Class 8 Maths important questions for chapter 6-Squares and Square Roots are available here with solutions. Our subject expert has also provided the solutions for these problems based on **CBSE latest syllabus (2019-2020)** and in reference with** NCERT book**. Therefore, it will be easy for students to revise the chapter with this material and score in the final examination.

Students can also reach to **BYJUâ€™S** to get Maths subject important questions for class 8 to get all the chapters. In chapter 6, we will learn to find the squares and square roots for different numbers.

**Also, check:**

- Important 2 Marks Question For Cbse Class 8 Maths
- Important 3 Marks Questions For Cbse Class 8 Maths
- Important 4 Marks Questions For Cbse Class 8 Maths

## Important Questions with Solutions For Class 8 Maths Chapter 6 (Squares and Square Roots)

**Q.1: How many numbers lie between squares of the following numbers? **

**(i) 12 and 13 **

**(ii) 25 and 26 **

**(iii) 99 and 100**

Solution: As we know, between n^{2} and (n+1)^{2}, the number ofÂ nonâ€“perfect square numbers are 2n.

(i) Between 12^{2} and 13^{2} there are 2Ã—12 = 24 natural numbers.

(ii) Between 25^{2} and 26^{2} there are 2Ã—25 = 50 natural numbers.

(iii) Between 99^{2} and 100^{2} there are 2Ã—99 =198 natural numbers.

**Q.2: Write a Pythagorean triplet whose one member is: **

**(i) 6 **

**(ii) 14 **

**(iii) 16 **

**(iv) 18 **

Solution:

We know, for any natural number m, 2m, m^{2}â€“1, m^{2}+1 is a Pythagorean triplet.

(i) 2m = 6

â‡’ m = 6/2 = 3

m^{2}â€“1= 3^{2} â€“ 1 = 9â€“1 = 8

m^{2}+1= 3^{2}+1 = 9+1 = 10

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

(ii) 2m = 14

â‡’ m = 14/2 = 7

m^{2}â€“1= 7^{2}â€“1 = 49â€“1 = 48

m^{2}+1 = 7^{2}+1 = 49+1 = 50

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

(iii) 2m = 16

â‡’ m = 16/2 = 8

m^{2}â€“1 = 8^{2}â€“1 = 64â€“1 = 63

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

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

(iv) 2m = 18

â‡’ m = 18/2 = 9

m^{2}â€“1 = 9^{2}â€“1 = 81â€“1 = 80

m^{2}+1 = 9^{2}+1 = 81+1 = 82

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

**Q.3: (n+1) ^{2}-n^{2} = ?**

Solution:

(n+1)^{2}-n^{2}

= (n^{2} + 2n + 1) â€“ n^{2}

= 2n + 1

**Q.4: Show that 121 is the sum of 11 odd natural numbers.**

Solution: As 121 = 11^{2}

Therefore, 121 = sum of first 11 odd natural numbers

= 1 + 3 + 5+ 7 + 9 + 11 +13 + 15 + 17 + 19 + 21

**Q.5: Show that the sum of two consecutive natural numbers is 13 ^{2}.**

Solution: Here, 2n + 1 = 13

So, n = 6

So, ( 2n + 1)^{2} = (2n^{2} + 2n) + (2n^{2} + 2n + 1)

Substitute n by 6,

(13)^{2} = ( 2 x 6^{2} + 2 x 6) + (2 x 6^{2} + 2 x 6 + 1)

= 84 + 85

**Q.6: Use the identity and find the square of 189?**

**(a – b) ^{2 } = a^{2} â€“ 2ab + b^{2}**

Solution: 189 = (200 â€“ 11)^{2}

= 40000 â€“ 2 x 200 x 11 + 11^{2}

= 40000 â€“ 4400 + 121

= 35721

**Q.7: What would be the square root of 625 using the identity (a +b) ^{2} = a^{2} + b^{2} + 2ab?**

Solution: (625)^{2}

= (600 + 25)^{2}

= 600^{2} + 2 x 600 x 25 +25^{2}

= 360000 + 30000 + 625

= 390625

**Q.8: Find the square roots of 100 and 169 by the method of repeated subtraction.**

Solution:

Let us find the square root of 100 first.

- 100 â€“ 1 = 99
- 99 â€“ 3 = 96
- 96 â€“ 5 = 91
- 91 â€“ 7 = 84
- 84 â€“ 9 = 75
- 75 â€“ 11 = 64
- 64 â€“ 13 = 51
- 51 â€“ 15 = 36
- 36 â€“ 17 = 19
- 19 â€“ 19 = 0

Here, we have performed subtraction ten times.

