# Class 8 NCERT Solutions – Chapter 6 Squares and Square Roots – Exercise 6.1

**Problem 1: **What will be the unit digit of the squares of the following numbers?

**Solution:**

To find the unit digit of the square of any number, find the square of the unit digit of that number and

its unit digit will be the unit digit of the square of the actual number.

eg:- if a number is 2007, the square of the unit digit of 2007 is 49 (7 square). and the unit digit of 49 is 9. Therefore, the unit digit of the square of 2007 is 9.

** (i) 81**

Ans:1

Reason:If a number has 1 or 9 in the units place, then its square ends in 1.

(or) Unit digit of 81 is 1. and 1^{2 }= 1, therefore unit digit of square of 81 is 1.

**(ii) 272**

Ans:4

Reason:Unit digit of 272 is 2 and 2^{2 }= 4. Therefore unit digit of the square of 272 is 4.

**(iii) 799**

Ans:1

Reason:If a number has 1 or 9 in the units place, then its square ends in 1. (or) Unit digit of 799 is 9 and 9^{2 }= 81, and unit digit of 81 is 1. Therefore unit digit of square of 799 is 1.

**(iv) 3853**

Ans: 9

Reason: Unit digit of 3853 is 3 and 3^{2 }= 9, therefore unit digit of square of 3853 is 9.

**(v) 1234**

Ans: 6

Reason: Unit digit of 1234 is 4 and 4^{2 }= 16, and the unit digit of 16 is 6. Therefore the unit digit of the square of 1234 is 6.

**(vi) 26387**

Ans: 9

Reason: Unit digit of 26387 is 7 and 7^{2 }= 49, and the unit digit of 49 is 9. Therefore the unit digit of the square of 26387 is 9.

**(vii) 52698**

Ans: 4

Reason: Unit digit of 52698 is 8 and 8^{2 }= 64, and the unit digit of 64 is 4. Therefore the unit digit of the square of 52698 is 4.

**(viii) 99880**

Ans: 0

Reason: If a number has 0 in its unit place then square of its number will also have 0 in its units place (since 0^{2 }= 0).

**(ix) 12796**

Ans: 6

Reason: Unit digit of 12796 is 6 and 6^{2 }= 36, and the unit digit of 36 is 6. Therefore the unit digit of the square of 12796 is 6.

**(x) 55555**

Ans: 5

Reason: The unit digit of 55555 is 5 and 5^{2 }= 25, and the unit digit of 25 is 5. Therefore the unit digit of the square of 55555 is 5.

**Problem 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:**

- If a number ends with 2 (or) 3 (or) 7 (or) 8 at units place then we can tell it is not a perfect square. From the above rule, we can tell
1057, 23453, 7928, 222222, 89722 are not perfect squares.- For a number to be a perfect square it should have an even number of zeros at the end. From rule 2 we can tell
64000, 222000, 505050 are not perfect squares as they are having an odd number of zeros at the end.

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

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

Ans:431 and 7779

Reason:The square of an odd number will be an odd number, and the square of an even number will be an even number.

**Problem 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 }= ………………………

Ans:100001

^{2 }= 1000020000110000001

^{2 }= 100000020000001

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

11^{2 }= 121

101^{2 }= 10201

10101^{2 }= 102030201

1010101^{2 }= ………………………

…………^{2 }= 10203040504030201

Ans:1010101

^{2 }= 1020304030201

101010101^{2 }= 10203040504030201

**Problem 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:**

To find pattern third number is related to first and second number. How?

If we multiply the first and second numbers we will get the third number.

The fourth number is related to the third number. How?

Forth number = third number + 1

4

^{2 }+ 5^{2 }+ __^{2 }= 21^{2}

Ans:4 * 5 = 205

^{2 }+ __^{2 }+ 30^{2 }= 31^{2}

Ans:5 * x = 30=> x = (30 / 5) = 6

6

^{2 }+ 7^{2 }+ x^{2 }= y^{2}

Ans:x = 6 * 7 = 42y = x + 1 = 43

**Problem 7: Without adding, find the sum.**

**Solution:**

The Sum of first n odd natural numbers is n^{2}

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

Ans:25

Sum of first 5 odd natural numbers is 5^{2}, and 5^{2 }= 25.

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

Ans:100Sum of first 10 odd natural numbers is 10

^{2}, and 10^{2 }= 100.

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

Ans:144Sum of first 12 odd natural numbers is 12

^{2}, and 12^{2 }= 144.

**Problem 8**

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

Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13Explanation: 49 = 7^{2}therefore the sum of the first 7 odd natural numbers is 49.

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

Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21Explanation: 121 = 11^{2 }therefore sum of first 11 odd natural numbers is 121.

**Problem 9: How many numbers lie between squares of the following numbers?**

In between n^{2} and (n + 1)^{2}, there will be 2n natural numbers

**(i) 12 and 13**

Ans: 24 (as n is 12, 2n will be 24)

**(ii) 25 and 26**

Ans: 50 (as n is 25, 2n will be 50)

**(iii) 99 and 100**

Ans: 198 (as n is 99, 2n will be 198)