = 20

**(iii) 1764**

1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2^{2} × 3^{2} × 7^{2}

√1764 = 2 × 3 × 7 = 42

**(iv) 4096**

4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2}

√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

**(v) Prime factorisation of 7744 is** 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

= 2^{2} × 2^{2} × 2^{2} × 11^{2}

√7744 = 2 × 2 × 2 × 11 = 88

**(vi) Prime factorisation of 9604 **

9604 = 2 × 2 × 7 × 7 × 7 × 7 = 2^{2} × 7^{2} × 7^{2}

√9604 = 2 × 7 × 7 = 98

**(vii) Prime factorisation of 5929 is **5929 = 7 × 7 × 11 × 11 = 7^{2} × 11^{2}

√5929 = 7 × 11 = 77

**(viii) Prime factorisation of 9216 **

9216 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3

= 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2} × 3^{2}

√9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96

**(ix) Prime factorisation of 529 is**

529 = 23 × 23 = 23^{2}

√529 = 23

**(x) Prime factorisation of 8100**

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2^{2} × 3^{2} × 3^{2} × 5^{2}

√8100 = 2 × 3 × 3 × 5 = 90

**Question 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) Prime factorisation of 252** is

252 = 2 × 2 × 3 × 3 × 7

Here, the prime factorisation is not in pair. 7 has no pair.

Thus, 7 is the smallest whole number by which the given number is multiplied to get a perfect square number.

The new square number is 252 × 7 = 1764

Square root of 1764 is

√1764 = 2 × 3 × 7 = 42

**(ii) Primp factorisation of 180** is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

New square number = 180 × 5 = 900

The square root of 900 is

√900 = 2 × 3 × 5 = 30

Thus, 5 is the smallest whole number by which the given number is multiplied to get a square number.

**(iii) Prime factorisation of 1008** is

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here, 7 has no pair.

New square number = 1008 × 7 = 7056

Thus, 7 is the required number.

Square root of 7056 is

√7056 = 2 × 2 × 3 × 7 = 84

**(iv) Prime factorisation of 2028 **is

2028 = 2 × 2 × 3 × 13 × 13

Here, 3 is not in pair.

3 is the required smallest whole number.

New square number = 2028 × 3 = 6084

Square root of 6084 is

√6084 = 2 × 13 × 3 = 78

**(v) Prime factorisation of 1458** is

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, 2 is not in pair.

2 is the required smallest whole number.

New square number = 1458 × 2 = 2916

Square root of 1458 is

√2916 = 3 × 3 × 3 × 2 = 54

**(vi) Prime factorisation of 768** is

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, 3 is not in pair.

Thus, 3 is the required whole number.

New square number = 768 × 3 = 2304

Square root of 2304 is

√2304 = 2 × 2 × 2 × 2 × 3 = 48

**Question 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) Prime factorisation of 252** is

252 = 2 × 2 × 3 × 3 × 7

Here 7 has no pair.

7 is the smallest whole number by which 252 is divided to get a square number.

New square number = 252 ÷ 7 = 36

Thus, √36 = 6

**(ii) Prime factorisation of 2925** is

2925 = 3 × 3 × 5 × 5 × 13

Here, 13 has no pair.

13 is the smallest whole number by which 2925 is divided to get a square number.

New square number = 2925 ÷ 13 = 225

Thus √225 = 15

**(iii) Prime factorisation of 396** is

396 = 2 × 2 × 3 × 3 × 11

Here 11 is not in pair.

11 is the required smallest whole number by which 396 is divided to get a square number.

New square number = 396 ÷ 11 = 36

Thus √36 = 6

**(iv) Prime factorisation of 2645** is

2645 = 5 × 23 × 23

Here, 5 is not in pair.

5 is the required smallest whole number.

By which 2645 is multiplied to get a square number

New square number = 2645 ÷ 5 = 529

Thus, √529 = 23

(v) Prime factorisation of 2800 is

2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

Here, 7 is not in pair.

7 is the required smallest number.

By which 2800 is multiplied to get a square number.

New square number = 2800 ÷ 7 = 400

Thus √400 = 20

**(vi) Prime factorisation of 1620** is

1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

Here, 5 is not in pair.

5 is the required smallest prime number.

By which 1620 is multiplied to get a square number = 1620 ÷ 5 = 324

Thus √324 = 18

**Question 7.** The students of class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

**Solution:**

Total amount of money donated = ₹ 2401

Total number of students in the class = √2401

= 72×72−−−−−−√

= 7×7×7×7−−−−−−−−−−√

= 7 × 7

= 49

**Question 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:**

Total number of rows = Total number of plants in each row = √2025

= 3×3×3×3×5×5−−−−−−−−−−−−−−−−√

= 32×32×52−−−−−−−−−−√

= 3 × 3 × 5

= 45

Thus the number of rows and plants = 45

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

**Solution:**

LCM of 4, 9, 10 = 180

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

Now prime factorisation of 180 is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

The required smallest square number = 180 × 5 = 900

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

**Solution:**

The smallest number divisible by 8, 15 and 20 is equal to their LCM.

LCM = 2 × 2 × 2 × 3 × 5 = 120

Here, 2, 3 and 5 have no pair.

The required smallest square number = 120 × 2 × 3 × 5 = 120 × 30 = 3600

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NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.4

**Additional Question and Answer on ****Squares and Square Roots**

**Squares and Square Roots**

**1. What is square root?**

The square root of any number is the number whose square is equal to the original number. For example, if we have a number x, then its square root is the number y such that x^2 y^2. If we want to find the square root of a number, we simply take the square root of both sides of the equation. So, if we wanted to find the square root for 12, we would do (12)^(1/2) 6.

**2. How does square roots work?**

Square roots are useful in many different situations. One of the most popular uses is to solve equations. If we have two numbers, say x and y, and we know their product, z xy, then we can use the following formula to find the square root:

z^(1/2)x+y

This means that if we know the sum of the square roots of two numbers, we can calculate the individual square roots of those numbers.

**3. Why should I learn about squares and square roots?**

Squares and square roots are often used in math problems. In fact, they are the basis for many mathematical operations. You might even encounter them in everyday life. For instance, if you were trying to figure out how much money you need to buy something, you could multiply the price times the quantity. Then you could divide the result by the cost per unit to get the total amount of money you need.

**4. How do I remember square roots?**

You can easily memorize the rules for finding square roots. Just think of the word “square” and the number 1. When you add the first letter of each word together, you get the number 2. Now, just keep adding the letters of the words together until you reach the number you are looking for.

**5. Where do I find square roots?**

There are several websites where you can find square roots online. However, some sites require you to download software to access the information. If you don’t have time to download software, you can always look at the answers to math questions on Khan Academy.