# NCERT Maths Class 8th Chapter 6 Squares And Square Roots Exercise 6.1 Problems

A square root of a number is the value that, when we multiply by itself, gives the number is known as square root. Ex: 4X4=16, so the square root of 16 is 4. The symbol of square root is which means a positive square root always. Always the positive square root a number gives the original number. Square root the radical symbol used to represent the root of a number.

The square root of any number is equal to the number when it is squared and gives the original number. If a number ends with even numbers or zeros then, it can have the . If the number is a perfect square number then, it exists a perfect square root. The two square roots can be multiplied for ex:3can be multiplied by2 then the result will be 6. The square root of any negative number is not defined because the perfect square can’t be negative. If a number ends with 1, 4, 5, 6, or 9 then the number will have the square root but, if the number ends with 2, 3, 7 or 8 then the perfect square root won’t be or can’t be.

There are two methods for finding the square roots: The Prime factorization and long division method. Square root is just the opposite process of squaring a number by using prime factorization, we can find the square root of perfect squares and by using the long division method, we can find the square root of imperfect squares. So we are going to discuss some of the problems related to squares and square roots.

## 6. 1 By using a given pattern, find the missing numbers?

12:+22+22=32
22+33+62=72
32+42+122=132
42+52+_2=212
52+_2+302=312
62+72+_2=_2

Sol: From the given pattern, we can observe that,
The first two numbers are the product of the third number.
By adding one to the third number, the fourth number can be obtained.

62+72+422=432

### So here we give you some questions try to solve them very easy questions

1. Find the missing digits by observing the following pattern?

112=121
1012=10201
10012=1002001
1000012=1_2_1
100000012=_?

2. Supply the missing number by observing the following pattern?
112=121
1012=10201
101012=102030201
10101012=_
_2=10203040504030201?

3. Without adding find the sum?

a). 1+3+5+7+9
b). 1+3+5+7+9+11+13+15+17+19
c) .1+3+5+7+9+11+13+15+17+19+21+23

4. What are the possibilities for expressing 49 as the sum of the 7 odd numbers?
Express 121as the sum of 11 odd numbers?

5. How many numbers lie in between the squares of the following numbers?

a) 12 and 13
b) 25 and 26
c) 99 and 100

### Solutions:

1. In the given pattern, we can observe that the squares of the given numbers have the same numbers of zeros before and after the digit 2 as it is in the original number

112=121
1012=10201
10012=1002001
1000012=10000200001
100000012=100000020000001

2. In the given pattern, we can observe that
112=121
1012=10201
101012=102030201
10101012=1020304030201
1010101012=10203040504030201

3. We know the first n odd natural numbers sum is n2
a) Here, we had to find the first five odd natural numbers and their sum
1+3+5+7+9=(5)2=25

b) Here, we had to find the first ten odd natural numbers and their sum
1+3+5+7+9+11+13+15+17+19=(10)2=100

c) Here, we had to find the first twelve odd natural numbers and their sum
1+3+5+7+9+11+13+15+17+19+21+23=(12)2=144

4. We know that the sum of the first n odd natural numbers is n2
a) 49=(7)2
49 is the sum of the first 7 odd natural numbers
49=1+3+5+7+9+11+13
b) 121=(11)2
The first 11 odd natural numbers sum is 121
121=1+3+5+7+9+11+13+15+17+19+21

5. We know that there will be 2n numbers in between the squares of the numbers n and (n+1)

a) 2×12=24 numbers will be in between 122 and 132
b) 2×25=50 numbers will be in between 252 and 262
c) 2×99=198 numbers will be in between 992 and 1002

#### Conclusion:

These are some of the problems of squares and square roots, very easy problems which will be helpful for higher studies and gain knowledge. It is very essential to learn them. It is a very important chapter and important topic.