**Notes**
**Worksheets**
**Ncert Solutions**
**Square Number**

if a natural number

*m *can be expressed as

*n*^{2}, where

*n *is also a natural

number, then

*m *is a

**square number**
**Example**

1=1^{2}

4=2^{2}

The numbers 1, 4, 9, 16 ... are square numbers. These numbers are also called

**perfect ****squares**.

**Important point to Note for Square Numbers**

**How to find the square of Number easily**

__Identity method__

We know that

(a+b)

^{2} =a

^{2} +2ab+b

^{2}
**Example**

23^{2}= (20+3)^{2} =400+9+120=529

__Special Case__

(

*a*5)

^{2}
=

* a*(

*a *+ 1) hundred + 25

**Example**

25^{2}=2(3) hundred +25=625

**Watch this tutorial for an example of square numbers**
**Pythagorean triplets**

For any natural number

*m *> 1, we have (2

*m*)

^{2} + (

*m*^{2} – 1)

^{2} = (

*m*^{2} + 1)

^{2}
So, 2

*m*,

*m*^{2} – 1 and

*m*^{2} + 1 forms a Pythagorean triplet

**Example**

6,8,10

6^{2} +8^{2} =10^{2}

Lets take some solved example on it

**Solved Example**

Find a Pythagorean triplet in which one member is 12.

**Solution:**

If we take *m*^{2} – 1 = 12

Then, *m*^{2} = 12 + 1 = 13

Then the value of *m *will not be an integer.

So, we try to take *m*^{2} + 1 = 12. Again *m*^{2} = 11 will not give an integer value for *m*.

So, let us take 2*m *= 12

then *m *= 6

Thus, *m*^{2} – 1 = 36 – 1 = 35 and *m*^{2} + 1 = 36 + 1 = 37

Therefore, the required triplet is 12, 35, 37

**Square Root**

Square root of a number is the number whose square is given number

So we know that

m=n

^{2}
Square root of m

√m =n

Square root is denoted by expression √

## How to Find Square root

**Finding square root through repeated subtraction**

We know sum of the first n odd natural numbers is n

^{2}. So in this method we subtract the odd number starting from 1 until we get the reminder as zero. The count of odd number will be the square root

Consider 36

Then,

(i) 36 – 1 = 35 (ii) 35 – 3 = 32 (iii) 32 – 5 = 27 (iv) 27 – 7 = 20

(v) 20 – 9 = 11 (vi)11 – 11 = 0

So 6 odd number, Square root is 6

**Finding square root through prime factorization**

This method, we find the prime factorization of the number.

We will get same prime number occurring in pair for perfect square number. Square root will be given by multiplication of prime factor occurring in pair

Consider

81

81= (3×3)×(3×3)

√81= 3×3=9

**Example**

Find the square root of 57600.

**Solution: **Write 57600 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5×3×3

Therefore 57600= 2 × 2 × 2 × 2 × 5×3 = 240

**Finding square root by division method**

This can be well explained with the example

**Step 1** Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar. So in the below example 6 and 25 will have separate bar

**Step 2** Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide and get the remainder

In the below example 4 < 6, So taking 2 as divisor and quotient and dividing, we get 2 as reminder

**Step 3** Bring down the number under the next bar to the right of the remainder.

In the below example we bring 25 down with the reminder, so the number is 225

**Step 4** Double the quotient and enter it with a blank on its right.

In the below example, it will be 4

**Step 5** Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.

In this case 45 × 5 = 225 so we choose the new digit as 5. Get the remainder.

**Step 6** Since the remainder is 0 and no digits are left in the given number, therefore the number on the top is square root

In case of Decimal Number, we count the bar on the integer part in the same manner as we did above, but for the decimal part, we start pairing the digit from first decimal part.

**Example**
Find the square root of 870.25

**Solution**
**Example**
Find the Square root of 841

**Solution**
**Estimating Digits in the Square Root**

if a perfect square is of

*n*-digits, then its square root will have n/2 digits if

*n *is even or (n+ 1)/2

if

*n *is odd

Or

We can create the pair in the number, the Number of pair gives the number of digit in the square root

**Example**
Find the number of digits in the square root of the following numbers.

(i) 25921

(ii) 37249

**Solution**
Here n is odd, so (n+1)/2 = 3

So three digits will be present in square root