Therefore, âˆš100 = 10

Now, the square root of 169:

- 169 â€“ 1 = 168
- 168 â€“ 3 = 165
- 165 â€“ 5 = 160
- 160 â€“ 7 = 153
- 153 â€“ 9 = 144
- 144 â€“ 11 = 133
- 133 â€“ 13 = 120
- 120 â€“ 15 = 105
- 105 â€“ 17 = 88
- 88 â€“ 19 = 69
- 69 â€“ 21 = 48
- 48 â€“ 23 = 25
- 25 â€“ 25 = 0

Here, we have performed subtraction thirteen times.

Therefore, âˆš169 = 13

**Q.10: Find the square root of 729 using factorisation method.**

Solution:

729 = 3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—1

â‡’ 729 = (3Ã—3)Ã—(3Ã—3)Ã—(3Ã—3)

â‡’ 729 = (3Ã—3Ã—3)Ã—(3Ã—3Ã—3)

â‡’ 729 = (3Ã—3Ã—3)^{2}

Therefore,

â‡’ âˆš729 = 3Ã—3Ã—3 = 27

**Q. 11: Find the smallest whole number by which 1008 should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.**

Solution:

Let us factorise the number 1008.

1008 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—7

= (2Ã—2)Ã—(2Ã—2)Ã—(3Ã—3)Ã—7

Here, 7 cannot be paired.

Therefore, we will multiply 1008 by 7 to get perfect square.

New number so obtained = 1008Ã—7 = 7056

Now, let us find the square root of 7056

7056 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—7Ã—7

â‡’ 7056 = (2Ã—2)Ã—(2Ã—2)Ã—(3Ã—3)Ã—(7Ã—7)

â‡’ 7056 = 22Ã—22Ã—32Ã—72

â‡’ 7056 = (2Ã—2Ã—3Ã—7)^{2}

Therefore;

â‡’ âˆš7056 = 2Ã—2Ã—3Ã—7 = 84

**Q. 12: Find the smallest whole number by which 2800 should be divided so as to get a perfect square. **

Solution:

Let us first factorise the number 2800.

2800 = 2Ã—2Ã—2Ã—2Ã—5Ã—5Ã—7

= (2Ã—2)Ã—(2Ã—2)Ã—(5Ã—5)Ã—7

Here, 7 cannot be paired.

Therefore, we will divide 2800 by 7 to get a perfect square.

New number = 2800 Ã· 7 = 400

**Q. 13: 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.**

Solution:

Let the number of rows be, x.

Therefore, the number of plants in each rows = x.

Total many contributed by all the students = xÃ—x = x^{2}

Given, x^{2} = Rs.2025

x^{2 }= 3Ã—3Ã—3Ã—3Ã—5Ã—5

â‡’ x^{2} = (3Ã—3)Ã—(3Ã—3)Ã—(5Ã—5)

â‡’ x^{2} = (3Ã—3Ã—5)Ã—(3Ã—3Ã—5)

â‡’ x^{2} = 45Ã—45 â‡’ x = âˆš(45Ã—45)

â‡’ x = 45

Therefore,

Number of rows = 45

Number of plants in each rows = 45

**Q.14: Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.**

Solution:

L.C.M of 8, 15 and 20 is (2Ã—2Ã—5Ã—2Ã—3) 120. 120 = 2Ã—2Ã—3Ã—5Ã—2 = (2Ã—2)Ã—3Ã—5Ã—2

Here, 3, 5 and 2 cannot be paired.

Therefore, we need to multiply 120 by (3Ã—5Ã—2) i.e. 30 to get a perfect square.

Hence, the smallest squared number which is divisible by numbers 8, 15 and 20 = 120Ã—30 = 3600

**Q.15: Find the square root of 7921 using long division method.**

Solution:

âˆ´ âˆš7921 = 89

**Q. 16: Find the square root of 42.25 using long division method.**

Solution:

âˆ´ âˆš42.25 = 6.5

**Q. 17: Find the least number which must be added to 1750 so as to get a perfect square. Also, find the square root of the obtained number.**

Solution:

Using long division method:

Here, (41)^{2} < 1750 > (42)^{2}

We can say 1750 is ( 164 â€“ 150 ) 14 less than (42)^{2}.

Therefore, if we add 14 to 1750, it will be a perfect square.

New number = 1750 + 14 = 1764

Therefore, the square root of 1764 is as follows:

âˆ´âˆš1764 = 42

### Class 8 Maths Chapter 6 Extra Questions

- In a right triangle ABC, âˆ B = 90Â°. a. If AB = 6 cm, BC = 8 cm, find AC b. If AC = 13 cm, BC = 5 cm, find AB.
- Find the length of the side of a square whose area is 441 m
^{2}. - 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?
- 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.
- Express 49 as the sum of 7 odd numbers.