Latest Updated : October 2023

If you are in search of **NCERT solutions for Class 8 Maths Chapter** **5 Squares and Square Roots**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths** of all the chapters, **exercise wise** with **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark **ncertforclass8.com** to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.1

**Class 8 Maths Chapter** **5 Squares and Square Roots** **Exercise 5.1**

**Exercise 5.1**

**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

Solution:

The unit digit of square of a number having ‘a’ at its unit place ends with **a×a.**

i. The unit digit of the square of a number having digit 1 as unit’s place is 1.

∴ Unit digit of the square of number 81 is equal to 1.

ii. The unit digit of the square of a number having digit 2 as unit’s place is 4.

∴ Unit digit of the square of number 272 is equal to 4.

iii. The unit digit of the square of a number having digit 9 as unit’s place is 1.

∴ Unit digit of the square of number 799 is equal to 1.

iv. The unit digit of the square of a number having digit 3 as unit’s place is 9.

∴ Unit digit of the square of number 3853 is equal to 9.

v. The unit digit of the square of a number having digit 4 as unit’s place is 6.

∴ Unit digit of the square of number 1234 is equal to 6.

vi. The unit digit of the square of a number having digit 7 as unit’s place is 9.

∴ Unit digit of the square of number 26387 is equal to 9.

vii. The unit digit of the square of a number having digit 8 as unit’s place is 4.

∴ Unit digit of the square of number 52698 is equal to 4.

viii. The unit digit of the square of a number having digit 0 as unit’s place is 01.

∴ Unit digit of the square of number 99880 is equal to 0.

ix. The unit digit of the square of a number having digit 6 as unit’s place is 6.

∴ Unit digit of the square of number 12796 is equal to 6.

x. The unit digit of the square of a number having digit 5 as unit’s place is 5.

∴ Unit digit of the square of number 55555 is equal to 5.

**2. The following numbers are obviously not perfect squares. Give reason.**

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222

(v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

Solution:

(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.

Therefore 1057 is not a perfect square because its unit’s place digit is 7.

(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 23453 is not a perfect square because its unit’s place digit is 3.

(iii)Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 7928 is not a perfect square because its unit’s place digit is 8.

(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 222222 is not a perfect square because its unit’s place digit is 2.

(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 64000 is not a perfect square because its unit’s place digit is single 0.

(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 89722 is not a perfect square because its unit’s place digit is 2.

(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.

(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 505050 is not a perfect square because its unit’s place digit is 0.

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

(i) 431 (ii) 2826 (iii) 7779 (iv) 82004

Solution:

We know that the square of an odd number is odd and the square of an even number is even.

i. The square of 431 is an odd number.

ii. The square of 2826 is an even number.

iii. The square of 7779 is an odd number.

iv. The square of 82004 is an even number.

**4. Observe the following pattern and find the missing digits.**

`\[101^{2\ }=\ 10201\]`

`\[1001^{2\ }=\ 1002001\]`

`\[100001^{2\ }=\ 1.....2......1\]`

`\[10000001^{2\ }=\ .............................\]`

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.

`\[∴\ 100001^2=\ 10000200001\]`

`\[∴\ 10000001^2=\ 100000020000001\]`

**5. Observe the following pattern and supply the missing numbers**.

`\[11^{2\ }=\ 121\]`

`\[101^{2\ }=\ 10201\]`

`\[10101^{2\ }=\ 102030201\]`

`\[1010101^{2\ }=\ ..........................\]`

`\[..................^{2\ }=\ 10203040504030201\]`

**Solution:**

`\[1010101^{2\ }=1020304030201\]`

`\[101010101^2\ =\ 10203040504030201\]`

**6. Using the given pattern, find the missing numbers.**

`\[1^{2\ }+\ 2^2\ +\ 2^{2\ }=\ 3^2\]`

`\[2^{2\ }+\ 3^2\ +\ 6^2\ =\ 7^2\]`

`\[3^{2\ }+\ 4^2\ +\ 12^{2\ }=\ 13^2\]`

`\[4^{2\ }+\ 5^2\ +\ .......^{2\ }=\ 21^2\]`

`\[5^{2\ }+\ ......^2\ +\ 30^{2\ }=\ 31^2\]`

`\[6^{2\ }+\ 7^2\ +\ .......^{2\ }=\ .........^2\]`

Solution:

`\[1^{2\ }+\ 2^2\ +\ 2^{2\ }=\ 3^2\]`

`\[2^{2\ }+\ 3^2\ +\ 6^{2\ }=\ 7^2\]`

`\[3^{2\ }+\ 4^2\ +\ 12^{2\ }=\ 13^2\]`

`\[4^{2\ }+\ 5^2\ +\ 20^{2\ }=\ 21^2\]`

`\[5^{2\ }+\ 6^2\ +\ 30^{2\ }=\ 31^2\]`

`\[6^{2\ }+\ 7^2\ +\ 42^{2\ }=\ 43^2\]`

**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**

Solution:

(i) Here, there are five odd numbers.

Therefore, square of 5 is 25 ∴ 1+3+5+7+9=52 = 25

(ii) Here, there are ten odd numbers.

Therefore, square of 10 is 100 ∴ 1+3+5+7+9+11 +13 + 15 +17+ 19 = 102 = 100

(iii) Here, there are twelve odd numbers.

Therefore, square of 12 is 144 ∴ 1+3+5+7+9+11 +13 + 15 +17 +19 + 21+ 23 = 122 = 144

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

Solution:

(i) 49 is the square of 7. Therefore, it is the sum of 7 odd numbers. 49=1+3+5+7+9+11 +13

(ii) 121 is the square of 11. Therefore, it is the sum of 11 odd numbers 121 1+3+5+7+9+11 +13+15+17 +19 +21

**9. How many numbers lie between squares of the following numbers?(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100**

Solution:

`\[Between\ n^2\ and\ (n+1)^{2\ },there\ are\ 2n\ non-perfect\ square\ numbers.\]`

`\[\left(i\right)\ Between\ 12^2\ and\ \ 13^{2\ },there\ are\ 2n\ =\ 2\ X\ 12\ =\ 24\ natural\ numbers.\]`

`\[\left(ii\right)\ Between\ 25^2\ and\ \ 26^{2\ },there\ are\ 2n\ =\ 2\ X\ 25\ =\ 50\ natural\ numbers.\]`

`\[\left(iii\right)\ Between\ 99^2\ and\ \ 100^{2\ },there\ are\ 2n\ =\ 2\ X\ 99\ =\ 198\ natural\ numbers.\]`

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